Chapter 4

Moving Charges and Magnetism

CBSE · 10 subtopics

Formula SheetEvery formula this chapter can be examined on, in one table.

Constants, units and sign conventions

Formula	Symbols, units and conditions
$\mu_0 = 4\pi \times 10^{-7}\ \text{T m A}^{-1}, \qquad \dfrac{\mu_0}{4\pi} = 10^{-7}\ \text{T m A}^{-1}$	$\mu_0$ is the permeability of free space. Every magnetic-field formula in this chapter carries it; the second form $\mu_0/4\pi = 10^{-7}$ is the one that makes CBSE numericals arithmetic-free.
$1\ \text{T} = \dfrac{1\ \text{N}}{(1\ \text{C})(1\ \text{m s}^{-1})} = 1\ \text{N A}^{-1}\,\text{m}^{-1}$	Tesla, the SI unit of magnetic field $B$ . Non-SI: $1\ \text{gauss} = 10^{-4}\ \text{T}$ , so $1\ \text{T} = 10^{4}\ \text{G}$ . Earth's field is about $3.6\times10^{-5}\ \text{T}$ . Other units: magnetic moment $m$ in $\text{A m}^2$ (equivalently $\text{J T}^{-1}$ ); torsional constant $k$ in $\text{N m rad}^{-1}$ ; $\mu_0$ in $\text{T m A}^{-1}$ . Useful constants for numericals: $e = 1.6\times10^{-19}\ \text{C}$ , $m_e = 9.1\times10^{-31}\ \text{kg}$ , $m_p = 1.67\times10^{-27}\ \text{kg}$ , $1\ \text{eV} = 1.6\times10^{-19}\ \text{J}$ . A dot means the current or field comes out of the page; a cross means it goes into the page.
$\mu_0\varepsilon_0 = \dfrac{1}{c^2}$	Links the two free-space constants to the speed of light; fixing one fixes the other.

Magnetic force

Formula	Symbols, units and conditions
$\vec{F} = q\left[\vec{E} + \vec{v}\times\vec{B}\right]$	Lorentz force on a charge $q$ moving with velocity $\vec{v}$ in fields $\vec{E}$ and $\vec{B}$ . The full statement; the magnetic part is the second term.
$\vec{F} = q(\vec{v}\times\vec{B}), \qquad F = qvB\sin\theta$	Magnetic force alone. $\theta$ is the angle between $\vec{v}$ and $\vec{B}$ . Zero if the charge is at rest, or if $\vec{v}$ is parallel or antiparallel to $\vec{B}$ ; maximum when $\theta = 90^\circ$ .
$B = \dfrac{F}{qv}$	Definition of the tesla: $B = 1\ \text{T}$ if a charge of $1\ \text{C}$ moving at $1\ \text{m s}^{-1}$ perpendicular to the field feels a force of $1\ \text{N}$ .
$\vec{F} = I(\vec{l}\times\vec{B}), \qquad F = BIl\sin\theta$	Force on a straight conductor of length $l$ carrying current $I$ in an external field $\vec{B}$ . $\vec{l}$ points along the current; $I$ itself is not a vector. $\theta$ is the angle between the wire and $\vec{B}$ .
$\vec{F} = I\displaystyle\int d\vec{l}\times\vec{B}$	For a wire of arbitrary shape: sum the strips $I\,d\vec{l}\times\vec{B}$ .
$BIl = mg \;\Rightarrow\; B = \dfrac{mg}{Il}, \qquad I = \dfrac{mg}{lB}$	Wire suspended in mid-air (or strings gone slack): magnetic force balances weight. Only $m/l$ , the mass per unit length, is really needed.

Motion in a magnetic field

Formula	Symbols, units and conditions
$qvB = \dfrac{mv^2}{r} \;\Rightarrow\; r = \dfrac{mv}{qB} = \dfrac{p}{qB}$	Radius of the circular path when $\vec{v}\perp\vec{B}$ . $p = mv$ is the linear momentum: equal momenta and equal charge magnitudes give equal radii.
$r = \dfrac{\sqrt{2mK}}{qB} = \dfrac{1}{B}\sqrt{\dfrac{2mV}{q}}$	Same radius rewritten for a particle of kinetic energy $K$ , or one accelerated from rest through a potential difference $V$ (so $K = qV$ ). Use it whenever the question compares proton, deuteron and alpha particle.
$\omega = 2\pi\nu = \dfrac{qB}{m}, \qquad \nu = \dfrac{qB}{2\pi m}, \qquad T = \dfrac{2\pi m}{qB}$	Angular frequency, cyclotron frequency and period of the circular motion. All three are independent of the speed, radius and energy of the particle.
$W = 0, \qquad K = \text{constant}$	The magnetic force is always perpendicular to $\vec{v}$ , so it does no work and the speed (and kinetic energy) never changes.
$v_\parallel = v\cos\theta, \qquad v_\perp = v\sin\theta, \qquad r = \dfrac{mv\sin\theta}{qB}$	Helical motion: $\vec{v}$ at angle $\theta$ to $\vec{B}$ . The parallel part is unaffected; the perpendicular part circles with this radius (the radius of the helix).
$\text{pitch } p = v_\parallel T = \dfrac{2\pi m\, v\cos\theta}{qB}$	Distance advanced along $\vec{B}$ in one full rotation.
$t = \dfrac{\pi r}{v} = \dfrac{\pi m}{qB}$	Time to traverse a semicircle inside a field region and come back out (independent of speed).

4.A Motion in combined electric and magnetic fields

Formula	Symbols, units and conditions
$qE = qvB \;\Rightarrow\; v = \dfrac{E}{B}$	Velocity-selector condition: in crossed $\vec{E}$ and $\vec{B}$ (both perpendicular to $\vec{v}$ ), only particles of this one speed pass undeflected. $E$ in $\text{V m}^{-1}$ , $B$ in T, $v$ in $\text{m s}^{-1}$ .
$E = vB$	Magnitude of the electric field needed to cancel the magnetic deflection; its direction must oppose $q\vec{v}\times\vec{B}$ .

