From the Lorentz force to the galvanometer — how moving charges make magnetic fields and how those fields push back
🔬 Interactive 3D · A moving charge curves as it feels $\vec{F}=q\,\vec{v}\times\vec{B}$, while a current in the wire wraps space in circular magnetic field lines.
Electricity and magnetism were studied as separate curiosities for over two thousand years, until in 1820 Hans Christian Oersted noticed that a steady current in a straight wire swung a nearby compass needle. That single deflection announced a deep truth: moving charge is the source of magnetism. This chapter turns that discovery into a working toolkit, and for JEE it is one of the highest-yield chapters in electromagnetism. It carries substantial weight on its own, and it almost never appears in isolation. A problem may start with a capacitor accelerating a charge (electrostatics), send that charge into a magnetic field to bend it in a circle (this chapter), and finish by asking for the flux swept out (electromagnetic induction). Learning to move fluently between these three chapters is where marks are won. 🔉⇢
Before any calculation, it helps to hold a small mental dictionary in your head, because every result in the chapter is one of three ideas wearing different clothes. First, a moving charge and an electric current are the same physical object seen at two scales, so the force on a single charge and the force on a current-carrying wire are two forms of one law. Second, a tiny current element is a source of magnetic field, and adding up such elements builds the field of any wire, loop or coil. Third, a closed current loop behaves, from far away and under torque, exactly like a magnetic dipole. Whenever a question looks unfamiliar, translate it into one of these three pictures and it becomes routine. 🔉⇢
The master equation of the chapter is the Lorentz force, $\vec{F} = q\,\vec{E} + q\,\vec{v}\times\vec{B}$. The whole chapter lives or dies on the magnetic part, $q\,\vec{v}\times\vec{B}$, and the single most common source of lost marks is rushing its direction. The disciplined habit is to fix direction before magnitude: settle the geometry of the cross product with the right-hand rule first, remembering that the force on a negative charge points opposite to the force on a positive one, and only then compute the size $F = qvB\sin\theta$. If velocity and field are parallel or antiparallel the magnetic force is exactly zero, and it is always perpendicular to both $\vec{v}$ and $\vec{B}$. 🔉⇢
A second fact about the magnetic force is subtle and endlessly examined: it does no work. Because $q\,\vec{v}\times\vec{B}$ is always perpendicular to the velocity, its dot product with the displacement vanishes, so it can never change the speed or the kinetic energy of a particle, only the direction of motion. An electric field can speed a charge up or slow it down; a magnetic field can only steer. Any question that assumes a magnetic field has added energy to a free charge is testing precisely this misconception, and recognising it instantly is worth easy marks. 🔉⇢
The immediate consequence of a steering-only force is beautiful. When a charge enters a uniform field perpendicular to it, the constant sideways push acts as a centripetal force and the charge runs in a circle of radius $r = mv/qB$. The time for one loop, $T = 2\pi m/qB$, and the corresponding cyclotron frequency $\nu_c = qB/2\pi m$, depend only on the mass, charge and field, and remarkably not at all on the speed or radius. If the velocity also has a component along the field, that component is untouched and the path becomes a helix whose forward step per turn is the pitch $p = v_\parallel T$. The speed-independence of the period is the exact principle that lets a cyclotron accelerate particles to high energy. 🔉⇢
Crossed electric and magnetic fields give a clean piece of apparatus that examiners love: the velocity selector. Arrange $\vec{E}$ and $\vec{B}$ perpendicular to each other and to the beam, and the electric force $qE$ opposes the magnetic force $qvB$. Only particles whose speed satisfies $v = E/B$ pass straight through undeflected; faster or slower ones are swept aside. This is the heart of mass spectrometers and of Thomson's measurement of the charge-to-mass ratio, and it is a favourite because it rewards a student who can balance two forces rather than memorise a formula. 🔉⇢
Having dealt with what a field does to a charge, the chapter turns to where the field comes from. The Biot-Savart law states that a current element $I\,d\vec{l}$ produces at a point a field $d\vec{B} = \dfrac{\mu_0}{4\pi}\dfrac{I\,d\vec{l}\times\hat{r}}{r^2}$, perpendicular to the plane containing the element and the line to the point, and falling off as the inverse square of distance. It is the magnetic cousin of Coulomb's law, but with two twists that trap the unwary: the source is a vector, not a scalar, and there is an angular factor $\sin\theta$, so an element contributes nothing along its own direction. Every field of a wire, arc or loop is built by superposing these contributions. 🔉⇢
The most-used result of the Biot-Savart integration is the field on the axis of a circular current loop of radius $R$, $B = \dfrac{\mu_0 I R^2}{2\,(R^2 + x^2)^{3/2}}$, whose special case at the centre, $B = \dfrac{\mu_0 I}{2R}$, appears in countless problems and scales to $\dfrac{\mu_0 N I}{2R}$ for $N$ tightly wound turns. Arcs are handled by taking the fraction of a full loop that the arc subtends, and straight radial segments pointing at the centre contribute nothing because $d\vec{l}$ and $\hat{r}$ are parallel there. Recognising which pieces of a bent wire contribute and which do not is a recurring skill. 🔉⇢
Ampere's circuital law, $\oint \vec{B}\cdot d\vec{l} = \mu_0 I_{\text{enc}}$, is the elegant shortcut, but it carries a loud warning that examiners exploit relentlessly: it is always true, yet it only lets you extract the field when the geometry has enough symmetry to pull $B$ outside the integral. For an infinite straight wire the amperian circle gives $B = \dfrac{\mu_0 I}{2\pi r}$ in one line; for a long solenoid or a toroid it is just as quick. But for a finite wire or a point on the axis of a loop the symmetry is absent and the law, though valid, is useless for finding the field, which is exactly why Biot-Savart still matters. Asking a student to say when the law helps is a standard trap. 🔉⇢
Applied to a densely wound coil, Ampere's law delivers the two field machines of the chapter. Inside a long solenoid the field is uniform and axial with magnitude $B = \mu_0 n I$, where $n$ is the number of turns per unit length, while the field outside is essentially zero. Bend the solenoid into a closed ring and it becomes a toroid, where the field is confined within the core, $B = \dfrac{\mu_0 N I}{2\pi r}$, and vanishes both inside the hole and outside the ring. These two devices are the standard way to produce a controlled, nearly uniform magnetic field, and the contrast between the uniform solenoid interior and the radius-dependent toroid field is a common comparison. 🔉⇢
Because a wire carrying current sits in the field of its neighbour, two parallel currents exert forces on each other: parallel currents attract and antiparallel currents repel, with a force per unit length $f = \dfrac{\mu_0 I_a I_b}{2\pi d}$. This is the opposite of the electrostatic rule for like charges, and it is the basis on which the ampere was historically defined, as the steady current that produces a force of $2\times 10^{-7}$ newton per metre between two wires one metre apart. The force on the wire itself follows from the single-charge law summed over all carriers, giving $\vec{F} = I\,\vec{L}\times\vec{B}$, the wire-scale twin of $q\,\vec{v}\times\vec{B}$. 🔉⇢
Finally, a current loop in a uniform field feels no net force but does feel a torque $\vec{\tau} = \vec{m}\times\vec{B}$, where the magnetic dipole moment is $\vec{m} = N I \vec{A}$. This is the direct analogue of an electric dipole in a uniform electric field, and it explains why a compass needle aligns with a field. The torque is what makes the moving coil galvanometer work: a radial field keeps the deflection proportional to current through $\varphi = \dfrac{N A B}{k}\,I$. From that one instrument the chapter builds practical meters, converting a galvanometer into an ammeter with a small shunt resistance in parallel and into a voltmeter with a large resistance in series. 🔉⇢
Approach: read the geometry before touching a formula. For a moving charge, decide the plane of the circle and the sense of $\vec{v}\times\vec{B}$ first, then use $r = mv/qB$ and the speed-independent period. For fields from currents, ask whether the symmetry is high enough for Ampere's law; if it is, use it, and if it is not, fall back to Biot-Savart and superposition. For loops in a field, split every problem into force (net zero) and torque ($\vec{m}\times\vec{B}$). Keep units honest, work with a clean right-hand rule every single time, and remember that examiners chain this chapter to electrostatics and induction, so a magnetism problem is often only the middle act of a longer story. 🔉⇢
This chapter breaks into the concepts below. Each has its own tab with interactive 3D, a full derivation, and a JEE worked example. The Contents tab has the complete prose.
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🗺️ Concept map — click any concept to open its tab. Each box shows the concept and its governing formula; the path is the recommended learning order.
A charge moving in a uniform magnetic field feels a force qv×B always perpendicular to its velocity, so its speed is unchanged and its path is a circle or, with a velocity component along B, a helix; the cyclotron exploits the speed-independent revolution frequency.
Ampere's law states that the line integral of B around any closed loop equals μ0 times the enclosed current; under enough symmetry it gives the field of a long wire and of a solenoid, whose interior field μ0 n I is uniform and whose exterior field nearly vanishes.
A current loop of moment m = N I A in a uniform field feels a torque τ = m×B that tries to align it with the field but no net force; this is the principle of the moving-coil galvanometer, and it makes a current loop behave exactly like a magnetic dipole.
The magnetic part of the Lorentz force is F = q v×B: its magnitude is qvB sinθ, its direction is perpendicular to both v and B, and because it is always perpendicular to v it can change a charge's direction but never its speed — a magnetic force does no work.
When a charged beam passes through crossed electric and magnetic fields, only particles with speed v = E/B pass undeflected because the electric force qE and magnetic force qvB cancel; this selects a single speed regardless of charge or mass.
A straight conductor carrying current I in a field B feels a force F = I L×B of magnitude BIL sinθ; it is the sum of the magnetic forces on all the drifting charges, and it is the bridge from single-charge dynamics to the forces on real wires.
Two long parallel wires a distance d apart exert a force per unit length μ0 I1 I2 / 2πd on each other — attractive for currents in the same direction, repulsive for opposite — and this force is what historically defined the ampere.
The Biot-Savart law gives the field of a current element, dB = (μ0/4π) I dl×r̂ / r²; integrating it over a long straight wire yields B = μ0 I / 2πr, the inverse-distance field whose circular field lines encircle the wire.
Integrating the Biot-Savart law around a circular loop of radius R gives the on-axis field B = μ0 I R² / 2(R²+x²)^{3/2}, which is μ0 I / 2R at the centre and falls off like an axial dipole far away.
A moving-coil galvanometer balances the deflecting torque N I A B on a coil against a restoring spring torque kφ, so its steady deflection φ = (NAB/k) I is proportional to current; a small shunt converts it to an ammeter and a large series resistance to a voltmeter.
The complete chapter in reading order. Each section is narratable; the concept tabs add interactive 3D and worked examples on top of this text.
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Every diagram below it is interactive: drag the controls and the numbers move with the drawing.
When a charge $q$ moves with velocity $\vec{v}$ through a magnetic field $\vec{B}$, it feels the magnetic part of the Lorentz force, $\vec{F}=q\,\vec{v}\times\vec{B}$. Because this force is built from a cross product, it points sideways -- always at right angles to both $\vec{v}$ and $\vec{B}$ -- and never along the line of motion. That single geometric fact drives everything in this topic. Since $\vec{F}$ is perpendicular to $\vec{v}$, it can do no work, so the particle's speed and kinetic energy stay exactly constant; only the direction of the velocity turns. A constant-magnitude force that continually turns the velocity without changing its length is precisely a centripetal force, so a charge launched perpendicular to a uniform field travels in a circle of radius $r=\dfrac{mv}{qB}$. Remarkably, the time for one loop, $T=\dfrac{2\pi m}{qB}$, and the corresponding frequency $f=\dfrac{qB}{2\pi m}$ do not depend on the speed at all: a faster particle simply sweeps a proportionally bigger circle in the same time. If the velocity also has a component along $\vec{B}$, that component sails through untouched and the motion becomes a helix whose pitch is $p=v_{\parallel}T$. The speed-independence of the period is the working principle of the cyclotron, a machine that accelerates ions to high energy using a fixed-frequency oscillator tuned to the resonance condition $f=\dfrac{qB}{2\pi m}$. 🔉⇢
Full derivation, worked example and interactive 3D on the Motion of a Charge in a Magnetic Field tab →
Ampere's circuital law states that the line integral of the magnetic field around any closed loop equals the permeability of free space times the net steady current that threads the loop: $\oint \vec{B}\cdot d\vec{l}=\mu_0 I_{\text{enc}}$. In words, if you walk once around a closed path and, at every step, add up the component of $\vec{B}$ along your direction of travel, the total you accumulate is fixed entirely by the current passing through any surface bounded by that path, and by nothing else. This is the exact magnetic counterpart of Gauss's law $\oint \vec{E}\cdot d\vec{A}=q_{\text{enc}}/\varepsilon_0$, which ties the flux of $\vec{E}$ through a closed surface to the charge inside. Both laws are always true for their respective steady sources, but, and this is the point a student must internalise, they hand you the field only when the geometry is symmetric enough to pull the field out of the integral. Where that symmetry is absent, the law remains perfectly correct yet completely useless for finding $\vec{B}$. 🔉⇢
Full derivation, worked example and interactive 3D on the Ampere's Circuital Law and the Solenoid tab →
A closed loop of wire carrying a steady current, when placed in a uniform magnetic field, experiences no net translational force, yet it is generally acted upon by a net torque that tries to rotate it. This single fact is the mechanical seed of one of the most productive ideas in magnetism: that a current loop behaves like a magnetic dipole. Just as an electric dipole $\mathbf{p}$ in a uniform electric field $\mathbf{E}$ feels the torque $\boldsymbol{\tau}=\mathbf{p}\times\mathbf{E}$ but no net force, a current loop of $N$ turns, area $A$ and current $I$ possesses a magnetic dipole moment $\mathbf{m}=NI A\,\hat{\mathbf{n}}$ and, in a uniform field $\mathbf{B}$, feels the torque $\boldsymbol{\tau}=\mathbf{m}\times\mathbf{B}$ whose magnitude is $\tau=NIAB\sin\theta$. The associated orientation energy is $U=-\mathbf{m}\cdot\mathbf{B}$, so the loop is in stable equilibrium when its moment is aligned with the field and in unstable equilibrium when it points against it. Exactly this torque, tamed by a soft-iron core that makes the field radial and by a spring that supplies a restoring couple, is what drives the pointer of a moving-coil galvanometer and gives the linear scale on which we read currents and voltages. 🔉⇢
Full derivation, worked example and interactive 3D on the Torque on a Current Loop and the Magnetic Dipole Moment tab →
A charge $q$ moving with velocity $\vec{v}$ through a magnetic field $\vec{B}$ experiences a magnetic force given by $\vec{F}=q\,\vec{v}\times\vec{B}$, and the single most important habit in handling it is to settle its direction before ever touching its magnitude. The direction is that of the cross product $\vec{v}\times\vec{B}$, read off with the right-hand rule, and it is always perpendicular to both $\vec{v}$ and $\vec{B}$ at once. Only afterwards do we attach the size $qvB\sin\theta$, where $\theta$ is the angle between $\vec{v}$ and $\vec{B}$. Because the force is forever perpendicular to the velocity, it can never do work: it bends the path of the particle without changing its speed or kinetic energy. This is the sharpest contrast with the electric force $q\vec{E}$, which points along the field and freely feeds energy to the charge. Combining the two gives the complete Lorentz force $\vec{F}=q\,(\vec{E}+\vec{v}\times\vec{B})$, and the SI unit that the magnetic term defines is the tesla. 🔉⇢
Consider first why direction must come first. The expression $\vec{v}\times\vec{B}$ is a vector product, and a vector product is meaningless until its direction is fixed; its magnitude alone tells you nothing about which way the particle will swerve. Geometrically, $\vec{v}\times\vec{B}$ is perpendicular to the plane that contains $\vec{v}$ and $\vec{B}$. If you lay your right hand so that the fingers point along $\vec{v}$ and then curl them through the smaller angle toward $\vec{B}$, your outstretched thumb points along $\vec{v}\times\vec{B}$. For a positive charge the force $\vec{F}$ is along this thumb direction; for a negative charge such as an electron the force is exactly opposite, because $q$ itself carries the minus sign. This sign-flip for negative carriers is not a detail to memorise separately, it falls straight out of the algebra. 🔉⇢
An equivalent statement is the screw rule: imagine turning a right-handed screw from $\vec{v}$ toward $\vec{B}$ through the angle between them; the direction in which the screw advances is the direction of $\vec{v}\times\vec{B}$. Whichever mental image you prefer, the payoff is the same discipline. You never guess the direction of the deflection from intuition about magnets, you construct it from the two vectors you were given. This is why examiners can pose the same physical situation with the field into the page or out of the page and expect a different answer: the geometry, not a remembered result, decides the sign. 🔉⇢
With direction settled, the magnitude follows cleanly. Writing $\vec{F}=q\,\vec{v}\times\vec{B}=qvB\sin\theta\,\hat{n}$, the scalar size is $qvB\sin\theta$, where $\hat{n}$ is the unit vector along $\vec{v}\times\vec{B}$ that we have just constructed. The factor $\sin\theta$ carries a great deal of physics. When $\vec{v}$ is perpendicular to $\vec{B}$, $\theta=90^{\circ}$ and $\sin\theta=1$, so the force is largest. When $\vec{v}$ is parallel or antiparallel to $\vec{B}$, $\theta=0^{\circ}$ or $180^{\circ}$ and $\sin\theta=0$, so the magnetic force vanishes entirely: a charge shot straight along the field line sails through undeflected. And if the charge is not moving at all, $v=0$ and the force is zero, so only a moving charge feels a magnetic force. These three limiting cases are worth reciting before any calculation. 🔉⇢
The magnitude relation also lets us define the unit of magnetic field. Dimensionally $[B]=[F/(qv)]$, so the unit of $B$ is $\text{N}\,\text{s}\,\text{C}^{-1}\,\text{m}^{-1}$, and this combination is named the tesla ($\text{T}$) after Nikola Tesla. Reading the definition physically, a field has magnitude $1\ \text{T}$ when a charge of $1\ \text{C}$ moving at $1\ \text{m}\,\text{s}^{-1}$ perpendicular to the field feels a force of $1\ \text{N}$. The tesla is a rather large unit, so a smaller non-SI unit, the gauss, equal to $10^{-4}\ \text{T}$, is also common; the Earth's own magnetic field is only about $3.6\times10^{-5}\ \text{T}$, which is why everyday magnetic forces on slow charges are so feeble. 🔉⇢
