Chapter 4 · Moving Charges and Magnetism

4.1 The Magnetic Force and Motion of a Charge in a Magnetic Field

A magnetic field pushes only on charges that move, always at right angles to their velocity: $\mathbf F=q\mathbf v\times\mathbf B$. The same force acts on a current-carrying wire, and it bends free charges into circles and helices.

Near the North and South Poles the night sky sometimes fills with moving curtains of green and red light — the aurora. The light comes from oxygen and nitrogen atoms high in the atmosphere, struck by fast electrons and protons that have streamed out from the Sun. Why only near the poles? Because those particles do not travel in straight lines once they reach the Earth's magnetic field: they are bent into tight spirals that wind along the field lines, and the field lines dip into the atmosphere near the magnetic poles. A hospital MRI scanner, a particle accelerator at CERN and the electric motor in a ceiling fan all rely on the same push — the force a magnetic field exerts on a moving charge.

In Chapter 1 charges at rest produced and felt electric fields. In 1820 Hans Christian Ørsted noticed that a compass needle swings when a current flows in a nearby wire: moving charges also produce a magnetic field, and — as this section shows — moving charges also feel one. We start from the force on a single moving charge, the Lorentz force, use it to define the field and its unit, extend it to the force on a current-carrying wire, and then follow a charged particle as the field bends its path into a circle or a helix. (Where the magnetic field itself comes from is the subject of Sections 4.2 and 4.3.)

The Magnetic Field and the Lorentz Force

Experiments with charged particles in the space near a magnet, or near a current, show a force with four unusual features. It is proportional to the charge $q$ and to its speed $v$: a charge at rest feels nothing. It depends on the direction of motion: there is one direction through each point along which a moving charge feels no force at all, and the force is largest for motion at right angles to that direction. And the force is always perpendicular both to the velocity and to that special direction. A vector product describes all four features at once, and the special direction is the direction of the magnetic field $\mathbf B$.

📘 Definition 4.1 — Magnetic field and the tesla
The magnetic field $\mathbf B$ at a point is the vector for which a charge $q$ moving through the point with velocity $\mathbf v$ experiences the magnetic force $$\mathbf F=q\,\mathbf v\times\mathbf B,\qquad F=qvB\sin\theta,$$ where $\theta$ is the angle between $\mathbf v$ and $\mathbf B$. A charge moving along $\mathbf B$ feels no magnetic force, which fixes the line of $\mathbf B$; the sign of the force on a positive charge fixes which way along that line $\mathbf B$ points. The SI unit of $B$ is the tesla (T): a field of $1$ T exerts a force of $1$ N on a charge of $1$ C moving at $1$ m/s at right angles to it, so $$1\ \text{T}=1\ \frac{\text{N}}{\text{C}\cdot\text{m/s}}=1\ \text{N A}^{-1}\text{m}^{-1}.$$

The tesla is a large unit. The Earth's field at the surface is a few times $10^{-5}$ T, a fridge magnet gives about $10^{-2}$ T, a strong laboratory electromagnet about $1$ T and an MRI scanner $1.5$–$3$ T. A smaller unit, the gauss ($1\ \text{G}=10^{-4}$ T), is still common in older books and in geophysics.

When an electric field is also present, the two forces simply add.

📐 Law 4.1 — The Lorentz force

A point charge $q$ moving with velocity $\mathbf v$ through a region where the electric field is $\mathbf E$ and the magnetic field is $\mathbf B$ experiences the force $$\mathbf F=q\left(\mathbf E+\mathbf v\times\mathbf B\right).$$ The magnetic part $q\mathbf v\times\mathbf B$:

1. is zero if the charge is at rest, or if it moves parallel or antiparallel to $\mathbf B$;

2. is perpendicular to both $\mathbf v$ and $\mathbf B$, with the direction given by the right-hand rule for $\mathbf v\times\mathbf B$ when $q\gt0$, and the opposite direction when $q\lt0$;

3. does no work on the charge, because it is always perpendicular to the velocity. It can change the direction of motion but never the speed or the kinetic energy.

The direction rule is the one for any cross product: point the fingers of the right hand along $\mathbf v$, curl them towards $\mathbf B$ through the smaller angle, and the thumb gives $\mathbf v\times\mathbf B$. For a negative charge, such as an electron, reverse the answer. In diagrams a field pointing out of the page is drawn as dots ⊙ (the tip of an arrow coming towards you) and a field into the page as crosses ⊗ (the tail feathers of an arrow going away). In components, with $\hat{\mathbf i}\times\hat{\mathbf j}=\hat{\mathbf k}$, $\hat{\mathbf j}\times\hat{\mathbf k}=\hat{\mathbf i}$, $\hat{\mathbf k}\times\hat{\mathbf i}=\hat{\mathbf j}$ and reversing the order changing the sign.