Biot-Savart law and the straight wire

Formula	Symbols, units and conditions
$d\vec{B} = \dfrac{\mu_0}{4\pi}\,\dfrac{I\,d\vec{l}\times\vec{r}}{r^3}$	Biot-Savart law in vector form: field of a current element $I\,d\vec{l}$ at position $\vec{r}$ from it. $d\vec{B}$ is perpendicular to the plane of $d\vec{l}$ and $\vec{r}$ .
$\lvert d\vec{B}\rvert = \dfrac{\mu_0}{4\pi}\,\dfrac{I\,dl\,\sin\theta}{r^2}$	Magnitude form, $\theta$ the angle between $d\vec{l}$ and $\vec{r}$ . Zero along the direction of $d\vec{l}$ itself.
$B = \dfrac{\mu_0 I}{2\pi r} = \dfrac{\mu_0}{4\pi}\cdot\dfrac{2I}{r}$	Infinitely long straight wire, $r$ the perpendicular distance. Field lines are concentric circles round the wire; direction by the right-hand thumb rule.
$B = \dfrac{\mu_0 I r}{2\pi a^2}\ (r<a), \qquad B = \dfrac{\mu_0 I}{2\pi r}\ (r>a)$	Thick straight wire of radius $a$ with the current spread uniformly over the cross-section: $B\propto r$ inside, $B\propto 1/r$ outside, maximum at the surface.
$B = \dfrac{\mu_0 I \phi}{4\pi R}$	Field at the centre of a circular arc of radius $R$ subtending angle $\phi$ (in radians) at the centre. Semicircle ( $\phi=\pi$ ): $B = \dfrac{\mu_0 I}{4R}$ . Full circle ( $\phi = 2\pi$ ): $B = \dfrac{\mu_0 I}{2R}$ .
$B_P = \dfrac{\mu_0 I}{2\pi}\left[\dfrac{1}{x} - \dfrac{1}{d-x}\right]$	Two long parallel wires distance $d$ apart carrying equal currents in the same direction; $P$ lies between them at distance $x$ from the first. Zero at $x = d/2$ . For antiparallel currents the two terms add instead of subtracting.

Magnetic field on the axis of a circular current loop

Formula	Symbols, units and conditions
$B = \dfrac{\mu_0 I R^2}{2\left(R^2+x^2\right)^{3/2}}$	On the axis of a single circular loop of radius $R$ , at distance $x$ from the centre. Directed along the axis, sense by the right-hand thumb rule (curl fingers with the current, thumb gives $\vec{B}$ ).
$B = \dfrac{\mu_0 N I R^2}{2\left(R^2+x^2\right)^{3/2}}$	Same, for a tightly wound coil of $N$ turns.
$B_{\text{centre}} = \dfrac{\mu_0 I}{2R} = \dfrac{\mu_0}{4\pi}\cdot\dfrac{2\pi I}{R}, \qquad B_{\text{centre}} = \dfrac{\mu_0 N I}{2R}$	Put $x=0$ : field at the centre of the loop or coil.
$B = \dfrac{\mu_0}{4\pi}\cdot\dfrac{2 m}{x^3} \quad (x\gg R), \qquad m = I A = I\pi R^2$	Axial far field of the loop, written as a magnetic dipole of moment $m$ . Compare $E = \dfrac{1}{4\pi\varepsilon_0}\dfrac{2p}{x^3}$ .
$B = \dfrac{\mu_0}{4\pi}\cdot\dfrac{m}{x^3} \quad (x\gg R)$	Field in the plane of the loop (equatorial position) far from it — half the axial value and oppositely directed.
$B_{\text{net}} = \sqrt{B_1^2 + B_2^2}, \qquad \tan\alpha = \dfrac{B_2}{B_1}$	Two coils with a common centre lying in perpendicular planes: their fields are perpendicular, so combine by Pythagoras. Equal radii with currents $I$ and $\sqrt{3}I$ give $B = \mu_0 I/R$ at $30^\circ$ to the stronger field.

Ampere's circuital law

Formula	Symbols, units and conditions
$\displaystyle\oint \vec{B}\cdot d\vec{l} = \mu_0 I$	Ampere's circuital law: the line integral of $\vec{B}$ round any closed loop equals $\mu_0$ times the total current threading the surface bounded by it. Sign of $I$ from the right-hand rule.
$B L = \mu_0 I_e$	Simplified form for an amperian loop chosen so that $\vec{B}$ is tangential and constant over a length $L$ and zero or normal elsewhere; $I_e$ is the enclosed current.
$\displaystyle\oint \vec{B}\cdot d\vec{l} = \mu_0\left(i_c + i_d\right)$	Ampere-Maxwell form: $i_c$ conduction current, $i_d$ displacement current through the loop.

The solenoid

Formula	Symbols, units and conditions
$B = \mu_0 n I, \qquad n = \dfrac{N}{L}$	Field well inside a long solenoid ( $L\gg$ radius): uniform, axial, independent of position. $n$ is turns per unit length in $\text{m}^{-1}$ . Field outside is taken as zero.
$B_{\text{in}} = \mu_0 I\,(n_2 - n_1), \qquad B_{\text{out}} = 0$	Two coaxial solenoids of equal length wound one over the other, same current in opposite senses.
$m = N I A$	Magnetic moment of a solenoid of $N$ turns and cross-section $A$ ; it behaves exactly like a bar magnet of this moment.

Force between two parallel currents, the ampere

Formula	Symbols, units and conditions
$B_a = \dfrac{\mu_0 I_a}{2\pi d}$	Field produced by wire $a$ at the position of wire $b$ , distance $d$ away.
$F_{ba} = I_b L B_a = \dfrac{\mu_0 I_a I_b}{2\pi d}\,L$	Force on a length $L$ of wire $b$ due to wire $a$ . Equal and opposite to $F_{ab}$ (Newton's third law).
$f = \dfrac{F}{L} = \dfrac{\mu_0 I_1 I_2}{2\pi d} = \dfrac{\mu_0}{4\pi}\cdot\dfrac{2 I_1 I_2}{d}$	Force per unit length, in $\text{N m}^{-1}$ . Parallel currents attract, antiparallel currents repel.
$I_1 = I_2 = 1\ \text{A},\ d = 1\ \text{m} \;\Rightarrow\; f = 2\times10^{-7}\ \text{N m}^{-1}$	The definition of the ampere.
$F_{\text{net}} = \dfrac{\mu_0 I_1 I_2 L}{2\pi}\left[\dfrac{1}{r_1} - \dfrac{1}{r_2}\right]$	Net force on a rectangular or square loop of side $L$ (sides parallel to the wire at distances $r_1$ and $r_2$ ) near a long straight wire. The two perpendicular sides contribute equal and opposite forces and cancel. Net force is towards the wire when the near side's current is parallel to the wire's.