Now to the deepest property: the magnetic force does no work. The rate at which any force delivers energy is $P=\vec{F}\cdot\vec{v}$. For the magnetic force $\vec{F}=q\,\vec{v}\times\vec{B}$, this power is $q\,(\vec{v}\times\vec{B})\cdot\vec{v}$, and the vector $\vec{v}\times\vec{B}$ is by construction perpendicular to $\vec{v}$, so the dot product is identically zero. No work means no change in kinetic energy, hence no change in the particle's speed. What the force does change is the direction of the momentum: it continually turns the velocity vector, curving the trajectory into a circle when $\vec{v}$ is perpendicular to a uniform $\vec{B}$, or a helix when $\vec{v}$ has a component along $\vec{B}$. The magnetic force steers, it never accelerates in the sense of speeding up. 🔉⇢
The electric force behaves in the opposite way, and holding the two side by side sharpens the understanding of both. The electric force $q\vec{E}$ points along the field (or against it, for a negative charge), so it generally has a component parallel to the velocity. That parallel component makes $\vec{F}\cdot\vec{v}$ nonzero, so the electric force does work, transferring energy to or from the charge and changing its speed. In short, an electric field can speed a particle up or slow it down, while a magnetic field can only redirect it. This distinction is why particle accelerators use electric fields to raise energy and magnetic fields to steer the beam around the ring. 🔉⇢
Both interactions occur together for a charge sitting in overlapping fields, and their combination is the full Lorentz force $\vec{F}=q\,(\vec{E}+\vec{v}\times\vec{B})$, first written in this compact form by H. A. Lorentz on the strength of the experiments of Ampere and others. The two contributions simply add as vectors, an instance of superposition, so one may compute the electric part $q\vec{E}$ and the magnetic part $q\,\vec{v}\times\vec{B}$ separately and sum them. Note that the electric part is present whether or not the charge moves, whereas the magnetic part switches off the moment the charge is at rest. Recognising which pieces are active in a given problem is often half the battle. 🔉⇢
The charge sign deserves one more careful look because it governs so many standard questions. Since the whole force scales with $q$, reversing the sign of the charge reverses the force for the same $\vec{v}$ and $\vec{B}$. Thus a proton and an electron launched with identical velocities into the same field are pushed in exactly opposite directions. This is the reasoning behind the classic result that, for a beam moving along one axis in a field along a second axis, the positive and negative species deflect toward opposite ends of the third axis. Always fix the direction of $\vec{v}\times\vec{B}$ first for a hypothetical positive charge, then flip it if the actual charge is negative. 🔉⇢
Because the force turns the velocity without changing its size, a charge entering a uniform field at right angles travels a circular arc. Setting the magnetic force equal to the centripetal requirement, $qvB=mv^{2}/r$, gives the orbit radius $r=mv/(qB)$, so a more energetic particle sweeps a larger circle. This is a direct consequence of the force law of this card and is the seed of later devices such as the cyclotron and the mass spectrometer; here it simply illustrates once more that a purely magnetic force produces steering, embodied in a fixed-speed circular motion, rather than any change of speed. 🔉⇢
For examinations the reliable procedure is therefore fixed and worth internalising. First, draw $\vec{v}$ and $\vec{B}$ and construct $\vec{v}\times\vec{B}$ with the right-hand rule to obtain the direction; second, if the charge is negative, reverse that direction; third, only then compute the magnitude $qvB\sin\theta$, checking the limiting cases where $\theta=0$ gives zero force and $\theta=90^{\circ}$ gives the maximum. Finally, remember that whatever the numbers, the magnetic force is perpendicular to $\vec{v}$, so it can never change the particle's speed or kinetic energy, and any answer suggesting otherwise signals a conceptual slip that should be caught before the marks are lost. 🔉⇢
A velocity selector is a region where a uniform electric field $\vec{E}$ and a uniform magnetic field $\vec{B}$ are arranged perpendicular to each other and perpendicular to the beam, so that the electric force and the magnetic force on a moving charge point along the same line but in opposite senses. A charge feels the full Lorentz force $\vec{F}=q\,(\vec{E}+\vec{v}\times\vec{B})$; when the two contributions exactly cancel the charge travels straight through undeflected. Setting the electric force $qE$ equal to the magnetic force $qvB$ gives the selection condition $v=E/B$. The remarkable feature is that this speed depends only on the field strengths, not on the charge $q$ or the mass $m$ of the particle, so the device transmits one sharply defined speed out of a beam containing many. This makes crossed fields the standard front end of a mass spectrometer, where a clean, single-speed beam must be prepared before masses are compared. 🔉⇢
To see the geometry, imagine the beam moving along the $x$-axis, the magnetic field along the $y$-axis, and the electric field along the $z$-axis. Following the direction-first discipline for the magnetic force, $\vec{v}\times\vec{B}$ for a positive charge with $\vec{v}=v\hat{i}$ and $\vec{B}=B\hat{j}$ points along $+\hat{k}$ (the $+z$ direction), so the magnetic force is $q v B$ along $+z$. If the electric field is arranged along $-z$, the electric force $q\vec{E}$ on the same positive charge is along $-z$. The two forces are then collinear and opposed, and the whole design rests on choosing the field orientations so that this opposition holds. 🔉⇢
In SI units the electric field is measured in volt per metre and the magnetic field in tesla, so the selected value $v=E/B$ emerges in metre per second; recall that one tesla is the field that exerts one newton on a coulomb of charge, and the smaller unit gauss is $10^{-4}$ tesla. Written as vectors, the transverse force $q\vec{E}+q\,\vec{v}\times\vec{B}$ lies in one plane normal to the beam, and it vanishes only when the two magnitudes are the same; the right hand rule fixes the sense of $\vec{v}\times\vec{B}$, while pointing $\vec{E}$ towards the other side makes the electric and magnetic forces collinear and opposed. 🔉⇢
The balance condition is now a one-line statement. The electric force has magnitude $qE$ and the magnetic force has magnitude $qvB$ (the charge moves perpendicular to $\vec{B}$, so $\sin\theta=1$). For the particle to go straight the net transverse force must be zero, so $qE=qvB$. The charge $q$ cancels from both sides, leaving $v=E/B$. Any particle whose speed equals this ratio experiences zero net force and crosses the region in a straight line; the fields have selected it. 🔉⇢
The independence from charge and mass is the property worth dwelling on. Both the electric and the magnetic forces are proportional to $q$, so when we demand that they balance, $q$ divides out and cannot influence the selected speed. Mass never enters the balance condition at all, because the condition is a statement about forces at the instant of crossing, not about accelerations or trajectories. Even the sign of the charge does not matter: reversing the sign of $q$ reverses both the electric force and the magnetic force together, so they still cancel. A velocity selector therefore transmits positive and negative particles of the correct speed alike, filtering purely on speed. 🔉⇢
What happens to particles of the wrong speed makes the filtering vivid. A particle faster than $E/B$ has an oversized magnetic force $qvB\gt qE$, so the magnetic term wins and it is deflected in the direction of the (uncancelled) magnetic force. A particle slower than $E/B$ has $qvB\lt qE$, so the electric term wins and it is deflected the opposite way. Only the particle with $v=E/B$ threads the needle. By placing a narrow exit slit downstream, one lets only the undeflected, correctly-speeded particles emerge, discarding the rest. 🔉⇢
This is precisely why the velocity selector is the entrance stage of many mass spectrometers and of the classic experiments on charged particles. In a Bainbridge-type spectrometer, ions first pass through crossed fields that fix their speed to $v=E/B$; they then enter a region of pure magnetic field $B'$, where the magnetic force alone bends them into a circular arc of radius $r=mv/(qB')$. Because $v$ is now known and identical for every transmitted ion, measuring the radius $r$ directly yields the mass-to-charge ratio $m/q=B' r/v=B B' r/E$. Without the speed-fixing first stage, ions of the same mass but different speeds would trace different radii and the mass measurement would blur. 🔉⇢
This crossed field arrangement was used in the experiments that measured the electron, and it survives as the front end of the mass spectrometer: once the two forces balance, the momentum of every transmitted ion is the same, so a following region of pure magnetic field can determine the mass. At the selected speed the electric and magnetic forces balance and their vector sum is zero, so no energy is given to the beam; the device passes the ions that already move at $E/B$ and turns the rest aside, and the whole result rests on the relation $v=E/B$. 🔉⇢
In the presence of both fields the net force on a moving charge is the familiar Lorentz expression, and the balance principle $qE=qvB$ shows that a proton and an electron of the same speed are transmitted alike; the selector fixes the speed, not the distance or the length of the path. 🔉⇢
A concrete calculation shows how tame the numbers are. Suppose the electric field between the plates of the selector is $E=2.0\times10^{4}\ \text{V}\,\text{m}^{-1}$ and the magnetic field is $B=0.10\ \text{T}$. The transmitted speed is $v=E/B=(2.0\times10^{4})/(0.10)=2.0\times10^{5}\ \text{m}\,\text{s}^{-1}$. A particle entering at exactly this speed, whatever its charge or mass, sails straight through; one entering at $3.0\times10^{5}\ \text{m}\,\text{s}^{-1}$ finds its magnetic force too strong and is swept aside, while one at $1.0\times10^{5}\ \text{m}\,\text{s}^{-1}$ is pushed the other way by the dominant electric force. 🔉⇢
For the device to work as advertised, both fields must be genuinely uniform and truly perpendicular over the region the beam traverses, and the beam must enter along the intended axis. Fringing fields at the edges of the capacitor plates and of the magnet pole faces slightly spoil the ideal picture, so real selectors use well-shaped electrodes and pole pieces and accept only a narrow cone of entry directions through collimating slits. These are engineering refinements; the physics remains the single clean balance $qE=qvB$. 🔉⇢
It is instructive to contrast the selector with a region of magnetic field alone. In a pure magnetic field the orbit radius $r=mv/(qB)$ depends on the speed, so a magnetic field spreads a beam of mixed speeds into a fan of different radii, it does not pick out one speed. The crossed-field selector, by pitting a speed-independent electric force against a speed-proportional magnetic force, converts that speed dependence into a sharp pass-or-deflect criterion. The two devices are complementary, and a spectrometer uses them in series precisely for that reason. 🔉⇢
A subtle but important point is that $v=E/B$ is exact only for the selected particles; it is not a formula for the speed of an arbitrary particle in the beam. Every particle feels the same fields, but only those already moving at $E/B$ experience zero net force. The selector does not change a particle's speed to $E/B$; it merely lets the ones that already have that speed pass while ejecting the others. Confusing selection with acceleration is a common conceptual slip. 🔉⇢
For examinations, the safe routine is to identify the three mutually perpendicular directions (beam, $\vec{E}$, $\vec{B}$), confirm using the right-hand rule that the electric and magnetic forces are collinear and opposed, and then write the balance $qE=qvB$ to obtain $v=E/B$. State explicitly that the result is independent of $q$ and $m$ and holds for either sign of charge, and, if the problem continues into a mass spectrometer, feed the selected $v$ into $r=mv/(qB')$ to extract $m/q$. Keeping the selection step and the deflection step conceptually separate prevents the most frequent mistakes. 🔉⇢
A current is nothing but charge carriers in ordered motion, and each moving carrier feels the magnetic force $q\,\vec{v}\times\vec{B}$. Adding these microscopic forces over all the carriers in a wire gives a single macroscopic force on the conductor, which for a straight segment of length $L$ carrying current $I$ in a uniform field $\vec{B}$ is $\vec{F}=I\,\vec{L}\times\vec{B}$. As always with a cross product, the direction comes first: it is that of $\vec{L}\times\vec{B}$, read with the right-hand rule, where $\vec{L}$ is a vector of length $L$ pointing along the direction of the current. The magnitude is then $BIL\sin\theta$, with $\theta$ the angle between the wire and the field, so the force is greatest when the wire is perpendicular to $\vec{B}$ and zero when it lies along $\vec{B}$. For a wire of arbitrary shape the force is found by summing over infinitesimal elements, $\vec{F}=\int I\,d\vec{l}\times\vec{B}$. Crucially, $\vec{B}$ here is the external field, not the field the wire itself produces. 🔉⇢
The derivation from the carriers makes the origin of the formula transparent. Take a straight rod of uniform cross-sectional area $A$ and length $l$, carrying one kind of mobile carrier (electrons, in a metal) of number density $n$. The total number of mobile carriers in the rod is $nlA$. For a steady current, each carrier drifts with an average velocity $\vec{v}_d$. In an external field $\vec{B}$, each carrier of charge $q$ feels $q\,\vec{v}_d\times\vec{B}$, and since all carriers share the same average drift, the forces add coherently. 🔉⇢
Summing the identical forces over all $nlA$ carriers gives the total force $\vec{F}=(nlA)\,q\,\vec{v}_d\times\vec{B}$. This is the whole force on the rod, expressed in microscopic quantities. The next step is to recognise the familiar combination $nq\vec{v}_d$, which is the current density $\vec{j}$, and $|nq\vec{v}_d|A$, which is the current $I$. Rewriting the total force in these terms is what converts an expression about drifting electrons into one about the measurable current. 🔉⇢
Carrying out that rewrite, $\vec{F}=[(nq\vec{v}_d)\,lA]\times\vec{B}=[\,\vec{j}\,Al\,]\times\vec{B}=I\,\vec{L}\times\vec{B}$, where $\vec{L}$ is a vector of magnitude $l$ whose direction is that of the current $I$. Two subtleties deserve emphasis. First, the current $I$ itself is a scalar, not a vector; the directional information has been transferred onto $\vec{L}$, which points along the flow of (conventional, positive) current. Second, the same formula holds whether the actual carriers are positive or negative, because reversing the sign of $q$ also reverses $\vec{v}_d$ for a given current direction, leaving $I\vec{L}$ unchanged. 🔉⇢
With the vector form in hand, the magnitude and direction follow the usual cross-product rules. The magnitude is $F=BIL\sin\theta$, where $\theta$ is the angle between the wire (the direction of $\vec{L}$) and $\vec{B}$. Thus a wire carrying current perpendicular to the field ($\theta=90^{\circ}$) feels the maximum force $BIL$, while a wire lying parallel to the field ($\theta=0^{\circ}$) feels no force at all. The direction of $\vec{F}$ is that of $\vec{L}\times\vec{B}$: point the right-hand fingers along the current and curl them toward $\vec{B}$; the thumb gives the force. This direction is perpendicular to both the wire and the field. 🔉⇢
For a conductor that is not straight, or a field that varies along it, we return to the differential statement. Treat the wire as a chain of tiny straight elements $d\vec{l}$, each pointing along the local current direction; the force on one element is $d\vec{F}=I\,d\vec{l}\times\vec{B}$, and the total force is the vector sum, $\vec{F}=\int I\,d\vec{l}\times\vec{B}$, taken along the whole conductor. In most problems this sum becomes an ordinary integral. A useful consequence is that a closed current loop placed in a uniform field experiences zero net force, because $\oint d\vec{l}=\vec{0}$; such a loop feels a torque, not a translation, a fact that underlies the current loop as a magnetic dipole. 🔉⇢
It cannot be stressed too often that the $\vec{B}$ in these formulae is the external magnetic field, the field set up by other magnets or currents, and not the field produced by the current in the wire itself. A wire does create its own magnetic field encircling it, but that self-field exerts no net force on the wire as a whole; only an externally imposed field pushes the conductor. Overlooking this and trying to include the wire's own field is a classic source of error. 🔉⇢
A clean numerical illustration is the wire that hangs in mid-air. A straight wire of mass $200\ \text{g}$ and length $1.5\ \text{m}$ carries a current of $2\ \text{A}$ and is suspended, without any support, in a uniform horizontal magnetic field. For the wire to float, the upward magnetic force $BIl$ must balance its weight $mg$: $BIl=mg$, so $B=mg/(Il)$. Substituting, $B=(0.200\times9.8)/(2\times1.5)=1.96/3.0\approx0.65\ \text{T}$. Note that only the mass per unit length actually matters, and that we have safely neglected the Earth's field, which at about $4\times10^{-5}\ \text{T}$ is far too small to matter here. 🔉⇢
The orientation dependence appears sharply in another standard case. A long straight conductor lying on a horizontal table carries a steady current in the Earth's horizontal magnetic field $B$. The force per unit length is $f=F/l=IB\sin\theta$. If the current runs east-to-west while the field points south-to-north, the wire is perpendicular to the field, $\theta=90^{\circ}$, and $f=IB$ is maximal. If instead the current runs south-to-north, parallel to the field, then $\theta=0^{\circ}$ and $f=0$: the conductor feels no force at all. The same current and same field give completely different forces depending only on the angle, which is the physical content of the $\sin\theta$ factor. 🔉⇢
A short numerical case reinforces the perpendicular maximum. A $3.0\ \text{cm}$ length of wire carrying $10\ \text{A}$ is placed inside a solenoid, perpendicular to its axis, where the uniform field is $0.27\ \text{T}$. Because the wire is perpendicular to the field, $\theta=90^{\circ}$, so the force is simply $F=BIL=0.27\times10\times0.030=0.081\ \text{N}$, directed perpendicular to both the wire and the solenoid axis. Had the wire instead been aligned with the axis, parallel to the field, the force would have been zero. The example also shows why the uniform interior field of a solenoid is a convenient, well-controlled setting in which to exert a known and calculable force on a conductor. 🔉⇢
For examinations the procedure is once again direction-first. Identify the direction of conventional current to fix $\vec{L}$ (or each $d\vec{l}$); construct $\vec{L}\times\vec{B}$ with the right-hand rule to get the direction of the force; then compute the magnitude $BIL\sin\theta$, checking the perpendicular case for the maximum and the parallel case for zero. For bent or curved wires, either integrate $I\,d\vec{l}\times\vec{B}$ or, in a uniform field, exploit the shortcut that only the straight-line displacement between the endpoints matters for the net force. And never forget that $\vec{B}$ is the external field acting on the wire, not the wire's own. 🔉⇢
Two long, straight, parallel wires carrying currents each sit in the magnetic field produced by the other, and since a current-carrying conductor feels a force in an external field, each wire pushes or pulls on its neighbour. The force per unit length between them is $f=\dfrac{\mu_0 I_1 I_2}{2\pi d}$, where $I_1$ and $I_2$ are the currents, $d$ is the separation, and $\mu_0=4\pi\times10^{-7}\ \text{T}\,\text{m}\,\text{A}^{-1}$ is the permeability of free space. The direction of the force follows a simple and memorable rule: parallel currents (flowing the same way) attract, while antiparallel currents (flowing opposite ways) repel. This is exactly the reverse of electrostatics, where like charges repel. The mutual force is the physical basis of the historic definition of the ampere, and it is a direct, verified consequence of combining the field of a long wire with the Lorentz force on a current. 🔉⇢
The starting point is the field of a long straight wire. Wire $a$, carrying current $I_a$, produces at the location of a parallel wire $b$ a distance $d$ away a magnetic field of magnitude $B_a=\dfrac{\mu_0 I_a}{2\pi d}$. By the right-hand grip rule, the field lines of wire $a$ are circles around it, and at the position of wire $b$ this field is perpendicular to $b$ and lies in the plane containing the two wires. This is the external field in which wire $b$ finds itself. 🔉⇢