✏️ Example 1 — Force on a proton and an electron
A proton moves at $2.0\times10^6$ m/s at $30^\circ$ to a uniform magnetic field of $0.50$ T. (a) Find the magnetic force on it and its acceleration. (b) An electron moves with the same velocity at the same point. How does the force on it compare? (c) What is the largest force the proton could feel at this speed? Take $m_p=1.67\times10^{-27}$ kg.
Concept $F=|q|vB\sin\theta$, perpendicular to the plane containing $\mathbf v$ and $\mathbf B$.
(a) Proton $F=(1.6\times10^{-19})(2.0\times10^6)(0.50)\sin30^\circ$ $=8.0\times10^{-14}$ N, along $\mathbf v\times\mathbf B$. $a=\dfrac{F}{m_p}=\dfrac{8.0\times10^{-14}}{1.67\times10^{-27}}$ $\approx4.8\times10^{13}\ \text{m/s}^2$, perpendicular to the velocity.
(b) Electron Same $|q|$, same $v$ and $\theta$, so the same magnitude $8.0\times10^{-14}$ N — but in the opposite direction, because $q$ is negative. (Its acceleration is $1836$ times larger, since its mass is that much smaller.)
(c) Maximum At $\theta=90^\circ$: $F_\text{max}=evB=1.6\times10^{-13}$ N.
✏️ Example 2 — The Lorentz force in components
An electron moves with velocity $\mathbf v=(3.0\,\hat{\mathbf i}+4.0\,\hat{\mathbf j})\times10^5$ m/s in a magnetic field $\mathbf B=0.20\,\hat{\mathbf k}$ T. (a) Find the magnetic force on it as a vector, and its magnitude. (b) What electric field $\mathbf E$, switched on at the same place, would make the total Lorentz force on the electron zero at this instant?
Cross product Using $\hat{\mathbf i}\times\hat{\mathbf k}=-\hat{\mathbf j}$ and $\hat{\mathbf j}\times\hat{\mathbf k}=\hat{\mathbf i}$: $$\begin{aligned}\mathbf v\times\mathbf B&=(2.0\times10^4)\left[3.0\,(\hat{\mathbf i}\times\hat{\mathbf k})+4.0\,(\hat{\mathbf j}\times\hat{\mathbf k})\right]\\&=(8.0\,\hat{\mathbf i}-6.0\,\hat{\mathbf j})\times10^4\ \text{T m/s}.\end{aligned}$$
(a) Force $q=-1.6\times10^{-19}$ C: $$\mathbf F=q\,\mathbf v\times\mathbf B=(-1.28\,\hat{\mathbf i}+0.96\,\hat{\mathbf j})\times10^{-14}\ \text{N},$$ with magnitude $\sqrt{1.28^2+0.96^2}\times10^{-14}=1.6\times10^{-14}$ N.
(b) Cancel it Zero total force needs $\mathbf E+\mathbf v\times\mathbf B=\mathbf 0$, i.e. $\mathbf E=-\mathbf v\times\mathbf B=(-8.0\,\hat{\mathbf i}+6.0\,\hat{\mathbf j})\times10^4$ N/C, of magnitude $1.0\times10^5$ N/C. The answer does not depend on the sign of the charge.
Check $\mathbf v\perp\mathbf B$ here, so $|\mathbf F|=evB$ $=(1.6\times10^{-19})(5.0\times10^5)(0.20)$ $=1.6\times10^{-14}$ N. ✓ Also $\mathbf F\cdot\mathbf v\propto(-1.28)(3.0)+(0.96)(4.0)$ $=0$: the force is perpendicular to the velocity, as it must be.

Force on a Current-Carrying Conductor

A current in a wire is a stream of moving charges (Chapter 3 — Current Electricity), so a wire in a magnetic field should feel the sum of the forces on all of them. Take a straight piece of wire of length $l$ and cross-sectional area $A$, in a uniform field $\mathbf B$. Suppose it contains $n$ mobile charge carriers per unit volume, each of charge $q$, drifting with the average velocity $\mathbf v_d$. The magnetic force on one carrier is $q\mathbf v_d\times\mathbf B$, and the piece contains $nAl$ carriers, so the total force is $$\mathbf F=(nAl)\,q\,\mathbf v_d\times\mathbf B=(nqAv_d)\,\mathbf l\times\mathbf B,$$ where $\mathbf l$ is a vector of length $l$ pointing along the wire in the direction of $\mathbf v_d$ — for positive carriers, the direction of the current. The bracket $nqAv_d$ is exactly the current $I$. (For electrons $q$ and $\mathbf v_d$ both change sign, so the product is unchanged: the force depends only on the conventional current.) The force on the carriers is passed on to the lattice of the wire, so the wire itself is pushed.

📐 Result 4.2 — Force on a current-carrying conductor

A straight conductor of length $l$ carrying a steady current $I$ in a uniform magnetic field $\mathbf B$ experiences the force $$\mathbf F=I\,\mathbf l\times\mathbf B,\qquad F=IlB\sin\theta,$$ where $\mathbf l$ points along the conductor in the direction of the current and $\theta$ is the angle between the conductor and $\mathbf B$. The force is perpendicular to both the wire and the field; it is zero when the wire lies along $\mathbf B$ and greatest, $IlB$, when the wire is perpendicular to $\mathbf B$. For a wire of any shape, or a non-uniform field, add the forces $I\,d\mathbf l\times\mathbf B$ on short pieces $d\mathbf l$.

Electric-motor engineers often use an equivalent finger rule, Fleming's left-hand rule: hold the thumb, first finger and second finger of the left hand mutually at right angles; point the First finger along the Field and the seCond finger along the Current; the thuMb then gives the direction of the force, i.e. of the Motion. It always agrees with $\mathbf F=I\mathbf l\times\mathbf B$; use whichever you find more reliable, but check the answer with the cross product when the geometry is three-dimensional.

A useful consequence: in a uniform field the forces on a closed loop of any shape add to zero, because the vectors $d\mathbf l$ round a closed loop add to zero. A loop in a uniform field can only be turned — the torque on it is the subject of Section 4.4. Similarly, a curved wire from $A$ to $B$ in a uniform field feels the same force as a straight wire from $A$ to $B$ carrying the same current.