Torque on a current loop, magnetic dipole

Formula	Symbols, units and conditions
$m = I A, \qquad \vec{m} = N I \vec{A}$	Magnetic moment of a plane current loop; $N$ turns multiply it. Direction of $\vec{A}$ (and $\vec{m}$ ) by the right-hand thumb rule. Unit $\text{A m}^2$ .
$\vec{\tau} = \vec{m}\times\vec{B}, \qquad \tau = m B\sin\theta = N I A B\sin\theta$	Torque on the loop, $\theta$ the angle between $\vec{m}$ (the normal to the loop) and $\vec{B}$ . Net force on the loop in a uniform field is zero.
$\tau_{\max} = N I A B \quad (\theta = 90^\circ), \qquad \tau = 0 \quad (\theta = 0^\circ \text{ or } 180^\circ)$	Maximum torque when $\vec{B}$ lies in the plane of the loop; zero when the plane is perpendicular to $\vec{B}$ .
$U = -\vec{m}\cdot\vec{B} = -mB\cos\theta$	Potential energy of the loop. Minimum at $\theta = 0^\circ$ (stable equilibrium, $\vec{m}\parallel\vec{B}$ ); maximum at $\theta = 180^\circ$ (unstable equilibrium, $\vec{m}$ antiparallel to $\vec{B}$ ).
$B \simeq \dfrac{\mu_0}{4\pi}\cdot\dfrac{2m}{x^3}$	A current loop at large distance behaves as a magnetic dipole; this is its axial field, the analogue of the electric dipole field with $1/\varepsilon_0 \to \mu_0$ , $\vec{p}\to\vec{m}$ , $\vec{E}\to\vec{B}$ .

The moving coil galvanometer

Formula	Symbols, units and conditions
$\tau = N I A B \qquad (\text{radial field, } \sin\theta = 1)$	Deflecting torque on the coil. The radial field guarantees $\theta = 90^\circ$ in every position.
$k\phi = N I A B \;\Rightarrow\; \phi = \left(\dfrac{N A B}{k}\right) I$	Equilibrium of deflecting torque against the spring's restoring torque $k\phi$ . Deflection is proportional to current, so the scale is linear. $k$ is the torsional constant in $\text{N m rad}^{-1}$ .
$I_S = \dfrac{\phi}{I} = \dfrac{N A B}{k}$	Current sensitivity: deflection per unit current, in $\text{rad A}^{-1}$ (or divisions per ampere).
$V_S = \dfrac{\phi}{V} = \dfrac{N A B}{k R} = \dfrac{I_S}{R}$	Voltage sensitivity: deflection per unit voltage; $R$ is the galvanometer's coil resistance.
$S = \left(\dfrac{I_g}{I_0 - I_g}\right) G, \qquad S = \dfrac{G}{n-1}$	Shunt for converting a galvanometer of resistance $G$ (full-scale current $I_g$ ) into an ammeter reading up to $I_0$ ; the second form applies when the range is increased $n = I_0/I_g$ times.
$R_{\text{eff}} = \dfrac{G S}{G + S} = \dfrac{G I_g}{I_0}$	Resistance of the finished ammeter — small, as an ammeter's must be.
$R = \dfrac{V}{I_g} - G, \qquad R = (n-1)G$	Series resistance for converting the same galvanometer into a voltmeter reading up to $V$ ; second form when the range is $n$ times the galvanometer's own $I_g G$ .
$R_{\text{voltmeter}} = G + R$	Resistance of the finished voltmeter — large, as a voltmeter's must be.

Short NotesThe chapter subtopic by subtopic, in the textbook's own order.10 subtopics

The whole chapter in brief

4.1 Oersted's experiment — where the chapter starts

In 1820 Oersted found that a current in a straight wire deflects a nearby compass needle. The needle sets itself tangential to a circle centred on the wire, in a plane perpendicular to the wire.
Reversing the current reverses the deflection; the deflection grows with the current and on bringing the needle closer.
Iron filings sprinkled round the wire arrange themselves in concentric circles.
Conclusion: moving charges (currents) produce a magnetic field in the space around them. This is the experimental root of the whole chapter.
Convention used throughout: a dot is a current or field coming out of the page, a cross is one going into the page.

Magnetic field and the Lorentz force

Just as a charge is the source of $\vec{E}$ , a current or moving charge is the source of $\vec{B}$ . Both are vector fields and both obey the principle of superposition — fields of several sources add vectorially.
Lorentz force: $\vec{F} = q\left[\vec{E} + \vec{v}\times\vec{B}\right]$ .
Magnetic part $\vec{F} = q(\vec{v}\times\vec{B})$ , magnitude $qvB\sin\theta$ . Key features: it depends on $q$ , $v$ and $B$ ; it is zero for a stationary charge and zero when $\vec{v}\parallel\vec{B}$ ; it is perpendicular to both $\vec{v}$ and $\vec{B}$ ; and it reverses for a negative charge.
One tesla: the field in which a charge of $1\ \text{C}$ moving perpendicular to the field at $1\ \text{m s}^{-1}$ experiences a force of $1\ \text{N}$ .
Force on a current-carrying rod: $\vec{F} = I(\vec{l}\times\vec{B})$ , $F = BIl\sin\theta$ , maximum when the rod is perpendicular to $\vec{B}$ , zero when parallel. Here $\vec{B}$ is the external field, not the field of the rod itself.

Motion of a charge in a magnetic field

The magnetic force is always perpendicular to $\vec{v}$ , so it does no work: speed and kinetic energy are unchanged, only the direction of momentum turns.
$\vec{v}\perp\vec{B}$ : the force is centripetal and the path is a circle of radius $r = \dfrac{mv}{qB} = \dfrac{p}{qB} = \dfrac{\sqrt{2mK}}{qB}$ .
$\vec{v}$ at an angle $\theta$ to $\vec{B}$ : the parallel component $v\cos\theta$ is untouched, the perpendicular component $v\sin\theta$ circles — the path is a helix with axis along $\vec{B}$ , radius $\dfrac{mv\sin\theta}{qB}$ and pitch $\dfrac{2\pi m v\cos\theta}{qB}$ .
$\vec{v}\parallel\vec{B}$ (or $v=0$ ): no force at all, straight line.
Cyclotron frequency $\nu = \dfrac{qB}{2\pi m}$ and period $T = \dfrac{2\pi m}{qB}$ are independent of speed, radius and energy — this is what makes a cyclotron possible.