Now apply the force on a current-carrying conductor. Wire $b$ carries current $I_b$ over a length $L$ in the field $B_a$, so it feels a force $F_{ba}=I_b L B_a$ (the wire is perpendicular to $B_a$, so $\sin\theta=1$). Substituting the field, $F_{ba}=I_b L\cdot\dfrac{\mu_0 I_a}{2\pi d}=\dfrac{\mu_0 I_a I_b}{2\pi d}L$. Dividing by the length gives the force per unit length $f_{ba}=\dfrac{\mu_0 I_a I_b}{2\pi d}$, which is symmetric in the two currents, as it must be. 🔉⇢
By the same reasoning applied to wire $a$ in the field of wire $b$, the force on $a$ has equal magnitude and points toward $b$: $\vec{F}_{ab}=-\vec{F}_{ba}$. The two forces are equal and opposite, consistent with Newton's third law, at least for steady currents in parallel wires. (For rapidly time-varying currents the naive third law can appear to fail, but momentum is still conserved once the momentum carried by the electromagnetic field is included; this subtlety lies beyond the steady-current picture used here.) 🔉⇢
The direction, attraction versus repulsion, is worth deriving rather than merely remembering. Take both currents flowing the same way. Wire $a$'s field at wire $b$ points in a definite direction; applying $\vec{F}=I\vec{L}\times\vec{B}$ to wire $b$ with the right-hand rule gives a force pointing from $b$ toward $a$. The symmetric calculation gives a force on $a$ pointing toward $b$. Hence parallel currents attract. If one current is reversed (antiparallel currents), both forces reverse and the wires repel. The mnemonic, parallel currents attract and antiparallel currents repel, is the opposite of the electrostatic rule for like and unlike charges, and this contrast is a favourite examination point. 🔉⇢
This mutual force provided the historic, pre-2019 definition of the ampere, one of the seven SI base units. The definition, adopted in 1946, reads: the ampere is that steady current which, when maintained in each of two very long, straight, parallel conductors of negligible cross-section placed one metre apart in vacuum, produces on each conductor a force of exactly $2\times10^{-7}\ \text{N}$ per metre of length. One can check the consistency: with $I_1=I_2=1\ \text{A}$ and $d=1\ \text{m}$, $f=\dfrac{\mu_0 (1)(1)}{2\pi(1)}=\dfrac{4\pi\times10^{-7}}{2\pi}=2\times10^{-7}\ \text{N}\,\text{m}^{-1}$, exactly as the definition demands. (In the modern SI, since 2019, the ampere is instead defined by fixing the numerical value of the elementary charge, but the force relation above remains the classic operational definition and is what most textbooks state.) 🔉⇢
Once the ampere is fixed, the coulomb follows immediately: when a steady current of $1\ \text{A}$ flows, the charge passing through a cross-section in $1\ \text{s}$ is one coulomb. Thus the mechanical force between wires, something one can literally weigh, ties the electrical units back to the mechanical units of newton, metre and second. In practice a very long pair of wires is impractical, so the force is measured with multiturn coils of well-defined geometry in an instrument called a current balance, with the Earth's field and stray fields carefully eliminated. 🔉⇢
A numerical example fixes the scale of these forces. Two long parallel wires $A$ and $B$ carry currents of $8.0\ \text{A}$ and $5.0\ \text{A}$ in the same direction and are separated by $4.0\ \text{cm}=0.040\ \text{m}$. The force per unit length is $f=\dfrac{\mu_0 I_A I_B}{2\pi d}=\dfrac{(4\pi\times10^{-7})(8.0)(5.0)}{2\pi(0.040)}$. Using $\dfrac{\mu_0}{2\pi}=2\times10^{-7}\ \text{T}\,\text{m}\,\text{A}^{-1}$, this is $f=\dfrac{(2\times10^{-7})(40)}{0.040}=2\times10^{-4}\ \text{N}\,\text{m}^{-1}$. Over a $10\ \text{cm}=0.10\ \text{m}$ section the force is $F=f\times0.10=2\times10^{-5}\ \text{N}$, and since the currents are parallel the force is attractive. 🔉⇢
Two features of the result are physically instructive. The force falls off as $1/d$, more slowly than the inverse-square electrostatic force between point charges, because a long wire's field itself decays only as $1/d$. And the force is proportional to the product of the currents, so doubling either current doubles the force, while doubling both quadruples it. These scalings, together with the attract/repel rule, let one predict the qualitative behaviour of bundles of wires, busbars and coils without recomputing from scratch. 🔉⇢
There is an appealing field-line picture behind these directions. Between two wires carrying current the same way, their circular fields point in opposite senses in the gap and partially cancel, so the field is weak between the wires and stronger on the outer sides; each conductor is pushed from the strong-field region toward the weak-field region, and the wires are drawn together. For antiparallel currents the fields instead reinforce in the gap and cancel outside, the crowded field lines between the wires behave like a compressed cushion, and the wires are forced apart. This qualitative reasoning agrees exactly with the sign obtained algebraically from $\vec{F}=I\,\vec{L}\times\vec{B}$, and it is a useful cross-check under examination conditions. 🔉⇢
For examinations, the reliable sequence is: write the field of one wire at the other, $B=\mu_0 I/(2\pi d)$; apply $F=BIL$ to the second wire to get $F=\dfrac{\mu_0 I_1 I_2}{2\pi d}L$, hence $f=\dfrac{\mu_0 I_1 I_2}{2\pi d}$; and determine the direction with the right-hand rule, remembering that parallel currents attract and antiparallel currents repel. Keep the constant handy as $\dfrac{\mu_0}{2\pi}=2\times10^{-7}\ \text{T}\,\text{m}\,\text{A}^{-1}$, and if asked for the total force on a segment, multiply the per-unit-length force by the segment length. Stating the attract-or-repel conclusion explicitly, and contrasting it with the electrostatic rule, secures the conceptual marks. 🔉⇢
The Biot–Savart law is the fundamental rule that tells us how much magnetic field a single infinitesimal element of a current-carrying conductor produces at a chosen point in space. If a short element of length $\mathrm{d}l$ carries a steady current $I$, and $\hat{r}$ is the unit vector pointing from the element to the field point a distance $r$ away, then the field contribution is $\mathrm{d}\vec{B}=\dfrac{\mu_0}{4\pi}\,\dfrac{I\,\mathrm{d}\vec{l}\times\hat{r}}{r^2}$. Three physical facts are packed into this one line: the field grows in direct proportion to the current and to the element length, it weakens as the inverse square of the distance, and it carries a $\sin\theta$ angular factor through the cross product. Because separate contributions add by the principle of superposition, integrating this elemental law along a complete circuit reconstructs the entire field — which is exactly how we shall obtain the field of a long straight wire, $B=\dfrac{\mu_0 I}{2\pi a}$. 🔉⇢
The central new object here is the current element $I\,\mathrm{d}\vec{l}$. It is a vector: its magnitude is the current multiplied by the tiny length of conductor, and its direction is the direction in which positive current flows through that stretch of wire. This is a sharp point of contrast with electrostatics. There the source of the field is the electric charge, which is a scalar, and the Coulomb field points straight along the line joining source and field point. In magnetism the source $I\,\mathrm{d}\vec{l}$ is a vector, and — as the cross product makes explicit — the field it produces does not point toward or away from the element at all. No isolated current element can exist on its own the way an isolated charge can; it is always part of a closed circuit, so the Biot–Savart law is ultimately a recipe for a quantity that only becomes physically complete after integration around the loop. 🔉⇢
Read the magnitude form to see the dependences cleanly. Writing the equation out gives $\mathrm{d}B=\dfrac{\mu_0}{4\pi}\,\dfrac{I\,\mathrm{d}l\,\sin\theta}{r^2}$, where $\theta$ is the angle between the element $\mathrm{d}\vec{l}$ and the displacement vector $\vec{r}$ drawn from the element to the point. The field is largest when the point lies off to the side of the element, where $\theta=90^\circ$ and $\sin\theta=1$. Crucially, the field vanishes along the line of the element itself: when the point lies straight ahead of or straight behind the current, $\theta=0$, $\sin\theta=0$, and $\mathrm{d}B=0$. This angular dependence has no counterpart in Coulomb's law and is one of the most heavily tested subtleties of the law in problems, because it means a straight segment contributes nothing to the field at points lying on its own extended line. 🔉⇢
The direction of $\mathrm{d}\vec{B}$ is fixed entirely by the cross product $\mathrm{d}\vec{l}\times\hat{r}$: the field is perpendicular to the plane that contains both the current element and the displacement vector. To read off which way it points, use the right-hand screw rule. Look at the plane containing $\mathrm{d}\vec{l}$ and $\vec{r}$ and rotate the first vector toward the second; if that rotation is anticlockwise as you view it, $\mathrm{d}\vec{B}$ points out toward you, and if clockwise, it points away. For a long straight wire this collapses into the familiar right-hand thumb rule: point the thumb along the current and the curled fingers trace the direction of the circular field lines that encircle the wire. 🔉⇢
The constant of proportionality $\dfrac{\mu_0}{4\pi}$ has, in SI units, the exact value $10^{-7}\ \mathrm{T\,m\,A^{-1}}$, so that $\mu_0=4\pi\times10^{-7}\ \mathrm{T\,m\,A^{-1}}$ is the permeability of free space. This exact numerical value is a great convenience in problem work, because $\dfrac{\mu_0}{4\pi}$ can be substituted directly as $10^{-7}$. The permeability is also tied to the other fundamental constants of electromagnetism through the relation $\varepsilon_0\mu_0=1/c^2$, where $\varepsilon_0$ is the permittivity of free space and $c$ is the speed of light in vacuum. Fixing $\mu_0$ therefore fixes $\varepsilon_0$, and the appearance of $c$ foreshadows the deep unity of electricity, magnetism, and light developed in the study of electromagnetic waves. 🔉⇢
It is worth setting the Biot–Savart law beside Coulomb's law to see what the two share and where they part company. Both are long-range laws that fall off as the inverse square of distance, and both obey the principle of superposition, so complicated sources can be handled by adding elemental contributions. But the differences are decisive. Coulomb's source is a scalar charge; the Biot–Savart source is the vector $I\,\mathrm{d}\vec{l}$. The electrostatic field lies along the line joining source and field point; the magnetic field lies perpendicular to the plane of $\vec{r}$ and $\mathrm{d}\vec{l}$. And the magnetic law carries the $\sin\theta$ angular factor that the electrostatic law entirely lacks. These structural differences are why magnetic field lines close on themselves rather than beginning and ending on sources. 🔉⇢
To turn the elemental law into something measurable we integrate it over a real conductor, and the archetype is the infinitely long straight wire. The result of that integration, carried out in the accompanying derivation, is a strikingly simple field whose magnitude at a perpendicular distance $a$ from the wire is $B=\dfrac{\mu_0 I}{2\pi a}$. Notice that the field of the extended wire falls off only as $1/a$, the first power of distance, even though each individual element contributed a $1/r^2$ term; the slower decay emerges from summing the contributions of the whole infinite line. This is a recurring theme in magnetostatics: the geometry of the source reshapes how the field decays with distance. 🔉⇢
The field lines of the straight wire are concentric circles lying in planes perpendicular to the wire, centred on the wire. Their sense is given by the right-hand thumb rule described earlier, and their spacing widens with distance because the magnitude drops as $1/a$. This circular, source-encircling pattern is qualitatively different from the radial, charge-centred pattern of an electric field, and it is the visual signature of the fact that magnetic field lines never terminate. The same integration machinery, applied to a bent conductor, yields the field of a circular loop on its axis, and applied to many stacked loops, the nearly uniform field of a solenoid — so the straight-wire result is the first member of a family. 🔉⇢
For examinations the Biot–Savart law is valuable both as a computational tool and as a source of quick qualitative judgements. Recognising that a straight segment contributes nothing along its own line, that the field of a finite arc depends on the angle it subtends at the centre, and that contributions from symmetric pieces of a loop can cancel, often lets one write down an answer with almost no algebra. When a full calculation is unavoidable, the standard route is always the same: choose the element, write $\mathrm{d}B$ with the correct $\sin\theta$, identify the direction from the cross product, exploit any symmetry to discard cancelling components, and finally integrate over the geometry. Mastery of this single procedure underlies essentially every steady-current field problem you will meet. 🔉⇢
On the axis of a circular loop of radius $R$ carrying a steady current $I$, the magnetic field points along the axis and has magnitude $B=\dfrac{\mu_0 I R^2}{2\big(R^2+x^2\big)^{3/2}}$, where $x$ is the distance from the centre of the loop measured along the axis. This one formula contains two important limits. At the centre, where $x=0$, it collapses to $B_0=\dfrac{\mu_0 I}{2R}$. Far away, where $x\gg R$, it falls as $1/x^3$ and reproduces exactly the field of a magnetic dipole of moment $m=I\pi R^2$. The result is obtained by summing Biot–Savart contributions around the loop and using the loop's symmetry to cancel every component perpendicular to the axis, so that only the axial part survives. 🔉⇢
Set up the calculation as follows. Place the loop in a plane with its centre at the origin $O$ and let the $x$-axis be the axis of the loop. The field point $P$ lies on this axis at distance $x$ from $O$. Take a small conducting element $\mathrm{d}\vec{l}$ of the loop; the displacement $\vec{r}$ from this element to $P$ has magnitude $r=\sqrt{R^2+x^2}$, the same for every element by symmetry. A key geometric fact is that the element $\mathrm{d}\vec{l}$, which lies in the plane of the loop, is always perpendicular to $\vec{r}$, so $|\mathrm{d}\vec{l}\times\vec{r}|=r\,\mathrm{d}l$ and the Biot–Savart magnitude simplifies to $\mathrm{d}B=\dfrac{\mu_0}{4\pi}\dfrac{I\,\mathrm{d}l}{R^2+x^2}$. 🔉⇢
Now consider the direction of each $\mathrm{d}\vec{B}$. It is perpendicular to the plane formed by $\mathrm{d}\vec{l}$ and $\vec{r}$, and as we move around the loop this direction sweeps around a cone about the axis. Resolve each $\mathrm{d}\vec{B}$ into a component $\mathrm{d}B_x$ along the axis and a component $\mathrm{d}B_\perp$ perpendicular to it. Here symmetry does the decisive work: for every element there is a diametrically opposite element whose perpendicular contribution is equal in magnitude but exactly opposite in direction. Summed around the loop, all the perpendicular components cancel to zero, and only the axial components add up. This cancellation is the single most important idea in the derivation, and recognising it saves an enormous amount of algebra. 🔉⇢
The axial component follows from a small piece of geometry. The angle $\theta$ between $\vec{r}$ and the axis satisfies $\cos\theta=\dfrac{R}{\big(R^2+x^2\big)^{1/2}}$, and the surviving contribution of each element is $\mathrm{d}B_x=\mathrm{d}B\,\cos\theta=\dfrac{\mu_0 I}{4\pi}\dfrac{R\,\mathrm{d}l}{\big(R^2+x^2\big)^{3/2}}$. Everything multiplying $\mathrm{d}l$ in this expression is the same for every element of the loop, so the integration reduces to summing $\mathrm{d}l$ around the loop, which simply gives its circumference $2\pi R$. Carrying out that sum yields the axial field quoted at the start, directed along the axis with a sense fixed by the right-hand rule. 🔉⇢
The special case $x=0$ deserves to be memorised on its own: at the centre of the loop $B_0=\dfrac{\mu_0 I}{2R}$. For a tightly wound coil of $N$ identical turns the field is simply $N$ times larger, because each turn contributes the same field, giving $B=\dfrac{\mu_0 N I R^2}{2\big(R^2+x^2\big)^{3/2}}$ on the axis and $B_0=\dfrac{\mu_0 N I}{2R}$ at the centre. It is worth noting a clean intermediate result: at the special point $x=R$, one axial radius from the centre, the field is $B=\dfrac{B_0}{2\sqrt{2}}$, a fact that makes a good quick check on any numerical answer. 🔉⇢
It helps to picture how the axial field varies as the point $P$ slides along the axis. The magnitude is greatest at the centre, $x=0$, where it equals $B_0=\dfrac{\mu_0 I}{2R}$, and it is symmetric about the centre, falling off in the same way whether $P$ moves to positive or negative $x$. Near the centre the field is broad and flat, changing only slowly with $x$; this gentle plateau is what the Helmholtz arrangement of two coaxial coils, separated by one radius, exploits to create a region of nearly uniform field. Far from the loop the curve steepens into the $1/x^3$ dipole tail. Because the profile has no kink or discontinuity, a single continuous expression describes the field everywhere on the axis, from the centre to infinity. 🔉⇢
The direction of the field is given by a right-hand thumb rule adapted to loops: curl the fingers of the right hand around the wire in the direction of the current, and the extended thumb points along the field on the axis. Equivalently, viewed from one face the current appears to circulate anticlockwise and that face behaves as a north pole, while the opposite face behaves as a south pole. In this sense a single current loop is the elementary magnet: its two faces play the roles of the two poles, and the whole external field pattern is that of a small bar magnet or, more precisely, a magnetic dipole. Stacking many such loops side by side along a common axis is exactly how a solenoid is built, and adding their axial fields is how the solenoid's strong, nearly uniform interior field arises. 🔉⇢
The dipole identification becomes exact in the far field. When $x\gg R$ we may neglect $R^2$ beside $x^2$ in the denominator, so $\big(R^2+x^2\big)^{3/2}\approx x^3$ and the field becomes $B\approx\dfrac{\mu_0 I R^2}{2x^3}$. Introduce the magnetic moment of the loop, $m=I A=I\pi R^2$, and rewrite this as $B\approx\dfrac{\mu_0}{4\pi}\dfrac{2m}{x^3}$. This is precisely the on-axis field of a magnetic dipole of moment $m$, and it has exactly the same form as the on-axis field of an electric dipole, $E=\dfrac{1}{4\pi\varepsilon_0}\dfrac{2p}{x^3}$, with $m$ playing the role of the electric dipole moment $p$ and $\mu_0$ replacing $1/\varepsilon_0$. 🔉⇢
This correspondence is conceptually rich. It tells us that from far enough away the detailed size and shape of the loop are invisible; all that survives is the single number $m=I\pi R^2$, the magnetic dipole moment, which acts as the effective source. The same moment governs the torque $\vec{\tau}=\vec{m}\times\vec{B}$ that an external field exerts on the loop and the energy of the loop in that field, tying the axial-field result to the mechanics of current loops treated elsewhere. The loop is thus the bridge between the microscopic Biot–Savart law and the macroscopic language of magnetic dipoles and materials. 🔉⇢
For examination purposes the axial-loop result is a workhorse. Problems commonly ask for the field at the centre, at a stated axial distance, at the midpoint between two coaxial coils (the Helmholtz arrangement, where the field is made especially uniform), or for the ratio of axial to central field. All of these follow from the single formula $B=\dfrac{\mu_0 N I R^2}{2\big(R^2+x^2\big)^{3/2}}$ together with the two limits. The reliable method is always to reduce the geometry to $R$ and $x$, apply the formula, and sanity-check against the centre value $\dfrac{\mu_0 N I}{2R}$ and the far-field $1/x^3$ dipole trend. Because the perpendicular components have already been cancelled once and for all in the derivation, no vector bookkeeping is needed at the point of use. 🔉⇢