✏️ Example 3 — Force on a wire at an angle to the field
A uniform horizontal magnetic field of $0.30$ T points due east. A straight horizontal wire $0.50$ m long carries a current of $4.0$ A in a direction $60^\circ$ north of east. Find the magnitude and direction of the force on the wire. What would the force be if the wire pointed due north? Due east?
Magnitude $F=IlB\sin\theta$ $=(4.0)(0.50)(0.30)\sin60^\circ$ $=0.60\times0.866\approx0.52$ N.
Direction Take $\hat{\mathbf i}$ east, $\hat{\mathbf j}$ north, $\hat{\mathbf k}$ vertically up. $\mathbf l=0.50(\cos60^\circ\,\hat{\mathbf i}+\sin60^\circ\,\hat{\mathbf j})$, $\mathbf B=0.30\,\hat{\mathbf i}$. Only the north part of $\mathbf l$ contributes: $\hat{\mathbf j}\times\hat{\mathbf i}=-\hat{\mathbf k}$, so the force is vertically downwards. (Fleming: first finger east, second finger roughly north, thumb points down.)
Other directions Due north ($\theta=90^\circ$): the maximum, $IlB=0.60$ N, downwards. Due east ($\theta=0^\circ$): zero — the wire is along the field.
Check Interactive 1 uses this wire ($I=4.0$ A, $B=0.30$ T): set $\theta=60^\circ$ and it reads $F=0.520$ N, into the page (the page is the horizontal plane seen from above, so "into the page" is downwards). ✓
✏️ Example 4 — Floating a rod on a magnetic field
A horizontal metal rod of length $40$ cm and mass $60$ g hangs from two light flexible leads in a uniform horizontal magnetic field of $0.50$ T that points due north. The rod lies east–west. What current, in which direction, makes the tension in the leads zero? ($g=9.8\ \text{m/s}^2$)
Condition The tension is zero when the magnetic force is upwards and equal to the weight: $IlB=mg$ (the rod is perpendicular to $\mathbf B$, so $\sin\theta=1$).
Current $I=\dfrac{mg}{lB}=\dfrac{(0.060)(9.8)}{(0.40)(0.50)}=\dfrac{0.588}{0.20}\approx2.9$ A.
Direction We need $\mathbf l\times\mathbf B$ upwards with $\mathbf B$ north: $\hat{\mathbf i}\times\hat{\mathbf j}=\hat{\mathbf k}$, so the current must flow east. A current flowing west would double the tension instead.
Interactive 1 — The Force on a Wire in a Magnetic Field

You look down on a horizontal wire $0.50$ m long in a uniform field $\mathbf B$ pointing to the right. The force $\mathbf F=I\mathbf l\times\mathbf B$ is perpendicular to the page: ⊙ means out of the page (up), ⊗ into the page (down), and the size of the symbol shows $|\mathbf F|$. Turn the wire with the slider or by dragging it, reverse the current, and watch the marker on the graph of $F/(IlB)=\sin\theta$. Press Play to rotate the wire.

4.00.30
angle $\theta$—
$F=IlB\sin\theta$ (N)—
direction of $\mathbf F$—
$F_\text{max}=IlB$ (N)—
⚠️ Common mistakes with the magnetic force

1. Expecting a force on a charge at rest. A magnetic field exerts no force on a stationary charge, however strong the field. Only moving charges (and currents) feel it.

2. Putting the force along the field. Unlike $q\mathbf E$, the magnetic force is never along $\mathbf B$; it is perpendicular to both $\mathbf B$ and $\mathbf v$ (or the wire).

3. Forgetting the sign of the charge. The right-hand rule gives $\mathbf v\times\mathbf B$; for an electron the force is opposite to it.

4. Using $\cos\theta$. The force is $qvB\sin\theta$ and $IlB\sin\theta$, with $\theta$ measured between $\mathbf v$ (or the wire) and $\mathbf B$: it vanishes for motion along the field.

5. Claiming the field speeds a particle up. The magnetic force does no work. Any change in kinetic energy must come from an electric field.

Motion of a Charge in a Uniform Magnetic Field

Because the magnetic force is always perpendicular to the velocity, it behaves exactly like the tension in the string of a whirling stone: it changes the direction of motion and leaves the speed alone. Suppose a particle of mass $m$ and charge $q$ moves with speed $v$ perpendicular to a uniform field $\mathbf B$. The force $qvB$ has constant size and always points sideways, so the particle moves in a circle in the plane perpendicular to $\mathbf B$. The magnetic force supplies the centripetal force (Class 11 Physics, Chapter 3 — circular motion): $$\frac{mv^2}{r}=qvB\quad\Longrightarrow\quad r=\frac{mv}{qB}=\frac{p}{qB},$$ where $p=mv$ is the momentum. The time for one revolution is $$T=\frac{2\pi r}{v}=\frac{2\pi m}{qB},$$ and the speed has cancelled. A faster particle moves on a bigger circle, but it takes exactly the same time to go round. This surprising fact is what makes the cyclotron possible (see the 🚀 box below).

If the velocity makes an angle $\theta$ with $\mathbf B$, split it into a component $v_\parallel=v\cos\theta$ along the field and a component $v_\perp=v\sin\theta$ perpendicular to it. The force $q\mathbf v\times\mathbf B$ involves only $v_\perp$, so $v_\parallel$ never changes, while $v_\perp$ turns round in a circle exactly as before. Steady drift along $\mathbf B$ plus circling round it gives a helix wound round a field line — the spiral of the aurora particles.

📐 Result 4.3 — Circular and helical motion in a uniform magnetic field

A particle of mass $m$ and charge $q$ in a uniform magnetic field $\mathbf B$ (no other forces) keeps a constant speed. If its velocity $\mathbf v$ makes an angle $\theta$ with $\mathbf B$ ($v_\perp=v\sin\theta$, $v_\parallel=v\cos\theta$), it moves on a helix whose axis is parallel to $\mathbf B$, with

$$r=\frac{mv_\perp}{|q|B},\qquad T=\frac{2\pi m}{|q|B},\qquad \nu=\frac1T=\frac{|q|B}{2\pi m},\qquad \omega=\frac{|q|B}{m},$$

and pitch (the distance moved along $\mathbf B$ in one revolution) $$p_\text{h}=v_\parallel T=\frac{2\pi m v\cos\theta}{|q|B}.$$ For $\theta=90^\circ$ the helix becomes a circle of radius $mv/(|q|B)$; for $\theta=0$ the particle moves undeflected along $\mathbf B$. The period and frequency do not depend on the speed. Seen with $\mathbf B$ pointing into the page, positive charges go round anticlockwise and negative charges clockwise.