Charges accelerated through a potential difference

$qV = \dfrac{1}{2}mv^2$ , so $r = \dfrac{1}{B}\sqrt{\dfrac{2mV}{q}}$ : double the accelerating voltage and $r$ grows by $\sqrt{2}$ .
Same $V$ , same $B$ : $r\propto\dfrac{\sqrt{m}}{q}$ (the standard proton–deuteron–alpha comparison).

Particle	Charge	Mass	$K$ at same $V$	$r$ at same $V$
Proton	$e$	$m$	$eV$	$1$
Deuteron	$e$	$2m$	$eV$	$\sqrt{2}$
Alpha	$2e$	$4m$	$2eV$	$1$

At the same kinetic energy instead of the same $V$ , $r\propto\dfrac{\sqrt{m}}{q}$ still, giving $r_\alpha : r_d : r_p = 1 : \sqrt{2} : 1$ .
Equal momenta and equal charge magnitude (a proton and an electron, say) give equal radii, because $r = p/(qB)$ .

Combined electric and magnetic fields

Velocity selector: with $\vec{E}$ , $\vec{B}$ and $\vec{v}$ mutually perpendicular, the particle goes undeflected when $qE = qvB$ , i.e. $v = \dfrac{E}{B}$ — independent of charge and mass, so it filters one speed out of a beam.
Fields parallel to each other, particle released from rest: $\vec{v}$ stays along $\vec{B}$ , $\vec{v}\times\vec{B} = 0$ always, and the particle simply accelerates in a straight line.

Biot-Savart law

Statement: for a current element $I\,d\vec{l}$ , the field at a point whose position vector from the element is $\vec{r}$ is $d\vec{B} = \dfrac{\mu_0}{4\pi}\dfrac{I\,d\vec{l}\times\vec{r}}{r^3}, \qquad \lvert d\vec{B}\rvert = \dfrac{\mu_0}{4\pi}\dfrac{I\,dl\,\sin\theta}{r^2}$
$d\vec{B}$ is perpendicular to the plane containing $d\vec{l}$ and $\vec{r}$ ; along $d\vec{l}$ itself ( $\theta = 0$ ) the field is zero.

Coulomb's law	Biot-Savart law
Inverse square in distance	Inverse square in distance
Superposition applies	Superposition applies
Long range	Long range
Source is a scalar, the charge $q$	Source is a vector, the element $I\,d\vec{l}$
$\vec{E}$ is along the line joining source and point	$\vec{B}$ is perpendicular to the plane of $\vec{r}$ and $I\,d\vec{l}$
No angle dependence	Carries a $\sin\theta$ factor

Standard field results worth memorising

Configuration	Field
Long straight wire, distance $r$	$\dfrac{\mu_0 I}{2\pi r}$
Thick wire radius $a$ , inside at $r<a$	$\dfrac{\mu_0 I r}{2\pi a^2}$
Centre of circular loop radius $R$	$\dfrac{\mu_0 I}{2R}$
Centre of a semicircular arc	$\dfrac{\mu_0 I}{4R}$
Centre of an arc of angle $\phi$ radians	$\dfrac{\mu_0 I \phi}{4\pi R}$
Axis of loop, distance $x$	$\dfrac{\mu_0 I R^2}{2(R^2+x^2)^{3/2}}$
Inside a long solenoid	$\mu_0 n I$
Outside a long solenoid	zero

Straight portions of a wire that pass through the point of interest contribute nothing, because $d\vec{l}\times\vec{r} = 0$ along them.
Field lines of a straight wire: concentric circles round the wire, direction by the right-hand thumb rule — grasp the wire with the right thumb along the current and the fingers curl the way $\vec{B}$ points.
Field lines of a circular loop: closed loops threading the coil, straight and dense along the axis near the centre, spreading and curving round outside — the pattern of a short bar magnet, with one face acting as N and the other as S. The other right-hand rule applies here: curl the fingers along the current, and the thumb points along $\vec{B}$ on the axis.

Ampere's circuital law

Statement: the line integral of $\vec{B}$ round a closed loop is $\mu_0$ times the net current threading the surface bounded by it, $\displaystyle\oint \vec{B}\cdot d\vec{l} = \mu_0 I$ . Sign convention: curl the right-hand fingers in the sense the loop is traversed, and the thumb gives the positive current direction.
Simplified working form $BL = \mu_0 I_e$ , valid when the amperian loop is chosen so that along a length $L$ the field is tangential and constant, and elsewhere it is normal to the loop or zero.
It is not independent physics — it follows from the Biot-Savart law, exactly as Gauss's law follows from Coulomb's law. It holds for steady currents only.
It always holds, but only helps where there is symmetry (straight wire, thick wire, solenoid, toroid). It cannot deliver $B = \mu_0 I/2R$ at the centre of a circular loop.
Ampere-Maxwell form, needed for the later chapter: $\displaystyle\oint \vec{B}\cdot d\vec{l} = \mu_0(i_c + i_d)$ .

The solenoid

A long wire wound as a closely spaced helix; each turn is effectively a circular loop and the fields add.
Between neighbouring turns the loop fields cancel. Deep inside, the field is strong, uniform and along the axis; just outside the curved surface it is essentially zero.
Result from a rectangular amperian loop $abcd$ with side $ab = h$ inside and parallel to the axis, $cd$ outside where $B=0$ , and $bc$ , $da$ perpendicular to the axis where the component of $\vec{B}$ along the path is zero: $Bh = \mu_0(nh)I$ , hence $B = \mu_0 n I$ .
Field lines: parallel to the axis inside, emerging from the face that acts as N, curving round outside and re-entering the S face — the field of a bar magnet.
Polarity from the end view: current anticlockwise as seen from a face makes that face N; clockwise makes it S.
A solenoid of $N$ turns and area $A$ has magnetic moment $m = NIA$ ; a soft iron core inside greatly increases the field.