A moving coil galvanometer is a coil of $N$ closely wound turns, free to rotate about a fixed axis in a uniform radial magnetic field, arranged so that the current through it produces a steady angular deflection that can be read on a scale. When current $I$ flows, the field exerts a deflecting torque of magnitude $\tau=NIAB$, where $A$ is the area of the coil and $B$ the field strength; a fine spring supplies a restoring torque $k\phi$ that grows with the twist $\phi$, and the pointer settles where the two balance. Setting $k\phi=NIAB$ gives the working relation $\phi=\Big(\dfrac{NAB}{k}\Big)I$, so the deflection is directly proportional to the current. From this single equation follow the instrument's current sensitivity $\dfrac{NAB}{k}$, its voltage sensitivity $\dfrac{NAB}{kR}$, and the rules for converting it into an ammeter with a small shunt or a voltmeter with a large series resistance. 🔉⇢
The construction is designed around one goal: a scale that reads linearly in current. A rectangular coil of many turns is wound on a light frame and pivoted so it can rotate between the poles of a permanent magnet. A cylindrical soft-iron core is fixed inside the coil. This core does two jobs. First, it concentrates and strengthens the field, so a given current gives a larger torque. Second, and more importantly, together with suitably shaped pole pieces it makes the field radial — the field lines everywhere point toward the axis — so that whatever the coil's angular position, the plane of the coil always lies parallel to the local field. Fine hairsprings at top and bottom both supply the restoring torque and serve as leads carrying current into and out of the coil, and a light pointer attached to the coil moves over a graduated scale. 🔉⇢
The role of the radial field is subtle and worth dwelling on. In a general uniform field the torque on a coil of magnetic moment $m=NIA$ is $\tau=NIAB\sin\theta$, where $\theta$ is the angle between the coil's normal and the field; because $\sin\theta$ changes as the coil turns, the scale would be badly non-linear. The radial geometry removes this problem: by design the field is always in the plane of the coil, so the angle between the field and the plane is zero, the coil normal is always perpendicular to $B$, and $\sin\theta=1$ at every deflection. The torque therefore reduces to the constant-coefficient form $\tau=NIAB$ independent of $\phi$, which is exactly what makes the equilibrium condition $k\phi=NIAB$ give a deflection strictly proportional to current and hence a uniform, evenly divided scale. 🔉⇢
The balance of torques is the heart of the instrument. The deflecting torque is $\tau=NIAB$; the derivation of this from the general vector law $\vec{\tau}=\vec{m}\times\vec{B}$ (with $\vec{m}=NI\vec{A}$) belongs to the treatment of torque on a current loop and is not repeated here — in the galvanometer we simply use its radial-field value with $\sin\theta=1$. Against this the spring exerts a restoring torque $k\phi$, where $k$ is the torsional constant, the restoring torque per unit angular twist. At the steady deflection the two are equal, $k\phi=NIAB$, and solving for the deflection gives $\phi=\dfrac{NAB}{k}\,I$. The quantity $\dfrac{NAB}{k}$ in the bracket is a fixed constant for a given galvanometer, so reading $\phi$ off the scale is equivalent to reading the current. 🔉⇢
The current sensitivity is defined as the deflection produced per unit current, $\dfrac{\phi}{I}=\dfrac{NAB}{k}$. A galvanometer is made more sensitive by increasing the number of turns $N$, the coil area $A$, or the field $B$, or by using a softer spring with a smaller torsional constant $k$. Increasing $N$ is the manufacturer's most convenient lever. Real galvanometers are extremely sensitive, giving a full-scale deflection for currents of only the order of microamperes, which is precisely why they must be modified before they can measure the far larger currents and voltages found in ordinary circuits. 🔉⇢
The voltage sensitivity is the deflection per unit voltage across the instrument. Since the current through a galvanometer of resistance $R$ is $I=V/R$, we have $\dfrac{\phi}{V}=\dfrac{NAB}{k}\dfrac{1}{R}$. A striking consequence is that improving current sensitivity does not automatically improve voltage sensitivity. Suppose we double the number of turns, $N\to 2N$; the current sensitivity $\dfrac{NAB}{k}$ doubles, but the coil now uses twice the length of wire, so its resistance also roughly doubles, $R\to 2R$. In the ratio $\dfrac{NAB}{kR}$ the two factors of two cancel, and the voltage sensitivity is essentially unchanged. This is why the modifications required to build a good ammeter differ from those required to build a good voltmeter. 🔉⇢
A galvanometer cannot be used directly as an ammeter for two reasons. It is far too sensitive, running off the scale at currents that ordinary circuits carry easily, and it has a comparatively large resistance, so inserting it in series would appreciably reduce the very current it is meant to measure. An ideal ammeter should have zero resistance and be able to carry the full circuit current. The fix is to connect a small resistance $r_s$, called a shunt, in parallel with the coil, so that the bulk of the current bypasses the sensitive coil through the shunt while a small, fixed fraction passes through the galvanometer. The parallel combination has resistance $\dfrac{R_G r_s}{R_G+r_s}\approx r_s$ when $R_G\gg r_s$, which is small enough to leave the circuit current almost undisturbed. 🔉⇢
To design the shunt, note that the coil and the shunt are in parallel and therefore share the same voltage. If $I_g$ is the current that gives full-scale deflection and $R_G$ is the coil resistance, then when the total current to be measured is $I$, the shunt carries $I-I_g$, and equating the voltages gives $I_g R_G=(I-I_g)\,r_s$, so $r_s=\dfrac{I_g R_G}{I-I_g}$. Because $r_s$ turns out to be very small, the combined meter resistance is tiny and the scale can be re-graduated to read the full current directly. The larger the range $I$ we want, the smaller the shunt required, so a range switch is really a bank of different shunts. 🔉⇢
The same galvanometer becomes a voltmeter when a large resistance $R$ is connected in series with the coil and the combination is placed in parallel with the circuit element whose voltage is wanted. An ideal voltmeter has infinite resistance so that it draws no current and does not disturb the circuit; the large series resistance approximates this, making the meter's resistance $R_G+R\approx R$ large and limiting the current it draws to a small value. For a full-scale current $I_g$ and desired full-scale voltage $V$, the series resistance follows from $V=I_g\,(R_G+R)$, giving $R=\dfrac{V}{I_g}-R_G$. Thus small shunts in parallel turn the galvanometer into a low-resistance ammeter, while large resistances in series turn it into a high-resistance voltmeter — opposite modifications for opposite purposes. 🔉⇢
For examinations the galvanometer is a compact source of standard questions: state the working equation $\phi=\dfrac{NAB}{k}I$ and explain why the radial field makes the scale linear; distinguish current sensitivity $\dfrac{NAB}{k}$ from voltage sensitivity $\dfrac{NAB}{kR}$ and explain why doubling $N$ helps the first but not the second; and design a shunt or series resistance for a stated range. The reliable approach is always to begin from the parallel-voltage condition for a shunt or the series-current condition for a multiplier, keep track of which resistance is small and which is large, and remember that an ideal ammeter has zero resistance while an ideal voltmeter has infinite resistance. 🔉⇢
🔬 Interactive 3D · A charge spirals between the dees with B into the page — change the speed and watch the radius grow while the revolution frequency stays fixed. speed v, magnetic field B, charge-to-mass ratio q/m
When a charge $q$ moves with velocity $\vec{v}$ through a magnetic field $\vec{B}$, it feels the magnetic part of the Lorentz force, $\vec{F}=q\,\vec{v}\times\vec{B}$. Because this force is built from a cross product, it points sideways -- always at right angles to both $\vec{v}$ and $\vec{B}$ -- and never along the line of motion. That single geometric fact drives everything in this topic. Since $\vec{F}$ is perpendicular to $\vec{v}$, it can do no work, so the particle's speed and kinetic energy stay exactly constant; only the direction of the velocity turns. A constant-magnitude force that continually turns the velocity without changing its length is precisely a centripetal force, so a charge launched perpendicular to a uniform field travels in a circle of radius $r=\dfrac{mv}{qB}$. Remarkably, the time for one loop, $T=\dfrac{2\pi m}{qB}$, and the corresponding frequency $f=\dfrac{qB}{2\pi m}$ do not depend on the speed at all: a faster particle simply sweeps a proportionally bigger circle in the same time. If the velocity also has a component along $\vec{B}$, that component sails through untouched and the motion becomes a helix whose pitch is $p=v_{\parallel}T$. The speed-independence of the period is the working principle of the cyclotron, a machine that accelerates ions to high energy using a fixed-frequency oscillator tuned to the resonance condition $f=\dfrac{qB}{2\pi m}$. 🔉⇢
Start from the full Lorentz force on a point charge $q$ located at $\vec{r}$ and moving with velocity $\vec{v}$ in the presence of an electric field $\vec{E}$ and a magnetic field $\vec{B}$: $\vec{F}=q\left[\vec{E}+\vec{v}\times\vec{B}\right]$. The electric part, $q\vec{E}$, is familiar and can point along the motion, so it changes speed. In this topic we set $\vec{E}=\vec{0}$ and isolate the purely magnetic force, $\vec{F}=q\,\vec{v}\times\vec{B}$. Before worrying about how big this force is, you must know which way it points, because the geometry is what makes magnetic motion so different from projectile motion under gravity. The vector product $\vec{v}\times\vec{B}$ is, by definition, perpendicular to the plane that contains $\vec{v}$ and $\vec{B}$. To find its direction use the right-hand rule: point the fingers of your right hand along $\vec{v}$, curl them towards $\vec{B}$ through the smaller angle, and the thumb points along $\vec{v}\times\vec{B}$. That thumb direction is the force on a positive charge. For a negative charge such as an electron, the sign of $q$ flips the result, so the force is exactly opposite. As a worked instance, if $\vec{v}$ points along $+x$ and $\vec{B}$ along $+y$, then $\vec{v}\times\vec{B}$ points along $+z$; a proton is pushed along $+z$ and an electron along $-z$. 🔉⇢
It helps to see this magnetic force as one part of the electromagnetic interaction between electricity and magnetism. Writing the two contributions as vectors, the electric force $q\vec{E}$ and the magnetic force $q\,\vec{v}\times\vec{B}$ add vectorially, and only the magnetic term depends on the velocity. The link between a current and the magnetic field it sets up was found by Oersted and measured by Ampere, so the field $\vec{B}$ that turns our charge is in turn produced by other moving charges -- a current in a straight wire, a solenoid, or a bar magnet. Because the force is measured in newtons and the field in tesla, the equation $F=qvB$ also fixes the units of the field. The direction in which the charge is deflected follows from the right hand rule, and reversing the field, or the sign of the charge, reverses the deflection; this directional rule is the very one used to explain the deflection of a compass needle placed near a current. 🔉⇢
Only once the direction is settled do we quantify the magnitude. Writing $\theta$ for the angle between $\vec{v}$ and $\vec{B}$, the size of the magnetic force is $F=qvB\sin\theta$, and in vector form $\vec{F}=qvB\sin\theta\,\hat{n}$, where $\hat{n}$ is the unit vector given by the right-hand rule. Two limits are worth memorising. When $\vec{v}$ is parallel or anti-parallel to $\vec{B}$, $\theta=0^\circ$ or $180^\circ$, $\sin\theta=0$, and the magnetic force vanishes entirely: a charge fired straight along the field lines feels nothing and moves in a straight line at constant speed. When $\vec{v}$ is perpendicular to $\vec{B}$, $\theta=90^\circ$, $\sin\theta=1$, and the force takes its maximum value $F=qvB$. A third fact follows directly from the formula: if the charge is at rest, $v=0$, so a stationary charge feels no magnetic force at all -- unlike the electric case, where even a motionless charge is pushed. This is also how the unit of $\vec{B}$ is defined. Dimensionally $[B]=[F/qv]$, so one tesla ($\text{T}$) is the field that exerts one newton on a charge of one coulomb moving at one metre per second perpendicular to the field. The tesla is a large unit; the gauss, equal to $10^{-4}\ \text{T}$, is common for weak fields, and the Earth's field is only about $3.6\times10^{-5}\ \text{T}$. 🔉⇢
The most important consequence of the perpendicularity is that a magnetic force does no work on the charge. Work is transferred only by a force component along (or against) the direction of motion; the rate of doing work is the power $P=\vec{F}\cdot\vec{v}$. But $\vec{F}=q\,\vec{v}\times\vec{B}$ is, by construction, perpendicular to $\vec{v}$, so their dot product is identically zero: $P=q(\vec{v}\times\vec{B})\cdot\vec{v}=0$ at every instant, because $\vec{v}\times\vec{B}$ can never have a component along $\vec{v}$. Zero power means zero work over any interval, and by the work-energy theorem the kinetic energy $\tfrac{1}{2}mv^2$ cannot change. Therefore the magnitude of the velocity -- the speed -- is a constant of the motion. What the force does change, continuously, is the direction of $\vec{v}$: it delivers momentum sideways while never adding to the energy budget. This is why a magnetic field can steer a charged beam, bend it, or trap it in a loop, yet can never, by itself, speed it up or slow it down. Any energy gain in a real accelerator must come from an electric field acting somewhere in the cycle; the magnetic field only shapes the trajectory. 🔉⇢
Now specialise to the cleanest case, a uniform field $\vec{B}$ with the velocity lying entirely in the plane perpendicular to it. We have just shown two things: the speed $v$ is constant, and the force $\vec{F}=q\,\vec{v}\times\vec{B}$ is always perpendicular to $\vec{v}$. Its magnitude, $F=qvB$, is therefore also constant. A force of fixed magnitude that stays permanently at right angles to the velocity is exactly the definition of a centripetal force: it never changes the speed, only bends the path, and it does so at a uniform rate. The natural response of a particle to such a force is uniform circular motion. Geometrically, the velocity vector rotates at a steady rate while its tip traces a circle, and the force always points from the particle towards a fixed centre. This is why the trajectory closes on itself into a perfect circle when $\vec{v}\perp\vec{B}$, rather than an ellipse or a spiral. Contrast this with a projectile in gravity, where the force has a fixed direction (downward) and a component along the motion, giving a parabola and a changing speed; here the force direction rotates with the particle and stays forever perpendicular, giving a circle at unchanging speed. 🔉⇢
Equating the required centripetal force $\dfrac{mv^2}{r}$ to the magnetic force $qvB$ and solving gives the radius $r=\dfrac{mv}{qB}=\dfrac{p}{qB}$, where $p=mv$ is the linear momentum. Reading this formula physically: the radius is proportional to momentum and inversely proportional to the field. A more energetic (faster or heavier) particle is harder to turn and so sweeps a larger circle; a stronger field turns the particle more sharply and shrinks the circle. This single relation underlies a family of instruments. In a mass spectrometer, ions of the same charge and energy separate by radius according to their mass, because $r\propto m$ for fixed $qB$ and speed. In a bubble chamber or cloud chamber, the curvature $1/r$ of a track in a known field reveals the particle's momentum and, from the sense of the curve, the sign of its charge. The same expression also tells you the field needed to hold a beam of given momentum on a ring of chosen radius, which is how the bending magnets of a synchrotron are specified. 🔉⇢
In this way the bending of a charged track becomes a clear measure of the momentum, and the same physics is used across many instruments: the mass spectrometer, the cyclotron, and the deflection of a beam in a magnetic region are all applications of the one relation $r=mv/(qB)$. These results describe a real phenomenon that experiments show directly: when a beam of electrons or protons moves through a magnetic region, its path can be observed, and the period stays fixed while the radius grows with the speed. The whole discussion can be summed up in a few equations that describe the motion; each is related to the single magnetic force, and the transverse push it gives never adds energy, so the speed cannot vanish or grow. The same behaviour is seen experimentally whenever charged particles move through the field of a magnet or a solenoid. 🔉⇢
The truly surprising result appears when we ask how long one revolution takes. The angular frequency is $\omega=\dfrac{v}{r}$, and substituting $r=\dfrac{mv}{qB}$ makes the speed cancel: $\omega=\dfrac{qB}{m}$. Hence the period $T=\dfrac{2\pi}{\omega}=\dfrac{2\pi m}{qB}$ and the frequency $f=\dfrac{qB}{2\pi m}$ depend only on the charge-to-mass ratio $q/m$ and on the field $B$ -- and not at all on the particle's speed, energy, or the radius of its orbit. The reason is a clean cancellation of two effects. A faster particle needs a bigger circle, and $r$ grows in exact proportion to $v$; the extra path length it must cover is precisely matched by its extra speed, so the time to go around stays the same. A slow particle crawls around a tight little circle in the same period that a fast one takes to race around a huge one. This constancy, $f=\dfrac{qB}{2\pi m}$, is called the cyclotron frequency, and its independence from energy is not a curiosity but the enabling fact behind an entire class of particle accelerators. 🔉⇢
Real beams are rarely launched exactly perpendicular to the field, so consider a general velocity making some angle with $\vec{B}$. Resolve it into a component $v_{\perp}$ perpendicular to $\vec{B}$ and a component $v_{\parallel}$ along $\vec{B}$. The cross product $\vec{v}\times\vec{B}$ involves only the perpendicular part, because the parallel part is anti-parallel to $\vec{B}$ and contributes nothing. Consequently the magnetic force acts entirely within the plane perpendicular to $\vec{B}$ and has no component along the field. The parallel motion is therefore completely unaffected -- the particle drifts along $\vec{B}$ at the steady speed $v_{\parallel}$ -- while the perpendicular motion is the uniform circle analysed above, now of radius $r=\dfrac{mv_{\perp}}{qB}$. Superposing a steady straight drift on a circle produces a helix wound around the field lines. The distance the particle advances along the field during one full revolution is called the pitch, $p=v_{\parallel}T=\dfrac{2\pi m\,v_{\parallel}}{qB}$. The radius of the circular part is the radius of the helix. When $v_{\parallel}=0$ the helix collapses to a flat circle; when $v_{\perp}=0$ it stretches into a straight line. This helical guiding of charges along field lines is exactly how charged particles from the Sun spiral down the Earth's field to produce the auroras. Notice too that the sense in which the helix winds -- clockwise or anticlockwise when viewed along $\vec{B}$ -- is fixed by the sign of the charge through the right-hand rule, so electrons and protons entering the same field with the same geometry coil in opposite senses. A useful sanity check for any oblique-entry problem is that as the entry angle shrinks towards zero the pitch grows without bound while the radius shrinks to zero, smoothly recovering the straight-line limit, and as the angle approaches $90^\circ$ the pitch tends to zero and the helix closes into the flat circle. 🔉⇢
The cyclotron turns the speed-independent period into a practical accelerator. Two hollow, D-shaped metal electrodes called dees are placed with a narrow gap between their straight edges, inside a uniform magnetic field perpendicular to their flat faces, and the whole assembly sits in a vacuum. A high-frequency alternating voltage from an oscillator is applied across the gap. Ions injected near the centre are accelerated across the gap by the electric field there, gaining energy each crossing. Once inside a dee the electric field is screened out (the dee is a hollow conductor), so the ion feels only the magnetic field and coasts along a semicircle at constant speed, taking a time $\tfrac{1}{2}T=\dfrac{\pi m}{qB}$ that is the same for every semicircle regardless of how fast the ion is now moving. The clever part is timing: because the half-period is fixed, the oscillator can flip the gap voltage at a constant rate and still catch the ion at the gap for a forward kick on every single crossing. The matching condition, that the oscillator frequency equals the cyclotron frequency, is the resonance condition $f_{\text{osc}}=f=\dfrac{qB}{2\pi m}$. Each crossing adds energy, so the ion spirals outward through ever larger semicircles at ever higher speed while its period stays put, until at the outer edge, radius $R$, it has kinetic energy $K=\dfrac{q^2B^2R^2}{2m}$ and is extracted as a beam. The scheme finally fails at very high energy: as the ion approaches relativistic speeds its effective mass increases, so its true revolution period lengthens, the ion arrives late at the gap, and it slips out of step with the fixed-frequency oscillator. This loss of resonance is why the simple cyclotron is limited to modest energies, and why relativistic machines instead sweep the frequency (the synchrocyclotron) or ramp the field (the synchrotron) to keep the timing matched. 🔉⇢