🧭 Problem-solving strategy — a charged particle in a uniform magnetic field
  1. Find the angle $\theta$ between $\mathbf v$ and $\mathbf B$ and split $v$ into $v_\perp=v\sin\theta$ and $v_\parallel=v\cos\theta$.
  2. If you are given the kinetic energy $K$ (or an accelerating potential difference $V$, with $K=|q|V$), use $p=mv=\sqrt{2mK}$, so that $r=\dfrac{\sqrt{2mK}}{|q|B}$ for perpendicular motion.
  3. Radius from $v_\perp$; period from $2\pi m/(|q|B)$ (no speed needed); pitch from $v_\parallel T$.
  4. Decide the sense of rotation from the force at one point: $q\mathbf v\times\mathbf B$ must point towards the centre of the circle.
✏️ Example 5 — An electron circling in a weak field
An electron moving at $2.0\times10^6$ m/s enters a uniform magnetic field of $1.0$ mT at right angles to the field. (a) Find the radius of its path. (b) Find the period and the frequency of revolution. (c) What is its kinetic energy in electron volts, and does it change? ($m_e=9.1\times10^{-31}$ kg)
(a) Radius $$r=\frac{m_ev}{eB}=\frac{(9.1\times10^{-31})(2.0\times10^6)}{(1.6\times10^{-19})(1.0\times10^{-3})}\approx1.1\times10^{-2}\ \text{m}=1.1\ \text{cm}.$$
(b) Period $T=\dfrac{2\pi m_e}{eB}=\dfrac{2\pi(9.1\times10^{-31})}{1.6\times10^{-22}}\approx3.6\times10^{-8}$ s, so $\nu=1/T\approx2.8\times10^{7}$ Hz $=28$ MHz.
(c) Energy $K=\tfrac12m_ev^2$ $=\tfrac12(9.1\times10^{-31})(2.0\times10^6)^2$ $=1.82\times10^{-18}$ J $=\dfrac{1.82\times10^{-18}}{1.6\times10^{-19}}$ eV $\approx11$ eV. It stays the same for ever: the magnetic force does no work.
Check Interactive 2, preset “Example 5”, draws this circle and reads $r=1.14$ cm and $T=3.57\times10^{-8}$ s. ✓
✏️ Example 6 — A proton on a helix
A proton moves at $4.0\times10^5$ m/s at $60^\circ$ to a uniform magnetic field of $0.10$ T. Find (a) the radius of its helical path, (b) the period of revolution, (c) the pitch of the helix, and (d) how many turns it makes while moving $1.0$ m along the field. ($m_p=1.67\times10^{-27}$ kg)
Split $v_\perp=v\sin60^\circ\approx3.46\times10^5$ m/s, $v_\parallel=v\cos60^\circ=2.0\times10^5$ m/s.
(a) Radius $r=\dfrac{m_pv_\perp}{eB}$ $=\dfrac{(1.67\times10^{-27})(3.46\times10^5)}{(1.6\times10^{-19})(0.10)}$ $\approx3.6\times10^{-2}$ m $=3.6$ cm.
(b) Period $T=\dfrac{2\pi m_p}{eB}=\dfrac{2\pi(1.67\times10^{-27})}{1.6\times10^{-20}}\approx6.6\times10^{-7}$ s.
(c) Pitch $p_\text{h}=v_\parallel T=(2.0\times10^5)(6.56\times10^{-7})\approx0.13$ m $=13$ cm.
(d) Turns $\dfrac{1.0}{0.131}\approx7.6$ turns. Check: Interactive 2, preset “Example 6”, shows $r=3.62$ cm, $T=6.56\times10^{-7}$ s, pitch $13.1$ cm. ✓
Interactive 2 — A Charged Particle in a Magnetic Field

Left: a 3-D view with $\mathbf B$ pointing to the right. Right: the same path seen looking along $\mathbf B$ (field into the page). Change the particle, its speed, the angle between $\mathbf v$ and $\mathbf B$ and the field strength; the drawing is rescaled to fit, and the readouts give the real radius, period and pitch. The last setting (🚀 beyond the board syllabus) adds an electric field at right angles to $\mathbf B$ — a velocity selector.

—90—
radius r—
period T—
pitch—
kinetic energy—
🚀 Beyond the board syllabus (JEE Main · NEET · JEE Advanced) — the velocity selector and the cyclotron

Not required by the CBSE syllabus. The force on a moving charge is listed by JEE Main, NEET and JEE Advanced (JEE Main and NEET name crossed electric and magnetic fields), and the velocity selector applies it directly. The cyclotron is not named in any 2026 syllabus; read that part as background.

Velocity selector. Let a positive charge move with velocity $\mathbf v$ through a region where a uniform $\mathbf E$ and a uniform $\mathbf B$ are perpendicular to each other and to $\mathbf v$, arranged so that $q\mathbf E$ and $q\mathbf v\times\mathbf B$ point in opposite directions. The forces balance, and the particle goes straight through, only if $qE=qvB$, i.e. $$v=\frac EB,$$ whatever its charge or mass. Faster particles are bent one way (the magnetic force wins), slower ones the other way. A slit at the far end therefore passes only particles of one speed — the first stage of a mass spectrometer, and the idea behind J. J. Thomson's 1897 measurement of $e/m$ for the electron.