Force between parallel currents, and the ampere

Wire $a$ makes a field $B_a = \dfrac{\mu_0 I_a}{2\pi d}$ at wire $b$ ; wire $b$ then feels $F_{ba} = I_b L B_a = \dfrac{\mu_0 I_a I_b L}{2\pi d}$ , and $\vec{F}_{ba} = -\vec{F}_{ab}$ , consistent with Newton's third law.
Parallel currents attract, antiparallel currents repel — the opposite of the electrostatic rule for like charges.
Force per unit length $f = \dfrac{\mu_0 I_1 I_2}{2\pi d}$ .
One ampere is the steady current which, maintained in each of two very long straight parallel conductors of negligible cross-section placed $1\ \text{m}$ apart in vacuum, produces a force of $2\times10^{-7}\ \text{N}$ per metre of length on each. One coulomb is then the charge crossing a section in one second when the current is $1\ \text{A}$ .
Loop near a straight wire: the two sides parallel to the wire feel unequal forces (nearer side stronger), the two perpendicular sides cancel, so there is a net force towards or away from the wire.

Torque on a current loop; the magnetic dipole

A current loop in a uniform field feels no net force but does feel a torque, exactly like an electric dipole in a uniform electric field.
$\vec{m} = N I \vec{A}$ (direction by the right-hand thumb rule), $\vec{\tau} = \vec{m}\times\vec{B}$ , $\tau = NIAB\sin\theta$ with $\theta$ measured between the normal to the coil and $\vec{B}$ .
$\vec{B}$ in the plane of the loop ( $\theta = 90^\circ$ ): torque is maximum, $NIAB$ . Plane perpendicular to $\vec{B}$ ( $\theta = 0^\circ$ ): torque is zero.
$\theta = 0^\circ$ is stable equilibrium ( $U$ minimum, flux through the loop maximum); $\theta = 180^\circ$ is unstable ( $U = -\vec{m}\cdot\vec{B}$ maximum).
A flexible loop in a field takes up a circular shape with its plane normal to $\vec{B}$ , because a circle encloses the greatest area for a given perimeter and so maximises the flux.
For the same length of wire and the same current, a circular loop has a larger area ( $L^2/4\pi$ ) than a square one ( $L^2/16$ ), so the circle has the larger magnetic moment and the larger torque; the ratio of moments square to circle is $\pi : 4$ .
At large distances a current loop is a magnetic dipole: axial field $\dfrac{\mu_0}{4\pi}\dfrac{2m}{x^3}$ , equatorial field $\dfrac{\mu_0}{4\pi}\dfrac{m}{x^3}$ . Substituting $1/\varepsilon_0\to\mu_0$ , $\vec{p}\to\vec{m}$ , $\vec{E}\to\vec{B}$ turns the electric dipole results into these.
A magnetic dipole (a current loop) is the elementary magnetic object; magnetic monopoles are not known to exist, which is why magnetic field lines are always closed loops.

The moving coil galvanometer

Principle: a current-carrying coil placed in a magnetic field experiences a torque whose magnitude depends on the current.
Construction, described: a coil of many turns wound on a light frame, suspended (or pivoted) between the concave pole pieces of a permanent magnet, with a cylindrical soft-iron core inside the coil; a phosphor-bronze suspension strip provides the restoring torque and carries the current in, a spring carries it out, and a pointer on a scale reads the deflection.
Working: deflecting torque $NIAB$ is balanced by the restoring torque $k\phi$ , so $\phi = \dfrac{NAB}{k}I$ — the deflection is proportional to the current and the scale is linear.
Radial field: the poles are cut concave and a soft-iron cylinder sits inside the coil, so that in every orientation the plane of the coil is parallel to $\vec{B}$ , $\theta = 90^\circ$ , $\sin\theta = 1$ . This keeps the torque at its maximum for all positions and is what makes the scale uniform.
Soft-iron core: high permeability, so it concentrates the field, making it both radial and stronger, which increases sensitivity.
The coil is wound on a metallic frame so that eddy currents damp its motion electromagnetically — a dead-beat galvanometer that settles without oscillating.
Current sensitivity $I_S = \dfrac{\phi}{I} = \dfrac{NAB}{k}$ ; increase it by increasing $N$ , increasing $A$ , using a stronger radial $B$ , or a suspension of smaller $k$ .
Voltage sensitivity $V_S = \dfrac{\phi}{V} = \dfrac{NAB}{kR} = \dfrac{I_S}{R}$ . Doubling $N$ doubles $I_S$ but also roughly doubles $R$ (twice the length of wire), so $V_S$ need not increase at all.
A galvanometer alone cannot serve as an ammeter: it gives full-scale deflection for currents of the order of microamperes, and its resistance is large enough to alter the current it is asked to measure.

Converting a galvanometer

Feature	Ammeter	Voltmeter
Measures	current	potential difference
Connected	in series	in parallel
Conversion	small shunt $S$ in parallel	large $R$ in series
Value needed	$S = \dfrac{I_g G}{I_0 - I_g}$	$R = \dfrac{V}{I_g} - G$
To multiply range $n$ times	$S = \dfrac{G}{n-1}$	$R = (n-1)G$
Resistance of the instrument	$\dfrac{GS}{G+S} = \dfrac{G I_g}{I_0}$ , small	$G + R$ , large
Ideal instrument	zero resistance	infinite resistance

Why a low-resistance ammeter: in series, a high resistance would reduce the very current being measured.
Why a high-resistance voltmeter: in parallel, it must draw almost no current so it does not disturb the potential difference it reads.
Higher range therefore means lower resistance for an ammeter and higher resistance for a voltmeter.

Traps that cost marks

$\theta$ in $\tau = NIAB\sin\theta$ is between the normal to the coil and $\vec{B}$ , not between the plane of the coil and $\vec{B}$ . "The plane makes $60^\circ$ with the field" means $\theta = 30^\circ$ .
Torque is $\vec{m}\times\vec{B}$ , a cross product — never a dot product. The dot product $-\vec{m}\cdot\vec{B}$ is the potential energy.
Torque zero when the plane of the loop is perpendicular to $\vec{B}$ ; torque maximum when $\vec{B}$ lies in the plane of the loop. Students routinely swap these.
Magnetic force does zero work, always. Kinetic energy and speed never change in a purely magnetic field, whatever the path.
In $\vec{F} = I(\vec{l}\times\vec{B})$ , the field $\vec{B}$ is the external field; a wire feels no force from its own field.
Parallel currents attract, antiparallel repel — the reverse of the like-charges rule. Say which, and why, when the question asks for the nature of the force.
Do not use $\dfrac{\mu_0 I}{2R}$ for a semicircular arc; it is $\dfrac{\mu_0 I}{4R}$ . And check whether a question's "loop" is a full circle, a semicircle, or an arc.
Straight segments in line with the centre contribute nothing to $B$ at the centre — say so explicitly, it carries a mark.
Ampere's law is not restricted to symmetric situations; it always holds, but it is only *useful* where symmetry lets you pull $B$ out of the integral. Write it that way.
Increasing $N$ increases current sensitivity but also the coil resistance, so voltage sensitivity may fall. Never claim one sensitivity follows the other.
Shunt goes in parallel and is small; the multiplier resistance goes in series and is large. Getting these the wrong way round loses the whole part.
Radius $r = mv/(qB)$ compares by momentum, not by mass alone: equal momenta give equal radii even for an electron and a proton.
When two coils are in perpendicular planes with a common centre, their fields are perpendicular — add by $\sqrt{B_1^2+B_2^2}$ , never arithmetically.
State the direction as well as the magnitude whenever a question says "find the magnitude and direction"; the direction is a separate mark.
Keep units straight: $n$ in $B = \mu_0 n I$ is turns per metre, not the total number of turns.