For problem solving, a compact checklist follows from all of the above. First decide the plane of the motion: a velocity purely perpendicular to $\vec{B}$ gives a circle, a velocity purely along $\vec{B}$ gives a straight line, and anything in between gives a helix -- so always resolve $\vec{v}$ into $v_{\perp}$ and $v_{\parallel}$ before reaching for a formula. Use $r=\dfrac{mv_{\perp}}{qB}$ for the radius, remembering it is the perpendicular speed that sets the radius. Reach for $T=\dfrac{2\pi m}{qB}$ and $f=\dfrac{qB}{2\pi m}$ whenever a question asks about time, frequency, or the pitch $p=v_{\parallel}T$, and recall these never contain the speed. When kinetic energy is involved, get it from the geometry: at the rim of a cyclotron $K=\dfrac{q^2B^2R^2}{2m}$. Finally, keep the energetics straight -- the magnetic field does no work, so any change in speed in a combined-field problem must be pinned entirely on the electric field. Holding these few relations, together with a firm grip on the right-hand rule for direction, resolves the great majority of JEE questions on this section without any further memorisation. Each result above can be written as a compact equation, and every one is a special case of the single vector relation $\vec{F}=q\,\vec{v}\times\vec{B}$; understanding that one expression, together with the right hand rule for its direction, is enough to give the radius, the period, and the pitch. As in the worked example and the figure for this section, a question is worked through by first choosing the plane of the motion and then applying the relation for the radius, just as in the exercises at the end of the chapter. 🔉⇢
Source: JEE-pattern
🔬 Interactive 3D · Current in the windings threads a field through the core — raise the turns per unit length and watch the interior field become uniform while the outside collapses. turns per unit length n, current I
Ampere's circuital law states that the line integral of the magnetic field around any closed loop equals the permeability of free space times the net steady current that threads the loop: $\oint \vec{B}\cdot d\vec{l}=\mu_0 I_{\text{enc}}$. In words, if you walk once around a closed path and, at every step, add up the component of $\vec{B}$ along your direction of travel, the total you accumulate is fixed entirely by the current passing through any surface bounded by that path, and by nothing else. This is the exact magnetic counterpart of Gauss's law $\oint \vec{E}\cdot d\vec{A}=q_{\text{enc}}/\varepsilon_0$, which ties the flux of $\vec{E}$ through a closed surface to the charge inside. Both laws are always true for their respective steady sources, but, and this is the point a student must internalise, they hand you the field only when the geometry is symmetric enough to pull the field out of the integral. Where that symmetry is absent, the law remains perfectly correct yet completely useless for finding $\vec{B}$. 🔉⇢
To state the law precisely, imagine an open surface with a boundary curve $C$. Break the boundary into small vector elements $d\vec{l}$, take at each the tangential component $B_t$ of the field, multiply by the length $dl$, and add these products all the way round; in the limit of vanishingly small elements the sum becomes the closed-loop integral $\oint \vec{B}\cdot d\vec{l}$. Ampere's law asserts this integral equals $\mu_0$ times the total current crossing the surface. A sign convention fixes what counts as positive current: curl the fingers of the right hand in the sense you traverse the loop, and the thumb points in the direction of current taken as positive. Two cautions matter for problems. First, $I_{\text{enc}}$ is the net current, so a wire carrying current one way and another carrying it the opposite way through the same loop partly or wholly cancel. Second, the law in this form holds only for steady currents that do not change with time; the situation must be magnetostatic. 🔉⇢
The parallel with Gauss's law runs deep and is worth dwelling on. Gauss's law relates a quantity on a boundary, the flux of the electric field through a closed surface, to a source in the interior, the enclosed charge. Ampere's law relates a quantity on a boundary, the circulation of the magnetic field around a closed loop, to a source threading the interior, the enclosed current. Just as Gauss's law contains no more physics than Coulomb's law but repackages it in a form that trivialises symmetric problems, Ampere's law contains no more physics than the Biot-Savart law; it merely repackages the same content so that highly symmetric current distributions become almost effortless. And just as Gauss's law applied to a lopsided charge distribution gives you one true scalar equation yet cannot resolve how $\vec{E}$ varies over the surface, Ampere's law applied to a lopsided current gives you one true equation and leaves the point-by-point behaviour of $\vec{B}$ undetermined. 🔉⇢
When does the law actually deliver $\vec{B}$? Only when you can find an Amperian loop along which, at every point, one of three things is true: (i) $\vec{B}$ is tangent to the loop and has the same constant magnitude $B$, or (ii) $\vec{B}$ is perpendicular to the loop so that $\vec{B}\cdot d\vec{l}=0$, or (iii) $\vec{B}$ is zero. On the stretches where (i) holds, $\vec{B}\cdot d\vec{l}=B\,dl$ and $B$ factors out of the integral; on the stretches where (ii) or (iii) holds, the contribution is simply zero. If $L$ is the total length of loop over which $\vec{B}$ is tangential and constant, the whole law collapses to the beautifully simple $BL=\mu_0 I_{\text{enc}}$. Everything then reduces to reading off $L$ and $I_{\text{enc}}$ from the geometry. Notice what has really happened: the single unknown scalar $B$ has been pulled outside the integral, so one equation now suffices to fix it. That is only legitimate when $B$ is genuinely constant along the chosen stretch, and the constancy is not something we assume for convenience, it is forced on us by the symmetry of the source. The entire art of applying Ampere's law is choosing a loop that meets one of these three conditions everywhere, and that is possible only when the current distribution itself has a matching symmetry. 🔉⇢
The cleanest example is the infinitely long, straight wire carrying steady current $I$. By symmetry the field can depend only on the perpendicular distance $r$ from the wire, since the problem looks identical if you slide along the wire or rotate about it, and the field must circle the wire, tangent to circles centred on it. So choose the Amperian loop to be a circle of radius $r$ coaxial with the wire. Along it $\vec{B}$ is everywhere tangential and of one constant magnitude $B$, condition (i) is met, $L=2\pi r$, and $I_{\text{enc}}=I$. Ampere's law gives $B\,(2\pi r)=\mu_0 I$, so $B=\dfrac{\mu_0 I}{2\pi r}$. This carries four lessons: the field has cylindrical symmetry, depending on the single coordinate $r$; the field lines are closed concentric circles, unlike electrostatic lines that start and end on charges; the field stays finite at any nonzero $r$ even though the wire is idealised as infinite; and the right-hand grip rule, thumb along the current and fingers curling in the sense of $\vec{B}$, fixes the direction. 🔉⇢
The notion of enclosed current becomes vivid when the wire has finite radius $a$ and carries its current $I$ spread uniformly over the cross-section. For a circular loop outside the wire, $r \gt a$, the whole current is enclosed and the earlier result $B=\dfrac{\mu_0 I}{2\pi r}$ holds, falling off as $1/r$. For a loop inside the conductor, $r \lt a$, only the fraction of current within radius $r$ is enclosed: since the current density is uniform, $I_{\text{enc}}=I\,\dfrac{\pi r^{2}}{\pi a^{2}}=\dfrac{I r^{2}}{a^{2}}$. Ampere's law then gives $B\,(2\pi r)=\mu_0\dfrac{I r^{2}}{a^{2}}$, so $B=\dfrac{\mu_0 I r}{2\pi a^{2}}$, which grows linearly with $r$. The field therefore rises in proportion to $r$ from zero at the axis to a maximum $\dfrac{\mu_0 I}{2\pi a}$ at the surface, then falls as $1/r$ outside. It is the enclosed current, not the total current, that governs the field at each radius. 🔉⇢
Now the crucial caution, and the reason this topic separates careful students from careless ones: Ampere's law is true for every closed loop and every steady current, but it lets you compute $\vec{B}$ only in the rare cases where symmetry supplies a loop meeting conditions (i) to (iii). Take the field on the axis of a single circular current loop, $B=\dfrac{\mu_0 I}{2R}$ at the centre, which we obtain from the Biot-Savart law. You cannot get this from Ampere's law: no closed loop through that point has $\vec{B}$ tangent and constant along it, so although $\oint\vec{B}\cdot d\vec{l}=\mu_0 I$ is a correct statement, it is one equation in a field that varies from place to place in an unknown way, that is, infinitely many unknowns and a single equation. The same impasse defeats the field of a finite straight wire, a bar magnet, or any point off the axis of a solenoid. The integral does not lie; it simply cannot be untangled. Recognising this is a skill: before reaching for Ampere's law, ask whether the current distribution has one of the three symmetries that make it work, namely an infinite straight wire (or coaxial cable), an infinitely long solenoid, or a toroid. If it does not, the law will still be true, but it will not hand you $\vec{B}$, and you must return to the Biot-Savart law. 🔉⇢
The solenoid is the workhorse that Ampere's law tames beautifully. A solenoid is a long wire wound into a closely spaced helix; when its length greatly exceeds its radius we call it a long solenoid and treat each closely packed turn as a circular loop carrying the same current $I$. The net field is the vector sum of the fields of all the turns. Between neighbouring turns the contributions from adjacent wires point oppositely and largely cancel, so the field in the region just outside the winding is weak. Inside, the fields of all the turns reinforce along the axis, giving a strong, uniform, axial field. As the solenoid is made longer and longer it comes to resemble an infinite cylindrical current sheet: the interior field becomes everywhere parallel to the axis and uniform, while the exterior field tends to zero. This idealised picture, a uniform axial field inside and zero field outside, is exactly what makes an Amperian loop usable here. 🔉⇢
Applying the law (derived step by step below) yields the interior field of a long solenoid, $B=\mu_0 n I$, where $n$ is the number of turns per unit length and $I$ the current. Three features deserve emphasis for problem solving. First, $B$ does not depend on where inside the solenoid you measure it, so the field is genuinely uniform across the bore, a fact students often doubt. Second, and more surprising, $B$ does not depend on the radius of the solenoid at all; a fat solenoid and a thin one with the same $n$ and $I$ produce the same interior field. Third, the field is set only by the product $nI$, the current per unit length along the winding. At the very end of a long solenoid, symmetry and superposition show the axial field drops to exactly half its central value, $B_{\text{end}}=\tfrac{1}{2}\mu_0 n I$, because the end point sees, in effect, only half of an infinite solenoid. 🔉⇢
Bending a long solenoid round into a closed doughnut gives a toroid, and it too has the symmetry Ampere's law needs. A toroid is a coil wound uniformly on a ring-shaped core; the field lines are circles concentric with the ring's axis, entirely confined within the core. Choose an Amperian loop to be one such circle of radius $r$ inside the windings. If the toroid carries $N$ turns in total, each threading the loop once, then $I_{\text{enc}}=N I$, the field is tangential and constant along the loop, and $B\,(2\pi r)=\mu_0 N I$ gives $B=\dfrac{\mu_0 N I}{2\pi r}$. Unlike the straight solenoid, the toroidal field is not perfectly uniform across the cross-section, since it falls as $1/r$ from inner edge to outer edge, but for a thin ring whose cross-section is small compared with $r$ it is nearly constant. A striking result is that the field both outside the toroid and in the empty central hole is zero, because an Amperian loop drawn there encloses either no current or equal and opposite currents that cancel. 🔉⇢
So the practical recipe for JEE is short. When you meet a magnetostatics problem, first test it for one of the three canonical high-symmetry geometries: infinite straight wire, infinite solenoid, or toroid, with the coaxial cable as a close relative of the wire. If it matches, choose the Amperian loop that follows the field lines, split the loop into stretches where $\vec{B}$ is tangential-and-constant, perpendicular, or zero, write $BL=\mu_0 I_{\text{enc}}$, and read off $L$ and $I_{\text{enc}}$. If it does not match, as with a finite wire, a single loop's centre, an off-axis point, or a bar magnet, accept that Ampere's law, though still exactly true, will not deliver the field, and use the Biot-Savart law instead. Knowing which tool the geometry permits is half the battle. 🔉⇢
Source: JEE-pattern
🔬 Interactive 3D · A current loop is free to turn in a uniform field — drag its orientation and read τ = mB sinθ pass through zero and maximum as the moment vector swings. orientation angle θ, magnetic moment m, field B
A closed loop of wire carrying a steady current, when placed in a uniform magnetic field, experiences no net translational force, yet it is generally acted upon by a net torque that tries to rotate it. This single fact is the mechanical seed of one of the most productive ideas in magnetism: that a current loop behaves like a magnetic dipole. Just as an electric dipole $\mathbf{p}$ in a uniform electric field $\mathbf{E}$ feels the torque $\boldsymbol{\tau}=\mathbf{p}\times\mathbf{E}$ but no net force, a current loop of $N$ turns, area $A$ and current $I$ possesses a magnetic dipole moment $\mathbf{m}=NI A\,\hat{\mathbf{n}}$ and, in a uniform field $\mathbf{B}$, feels the torque $\boldsymbol{\tau}=\mathbf{m}\times\mathbf{B}$ whose magnitude is $\tau=NIAB\sin\theta$. The associated orientation energy is $U=-\mathbf{m}\cdot\mathbf{B}$, so the loop is in stable equilibrium when its moment is aligned with the field and in unstable equilibrium when it points against it. Exactly this torque, tamed by a soft-iron core that makes the field radial and by a spring that supplies a restoring couple, is what drives the pointer of a moving-coil galvanometer and gives the linear scale on which we read currents and voltages. 🔉⇢
To see why the field exerts a torque but no net force, picture a flat rectangular loop $ABCD$ carrying current $I$, with sides of length $a$ and $b$, immersed in a uniform field $\mathbf{B}$. Each straight arm of length $\ell$ feels the magnetic force $\mathbf{F}=I\boldsymbol{\ell}\times\mathbf{B}$. Because the field is the same everywhere, the forces on any pair of opposite arms are equal in magnitude and opposite in direction: the loop is pulled equally in opposite ways, so the four forces add vectorially to zero and there is no acceleration of the centre of mass. What survives is a couple. Opposite forces that do not share the same line of action cannot cancel their turning effects, and this residual couple is the torque on the loop. It is essential to appreciate that the vanishing of the net force is a special property of a uniform field; in a non-uniform field the forces on opposite arms differ and a genuine net force can appear, which is precisely how a bar magnet is pulled into a region of stronger field. 🔉⇢
Consider first the simplest orientation, in which the plane of the loop contains the field, so that $\mathbf{B}$ lies in the plane of the loop. Two of the arms are parallel or antiparallel to $\mathbf{B}$ and feel no force; the other two, each of length $b$ and perpendicular to $\mathbf{B}$, feel forces of magnitude $F=IbB$, one directed into the plane of the loop and the other out of it. These two forces are equal and opposite but are separated by the perpendicular distance $a$, so they form a couple. The magnitude of the torque of this couple is force times the perpendicular separation between the two lines of action, which works out to $\tau=IbB\times a=I(ab)B=IAB$, where $A=ab$ is the area of the loop. This is the maximum torque the field can exert, obtained precisely when the field lies in the plane of the loop, that is, when the normal to the loop is perpendicular to $\mathbf{B}$. 🔉⇢
Now let the loop be tilted so that its plane no longer contains the field. It is far cleaner to describe the orientation not by the angle the plane makes with $\mathbf{B}$ but by the angle $\theta$ between the field and the normal $\hat{\mathbf{n}}$ to the loop. The two arms that lie along the axis of rotation now carry equal and opposite forces that are collinear, so they cancel completely and contribute neither force nor torque. The other pair still carries equal and opposite forces of magnitude $F=IbB$, but the perpendicular distance between their lines of action shrinks from $a$ to $a\sin\theta$. The torque therefore becomes $\tau=IbB\,(a\sin\theta)=IAB\sin\theta$. When the loop has $N$ closely wound turns the current-carrying arm is effectively traversed $N$ times, so every force is multiplied by $N$ and the torque becomes $\tau=NIAB\sin\theta$. This is the master result for the magnitude of the torque on a planar current loop in a uniform field, and every later formula in this concept is a corollary of it. 🔉⇢
The recurring combination $NIA$ is given a name and a symbol of its own: the magnetic dipole moment of the loop, $\mathbf{m}=NI A\,\hat{\mathbf{n}}$. Its magnitude is $m=NIA$, measured in ampere metre-squared ($\mathrm{A\,m^{2}}$), and its direction is that of the area vector $\hat{\mathbf{n}}$, fixed by the right-hand rule: curl the fingers of the right hand along the direction of the conventional current around the loop and the extended thumb points along $\mathbf{m}$. Equivalently, if you look at the face of the loop through which the current appears to circulate anticlockwise, the moment points out towards you. The magnetic moment compresses everything about how strongly a given loop couples to an external field into a single vector, in exact analogy with the electric dipole moment $\mathbf{p}$ of two equal and opposite charges. The larger the number of turns, the larger the current, or the greater the enclosed area, the stronger the loop responds to a field. 🔉⇢
Armed with $\mathbf{m}$, the torque acquires its compact vector form $\boldsymbol{\tau}=\mathbf{m}\times\mathbf{B}$, whose magnitude $\tau=mB\sin\theta=NIAB\sin\theta$ reproduces the master result and whose direction, given by the right-hand rule for the cross product, correctly points along the axis about which the loop turns. Reading this expression tells us the whole story of equilibrium. The torque vanishes when $\sin\theta=0$, that is at $\theta=0$ and $\theta=\pi$, the two orientations in which the moment is parallel or antiparallel to the field. At $\theta=0$, with $\mathbf{m}$ aligned along $\mathbf{B}$, the equilibrium is stable: any small tilt produces a restoring torque that drives the loop back. At $\theta=\pi$, with $\mathbf{m}$ opposed to $\mathbf{B}$, the equilibrium is unstable: the slightest disturbance produces a torque that grows and flips the loop over. The torque is greatest, equal to $mB$, at $\theta=\pi/2$, when the moment is perpendicular to the field. This tendency of a magnetic moment to swing into alignment with an applied field is exactly why a compass needle, itself a magnetic dipole, turns to point along the Earth's field. 🔉⇢