Cyclotron. Because the period $T=2\pi m/(qB)$ does not depend on speed, ions circling in a uniform field between two hollow D-shaped electrodes ("dees") arrive at the gap between them at regular intervals. An alternating voltage of the cyclotron frequency $\nu_c=\dfrac{qB}{2\pi m}$ then gives them a kick every half-turn; they spiral outwards on ever larger semicircles and leave at the rim, radius $R$, with $v=qBR/m$ and $$K_\text{max}=\frac{q^2B^2R^2}{2m}.$$ For protons with $B=1.5$ T and $R=0.50$ m: $\nu_c\approx23$ MHz and $K_\text{max}\approx4.3\times10^{-12}$ J $\approx27$ MeV. (At much higher energies the mass increases with speed — relativity — and the timing fails.) Cyclotrons produce the short-lived isotopes used in PET scans in hospitals. Neither device is in the current NCERT syllabus; JEE Main and NEET may still set problems on them.

Exercises

I
The force on a moving charge. Use $e=1.6\times10^{-19}$ C; give directions where asked.
  • 1 A proton moves at $3.0\times10^6$ m/s perpendicular to a field of $0.40$ T. Find the magnetic force on it.
  • 2 An electron moves at $5.0\times10^6$ m/s at $30^\circ$ to a field of $2.0$ mT. Find the magnitude of the force.
  • 3 A charged particle moves through a magnetic field and feels no force. Give two possible reasons.
  • 4 An alpha particle ($q=2e$) moving at $1.5\times10^7$ m/s feels a maximum magnetic force of $4.8\times10^{-12}$ N. Find $B$.
  • 5 A charge $q=+2.0\ \mu\text{C}$ has velocity $\mathbf v=2.0\times10^5\,\hat{\mathbf i}$ m/s in $\mathbf B=0.50\,\hat{\mathbf j}$ T. Find the force vector.
  • 6 A charge $q=-3.0\ \mu\text{C}$ has $\mathbf v=(2.0\,\hat{\mathbf i}+3.0\,\hat{\mathbf j})\times10^4$ m/s in $\mathbf B=0.40\,\hat{\mathbf i}$ T. Find the force vector.
  • 7 Show that a magnetic force can never change the kinetic energy of a charged particle.
  • 8 A proton moves with $\mathbf v=1.0\times10^5\,\hat{\mathbf i}$ m/s where $\mathbf E=3.0\times10^4\,\hat{\mathbf j}$ N/C and $\mathbf B=0.20\,\hat{\mathbf k}$ T. Find the Lorentz force on it.
1 $F=evB=(1.6\times10^{-19})(3.0\times10^6)(0.40)\approx1.9\times10^{-13}$ N, perpendicular to $\mathbf v$ and $\mathbf B$.
2 $F=evB\sin30^\circ=(1.6\times10^{-19})(5.0\times10^6)(2.0\times10^{-3})(0.5)=8.0\times10^{-16}$ N.
3 (i) It is moving parallel or antiparallel to $\mathbf B$ ($\sin\theta=0$); (ii) it is at rest — or, of course, it is uncharged.
4 Maximum force means $\theta=90^\circ$: $B=\dfrac{F}{qv}=\dfrac{4.8\times10^{-12}}{(3.2\times10^{-19})(1.5\times10^7)}=1.0$ T.
5 $\mathbf F=q\mathbf v\times\mathbf B=(2.0\times10^{-6})(2.0\times10^5)(0.50)\,(\hat{\mathbf i}\times\hat{\mathbf j})=0.20\,\hat{\mathbf k}$ N.
6 Only $v_y$ contributes: $\mathbf v\times\mathbf B=(3.0\times10^4)(0.40)(\hat{\mathbf j}\times\hat{\mathbf i})=-1.2\times10^4\,\hat{\mathbf k}$ T m/s. $\mathbf F=(-3.0\times10^{-6})(-1.2\times10^4\,\hat{\mathbf k})=3.6\times10^{-2}\,\hat{\mathbf k}$ N.
7 The power delivered by a force is $\mathbf F\cdot\mathbf v$. For $\mathbf F=q\mathbf v\times\mathbf B$, the vector $\mathbf v\times\mathbf B$ is perpendicular to $\mathbf v$, so $\mathbf F\cdot\mathbf v=0$ at every instant: no work is done and $K$ stays constant (work–energy theorem).
8 $\mathbf v\times\mathbf B=(1.0\times10^5)(0.20)(\hat{\mathbf i}\times\hat{\mathbf k})=-2.0\times10^4\,\hat{\mathbf j}$. $\mathbf E+\mathbf v\times\mathbf B=1.0\times10^4\,\hat{\mathbf j}$ N/C, so $\mathbf F=e(1.0\times10^4)\,\hat{\mathbf j}=1.6\times10^{-15}\,\hat{\mathbf j}$ N — the electric force is only partly cancelled.
II
Forces on current-carrying conductors. $g=9.8\ \text{m/s}^2$.
  • 1 A $2.0$ m wire carrying $5.0$ A lies perpendicular to a field of $0.15$ T. Find the force on it.
  • 2 A $25$ cm wire carrying $8.0$ A makes $45^\circ$ with a field of $0.60$ T. Find the force.
  • 3 A $1.2$ m wire perpendicular to a field of $0.050$ T feels a force of $0.18$ N. What current does it carry?
  • 4 A horizontal power line carries $500$ A from west to east through a uniform horizontal magnetic field of $4.0\times10^{-5}$ T pointing north. Find the force on each metre of the line and its direction.
  • 5 A wire segment $\mathbf l=0.20\,\hat{\mathbf i}$ m carries $3.0$ A in a field $\mathbf B=(0.30\,\hat{\mathbf i}+0.40\,\hat{\mathbf j})$ T. Find the force vector.
  • 6 Explain why the net magnetic force on a closed rectangular loop carrying a current in a uniform field is zero. Is the net torque necessarily zero?
  • 7 A wire bent into a semicircle of radius $10$ cm carries $2.0$ A and lies in a plane perpendicular to a uniform field of $0.50$ T. Find the net force on it.
  • 8 A straight horizontal wire has a mass of $50$ g per metre. What current in it, at right angles to a horizontal field of $0.70$ T, would support its weight?
1 $F=IlB=(5.0)(2.0)(0.15)=1.5$ N.
2 $F=IlB\sin45^\circ=(8.0)(0.25)(0.60)(0.707)\approx0.85$ N.
3 $I=\dfrac{F}{lB}=\dfrac{0.18}{(1.2)(0.050)}=3.0$ A.
4 $\dfrac Fl=IB=(500)(4.0\times10^{-5})=2.0\times10^{-2}$ N/m. East $\times$ north $=$ up ($\hat{\mathbf i}\times\hat{\mathbf j}=\hat{\mathbf k}$): the force is vertically upwards — tiny compared with the weight of the cable.
5 $\mathbf F=I\mathbf l\times\mathbf B=(3.0)(0.20)\left[0.30\,(\hat{\mathbf i}\times\hat{\mathbf i})+0.40\,(\hat{\mathbf i}\times\hat{\mathbf j})\right]=0.24\,\hat{\mathbf k}$ N. The field component along the wire contributes nothing.
6 Opposite sides carry the current in opposite directions, so their forces $I\mathbf l\times\mathbf B$ are equal and opposite; the four forces add to zero (generally, $\oint d\mathbf l=\mathbf 0$). The torque need not be zero: the forces on opposite sides may act along different lines and form a couple (Section 4.4).