Solved Numerical ExamplesWorked numericals to attempt first, each with a hint before the full solution.Coming soon

These are being added chapter by chapter. They are solved numericals from the prescribed book and the reference guides — not previous board questions, which live under Practice — so that you can work through the method before meeting it in a real paper.

Important DerivationsEvery derive / obtain / prove question in this chapter.10

1
Derivation 1
(a) A particle of charge ' $q$ ' and mass ' $m$ ', moving with velocity ' $\vec{v}$ ' is subjected to a uniform magnetic field $\vec{B}$ perpendicular to its velocity. Show that the particle describes a circular path. Obtain expression for the radius of the circular path of the particle. (b) Explain, how its path will be affected if the velocity $\vec{v}$ makes an angle $\theta$ ( $\ne 90°$ ) with the direction of the magnetic field.
Asked in: CBSE Class 12 Physics — Long Answer · AI 2019C · Section E, 5 marks
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(a) The magnetic force on the moving charge is $\vec{F} = q(\vec{v}\times\vec{B})$ . This force acts perpendicular to the velocity vector at all instants; it therefore only changes the direction of velocity, not its magnitude. Since the magnitude of speed remains constant while the direction of the force is always perpendicular to velocity (i.e. towards a fixed centre), the particle moves in a circular path, with the magnetic force providing the necessary centripetal force. So, $F = qvB\sin\theta$ with $\theta = 90°$ : $F = qvB$ Equating to the centripetal force: $\dfrac{mv^2}{r} = qvB \Rightarrow r = \dfrac{mv}{qB}$ (b) If $\theta \ne 90°$ , the velocity has a component along $\vec{B}$ (which experiences no force) and a component perpendicular to $\vec{B}$ (which executes circular motion). The particle then moves along $B$ with uniform velocity while circling around it, so its motion becomes helical.
2
Derivation 2
(a) Write the expression for the force $\vec{F}$ acting on a particle of mass $m$ and charge $q$ moving with velocity $\vec{v}$ in a magnetic field $\vec{B}$ . Under what conditions will it move in (i) a circular path and (ii) a helical path? (b) Show that the kinetic energy of the particle moving in magnetic field remains constant.
Asked in: CBSE Class 12 Physics — SA · AI 2017 · Section C, 3 marks
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(a) $\vec{F} = q(\vec{v}\times\vec{B})$ (i) The particle moves in a circular path when $\vec{v}$ is perpendicular to $\vec{B}$ (i.e. the entire velocity provides for circular motion, with no component along $B$ ). (ii) The particle moves in a helical path when $\vec{v}$ makes an angle $\theta$ ( $\ne 90°$ , $\ne 0°$ ) with $\vec{B}$ — the component of $\vec{v}$ along $\vec{B}$ produces uniform linear motion while the perpendicular component produces circular motion, and together they trace a helix. (b) The magnetic force $\vec{F} = q(\vec{v}\times\vec{B})$ always acts perpendicular to $\vec{v}$ , so $\vec{F}\cdot\vec{v} = q(\vec{v}\times\vec{B})\cdot\vec{v} = 0$ . By Newton's second law, $\vec{F} = m\dfrac{d\vec{v}}{dt}$ , so $m\dfrac{d\vec{v}}{dt}\cdot\vec{v} = 0$ , i.e. $\dfrac{m}{2}\left[\dfrac{d\vec{v}}{dt}\cdot\vec{v} + \vec{v}\cdot\dfrac{d\vec{v}}{dt}\right] = 0$ $\Rightarrow \dfrac{m}{2}\dfrac{d}{dt}(\vec{v}\cdot\vec{v}) = 0 \Rightarrow \dfrac{d}{dt}\left(\dfrac{1}{2}mv^2\right) = 0 \Rightarrow \dfrac{dK}{dt} = 0$ Hence $K =$ constant, i.e. the kinetic energy remains constant.
3
Derivation 3
Deduce an expression for the frequency of revolution of a charged particle in a magnetic field and show that it is independent of velocity or energy of the particle.
Asked in: CBSE Class 12 Physics — Long Answer · AI 2014 · Section E, 5 marks
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The charged particle moves in a circular path with a constant speed and is acted upon only by the magnetic field. The radius of the circular path is given by $qvB = \dfrac{mv^2}{r}$ , i.e. $r = \dfrac{mv}{qB}$ . Period of revolution, $T = \dfrac{2\pi r}{v} = \dfrac{2\pi}{v}\cdot\dfrac{mv}{qB} = \dfrac{2\pi m}{qB}$ Frequency of revolution, $f = \dfrac{1}{T} = \dfrac{qB}{2\pi m}$ Clearly, the frequency $f$ is independent of both $v$ and $r$ (they cancelled out of the expression), and since it depends only on $q$ , $B$ and $m$ , it is also independent of the energy of the particle.
4
Derivation 4
Use Biot-Savart law to derive the expression for the magnetic field on the axis of a current carrying circular loop of radius $R$ . Draw the magnetic field lines due to circular wire carrying current $I$ .
Asked in: CBSE Class 12 Physics — Long Answer · AI 2016 · Section E, 5 marks
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Consider a circular loop of radius $R$ carrying current $I$ , and a point $P$ on its axis at distance $x$ from the centre $O$ . By Biot-Savart's law, small field due to element $d\vec{\ell}$ at $P$ : $dB = \dfrac{\mu_0}{4\pi}\dfrac{I\,d\ell\sin90°}{s^2} = \dfrac{\mu_0}{4\pi}\dfrac{I\,d\ell}{(R^2+x^2)}$ , since $d\vec{\ell} \perp \vec{s}$ everywhere on the loop (where $s = \sqrt{R^2+x^2}$ ). The component $dB\cos\phi$ (perpendicular to the axis) due to a current element is cancelled by the equal and opposite component $dB\cos\phi$ of the diametrically opposite current element, whereas the components $dB\sin\phi$ (along the axis) add up. So, net field at $P$ due to entire loop is $B = \oint dB\sin\phi = \displaystyle\int_0^{2\pi R}\dfrac{\mu_0}{4\pi}\dfrac{I\,d\ell}{(R^2+x^2)}\cdot\dfrac{R}{(R^2+x^2)^{1/2}}$ $B = \dfrac{\mu_0 IR}{4\pi(R^2+x^2)^{3/2}}\int_0^{2\pi R}d\ell = \dfrac{\mu_0 IR}{4\pi(R^2+x^2)^{3/2}}\times2\pi R$ $B = \dfrac{\mu_0 IR^2}{2(R^2+x^2)^{3/2}}$ , directed along the axis: (a) towards the coil if current in it is anticlockwise, or (b) away from the coil if current is clockwise (as seen from that side) — [book states: towards the coil if current is clockwise, away if anticlockwise, per the sign convention used]. The magnetic field lines due to a circular wire carrying current $I$ form closed loops that pass through the centre along the axis (straight and dense near the centre) and curve around and close up outside the loop, resembling those of a bar magnet's field, being directed by the right-hand thumb rule.