Because the field does work as the loop rotates, we can package the same information as an orientation-dependent potential energy. Taking the reference where $\mathbf{m}$ is perpendicular to $\mathbf{B}$ to have zero energy, the work done against the magnetic torque in turning the moment to an angle $\theta$ integrates to the potential energy $U(\theta)=-mB\cos\theta=-\mathbf{m}\cdot\mathbf{B}$. This function is a minimum, $U=-mB$, at $\theta=0$, confirming that alignment is the stable, lowest-energy state; it is a maximum, $U=+mB$, at $\theta=\pi$, confirming that anti-alignment is unstable; and it passes through zero at $\theta=\pi/2$. The work an external agent must do to rotate the loop from an angle $\theta_1$ to $\theta_2$ at constant current is simply the change in this energy, $W=U(\theta_2)-U(\theta_1)=-mB(\cos\theta_2-\cos\theta_1)$. The energy picture and the torque picture are two faces of the same physics, related by $\tau=-\,\mathrm{d}U/\mathrm{d}\theta$ for the magnitude of the aligning torque. 🔉⇢
The results derived for a rectangle hold for a loop of any shape, because any planar loop can be tiled by infinitesimal rectangles whose internal current-carrying edges cancel in pairs, leaving only the current along the outer boundary. In particular, a circular loop of radius $R$ carrying current $I$ has magnetic moment $m=I\,\pi R^{2}$, or $m=NI\,\pi R^{2}$ for $N$ turns, directed along the axis of the loop by the same right-hand rule. Viewed from far away, such a current loop is indistinguishable from a tiny bar magnet: the magnetic field it produces on its axis at a large distance $x\gg R$ falls off as $B=\dfrac{\mu_{0}}{2\pi}\dfrac{m}{x^{3}}$, the signature inverse-cube law of a magnetic dipole, with the same $\mathbf{m}=NI\pi R^{2}\,\hat{\mathbf{n}}$ appearing in the numerator. This equivalence is what allows us to treat atoms, in which electrons circulate in tiny current loops, as elementary magnetic dipoles, and it underlies the entire language of magnetism in matter. 🔉⇢
The moving-coil galvanometer is the most direct engineering application of the torque $\boldsymbol{\tau}=\mathbf{m}\times\mathbf{B}$. It consists of a rectangular coil of many turns wound on a light frame, free to rotate about a fixed axis, suspended in the gap of a permanent magnet whose pole pieces are curved and which surrounds a cylindrical soft-iron core. The soft-iron core does two jobs: it concentrates and strengthens the field, and, crucially, it shapes the field so that in the gap the lines of $\mathbf{B}$ are always radial, that is, always lying in the plane of the coil and always perpendicular to its side arms. Because the field is radial, the normal to the coil is at all times perpendicular to $\mathbf{B}$, so the angle between $\mathbf{m}$ and $\mathbf{B}$ stays at $90^\circ$ no matter how far the coil has turned, and $\sin\theta=1$ throughout the swing. The deflecting torque is therefore $\tau=NIAB$, independent of the deflection angle. 🔉⇢
A fine spiral spring attached to the coil supplies a restoring torque proportional to the twist, $\tau_{\text{restoring}}=k\varphi$, where $k$ is the torsional constant of the spring and $\varphi$ the angular deflection. The coil settles at the deflection where the deflecting and restoring torques balance, giving the reading relation $\varphi=\left(\dfrac{NAB}{k}\right)I$. Because $N$, $A$, $B$ and $k$ are all fixed for a given instrument, the deflection is directly proportional to the current, and it is precisely the radial-field trick that makes the scale linear and uniformly divided rather than crowded like a $\sin\theta$ scale. The current sensitivity, the deflection per unit current, is $\varphi/I=NAB/k$, and it can be raised by increasing the number of turns, the coil area, or the field, or by using a softer spring. The voltage sensitivity, the deflection per unit voltage, is $\varphi/V=(NAB/k)(1/R)$, where $R$ is the total resistance of the coil circuit; note that simply doubling $N$ doubles the current sensitivity but usually also doubles the resistance, so the voltage sensitivity need not improve. By adding a small shunt resistance in parallel the galvanometer becomes an ammeter, and by adding a large resistance in series it becomes a voltmeter. 🔉⇢
The parallel with the electric dipole is worth drawing out fully, because it lets us carry over intuition and formulae wholesale. An electric dipole of moment $\mathbf{p}$ in a uniform field $\mathbf{E}$ feels torque $\boldsymbol{\tau}=\mathbf{p}\times\mathbf{E}$ and has energy $U=-\mathbf{p}\cdot\mathbf{E}$; a magnetic dipole of moment $\mathbf{m}$ in a uniform field $\mathbf{B}$ feels torque $\boldsymbol{\tau}=\mathbf{m}\times\mathbf{B}$ and has energy $U=-\mathbf{m}\cdot\mathbf{B}$. The correspondence $\mathbf{p}\leftrightarrow\mathbf{m}$ and $\mathbf{E}\leftrightarrow\mathbf{B}$ is exact for these expressions, which is why both dipoles align with their respective fields, both store the least energy when aligned, and both radiate a field that falls off as the inverse cube of distance. The dimensions of the magnetic moment follow directly from $m=NIA$ as $[\text{A}][\text{L}^2]$, giving the SI unit $\mathrm{A\,m^{2}}$; equivalently, from $U=-mB$, one ampere metre-squared times one tesla is one joule, so $1\ \mathrm{A\,m^{2}}=1\ \mathrm{J\,T^{-1}}$, the unit in which atomic and nuclear moments are usually quoted. 🔉⇢
One more consequence of the aligning torque deserves emphasis because it recurs throughout physics: a magnetic dipole displaced slightly from its stable orientation oscillates. Near $\theta=0$ the restoring torque is $\tau=-mB\sin\theta\approx -mB\,\theta$ for small $\theta$, which is a linear restoring torque, the rotational analogue of a spring. If the loop has moment of inertia $J$ about its rotation axis, Newton's law for rotation gives $J\,\ddot{\theta}=-mB\,\theta$, the equation of angular simple harmonic motion with angular frequency $\omega=\sqrt{mB/J}$ and period $T=2\pi\sqrt{J/(mB)}$. This is exactly how a compass needle wobbles about magnetic north before settling, and measuring such an oscillation period is a standard laboratory route to determining an unknown magnetic moment or an unknown field. It is the same mathematics that in the galvanometer, once damping is added, governs how quickly the pointer swings to its steady reading. 🔉⇢
Source: JEE-pattern
Source: JEE-pattern (NCERT Ch 4)
Source: JEE-pattern (NCERT Ch 4)
Source: JEE-pattern (NCERT Ch 4, Example 4.1)
Source: JEE-pattern (NCERT Ch 4, Exercise 4.7)
Source: JEE-pattern (NCERT Ch 4)
Source: JEE-pattern (NCERT Ch 4)
Source: JEE-pattern (NCERT Ch 4)
Source: JEE-pattern
Source: JEE-pattern
Source: JEE-pattern
Source: JEE-pattern
Source: JEE-pattern
Source: JEE-pattern
Source: JEE-pattern
Source: JEE-pattern
Source: JEE-pattern
Source: JEE-pattern
Source: JEE-pattern
Source: JEE-pattern
Source: JEE-pattern
Source: JEE-pattern
Source: JEE-pattern
Source: JEE-pattern
Source: JEE-pattern
Source: JEE-pattern
Source: JEE-pattern
Source: JEE-pattern
Source: JEE-pattern
Source: JEE-pattern
Source: JEE-pattern
Source: JEE-pattern
Source: JEE-pattern
These worked examples are taught in full alongside their interactive scene:
| Quantity | Formula | What it means / when to use | Source |
|---|---|---|---|
| Lorentz force 🔉⇢ | $\vec{F} = q\,\vec{E} + q\,\vec{v}\times\vec{B}$ | Here $q$ is the charge, $\vec{v}$ its velocity, $\vec{E}$ the electric field and $\vec{B}$ the magnetic field. The total force is the vector sum of an electric part $q\vec{E}$ and a magnetic part $q\,\vec{v}\times\vec{B}$; applies to any point charge in combined fields. | NCERT Class XII Physics, Ch. 4 |
| Magnitude of the magnetic force 🔉⇢ | $F = qvB\sin\theta$ | Here $\theta$ is the angle between $\vec{v}$ and $\vec{B}$. Fix the direction of $\vec{v}\times\vec{B}$ with the right-hand rule first; the force is zero when $\vec{v}$ is parallel or antiparallel to $\vec{B}$ ($\theta=0$ or $\pi$) and maximum when they are perpendicular. | NCERT Class XII Physics, Ch. 4 |
| Magnetic force does no work 🔉⇢ | $\vec{F}_{\text{mag}}\cdot\vec{v} = 0$ | Because $q\,\vec{v}\times\vec{B}$ is always perpendicular to $\vec{v}$, the magnetic force can change the direction of motion but never the speed or kinetic energy; applies to a free charge moving in any magnetic field. | NCERT Class XII Physics, Ch. 4 |
| Definition of the tesla 🔉⇢ | $1\ \text{T} = 1\ \text{N}\,\text{A}^{-1}\,\text{m}^{-1}$ | The SI unit of $B$ is the tesla: a field of $1$ T exerts $1$ N on a $1$ C charge moving at $1$ m s$^{-1}$ perpendicular to it. The non-SI gauss is $1\ \text{G}=10^{-4}$ T; applies as the unit definition of magnetic field. | NCERT Class XII Physics, Ch. 4 |
| Quantity | Formula | What it means / when to use | Source |
|---|---|---|---|
| Radius of circular motion 🔉⇢ | $r = \dfrac{mv}{qB}$ | Here $m$ is mass, $v$ speed, $q$ charge and $B$ the field, with $\vec{v}$ perpendicular to $\vec{B}$. The magnetic force supplies the centripetal force $mv^2/r=qvB$; applies to a charge moving in a plane perpendicular to a uniform field. | NCERT Class XII Physics, Ch. 4 |
| Period of revolution 🔉⇢ | $T = \dfrac{2\pi m}{qB}$ | Here $T$ is the time for one full circle. It depends only on $m$, $q$ and $B$ and is independent of the speed or radius; applies to any charge in a uniform magnetic field. | NCERT Class XII Physics, Ch. 4 |
| Cyclotron frequency 🔉⇢ | $\nu_c = \dfrac{qB}{2\pi m}\qquad \omega_c = \dfrac{qB}{m}$ | Here $\nu_c$ is the number of revolutions per second and $\omega_c$ the angular frequency. Its independence from speed lets a cyclotron accelerate particles with a fixed-frequency oscillating voltage; applies to circular motion in a uniform field. | NCERT Class XII Physics, Ch. 4 |
| Pitch of the helix 🔉⇢ | $p = v_\parallel T = \dfrac{2\pi m\,v_\parallel}{qB}$ | Here $v_\parallel$ is the velocity component along $\vec{B}$, which is unaffected by the field. The perpendicular component gives circular motion while $v_\parallel$ carries the particle forward one pitch per turn; applies when velocity is oblique to the field. | NCERT Class XII Physics, Ch. 4 |
| Velocity selector 🔉⇢ | $v = \dfrac{E}{B}$ | For crossed electric and magnetic fields perpendicular to the beam, only particles with this speed pass undeflected because $qE=qvB$. Faster or slower particles are deflected; applies in mass spectrometers and charge-to-mass measurements. | NCERT-derived / JEE-pattern (crossed $E$ and $B$ fields) |
| Quantity | Formula | What it means / when to use | Source |
|---|---|---|---|
| Force on a current-carrying conductor 🔉⇢ | $\vec{F} = I\,\vec{L}\times\vec{B}$ | Here $I$ is the current and $\vec{L}$ a vector of magnitude equal to the wire length pointing along the current. This is the sum of $q\,\vec{v}\times\vec{B}$ over all carriers; applies to a straight wire in a uniform external field. | NCERT Class XII Physics, Ch. 4 |
| Magnitude of the force on a wire 🔉⇢ | $F = BIL\sin\theta$ | Here $\theta$ is the angle between the wire and $\vec{B}$. The force is maximum when the wire is perpendicular to the field and zero when it is parallel; applies to a straight current-carrying conductor. | NCERT Class XII Physics, Ch. 4 |
| Force per unit length between parallel currents 🔉⇢ | $f = \dfrac{\mu_0 I_a I_b}{2\pi d}$ | Here $I_a$ and $I_b$ are the two currents and $d$ their separation. Parallel currents attract and antiparallel currents repel; applies to two long straight parallel wires. | NCERT Class XII Physics, Ch. 4 |
| Definition of the ampere 🔉⇢ | $f = 2\times 10^{-7}\ \text{N}\,\text{m}^{-1}\ \text{when}\ I_a=I_b=1\ \text{A},\ d=1\ \text{m}$ | The ampere is the steady current that, in each of two very long parallel wires one metre apart in vacuum, produces a force of $2\times 10^{-7}$ N per metre of length; applies as the historical SI definition of current. | NCERT Class XII Physics, Ch. 4 |
| Quantity | Formula | What it means / when to use | Source |
|---|---|---|---|
| Biot-Savart law (vector form) 🔉⇢ | $d\vec{B} = \dfrac{\mu_0}{4\pi}\dfrac{I\,d\vec{l}\times\hat{r}}{r^2}$ | Here $I\,d\vec{l}$ is a current element, $\hat{r}$ the unit vector from the element to the field point and $r$ their distance, with $\mu_0/4\pi = 10^{-7}$ T m A$^{-1}$. The field is perpendicular to the plane of $d\vec{l}$ and $\hat{r}$; applies to any steady current by integration. | NCERT Class XII Physics, Ch. 4 |
| Biot-Savart law (magnitude) 🔉⇢ | $dB = \dfrac{\mu_0}{4\pi}\dfrac{I\,dl\,\sin\theta}{r^2}$ | Here $\theta$ is the angle between $d\vec{l}$ and $\hat{r}$; the element contributes nothing along its own direction ($\theta=0$). Add contributions by superposition; applies to computing the field of wires and arcs. | NCERT Class XII Physics, Ch. 4 |
| Field of a long straight wire 🔉⇢ | $B = \dfrac{\mu_0 I}{2\pi r}$ | Here $r$ is the perpendicular distance from an infinitely long straight wire. The field lines are circles concentric with the wire; applies outside a long straight current-carrying conductor and follows from Ampere's law by symmetry. | NCERT Class XII Physics, Ch. 4 |
| Quantity | Formula | What it means / when to use | Source |
|---|---|---|---|
| Field on the axis of a circular loop 🔉⇢ | $B = \dfrac{\mu_0 I R^2}{2\,(R^2 + x^2)^{3/2}}$ | Here $R$ is the loop radius and $x$ the axial distance from the centre. The field points along the axis by the right-hand thumb rule; applies on the axis of a single circular current loop. | NCERT Class XII Physics, Ch. 4 |
| Field at the centre of a loop 🔉⇢ | $B = \dfrac{\mu_0 I}{2R}$ | The special case $x=0$ of the on-axis field. For $N$ tightly wound turns it becomes $B=\dfrac{\mu_0 N I}{2R}$; applies at the centre of a circular coil. | NCERT Class XII Physics, Ch. 4 |
| Far-axial field (dipole limit) 🔉⇢ | $B \simeq \dfrac{\mu_0}{4\pi}\dfrac{2m}{x^3},\quad m=IA$ | For $x \gg R$ the loop behaves like a magnetic dipole of moment $m=I\pi R^2$, and the axial field falls off as $1/x^3$; applies far from a small current loop. | NCERT-derived / JEE-pattern (dipole limit of a loop) |
| Quantity | Formula | What it means / when to use | Source |
|---|---|---|---|
| Ampere's circuital law 🔉⇢ | $\oint \vec{B}\cdot d\vec{l} = \mu_0 I_{\text{enc}}$ | Here $I_{\text{enc}}$ is the net current threading the amperian loop, with sign fixed by the right-hand rule. The law is always true but only yields $B$ when symmetry lets $B$ be pulled out of the integral; applies to highly symmetric current distributions. | NCERT Class XII Physics, Ch. 4 |
| Field inside a long solenoid 🔉⇢ | $B = \mu_0 n I$ | Here $n$ is the number of turns per unit length. The interior field is uniform and axial while the exterior field is nearly zero; applies well inside a long, tightly wound solenoid. | NCERT Class XII Physics, Ch. 4 |
| Field inside a toroid 🔉⇢ | $B = \dfrac{\mu_0 N I}{2\pi r}$ | Here $N$ is the total number of turns and $r$ the distance from the centre to a point in the core. The field is confined to the core and is zero in the central hole and outside; applies within the core of a toroidal coil. | NCERT Class XII Physics, Ch. 4 |
| Quantity | Formula | What it means / when to use | Source |
|---|---|---|---|
| Magnetic dipole moment of a loop 🔉⇢ | $\vec{m} = N I \vec{A}$ | Here $N$ is the number of turns, $I$ the current and $\vec{A}$ the area vector (magnitude equal to the loop area, direction by the right-hand thumb rule). Its unit is A m$^2$; applies to any planar current loop treated as a dipole. | NCERT Class XII Physics, Ch. 4 |
| Torque on a current loop 🔉⇢ | $\vec{\tau} = \vec{m}\times\vec{B},\quad \tau = NIAB\sin\theta$ | Here $\theta$ is the angle between $\vec{m}$ and $\vec{B}$. The net force is zero but the couple rotates the loop toward alignment; equilibrium is stable when $\vec{m}\parallel\vec{B}$; applies to a loop in a uniform field. | NCERT Class XII Physics, Ch. 4 |
| Galvanometer deflection 🔉⇢ | $\varphi = \dfrac{N A B}{k}\,I$ | Here $k$ is the torsional constant of the suspension spring and $\varphi$ the steady deflection. A radial field keeps $\sin\theta=1$ so deflection is proportional to current; applies to a moving coil galvanometer. | NCERT Class XII Physics, Ch. 4 |
| Ammeter shunt resistance 🔉⇢ | $I_g R_G = (I - I_g)\,r_s$ | Here $I_g$ is the full-scale galvanometer current, $R_G$ its resistance, $I$ the line current and $r_s$ a small shunt in parallel. The shunt diverts most of the current; applies when converting a galvanometer to an ammeter. | NCERT Class XII Physics, Ch. 4 |
| Voltmeter series resistance 🔉⇢ | $R = \dfrac{V}{I_g} - R_G$ | Here $V$ is the full-scale voltage and $R$ a large resistance in series with the galvanometer of resistance $R_G$. The large resistance keeps the drawn current tiny; applies when converting a galvanometer to a voltmeter. | NCERT Class XII Physics, Ch. 4 |
Every PYQ traces to an official NTA/JAB paper (year + shift + paper). Answer keys match the official key. OCR-extracted questions are flagged for SME verification.
Two ions of masses 4 amu and 16 amu have charges +2e and +3e respectively. These ions pass through the region of constant perpendicular magnetic field. The kinetic energy of both ions is same. Then :
A coil having N turns is wound tightly in the form of a spiral with inner and outer radii 'a' and 'b' respectively. Find the magnetic field at centre, when a current I passes through coil:
Two long straight wires P and Q carrying equal current 10A each were kept parallel to each other at 5 cm distance. Magnitude of magnetic force experienced by 10 cm length of wire P is F$_1$. If distance between wires is halved and currents on them are doubled, force F$_2$ on 10 cm length of wire P will be:
A circular loop of radius $r$ is carrying current I A. The ratio of magnetic field at the center of circular loop and at a distance r from the center of the loop on its axis is :
The electric current in a circular coil of four turns produces a magnetic induction 32 T at its centre. The coil is unwound and is rewound into a circular coil of single turn, the magnetic induction at the centre of the coil by the same current will be :
A rod with circular cross-section area $2 \mathrm{~cm}^{2}$ and length $40 \mathrm{~cm}$ is wound uniformly with 400 turns of an insulated wire. If a current of $0.4 \mathrm{~A}$ flows in the wire windings, the total magnetic flux through the rod's cross-section (single-turn flux, not flux linkage) is $4 \pi \times 10^{-6} \mathrm{~Wb}$. The relative permeability of the rod is (Given : Permeability of vacuum $\mu_{0}=4 \pi \times 10^{-7} \mathrm{NA}^{-2}$)
A single turn current loop in the shape of a right angle triangle with sides 5 cm, 12 cm, 13 cm is carrying a current of 2 A. The loop is in a uniform magnetic field of magnitude 0.75 T whose direction is parallel to the current in the 13 cm side of the loop. The magnitude of the magnetic force on the 5 cm side will be $\frac{x}{130}$ N. The value of $x$ is ____________.
For a moving coil galvanometer, the deflection in the coil is 0.05 rad when a current of 10 mA is passes through it. If the torsional constant of suspension wire is $4.0\times10^{-5}\mathrm{N~m~rad^{-1}}$, the magnetic field is 0.01T and the number of turns in the coil is 200, the area of each turn (in cm$^2$) is :
The magnetic moments associated with two closely wound circular coils $\mathrm{A}$ and $\mathrm{B}$ of radius $\mathrm{r}_{\mathrm{A}}=10 \mathrm{cm}$ and $\mathrm{r}_{\mathrm{B}}=20 \mathrm{~cm}$ respectively are equal if : (Where $\mathrm{N}_{\mathrm{A}}, \mathrm{I}_{\mathrm{A}}$ and $\mathrm{N}_{\mathrm{B}}, \mathrm{I}_{\mathrm{B}}$ are number of turn and current of $\mathrm{A}$ and $\mathrm{B}$ respectively)
A long conducting wire having a current I flowing through it, is bent into a circular coil of $\mathrm{N}$ turns. Then it is bent into a circular coil of $\mathrm{n}$ turns. The magnetic field is calculated at the centre of coils in both the cases. The ratio of the magnetic field in first case to that of second case is :
A long solenoid is formed by winding 70 turns cm$^{-1}$. If 2.0 A current flows, then the magnetic field produced inside the solenoid is ____________ ($\mu_0=4\pi\times10^{-7}$ TmA$^{-1}$)
A particle of mass $m$ and charge $q$, moving with velocity $V$ enters Region II normal to the boundary. Region II has a uniform magnetic field $B$ perpendicular to the plane of the paper. The length of the Region II is $\ell$. Regions I and III are field free. Choose the correct choice(s).
STATEMENT-1: The sensitivity of a moving coil galvanometer is increased by placing a suitable magnetic material as a core inside the coil. and STATEMENT-2: Soft iron has a high magnetic permeability and cannot be easily magnetized or demagnetized.