7 In a uniform field a curved wire feels the same force as the straight chord joining its ends, here of length $2R=0.20$ m perpendicular to $\mathbf B$: $F=I(2R)B=(2.0)(0.20)(0.50)=0.20$ N, perpendicular to the chord in the plane of the semicircle.
8 Per metre: $IB=\lambda g$, so $I=\dfrac{(0.050)(9.8)}{0.70}=0.70$ A (in the direction that makes $\mathbf l\times\mathbf B$ point up).
III
Circular and helical paths. $m_p=1.67\times10^{-27}$ kg, $m_e=9.1\times10^{-31}$ kg; take the alpha-particle mass as $4m_p$ and its charge as $2e$.
  • 1 A proton moves at $1.0\times10^6$ m/s perpendicular to a field of $0.20$ T. Find the radius of its path.
  • 2 Find the period and frequency of revolution of this proton.
  • 3 The proton's speed is doubled. What happens to the radius and to the period?
  • 4 An electron moves on a circle of radius $2.0$ cm in a field of $5.0\times10^{-4}$ T. Find its speed.
  • 5 A proton and an alpha particle enter the same field perpendicularly with the same speed. Find the ratio of the radii $r_\alpha/r_p$.
  • 6 Repeat part 5 for the same kinetic energy.
  • 7 An electron and a proton have the same momentum and move perpendicular to the same field. Compare their radii and their senses of rotation.
  • 8 An electron moves at $3.0\times10^6$ m/s at $30^\circ$ to a field of $2.0$ mT. Find the radius, the period and the pitch of its helix.
  • 9 How much work does the field do on the electron of part 8 in one complete turn? In $1$ s?
1 $r=\dfrac{m_pv}{eB}=\dfrac{(1.67\times10^{-27})(1.0\times10^6)}{(1.6\times10^{-19})(0.20)}\approx5.2\times10^{-2}$ m $=5.2$ cm.
2 $T=\dfrac{2\pi m_p}{eB}=\dfrac{2\pi(1.67\times10^{-27})}{3.2\times10^{-20}}\approx3.3\times10^{-7}$ s; $\nu=1/T\approx3.0\times10^{6}$ Hz.
3 $r\propto v$, so the radius doubles to $10.4$ cm; $T$ does not depend on $v$, so the period is unchanged.
4 $v=\dfrac{eBr}{m_e}=\dfrac{(1.6\times10^{-19})(5.0\times10^{-4})(0.020)}{9.1\times10^{-31}}\approx1.8\times10^{6}$ m/s.
5 $r=\dfrac{mv}{qB}\propto\dfrac mq$: $\dfrac{r_\alpha}{r_p}=\dfrac{4m_p/2e}{m_p/e}=2$.
6 $r=\dfrac{\sqrt{2mK}}{qB}\propto\dfrac{\sqrt m}{q}$: $\dfrac{r_\alpha}{r_p}=\dfrac{\sqrt4/2}{1}=1$ — equal radii.
7 $r=\dfrac{p}{|q|B}$ with the same $p$ and the same $|q|=e$: equal radii. They go round in opposite senses because their charges have opposite signs.
8 $v_\perp=1.5\times10^6$ m/s, $v_\parallel=2.60\times10^6$ m/s. $r=\dfrac{(9.1\times10^{-31})(1.5\times10^6)}{(1.6\times10^{-19})(2.0\times10^{-3})}\approx4.3\times10^{-3}$ m $=4.3$ mm; $T=\dfrac{2\pi m_e}{eB}\approx1.79\times10^{-8}$ s; pitch $=v_\parallel T\approx4.6\times10^{-2}$ m $=4.6$ cm.
9 Zero in both cases: the magnetic force is always perpendicular to the velocity, so it never does work.
IV
Applications and longer problems.
  • 1 Singly charged neon ions of masses $20$ u and $22$ u ($1\ \text{u}=1.66\times10^{-27}$ kg), all moving at $1.0\times10^5$ m/s, enter a field of $0.20$ T at right angles and travel half a circle before striking a detector. Find the two radii and the separation of the two spots.
  • 2 Modelling. A solar-wind proton moving at $4.0\times10^5$ m/s meets a region of space where the magnetic field is $5.0\times10^{-8}$ T, perpendicular to its velocity. Find the radius of its circular path.
  • 3 An electron is accelerated from rest through $2.0$ kV and then moves perpendicular to a field of $1.0$ mT. Find its speed and the radius of its path.
  • 4 A proton moving at $2.0\times10^6$ m/s enters, at right angles, a region $3.0$ cm wide in which there is a uniform field of $0.50$ T parallel to the boundaries. Does it get through? If so, through what angle is it deflected?
  • 5 Find the frequency of revolution of an electron in a field of $0.10$ T. Why does it not depend on the electron's speed?
  • 6 A beam of protons passes undeflected through a region with $\mathbf E=2.0\times10^4\,\hat{\mathbf j}$ N/C and $\mathbf B$ along $\hat{\mathbf k}$, moving along $+\hat{\mathbf i}$ at $1.0\times10^5$ m/s. Using $\mathbf F=q(\mathbf E+\mathbf v\times\mathbf B)=\mathbf 0$, find $\mathbf B$.
1 $r=\dfrac{mv}{eB}$: $r_{20}=\dfrac{(20)(1.66\times10^{-27})(1.0\times10^5)}{(1.6\times10^{-19})(0.20)}\approx0.104$ m, $r_{22}\approx0.114$ m. After a semicircle each spot is a diameter from the entry point, so the separation is $2(r_{22}-r_{20})\approx2.1\times10^{-2}$ m $=2.1$ cm.
2 $r=\dfrac{m_pv}{eB}=\dfrac{(1.67\times10^{-27})(4.0\times10^5)}{(1.6\times10^{-19})(5.0\times10^{-8})}\approx8.4\times10^{4}$ m $\approx84$ km — small compared with the size of the Earth's magnetic field region, so the proton is guided along field lines.
3 $eV=\tfrac12m_ev^2\Rightarrow v=\sqrt{\dfrac{2(1.6\times10^{-19})(2000)}{9.1\times10^{-31}}}\approx2.65\times10^7$ m/s. $r=\dfrac{m_ev}{eB}=\dfrac{(9.1\times10^{-31})(2.65\times10^7)}{(1.6\times10^{-19})(1.0\times10^{-3})}\approx0.15$ m.
4 $r=\dfrac{(1.67\times10^{-27})(2.0\times10^6)}{(1.6\times10^{-19})(0.50)}\approx4.2$ cm, larger than the $3.0$ cm width, so it gets through. The centre of the arc lies on the entry boundary, so when it leaves, $\sin\phi=\dfrac{d}{r}=\dfrac{3.0}{4.18}\approx0.72$: deflection $\phi\approx46^\circ$.
5 $\nu=\dfrac{eB}{2\pi m_e}=\dfrac{(1.6\times10^{-19})(0.10)}{2\pi(9.1\times10^{-31})}\approx2.8\times10^{9}$ Hz. A faster electron goes round a larger circle in proportion ($r\propto v$), so the time per turn $2\pi r/v$ is the same.
6 Need $\mathbf v\times\mathbf B=-\mathbf E=-2.0\times10^4\,\hat{\mathbf j}$. With $\mathbf B=B_z\hat{\mathbf k}$: $\mathbf v\times\mathbf B=vB_z(\hat{\mathbf i}\times\hat{\mathbf k})=-vB_z\,\hat{\mathbf j}$, so $vB_z=2.0\times10^4$ and $B_z=\dfrac{2.0\times10^4}{1.0\times10^5}=0.20$ T: $\mathbf B=0.20\,\hat{\mathbf k}$ T.
🎯 Exam corner — the magnetic force BoardJEE MainJEE AdvancedNEET