5
Derivation 5
A circular loop of radius $R$ carries a current $I$ . Obtain an expression for the magnetic field at a point on its axis at a distance $x$ from its centre. OR Write, using Biot-Savart law, the expression for the magnetic field $\vec{B}$ due to an element $d\vec{\ell}$ carrying current $I$ at a distance $\vec{r}$ from it in a vector form. Hence derive the expression for the magnetic field due to a current carrying loop of radius $R$ at a point $P$ distant $x$ from its centre along the axis of the loop.
Asked in: CBSE Class 12 Physics — Long Answer · 3/5, 2020 · Section E, 5 marks
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By Biot-Savart's law, the small magnetic field due to a current element of the circular coil of radius $R$ at point $P$ (on the axis, distance $x$ from centre) is $dB = \dfrac{\mu_0}{4\pi}\dfrac{I\,d\ell\sin90°}{s^2} = \dfrac{\mu_0}{4\pi}\dfrac{I\,d\ell}{(R^2+x^2)}$ , since $d\vec{\ell} \perp \vec{s}$ , where $s = \sqrt{R^2+x^2}$ . The component $dB\cos\phi$ (perpendicular to axis) cancels for diametrically opposite elements; the axial components $dB\sin\phi$ add up. $B = \oint dB\sin\phi = \displaystyle\int_0^{2\pi R}\dfrac{\mu_0}{4\pi}\dfrac{I\,d\ell}{(R^2+x^2)}\cdot\dfrac{R}{(R^2+x^2)^{1/2}} = \dfrac{\mu_0 IR}{4\pi(R^2+x^2)^{3/2}}\times2\pi R$ $B = \dfrac{\mu_0 IR^2}{2(R^2+x^2)^{3/2}}$ , directed along the axis (towards the coil if current is clockwise, away if anticlockwise, as seen from the point of observation).
6
Derivation 6
Draw the magnetic field lines due to a current passing through a long solenoid. Use Ampere's circuital law, to obtain the expression for the magnetic field due to the current $I$ in a long solenoid having $n$ number of turns per unit length.
Asked in: CBSE Class 12 Physics — SA · 2014C, Ev · Section B, 2 marks
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The magnetic field lines inside a long solenoid run parallel to its axis (nearly uniform), emerging from one end (acting as north pole) and curving back around outside to enter the other end (acting as south pole), resembling the field of a bar magnet. Consider a rectangular Amperian loop $abcd$ near the middle of the solenoid, where $PQ = l$ (the length of side $ab$ , parallel to the axis, lying inside the solenoid). Let the magnetic field along path $ab$ be $B$ , and it is zero along $cd$ (well outside the solenoid). As paths $bc$ and $da$ are perpendicular to the axis, the field component along these paths is zero (field is either axial-only inside or negligible outside), so they do not contribute to the line integral. Total number of turns in length $l$ : $N = nl$ $\displaystyle\int_{abcd}\vec{B}\cdot d\vec{l} = \int_a^b\vec{B}\cdot d\vec{l} + \int_b^c\vec{B}\cdot d\vec{l} + \int_c^d\vec{B}\cdot d\vec{l} + \int_d^a\vec{B}\cdot d\vec{l}$ $= Bl + 0 + 0 + 0 = Bl$ By Ampere's circuital law: $\displaystyle\int_{abcd}\vec{B}\cdot d\vec{l} = \mu_0\times$ (total current through rectangle) $= \mu_0\times nl\times I = \mu_0 nlI$ From these two results: $Bl = \mu_0 nlI \Rightarrow B = \mu_0 nI$ This gives the magnetic field strength inside a straight current-carrying solenoid, directed along the axis of the solenoid. (The field just outside the curved surface, near the middle of the solenoid, is essentially zero.)
7
Derivation 7
An observer to the left of a solenoid of $N$ turns each of cross section area ' $A$ ' observes that a steady current $I$ in it flows in the clockwise direction. Depict the magnetic field lines due to the solenoid specifying its polarity and show that it acts as a bar magnet of magnetic moment $m = NIA$ .
Asked in: CBSE Class 12 Physics — Long Answer · 2015 · Section E, 5 marks
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Given that the current flows in the clockwise direction for an observer on the left side of the solenoid. This means that the left face of the solenoid acts as a south pole and the right face acts as a north pole (using the right-hand thumb rule / end-view convention: clockwise current as seen means field lines go away from the observer through the coil, emerging as N on the far/right side). Inside a bar magnet, the magnetic field lines are directed from south to north. Therefore, the magnetic field lines inside the solenoid are directed from left to right (analogous to the field inside a bar magnet), and outside they curve around from the north (right) face back to the south (left) face, exactly like a bar magnet's external field lines. Magnetic moment of a single current loop is given by $m' = IA$ , where $I$ = current flowing through the loop, $A$ = area of the loop. So, magnetic moment of the whole solenoid (with $N$ such turns, each contributing equally and coherently since they carry the same current in the same sense) is given by $m = Nm' = N(IA)$ This is exactly analogous to the magnetic moment of a bar magnet, confirming that the current-carrying solenoid behaves as a bar magnet of moment $m = NIA$ , with the polarity as identified above (south on the observer's left face, north on the right face).