A steady current $I$ goes through a wire loop PQR having shape of a right angle triangle with PQ $= 3x$, PR $= 4x$ and QR $= 5x$. If the magnitude of the magnetic field at P due to this loop is $k\left(\dfrac{\mu_0 I}{48\pi x}\right)$, find the value of $k$.
A thin flexible wire of length $L$ is connected to two adjacent fixed points and carries a current $I$ in the clockwise direction, as shown in the figure. When the system is put in a uniform magnetic field of strength $B$ going into the plane of the paper, the wire takes the shape of a circle. The tension in the wire is
A long insulated copper wire is closely wound as a spiral of $N$ turns. The spiral has inner radius $a$ and outer radius $b$. The spiral lies in the X-Y plane and a steady current $I$ flows through the wire. The Z-component of the magnetic field at the center of the spiral is
Consider the motion of a positive point charge in a region where there are simultaneous uniform electric and magnetic fields $\vec{E}=E_0\hat{j}$ and $\vec{B}=B_0\hat{j}$. At time $t=0$, this charge has velocity $\vec{v}$ in the $x$-$y$ plane, making an angle $\theta$ with the $x$-axis. Which of the following option(s) is(are) correct for time $t>0$?
A particle of mass $M$ and positive charge $Q$, moving with a constant velocity $\vec{u}_1=4\hat{i}\ \text{m s}^{-1}$, enters a region of uniform static magnetic field normal to the $x$-$y$ plane. The region of the magnetic field extends from $x=0$ to $x=L$ for all values of $y$. After passing through this region, the particle emerges on the other side after $10$ milliseconds with a velocity $\vec{u}_2=2\left(\sqrt{3}\,\hat{i}+\hat{j}\right)\ \text{m s}^{-1}$. The correct statement(s) is (are)
A steady current $I$ flows along an infinitely long hollow cylindrical conductor of radius $R$. This cylinder is placed coaxially inside an infinite solenoid of radius $2R$. The solenoid has $n$ turns per unit length and carries a steady current $I$. Consider a point $P$ at a distance $r$ from the common axis. The correct statement(s) is (are)
Two parallel wires in the plane of the paper are distance $X_0$ apart. A point charge is moving with speed $u$ between the wires in the same plane at a distance $X_1$ from one of the wires. When the wires carry current of magnitude $I$ in the same direction, the radius of curvature of the path of the point charge is $R_1$. In contrast, if the currents $I$ in the two wires have directions opposite to each other, the radius of curvature of the path is $R_2$. If $\dfrac{X_0}{X_1} = 3$, the value of $\dfrac{R_1}{R_2}$ is
A circular loop of radius $a$ lies between two long parallel wires (numbered 1 and 2), all in the plane of the paper. Wire 1 is the vertical line on the left through the points $P$ (lower) and $Q$ (upper); wire 2 is the vertical line on the right through the points $R$ (lower) and $S$ (upper). The distance of each wire from the centre of the loop is $d$. The loop and the wires are carrying the same current $I$. The current in the loop is in the counterclockwise direction if seen from above. When $d \approx a$ but wires are not touching the loop, it is found that the net magnetic field on the axis of the loop is zero at a height $h$ above the loop. In that case
A circular loop of radius $a$ lies between two long parallel wires (numbered 1 and 2), all in the plane of the paper. The distance of each wire from the centre of the loop is $d$. The loop and the wires are carrying the same current $I$. The current in the loop is in the counterclockwise direction if seen from above. Consider $d \gg a$, and the loop is rotated about its diameter parallel to the wires by $30^\circ$ from the position shown in the figure. If the currents in the wires are in the opposite directions, the torque on the loop at its new position will be (assume that the net field due to the wires is constant over the loop)
PARAGRAPH: In a thin rectangular metallic strip a constant current $I$ flows along the positive $x$-direction. The length, width and thickness of the strip are $l$, $w$ and $d$, respectively. A uniform magnetic field $\vec{B}$ is applied on the strip along the positive $y$-direction. Due to this, the charge carriers experience a net deflection along the $z$-direction. This results in accumulation of charge carriers on the surface $PQRS$ and appearance of equal and opposite charges on the face opposite to $PQRS$. A potential difference along the $z$-direction is thus developed. Charge accumulation continues until the magnetic force is balanced by the electric force. The current is assumed to be uniformly distributed on the cross section of the strip and carried by electrons. Consider two different metallic strips (1 and 2) of same dimensions (length $l$, width $w$ and thickness $d$) with carrier densities $n_1$ and $n_2$, respectively. Strip 1 is placed in magnetic field $B_1$ and strip 2 is placed in magnetic field $B_2$, both along positive $y$-directions. Then $V_1$ and $V_2$ are the potential differences developed between $K$ and $M$ in strips 1 and 2, respectively. Assuming that the current $I$ is the same for both the strips, the correct option(s) is(are)
Consider two identical galvanometers and two identical resistors with resistance $R$. If the internal resistance of the galvanometers $R_C < R/2$, which of the following statement(s) about any one of the galvanometers is(are) true?
A symmetric star shaped conducting wire loop (a regular six-pointed star, as shown in the figure) is carrying a steady state current $I$. The distance between the diametrically opposite vertices of the star is $4a$. The magnitude of the magnetic field at the center of the loop is
A uniform magnetic field $B$ exists in the region between $x = 0$ and $x = \dfrac{3R}{2}$ (region 2) pointing normally into the plane of the paper. Region 1 is $x < 0$ and region 3 is $x > \dfrac{3R}{2}$. A particle with charge $+Q$ and momentum $p$ directed along $x$-axis enters region 2 from region 1 at point $P_1$ $(y = -R)$. The point $P_2$ is the point where the boundary $x = \dfrac{3R}{2}$ meets the $x$-axis. Which of the following option(s) is/are correct?
A moving coil galvanometer has 50 turns and each turn has an area $2 \times 10^{-4}\ \mathrm{m^2}$. The magnetic field produced by the magnet inside the galvanometer is $0.02\ \mathrm{T}$. The torsional constant of the suspension wire is $10^{-4}\ \mathrm{N\,m\,rad^{-1}}$. When a current flows through the galvanometer, a full scale deflection occurs if the coil rotates by $0.2\ \mathrm{rad}$. The resistance of the coil of the galvanometer is $50\ \Omega$. This galvanometer is to be converted into an ammeter capable of measuring current in the range $0 - 1.0\ \mathrm{A}$. For this purpose, a shunt resistance is to be added in parallel to the galvanometer. The value of this shunt resistance, in $ohms$, is __________.
Two infinitely long straight wires lie in the $xy$-plane along the lines $x = \pm R$. The wire located at $x = +R$ carries a constant current $I_1$ and the wire located at $x = -R$ carries a constant current $I_2$. A circular loop of radius $R$ is suspended with its centre at $(0, 0, \sqrt{3}R)$ and in a plane parallel to the $xy$-plane. This loop carries a constant current $I$ in the clockwise direction as seen from above the loop. The current in the wire is taken to be positive if it is in the $+\hat{j}$ direction. Which of the following statements regarding the magnetic field $\vec{B}$ is (are) true?
A circular coil of radius $R$ and $N$ turns has negligible resistance. As shown in the schematic figure, its two ends are connected to two wires and it is hanging by those wires with its plane being vertical. The wires are connected to a capacitor with charge $Q$ through a switch. The coil is in a horizontal uniform magnetic field $B_o$ parallel to the plane of the coil. When the switch is closed, the capacitor gets discharged through the coil in a very short time. By the time the capacitor is discharged fully, magnitude of the angular momentum gained by the coil will be (assume that the discharge time is so short that the coil has hardly rotated during this time)
An $\alpha$-particle (mass 4 amu) and a singly charged sulfur ion (mass 32 amu) are initially at rest. They are accelerated through a potential $V$ and then allowed to pass into a region of uniform magnetic field which is normal to the velocities of the particles. Within this region, the $\alpha$-particle and the sulfur ion move in circular orbits of radii $r_\alpha$ and $r_S$, respectively. The ratio $(r_S/r_\alpha)$ is ___.
Two concentric circular loops, one of radius $R$ and the other of radius $2R$, lie in the xy-plane with the origin as their common center, as shown in the figure. The smaller loop carries current $I_1$ in the anti-clockwise direction and the larger loop carries current $I_2$ in the clockwise direction, with $I_2 > 2I_1$. $\vec{B}(x, y)$ denotes the magnetic field at a point $(x, y)$ in the xy-plane. Which of the following statement(s) is(are) correct?
An infinitely long wire, located on the $z$-axis, carries a current $I$ along the $+z$-direction and produces the magnetic field $\vec{B}$. The magnitude of the line integral $\int \vec{B}\cdot \vec{dl}$ along a straight line from the point $(-\sqrt{3}a,\, a,\, 0)$ to $(a,\, a,\, 0)$ is given by [$\mu_0$ is the magnetic permeability of free space.]
A thin stiff insulated metal wire is bent into a circular loop with its two ends extending tangentially from the same point of the loop. The wire loop has mass $m$ and radius $r$ and it is in a uniform vertical magnetic field $B_0$, as shown in the figure. Initially, it hangs vertically downwards, because of acceleration due to gravity $g$, on two conducting supports at P and Q. When a current $I$ is passed through the loop, the loop turns about the line PQ by an angle $\theta$ given by
A conducting solid sphere of radius $R$ and mass $M$ carries a charge $Q$. The sphere is rotating about an axis passing through its center with a uniform angular speed $\omega$. The ratio of the magnitudes of the magnetic dipole moment to the angular momentum about the same axis is given as $\alpha\dfrac{Q}{2M}$. The value of $\alpha$ is ___
A hollow, right circular cone of base radius $R$ and height $h$, with its tip at the origin is rotating about the $Z$-axis with an angular velocity $\omega$, as shown in the figure. The cone carries a total charge $Q$ uniformly distributed on its curved surface. The magnitude of magnetic field at a point $(0,0,z)$, where $z \gg R$ and $z \gg h$, is $\dfrac{n\mu_0}{4\pi}\dfrac{QR^2\omega}{z^3}$. The value of $n$ is:
List-I contains four conducting loops lying in the 𝑋𝑌 plane, as shown in the figures. The loops are rotating about 𝑍 axis passing through the point 𝑂 with time period 𝑇 in clockwise direction. The region 𝑥> 0 contains a uniform magnetic field 𝐵 in the +𝑧 direction. List-II contains the qualitative variation of the induced current 𝑖(𝑡) for each of these loops. Choose the option which describes the correct match between the entries in List-I to those in List-II. List-I List-II (P) (1) (Q) (2) (R) (3) (S) (4) (5)
A particle having charge $10^{-9}$ C moving in $x$-$y$ plane in fields of $0.4 \hat{j}$ N/C and $4 \times 10^{-3} \hat{k}$ T experiences a force of $(4 \hat{i} + 2 \hat{j}) \times 10^{-10}$ N. The velocity of the particle at that instant is _________ m/s.
A current carrying circular loop of radius 2 cm with unit normal $\hat{n}=\frac{\hat{k}+\hat{i}}{\sqrt{2}}$ is placed in a magnetic field, $\vec{B}=B_o(3 \hat{i}+2 \hat{k})$. If $B_o=4 \times 10^{-3} \mathrm{~T}$ and current $I=100 \sqrt{2} \mathrm{~A}$, the torque experienced by the loop is $\_\_\_\_$ Wb.A. ( $\pi=3.14$ )
An insulated wire is wound so that it forms a flat coil with $N=200$ turns. The radius of the innermost turn is $r_1=3 \mathrm{~cm}$, and of the outermost turn $r_2=6 \mathrm{~cm}$. If 20 mA current flows in it then the magnetic moment will be $\alpha \times 10^{-2} \mathrm{~A} . \mathrm{m}^2$. The value of $\alpha$ is $\_\_\_\_$ .
A small cube of side 1 mm is placed at the centre of a circular loop of radius 10 cm carrying a current of 2 A . The magnetic energy stored inside the cube is $\alpha \times 10^{-14} \mathrm{~J}$. The value of $\alpha$ is $\_\_\_\_$ . $\left(\mu_{\mathrm{o}}=4 \pi \times 10^{-7} \mathrm{Tm} / \mathrm{A}, \pi=3.14\right)$
In a vacuum chamber, a particle of charge $1\ \mu\mathrm{C}$ and mass $1\ \mathrm{mg}$ is projected with a velocity $(\hat{i} + 2\hat{j})\ \mathrm{ms^{-1}}$ from the $XZ$ plane at time $t = 0$ in an electric field of $1\hat{i}\ \mathrm{Vm^{-1}}$. At $t = 0.2\ \mathrm{s}$, the electric field is switched off and a magnetic field of $6\hat{j}\ \mathrm{T}$ is switched on. The acceleration due to gravity is $-10\hat{j}\ \mathrm{ms^{-2}}$. Correct option(s) is/are:
An electron and a proton are moving on straight parallel paths with same velocity. They enter a semi- infinite region of uniform magnetic field perpendicular to the velocity. Which of the following statement(s) is / are true?
A particle of mass M and positive charge Q, moving with a constant velocity 1 1 ˆ u 4ims , enters a region of uniform static magnetic field normal to the x-y plane. The region of the magnetic field extends from x = 0 to x = L for all values of y. After passing through this region, the particle emerges on the other side after 10 milliseconds with a velocity 2 ˆ ˆ u 2 3i j m/s1. The correct statement(s) is (are)
A steady current I flows along an infinitely long hollow cylindrical conductor of radius R. This cylinder is placed coaxially inside an infinite solenoid of radius 2R. The solenoid has n turns per unit length and carries a steady current I. Consider a point P at a distance r from the common axis. The correct statement(s) is (are)
A charge particle of $2 ~\mu \mathrm{C}$ accelerated by a potential difference of $100 \mathrm{~V}$ enters a region of uniform magnetic field of magnitude $4 ~\mathrm{mT}$ at right angle to the direction of field. The charge particle completes semicircle of radius $3 \mathrm{~cm}$ inside magnetic field. The mass of the charge particle is __________ $\times 10^{-18} \mathrm{~kg}$
Two long parallel wires carrying currents 8A and 15A in opposite directions are placed at a distance of 7 cm from each other. A point P is at equidistant from both the wires such that the lines joining the point P to the wires are perpendicular to each other. The magnitude of magnetic field at P is _____________ $\times~10^{-6}$ T. (Given : $\sqrt2=1.4$)
A long straight wire of circular cross-section (radius a) is carrying steady current I. The current I is uniformly distributed across this cross-section. The magnetic field is
A charge particle moving in magnetic field B, has the components of velocity along B as well as perpendicular to B. The path of the charge particle will be
An electron is allowed to move with constant velocity along the axis of current carrying straight solenoid. A. The electron will experience magnetic force along the axis of the solenoid. B. The electron will not experience magnetic force. C. The electron will continue to move along the axis of the solenoid. D. The electron will be accelerated along the axis of the solenoid. E. The electron will follow parabolic path-inside the solenoid. Choose the correct answer from the options given below:
An electron is moving along the positive $\mathrm{x}$-axis. If the uniform magnetic field is applied parallel to the negative z-axis, then A. The electron will experience magnetic force along positive y-axis B. The electron will experience magnetic force along negative y-axis C. The electron will not experience any force in magnetic field D. The electron will continue to move along the positive $\mathrm{x}$-axis E. The electron will move along circular path in magnetic field Choose the correct answer from the options given below:
A proton with a kinetic energy of $2.0 ~\mathrm{eV}$ moves into a region of uniform magnetic field of magnitude $\frac{\pi}{2} \times 10^{-3} \mathrm{~T}$. The angle between the direction of magnetic field and velocity of proton is $60^{\circ}$. The pitch of the helical path taken by the proton is __________ $\mathrm{cm}$. (Take, mass of proton $=1.6 \times 10^{-27} \mathrm{~kg}$ and Charge on proton $=1.6 \times 10^{-19} \mathrm{C}$ ).
The ratio of magnetic field at the centre of a current carrying coil of radius $r$ to the magnetic field at distance $r$ from the centre of coil on its axis is $\sqrt{x}: 1$. The value of $x$ is __________
A particle of charge $q$ and mass $m$ is projected from origin with an initial velocity $\vec{v}=\left(\frac{v_0}{\sqrt{2}} \hat{x}+\frac{v_0}{\sqrt{2}} \hat{y}\right)$. There exists a uniform magnetic field $\vec{B}=B_0 \hat{z}$ and a space varying electric field $\vec{E}=E_{\mathrm{o}} \mathrm{e}^{-\lambda x} \hat{x}$ within the region $0 \leqslant x \leqslant L$. After travelling a distance such that $x$-coordinate has changed from $x=0$ to $x=L$, the change in the kinetic energy is $\_\_\_\_$ .
A current of 30 A each flows in opposite directions in two conducting wires, placed parallel to each other at a distance of 8 cm . The magnetic field at the mid point between the two wires is $\_\_\_\_ \mu \mathrm{T}$. $\left(\frac{\mu_{\mathrm{o}}}{4 \pi}=10^{-7} \mathrm{~N} / \mathrm{A}^2\right)$
1 $\mu$C charge moving with velocity $\vec{v} = (\hat{i} - 2\hat{j} + 3\hat{k})$ m/s in the region of magnetic field $\vec{B} = (2\hat{i} + 3\hat{j} - 5\hat{k})$ T. The magnitude of force acting on it is $\sqrt{\alpha} \times 10^{-6}$ N. The value of $\alpha$ is _______.
A moving coil of galvanometer when shunted with $2 \Omega$ resistance gives a full scale deflection for a current of 500 mA . When a resistance of $470 \Omega$ is connected in series it gives a full scale deflection for 10 V potential applied on it. The value of resistance of galvanometer coil is $\_\_\_\_ \Omega$.
A 5 mg particle carrying a charge of $5 \pi \times 10^{-6} \mathrm{C}$ is moving with velocity of $(3 \hat{i}+2 \hat{k}) \times 10^{-2} \mathrm{~m} / \mathrm{s}$ in a region having magnetic field $\vec{B}=0.1 \hat{k} \mathrm{~Wb} / \mathrm{m}^2$. It moves a distance of $\alpha$ meter along $\hat{k}$ when it completes 5 revolutions. The value of $\alpha$ is $\_\_\_\_$.
A circular coil of radius 2 cm and 125 turns carries a current of 1 A . The coil is placed in a uniform magnetic field of magnitude 0.4 T . The axis of the coil makes an angle of $30^{\circ}$ with the direction of the magnetic field. The torque acting on the coil is $\alpha \times 10^{-4} \mathrm{~N} . \mathrm{m}$. The value of $\alpha$ is $\_\_\_\_$ . $(\pi=3.14)$
The charged particle moving in a uniform magnetic field of $(3 \hat{i}+2 \hat{j}) \mathrm{T}$ has an acceleration $\left(4 \hat{i}-\frac{x}{2} \hat{j}\right) \mathrm{m} / \mathrm{s}^2$. The value of $x$ is
Distribution — advanced: 13 · easy: 29 · hard: 18 · medium: 40. Every question carries a source trace; each ends in an SME-verify solution.