E1. Board · 3 marks Starting from the force on a single moving charge, derive the expression for the force on a straight conductor of length $l$ carrying a current $I$ in a uniform magnetic field $\mathbf B$. A $0.60$ m conductor carrying $3.0$ A is placed at $30^\circ$ to a field of $0.40$ T. Find the force on it.

Setup Conductor of length $l$, area $A$, $n$ carriers per unit volume, each of charge $q$, drift velocity $\mathbf v_d$. Force on one carrier: $q\mathbf v_d\times\mathbf B$.
Sum Number of carriers $=nAl$, so $\mathbf F=nAl\,q\,\mathbf v_d\times\mathbf B$. Since $I=nqAv_d$ and $\mathbf l$ points along $\mathbf v_d$ (direction of current), $\mathbf F=I\,\mathbf l\times\mathbf B$, magnitude $IlB\sin\theta$.
Numbers $F=(3.0)(0.60)(0.40)\sin30^\circ=0.36$ N, perpendicular to both the conductor and $\mathbf B$. Exam tip: one mark goes for stating that the force on the carriers is transferred to the conductor, and one for $I=nqAv_d$.

E2. Board · 1 mark Assertion (A): A charged particle moving in a uniform magnetic field keeps a constant kinetic energy. Reason (R): The magnetic force on a moving charge is always perpendicular to its velocity. (A) Both A and R are true and R is the correct explanation of A. (B) Both are true but R is not the correct explanation of A. (C) A is true, R is false. (D) A is false, R is true.

Concept R is true ($q\mathbf v\times\mathbf B\perp\mathbf v$). A force perpendicular to the velocity has zero power $\mathbf F\cdot\mathbf v$, so it does no work and $K$ is constant: A is true because of R.
Answer (A). Exam tip: velocity and momentum do change (direction); only speed and kinetic energy stay fixed.

E3. JEE Main A proton and a deuteron (same charge, twice the mass) with equal kinetic energies enter a uniform magnetic field at right angles. The ratio of the radius of the deuteron's path to that of the proton is (A) $1/\sqrt2$ (B) $1$ (C) $\sqrt2$ (D) $2$.

Concept $r=\dfrac{mv}{qB}=\dfrac{\sqrt{2mK}}{qB}$; with equal $K$ and $q$, $r\propto\sqrt m$.
Ratio $\dfrac{r_d}{r_p}=\sqrt{\dfrac{2m_p}{m_p}}=\sqrt2$.
Answer (C). Exam tip: (D) is the answer for equal speeds; always ask which quantity is equal — speed, momentum or kinetic energy.