8
Derivation 8
Two long straight parallel conductors carry steady current $I_1$ and $I_2$ separated by a distance $d$ . If the currents are flowing in the same direction, show how the magnetic field set up in one produces an attractive force on the other. Obtain the expression for this force. Hence define one ampere.
Asked in: CBSE Class 12 Physics — SA · Delhi 2016, An · Section C, 3 marks
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When two parallel infinite straight wires carrying currents $I_1$ and $I_2$ are placed at distance $d$ from each other, then current $I_1$ produces magnetic field, which at any point on the second current carrying wire is $B_1 = \dfrac{\mu_0 I_1}{2\pi d}$ , directed inwards, perpendicular to the plane of wires (by the right hand rule, for the given current directions). So, this current-carrying wire (carrying $I_2$ ) then experiences a force due to this magnetic field which on its length $l$ is given by $\vec{F}_{21} = I_2(\vec{l}\times\vec{B}_1)$ $F_{21} = F_{12} = I_2 l B_1\sin90° = I_2 l\times\dfrac{\mu_0 I_1}{2\pi d}$ or $F_{21} = F_{12} = \dfrac{\mu_0 I_1 I_2}{2\pi d}l$ The vector product $(\vec{l}\times\vec{B}_1)$ has a direction towards the wire carrying current $I_1$ . Hence, both the wires attract each other. So, force per unit length that each wire exerts on the other is $F = \dfrac{\mu_0 I_1 I_2}{2\pi d}$ If $I_1 = I_2 = 1$ A and $d = 1$ m and $l = 1$ m, then $F = \dfrac{\mu_0}{2\pi} = 2\times10^{-7}$ N m $^{-1}$ Thus, electric current through each of two parallel long wires placed at distance of 1 m from each other is said to be 1 ampere, if they exert a force of $2\times10^{-7}$ N m $^{-1}$ on each other.
9
Derivation 9
(a) Write an expression of magnetic moment associated with a current ( $I$ ) carrying circular coil of radius $r$ having $N$ turns. (b) Consider the above mentioned coil placed in YZ plane with its centre at the origin. Derive expression for the value of magnetic field due to it at point $(x, 0, 0)$ .
Asked in: CBSE Class 12 Physics — SA · 2020, Ev · Section C, 3 marks
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(a) The magnetic moment associated with a current ( $I$ ) carrying circular coil of radius $r$ having $N$ turns is given by, $M = NIA = NI\pi r^2$ . (b) Magnetic field at a distance $x$ from the centre of the ring due to element $d\vec{l}$ : $dB = \dfrac{\mu_0}{4\pi}\dfrac{i\,dl\sin90°}{r_1^2}$ (where $r_1 = \sqrt{r^2+x^2}$ is the distance from element to the point), with the perpendicular components ( $dB\cos\theta$ ) cancelling by symmetry and the axial components ( $dB\sin\theta$ ) adding up, exactly as in the standard on-axis field derivation. Since the angle between $d\vec{l}$ and $\vec{r}_1$ is 90°, the component $dB\cos\theta$ gets cancelled due to symmetry, leaving $B = \displaystyle\int dB\sin\theta = \int\left(\dfrac{\mu_0\,i\,dl}{4\pi r_1^2}\right)\sin\theta$ Here, $r_1$ and $\theta$ are constants along the loop, and $\sin\theta = \dfrac{r}{r_1}$ , so $B = \displaystyle\int\dfrac{\mu_0\,i\,dl}{4\pi r_1^2}\left(\dfrac{r}{r_1}\right) = \dfrac{\mu_0\,ir}{4\pi r_1^3}\int dl = \dfrac{\mu_0\,iR}{4\pi r^3}\int dl$ $= \dfrac{\mu_0\,iR}{4\pi r^3}(2\pi R) = \dfrac{\mu_0\,iR^2}{2r^3}$ Putting $r = (R^2+x^2)^{1/2}$ , we get $B = \dfrac{\mu_0\,iR^2}{2(R^2+x^2)^{3/2}}$ For $N$ turns: $B = \dfrac{\mu_0 NiR^2}{2(R^2+x^2)^{3/2}}$ , directed along the axis (the $x$ -axis).
10
Derivation 10
Derive the expression for the torque acting on the rectangular current carrying coil of a galvanometer. Why is the magnetic field made radial? OR Obtain the expression for the deflecting torque acting on the current carrying rectangular coil of a galvanometer in a uniform magnetic field. Why is a radial magnetic field employed in the moving coil galvanometer?
Asked in: CBSE Class 12 Physics — Long Answer · 3/5, 2020 · Section E, 5 marks
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Consider a rectangular coil of $N$ turns, area $A = l\times b$ , carrying current $I$ , placed in a uniform magnetic field $B$ such that the normal to its plane makes angle $\theta$ with $B$ . The forces on the two sides of length $b$ (parallel to the axis of rotation) are equal, opposite and collinear (they act along the coil's rotation axis), so they cancel out (produce no net force or torque contribution about the axis). The forces on the two sides of length $l$ (perpendicular to the axis) are equal and opposite, each of magnitude $F = BIl$ , but not collinear — they form a couple. Deflecting torque $\tau =$ Force $\times$ perpendicular distance $= BIl\times b\sin\theta = BIA\sin\theta$ (for one turn) For $N$ turns: $\tau = NIAB\sin\theta = MB\sin\theta$ , where $M = NIA$ is the magnetic moment of the coil. If $C$ is the torsional rigidity of the wire and $\theta$ is the twist of the suspension strip, then restoring torque $= C\theta$ . For equilibrium, deflecting torque = restoring torque, i.e. $NIAB\sin\theta = C\theta$ ; when the field is radial, $\sin\theta = 1$ at every position, so $NIAB = C\theta \Rightarrow \theta = \left(\dfrac{NAB}{C}\right)I$ , i.e. $\theta \propto I$ . The magnetic field is made radial so that the plane of the coil remains parallel to the field at every angular position, ensuring the deflection is always directly proportional to the current, giving a linear scale that can be calibrated uniformly.

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