Source: NCERT Ch 4 (§4.3)
Source: NCERT Ch 4 (§4.3, Eq. 4.5)
Source: NCERT Ch 4 (§4.3, Eq. 4.6a)
Source: NCERT Ch 4 (Example 4.3)
Source: NCERT Ch 4 (§4.3, Fig. 4.6)
Source: NCERT Ch 4 (§4.3, Eq. 4.6b)
Source: JEE pattern — Moving Charges
Source: JEE pattern — Moving Charges
Source: NCERT Ch 4 (§4.3)
Source: NCERT Ch 4 (Summary, cyclotron frequency)
Source: NCERT Ch 4 (§4.6, Eq. 4.13a)
Source: NCERT Ch 4 (§4.7, Eq. 4.16)
Source: NCERT Ch 4 (§4.7)
Source: NCERT Ch 4 (Example 4.8)
Source: NCERT Ch 4 (Example 4.7, Eq. 4.15b)
Source: NCERT Ch 4 (Example 4.7, Eq. 4.15a)
Source: NCERT Ch 4 (§4.6)
Source: JEE pattern — Ampere's law
Source: NCERT Ch 4 (§4.7)
Source: JEE pattern — Solenoid
Source: NCERT Ch 4 (§4.9, Eq. 4.21/4.24)
Source: NCERT Ch 4 (§4.9, Eq. 4.24)
Source: NCERT Ch 4 (§4.9.1)
Source: NCERT Ch 4 (§4.9.1)
Source: NCERT Ch 4 (§4.9.1)
Source: JEE pattern — Torque on loop
Source: JEE pattern — Magnetic dipole
Source: NCERT Ch 4 (§4.9, dimensions table)
Source: JEE pattern — Magnetic dipole
Source: NCERT Ch 4 (Example 4.10)
Source: NCERT Ch 4 (§4.2.2, Eq. 4.3)
Source: NCERT Ch 4 (§4.2.2)
Source: NCERT Ch 4 (Example 4.2)
Source: NCERT Ch 4 (§4.2.2)
Source: JEE pattern — Lorentz force
Source: NCERT Ch 4 (§4.2.2)
Source: NCERT Ch 4 (§4.2.2)
Source: NCERT Ch 4 (§4.3)
Source: NCERT Ch 4 (§4.2.2, Lorentz force)
Source: NCERT Ch 4 (§4.2.2)
Source: NCERT Ch 4 (Lorentz force application)
Source: NCERT Ch 4 (Lorentz force)
Source: JEE pattern — Velocity selector
Source: JEE pattern — Velocity selector
Source: JEE pattern — Applications
Source: JEE pattern — Velocity selector
Source: JEE pattern — Crossed fields
Source: JEE pattern — Velocity selector
Source: JEE pattern — Velocity selector
Source: JEE pattern — Velocity selector
Source: NCERT Ch 4 (§4.2.3, Eq. 4.4)
Source: NCERT Ch 4 (§4.2.3)
Source: JEE pattern — F = BIL
Source: NCERT Ch 4 (Example 4.1)
Source: NCERT Ch 4 (§4.2.3)
Source: NCERT Ch 4 (Exercise 4.6)
Source: NCERT Ch 4 (§4.2.3)
Source: JEE pattern — F = BIL sinθ
Source: NCERT Ch 4 (§4.2.3)
Source: NCERT Ch 4 (Example 4.9)
Source: NCERT Ch 4 (§4.8)
Source: NCERT Ch 4 (§4.8, Eq. 4.19)
Source: NCERT Ch 4 (§4.8)
Source: NCERT Ch 4 (Exercise 4.7)
Source: NCERT Ch 4 (§4.8, Eq. 4.19)
Source: NCERT Ch 4 (§4.8, Eq. 4.18)
Source: NCERT Ch 4 (§4.8)
Source: JEE pattern — Parallel currents
Source: JEE pattern — Parallel currents
Source: JEE pattern — Parallel currents
Source: NCERT Ch 4 (§4.4, Eq. 4.7a)
Source: NCERT Ch 4 (§4.4, Eq. 4.7b)
Source: NCERT Ch 4 (§4.4, Eq. 4.7c)
Source: NCERT Ch 4 (§4.4)
Source: NCERT Ch 4 (Example 4.5)
Source: NCERT Ch 4 (Example 4.5a)
Source: NCERT Ch 4 (§4.4)
Source: NCERT Ch 4 (Example 4.4)
Source: NCERT Ch 4 (§4.4, Eq. 4.7b)
Source: NCERT Ch 4 (§4.6, Eq. 4.14)
Source: NCERT Ch 4 (§4.5, Eq. 4.11)
Source: NCERT Ch 4 (§4.5, Eq. 4.12)
Source: NCERT Ch 4 (Example 4.6)
Source: NCERT Ch 4 (Example 4.6)
Source: NCERT Ch 4 (§4.9.2, Eq. 4.25a)
Source: JEE pattern — Circular loop
Source: JEE pattern — Arc field
Source: JEE pattern — Loop shapes
Source: NCERT Ch 4 (§4.5, Fig. 4.10)
Source: JEE pattern — Helmholtz coils
Source: NCERT Ch 4 (§4.10, Eq. 4.26)
Source: NCERT Ch 4 (§4.10)
Source: NCERT Ch 4 (§4.10, Eq. 4.27)
Source: NCERT Ch 4 (§4.10)
Source: NCERT Ch 4 (§4.10)
Source: NCERT Ch 4 (§4.10, Eq. 4.28)
Source: NCERT Ch 4 (§4.10)
Source: NCERT Ch 4 (Example 4.12)
Source: NCERT Ch 4 (§4.10)
Source: NCERT Ch 4 (Exercise 4.10)
MIT OCW / Walter Lewin, 3Blue1Brown, Khan, Veritasium. Each video is broken into clips — click a clip to play just that segment.
👁 Observe: The F = qv x B cross-product: why the force is perpendicular to both velocity and field, and does no work.
📚 Teaches: magnetic-force-on-a-charge
👁 Observe: Why a charge moves in a circle (or helix), and how radius r = mv/qB and the period are independent of speed.
📚 Teaches: motion-in-magnetic-field
👁 Observe: How F = I L x B arises from summing the magnetic force on the moving charges in a wire.
📚 Teaches: force-on-current-carrying-conductor
👁 Observe: Why like currents attract and unlike repel, and how this defines the ampere.
📚 Teaches: force-between-parallel-currents
👁 Observe: How each current element dl contributes dB, with the 1/r^2 and sin-theta dependence.
📚 Teaches: biot-savart-law
👁 Observe: Choosing an Amperian loop and deriving B = mu0 n I inside a long solenoid.
📚 Teaches: amperes-law-and-solenoid
👁 Observe: How tau = m x B acts on a current loop and links to the magnetic dipole moment.
📚 Teaches: torque-on-current-loop
👁 Observe: Why a charge moves in a circle (or helix), and how radius r = mv/qB and the period are independent of speed.
📚 Teaches: motion-in-magnetic-field
👁 Observe: How balancing qE against qvB lets only one speed v = E/B pass straight through.
📚 Teaches: velocity-selector
👁 Observe: Why like currents attract and unlike repel, and how this defines the ampere.
📚 Teaches: force-between-parallel-currents
👁 Observe: Setting up the axial-field integral and the result B = mu0 I R^2 / 2(R^2+x^2)^(3/2).
📚 Teaches: field-on-axis-of-circular-loop
👁 Observe: Choosing an Amperian loop and deriving B = mu0 n I inside a long solenoid.
📚 Teaches: amperes-law-and-solenoid
👁 Observe: How tau = m x B acts on a current loop and links to the magnetic dipole moment.
📚 Teaches: torque-on-current-loop
👁 Observe: How torque balances the restoring spring so deflection is proportional to current; sensitivity.
📚 Teaches: moving-coil-galvanometer
👁 Observe: The F = qv x B cross-product: why the force is perpendicular to both velocity and field, and does no work.
📚 Teaches: magnetic-force-on-a-charge
👁 Observe: Why a charge moves in a circle (or helix), and how radius r = mv/qB and the period are independent of speed.
📚 Teaches: motion-in-magnetic-field
👁 Observe: How balancing qE against qvB lets only one speed v = E/B pass straight through.
📚 Teaches: velocity-selector
Full lecture — no clip index.
👁 Observe: Why like currents attract and unlike repel, and how this defines the ampere.
📚 Teaches: force-between-parallel-currents
👁 Observe: How each current element dl contributes dB, with the 1/r^2 and sin-theta dependence.
📚 Teaches: biot-savart-law
👁 Observe: Setting up the axial-field integral and the result B = mu0 I R^2 / 2(R^2+x^2)^(3/2).
📚 Teaches: field-on-axis-of-circular-loop
Full lecture — no clip index.
👁 Observe: Choosing an Amperian loop and deriving B = mu0 n I inside a long solenoid.
📚 Teaches: amperes-law-and-solenoid
👁 Observe: How torque balances the restoring spring so deflection is proportional to current; sensitivity.
📚 Teaches: moving-coil-galvanometer
👁 Observe: How balancing qE against qvB lets only one speed v = E/B pass straight through.
📚 Teaches: velocity-selector
👁 Observe: Setting up the axial-field integral and the result B = mu0 I R^2 / 2(R^2+x^2)^(3/2).
📚 Teaches: field-on-axis-of-circular-loop
👁 Observe: How torque balances the restoring spring so deflection is proportional to current; sensitivity.
📚 Teaches: moving-coil-galvanometer
👁 Observe: The F = qv x B cross-product: why the force is perpendicular to both velocity and field, and does no work.
📚 Teaches: magnetic-force-on-a-charge
👁 Observe: How F = I L x B arises from summing the magnetic force on the moving charges in a wire.
📚 Teaches: force-on-current-carrying-conductor
👁 Observe: How each current element dl contributes dB, with the 1/r^2 and sin-theta dependence.
📚 Teaches: biot-savart-law
👁 Observe: How F = I L x B arises from summing the magnetic force on the moving charges in a wire.
📚 Teaches: force-on-current-carrying-conductor
👁 Observe: How tau = m x B acts on a current loop and links to the magnetic dipole moment.
📚 Teaches: torque-on-current-loop
Answers are grounded in tier-1 material (NCERT / JEE PYQ) and cited — the AI will say so if a doubt isn't in the indexed syllabus.
Students read $\theta$ in $\tau = mB\sin\theta$ as the angle between the plane of the loop and $\vec{B}$, when it is actually the angle between the area vector (normal) $\vec{m}$ and $\vec{B}$. Because the two angles are complementary, they end up computing $mB\cos\theta$ and get maximum torque and zero torque exactly backwards.
Fix: Always draw the normal to the loop first, then measure the angle to $\vec{B}$ from that normal. Remember the check: torque is maximum when $\vec{B}$ lies IN the plane of the loop ($\theta = 90^{\circ}$) and zero when $\vec{B}$ is perpendicular to the plane ($\theta = 0$).
The right-hand rule gives the direction of $\vec{v} \times \vec{B}$, but students forget that for a negative charge the force is opposite to that. They then send an electron curving the same way as a proton, flipping the whole trajectory.
Fix: Split the work: get $\vec{v} \times \vec{B}$ by the hand rule with zero regard for the charge, then multiply by $q$ including its sign. For any negative charge, reverse the arrow you just found.
Because a magnetic field clearly bends particle paths, students assume it also changes their kinetic energy or speed, and try to compute work done by the magnetic force on a free charge.
Fix: The magnetic force is always perpendicular to velocity, so $\vec{F} \cdot \vec{v} = 0$ and it does zero work; speed and $|\vec{v}|$ stay constant while only direction changes. Any change in kinetic energy must come from an electric field or an external source, never from $q\,\vec{v} \times \vec{B}$ itself.
Seeing that $\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{\text{enc}}$ is always true, students try to use it to find $B$ at the centre of a single loop or near a finite wire, and get stuck or invent a wrong constant $B$.
Fix: The law only yields $B$ when symmetry lets you pull $B$ out of the integral: infinite straight wire, long solenoid, toroid. For a finite wire or a single loop, abandon Ampère's law and integrate the Biot-Savart law instead.
Students expect a fatter solenoid to give a different interior field, and try to insert the cross-sectional area or radius into $B = \mu_0 n I$.
Fix: Inside a long solenoid $B = \mu_0 n I$ depends only on turns per unit length $n$ and current $I$, never on the radius. Radius enters the flux $\Phi = BA$ and the inductance, but the field itself is uniform and radius-free across the interior.
Under exam pressure students put a large resistance in parallel to make an ammeter, or a small resistance in series to make a voltmeter, exactly reversing the two conversions.
Fix: Ammeter needs low resistance, so a SMALL shunt goes in PARALLEL with the galvanometer. Voltmeter needs high resistance, so a LARGE resistor goes in SERIES. Tie it to purpose: an ammeter must not throttle the current it measures; a voltmeter must not drain the branch it reads.
When a charge enters a field at an angle, students plug the full speed $v$ into $r = mv/(qB)$ and forget that the parallel velocity component produces forward drift, so they cannot separate the circular size from the axial advance.
Fix: Resolve the velocity first. Use only $v_{\perp}$ for the radius $r = m v_{\perp}/(qB)$, and only $v_{\parallel}$ for the pitch $p = 2\pi m v_{\parallel}/(qB)$. The full speed appears in neither unless the entry is exactly perpendicular.
Since the orbit radius grows with speed, students conclude the revolution frequency must change too, and worry that the accelerating voltage will fall out of step immediately.
Fix: Non-relativistically, $f_c = qB/(2\pi m)$ is independent of speed and radius, which is exactly why a fixed-frequency cyclotron works. Frequency only starts to drop when relativistic mass increase becomes significant, which is when a synchrocyclotron is required.
A cyclotron with dee radius $R=0.50\,\mathrm{m}$ operates in a uniform field $B=1.5\,\mathrm{T}$ and accelerates protons ($m=1.67\times10^{-27}\,\mathrm{kg}$, $q=1.6\times10^{-19}\,\mathrm{C}$). Ignoring relativistic effects, find the maximum kinetic energy of the emerging protons, in MeV.
JEE-Advanced pattern (NCERT Ch 4)
Singly charged ions first pass a velocity selector with $E=2.0\times10^{5}\,\mathrm{V/m}$ and $B_1=0.10\,\mathrm{T}$ (crossed). They then enter a region of field $B_2=0.20\,\mathrm{T}$ perpendicular to their velocity and bend into a circle of radius $R=0.50\,\mathrm{m}$. Find the mass of an ion.
JEE-Advanced pattern (NCERT Ch 4)
A proton moves with speed $v=4.0\times10^{6}\,\mathrm{m/s}$ making an angle of $30^{\circ}$ with a uniform field $B=0.30\,\mathrm{T}$. Find the radius of the helical path and its pitch. Take $m=1.67\times10^{-27}\,\mathrm{kg}$, $q=1.6\times10^{-19}\,\mathrm{C}$.
JEE-Advanced pattern (NCERT Ch 4)
Two long straight parallel wires $5.0\,\mathrm{cm}$ apart carry currents $10\,\mathrm{A}$ and $15\,\mathrm{A}$ in the same direction. Find the magnitude and nature of the force per unit length between them.
JEE-Advanced pattern (NCERT Ch 4)
A conducting rod of mass $m=0.10\,\mathrm{kg}$ rests on two frictionless rails separated by $L=0.50\,\mathrm{m}$ on a plane inclined at $\theta=30^{\circ}$ to the horizontal. A uniform vertical field $B=0.50\,\mathrm{T}$ acts everywhere. A current $I$ flows along the rod. Find the current that keeps the rod in equilibrium. Take $g=9.8\,\mathrm{m/s^{2}}$.
JEE-Advanced pattern (NCERT Ch 4)
A proton enters, moving perpendicular to the boundary, a slab-shaped region of uniform field $B=0.20\,\mathrm{T}$ of width $w=5.0\,\mathrm{cm}$, with speed $v=1.0\times10^{6}\,\mathrm{m/s}$. Does it emerge from the far side, and if so at what angle to its original direction? Take $m=1.67\times10^{-27}\,\mathrm{kg}$, $q=1.6\times10^{-19}\,\mathrm{C}$.
JEE-Advanced pattern (NCERT Ch 4)
A circular coil of $N=100$ turns and radius $R=0.10\,\mathrm{m}$ carries current $I=2.0\,\mathrm{A}$. Find the magnetic field on its axis at a distance $x=0.10\,\mathrm{m}$ from the centre.
JEE-Advanced pattern (NCERT Ch 4)
A rectangular coil of sides $0.10\,\mathrm{m}\times0.20\,\mathrm{m}$ has $N=50$ turns and carries $I=2.0\,\mathrm{A}$ in a uniform field $B=0.50\,\mathrm{T}$, with the plane of the coil parallel to the field. Find the magnetic moment of the coil and the torque on it.
JEE-Advanced pattern (NCERT Ch 4)
An electron passes undeflected through a region of crossed fields with $E=3.0\times10^{4}\,\mathrm{V/m}$ and $B=1.0\times10^{-2}\,\mathrm{T}$ perpendicular to each other and to the velocity. Find the electron's speed. If the electric field is then switched off, find the radius of its circular path. Take $m=9.1\times10^{-31}\,\mathrm{kg}$, $e=1.6\times10^{-19}\,\mathrm{C}$.
JEE-Advanced pattern (NCERT Ch 4)
A toroid of mean radius $r=0.15\,\mathrm{m}$ has $N=1000$ turns and carries a current $I=3.0\,\mathrm{A}$. Find the magnetic field along the central circle of the core.
JEE-Advanced pattern (NCERT Ch 4)
A wire carrying current $I=10\,\mathrm{A}$ is bent into an arc subtending three-quarters of a full circle (a $270^{\circ}$ arc) of radius $R=0.050\,\mathrm{m}$. Find the magnetic field at the centre of the arc.
JEE-Advanced pattern (NCERT Ch 4)
A proton is accelerated to a kinetic energy of $50\,\mathrm{MeV}$; its rest energy is $938\,\mathrm{MeV}$. By what factor does its cyclotron frequency in a fixed field differ from the low-energy value, and what does this imply for a simple cyclotron?
JEE-Advanced pattern (relativistic cyclotron)
Two protons of different speeds both enter a uniform field $B=0.50\,\mathrm{T}$ perpendicular to their velocities and each traverses a semicircle before leaving. Show that they spend equal times inside, and compute that time. Take $m=1.67\times10^{-27}\,\mathrm{kg}$, $q=1.6\times10^{-19}\,\mathrm{C}$.
JEE-Advanced pattern (NCERT Ch 4)
A flat copper strip of thickness $t=0.20\,\mathrm{mm}$ carries a current $I=5.0\,\mathrm{A}$ in a perpendicular field $B=0.50\,\mathrm{T}$. The carrier density is $n=8.5\times10^{28}\,\mathrm{m^{-3}}$ and each carrier has charge $e=1.6\times10^{-19}\,\mathrm{C}$. Find the Hall voltage across the strip.
JEE-Advanced pattern (Hall effect)
A straight wire of mass per unit length $\lambda=50\,\mathrm{g/m}$ is to be suspended in mid-air by passing a current through it in a horizontal uniform field $B=0.40\,\mathrm{T}$ perpendicular to the wire. Find the required current. Take $g=9.8\,\mathrm{m/s^{2}}$.
JEE-Advanced pattern (NCERT Ch 4, Example 4.1 style)
| Chapter-mock score | Percentile band | Projected AIR band |
|---|---|---|
| 90-100% | 99.5+ percentile | $\lt 1000$ |
| 80-89% | 99.0-99.5 percentile | $1000-3000$ |
| 70-79% | 98.0-99.0 percentile | $3000-8000$ |
| 55-69% | 95.0-98.0 percentile | $8000-20000$ |
| 40-54% | 90.0-95.0 percentile | $20000-50000$ |
| 25-39% | 80.0-90.0 percentile | $50000-120000$ |
| 0-24% | $\lt 80$ percentile | $\gt 120000$ |
Indicative - based on historical JEE marks→percentile→JoSAA closing-rank trends
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Authoritative & comprehensive JEE Main + Advanced resource · sources traced Tier 1–3 · SME-review state (append ?review=1)