E4. NEET A charged particle moves in a uniform magnetic field with its velocity at $60^\circ$ to the field. The ratio of the pitch of its helical path to the radius is (A) $2\pi/\sqrt3$ (B) $2\pi\sqrt3$ (C) $\pi/\sqrt3$ (D) $\sqrt3/(2\pi)$.

Concept Pitch $=v\cos\theta\cdot\dfrac{2\pi m}{qB}$, radius $=\dfrac{mv\sin\theta}{qB}$, so $\dfrac{\text{pitch}}{r}=2\pi\cot\theta$.
Numbers $2\pi\cot60^\circ=\dfrac{2\pi}{\sqrt3}\approx3.6$.
Answer (A). Exam tip: the mass, charge, speed and field all cancel — only the angle matters.

E5. JEE Advanced (One or more correct.) A charged particle moves in a region containing only a uniform magnetic field $\mathbf B$, with its velocity making an angle $\theta$ ($0\lt\theta\lt90^\circ$) with $\mathbf B$. Which statements are correct? (A) Its kinetic energy is constant. (B) Its momentum is constant. (C) Its period of revolution does not depend on its speed. (D) Its path is a helix whose axis is parallel to $\mathbf B$.

Options (A) No work is done: true. (B) The direction of $\mathbf v_\perp$ keeps turning, so $\mathbf p$ changes: false (only $|\mathbf p|$ and $p_\parallel$ are constant). (C) $T=2\pi m/(|q|B)$: true. (D) $v_\parallel$ constant plus circular motion of $v_\perp$: true.
Answer (A), (C), (D). Exam tip: "momentum" in an option means the vector unless the question says "magnitude".

E6. JEE Main NEET In a region, a uniform electric field of $3.0\times10^4$ V/m and a uniform magnetic field of $0.15$ T are perpendicular to each other. A beam of ions moving perpendicular to both fields passes through undeflected. Their speed is (A) $4.5\times10^3$ m/s (B) $2.0\times10^4$ m/s (C) $2.0\times10^5$ m/s (D) $4.5\times10^5$ m/s.

Concept This is a velocity selector — beyond the NCERT syllabus, see the 🚀 box. Undeflected means $qE=qvB$.
Numbers $v=\dfrac EB=\dfrac{3.0\times10^4}{0.15}=2.0\times10^5$ m/s, for any charge or mass.
Answer (C). Exam tip: (A) is $EB$ — check units: $\text{(V/m)}/\text{T}=\text{m/s}$. Interactive 2, last setting, opens on exactly these values.
🔬 Research–Teaching NexusResearch-tutored · optional, about 30 minutes

Research question. How tightly does the Earth's magnetic field wind up the charged particles of the solar wind, and why does that steer them towards the polar regions?

Why it matters. Charged particles trapped and guided by the Earth's field produce the aurora, but they also damage satellite electronics, disturb GPS and radio signals, and during strong solar storms induce currents in power grids — a 1989 storm blacked out the Canadian province of Quebec for nine hours. Space-weather centres in the USA, Europe, India and Japan forecast these events using exactly the gyration physics of this section.

What to do.

  1. Model the field strength above the magnetic equator as $B=B_0\left(\dfrac{R_E}{R}\right)^3$ with $B_0=3.0\times10^{-5}$ T and $R_E=6400$ km. Find $B$ at $R=1,\,2,\,5$ and $10\,R_E$.
  2. For a solar-wind proton with $v_\perp=4.0\times10^5$ m/s, and for an electron of kinetic energy $1.0$ keV, compute the radius of gyration $r=mv_\perp/(eB)$ and the period $T=2\pi m/(eB)$ at each distance. Tabulate your results.
  3. Compare each radius with the distance over which the field changes appreciably (roughly $R$ itself). What does the ratio $r/R$ tell you about whether a particle "follows" the field line?
  4. Field lines from the equatorial region far out in space reach the ground at high latitudes. Use this to explain, in a paragraph, why aurorae appear in rings round the magnetic poles.

Tools: Calculator or spreadsheet; optionally a plotting program for $r$ against $R$ on log–log axes.

Hand in: Your table, the graph with its slope, and the paragraph explaining the location of aurorae.

Going further. Particles spiralling into a region of stronger field are often turned back ("magnetic mirror"), so they bounce between the two hemispheres — the Van Allen radiation belts. Find out how satellite designers protect electronics that must cross these belts.

Instructor note: $B=3.0\times10^{-5},\ 3.75\times10^{-6},\ 2.4\times10^{-7},\ 3.0\times10^{-8}$ T. Proton ($mv/e=4.18\times10^{-3}$ T m): $r\approx140$ m, $1.1$ km, $17$ km, $140$ km; $T\approx2.2$ ms, $17$ ms, $0.27$ s, $2.2$ s. Electron at $1$ keV ($v\approx1.9\times10^7$ m/s, $mv/e\approx1.07\times10^{-4}$ T m): $r\approx3.6$ m, $28$ m, $0.44$ km, $3.6$ km. Slope $+3$ on log–log axes. In every case $r/R\lt10^{-3}$, so particles are tightly tied to field lines and are guided along them to high latitudes, where the lines enter the atmosphere — the auroral ovals.

🧪 Self-check quiz

0/5 correct

Q1. The tesla can be written as

Q2. An electron moves due north through a magnetic field that points vertically upwards. The magnetic force on it points

Q3. A proton moves on a circle in a uniform magnetic field. Which change doubles the radius of the circle?

Q4. The period of revolution of a charged particle moving perpendicular to a uniform magnetic field

Q5. A straight wire carrying a current lies along the direction of a uniform magnetic field. The force on it is

← Chapter 4 — Moving Charges and Magnetism4.2 The Biot–Savart Law and the Field of a Circular Loop →