Chapter 6

Electromagnetic Induction

CBSE · 8 subtopics

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

Magnetic Flux

Formula	Symbols, units and conditions
$\Phi_B = \vec{B}\cdot\vec{A} = BA\cos\theta$	$\Phi_B$ magnetic flux through a plane area $A$ (in $\text{m}^2$ ) in a uniform field $B$ (in tesla, T); $\theta$ is the angle between $\vec{B}$ and the area vector $\vec{A}$ (the normal to the surface). SI unit of flux: weber (Wb) $= \text{T m}^2$ . Flux is a scalar.
$\Phi_B = \sum_{\text{all}} \vec{B}_i\cdot d\vec{A}_i$	For a curved surface or a non-uniform field: sum over all area elements $d\vec{A}_i$ .
$\Phi_B = BA\sin\alpha$	Use this when the question gives the angle $\alpha$ that the PLANE of the coil makes with $\vec{B}$ , since then $\theta = 90^{\circ}-\alpha$ .
$N\Phi_B = \text{flux linkage}$	For a closely wound coil of $N$ turns; unit weber-turn. This, not $\Phi_B$ , is what appears in $L$ and $M$ .

Faraday's Law of Induction

Formula	Symbols, units and conditions
$\varepsilon = -\dfrac{d\Phi_B}{dt}$	Induced emf in a single loop, in volt (V). The minus sign is Lenz's law. Applies whenever the flux linked with the circuit changes, by any cause.
$\varepsilon = -N\dfrac{d\Phi_B}{dt}$	Closely wound coil of $N$ turns; $\Phi_B$ is the flux through ONE turn.
$\varepsilon = -NA\cos\theta\,\dfrac{dB}{dt}$	Only the field is changing; $A$ and $\theta$ fixed. With the coil normal to $\vec{B}$ , $\varepsilon = -NA\dfrac{dB}{dt}$ .
$I = \dfrac{\varepsilon}{R}$	Induced current, in ampere (A); $R$ is the total resistance of the closed circuit, in ohm ( $\Omega$ ).
$q = \int I\,dt = \dfrac{N\,\Delta\Phi_B}{R}$	Induced charge in coulomb (C). Note it depends only on the TOTAL flux change, not on how fast the change happens. On an $I$ - $t$ graph, $q$ is the area under the curve.
$\varepsilon = -\dfrac{d}{dt}\left(at^2+bt+c\right) = -(2at+b)$	Pattern for the standard MCQ where the flux is given as a polynomial in $t$ (in Wb) and the emf is wanted at a stated instant.

Lenz's Law and Conservation of Energy

Formula	Symbols, units and conditions
$\varepsilon = -\dfrac{d\Phi_B}{dt}$	The negative sign IS Lenz's law: the induced emf drives a current whose own flux opposes the change that produced it. No separate formula; it fixes the sign or sense of every answer above.

Motional Electromotive Force

Formula	Symbols, units and conditions
$\varepsilon = Blv$	Rod of length $l$ (m) moving with speed $v$ ( $\text{m s}^{-1}$ ) perpendicular to itself and to a uniform field $B$ ; $\vec{v}$ , $\vec{B}$ , $\vec{l}$ mutually perpendicular.
$\varepsilon = Blv\sin\theta$	$\theta$ is the angle between $\vec{v}$ and $\vec{B}$ (rod still perpendicular to both). Zero if the rod moves along $\vec{B}$ .
$\varepsilon = \int (\vec{v}\times\vec{B})\cdot d\vec{l}$	General form; the one to quote when the rod is not uniformly moving, e.g. a rotating rod.
$\varepsilon = \dfrac{1}{2}B\omega l^2 = B\pi\nu l^2$	Rod of length $l$ rotating with angular speed $\omega$ ( $\text{rad s}^{-1}$ ), or frequency $\nu$ (rev per second), about one end, in a field $B$ parallel to the axis of rotation. Use $\omega = 2\pi\nu$ .
$\varepsilon = \dfrac{B\omega r^2}{2}$	Between the centre and the rim of a conducting disc of radius $r$ spinning about its axis in a field $B$ normal to its plane. Independent of the material of the disc.
$\varepsilon_{PQ} = 0,\qquad \varepsilon_{\text{half}} = \dfrac{B\omega l^2}{8}$	Rod of length $l$ rotating about its MID-POINT: the two halves give equal and opposite emfs, so the potential difference between the ends P and Q is zero, though each half separately has $B\omega l^2/8$ .
$\varepsilon = A v\,\dfrac{dB}{dx}$	Loop of area $A$ moving with speed $v$ along $x$ through a field with a uniform gradient $dB/dx$ (in $\text{T m}^{-1}$ ; convert from $\text{T cm}^{-1}$ by multiplying by $10^{2}$ ).
$\Phi_B = Blx,\qquad \lvert\varepsilon\rvert = Blv$	Rectangular loop of width $l$ entering or leaving a field region, $x$ = length already inside. Flux ramps linearly, emf is a constant block, and both are zero while the loop is wholly inside or wholly outside.

6.A Energy Consideration: A Quantitative Study

Formula	Symbols, units and conditions
$I = \dfrac{Blv}{R}$	Current in the rod-on-rails circuit; $R$ is the total resistance of the loop (usually just the rod's resistance $r$ , the rails being of negligible resistance).
$F = BIl = \dfrac{B^2l^2v}{R}$	Retarding magnetic force on the moving rod, in newton (N). Directed opposite to $\vec{v}$ (Lenz's law). The external agent must apply an equal and opposite force for uniform velocity.
$P_{\text{mech}} = Fv = \dfrac{B^2l^2v^2}{R}$	Mechanical power the external agent must supply, in watt (W).
$P_{\text{Joule}} = I^2R = \dfrac{B^2l^2v^2}{R}$	Heat dissipated per second. $P_{\text{mech}} = P_{\text{Joule}}$ — this equality is the statement of energy conservation for the chapter.
$q = \dfrac{\Delta\Phi_B}{R} = \dfrac{Bl\,\Delta x}{R}$	Charge that flows while the rod sweeps a distance $\Delta x$ .
$i = \dfrac{B\omega l^2}{2R},\qquad F = Bil = \dfrac{\pi\nu B^2l^3}{R},\qquad P = Fv$	Rotating rod of length $l$ , resistance $R$ , frequency $\nu$ ; here $v = l\omega = 2\pi l\nu$ is the speed of the free (outer) end, so $P = \dfrac{\pi\nu B^2l^3v}{R}$ .

Inductance — Mutual Inductance

Formula	Symbols, units and conditions
$N_1\Phi_1 = M_{12}I_2,\qquad M_{12}=M_{21}=M$	$M$ is the mutual inductance, SI unit henry (H); $1\ \text{H} = 1\ \text{Wb A}^{-1}$ . Dimensions $[\text{M L}^2\text{T}^{-2}\text{A}^{-2}]$ . Reciprocity $M_{12}=M_{21}$ is general.
$\varepsilon_1 = -M\dfrac{dI_2}{dt}$	Emf induced in coil 1 by a changing current in coil 2. Numerically, $M = \dfrac{\lvert\varepsilon_1\rvert}{\lvert dI_2/dt\rvert}$ .
$\Delta\Phi_{\text{linkage}} = M\,\Delta I$	Change of flux linkage with the second coil, in Wb (weber-turns).
$M = \mu_0 n_1 n_2 \pi r_1^2\, l = \dfrac{\mu_0 N_1 N_2 \pi r_1^2}{l}$	Two long coaxial solenoids of the same length $l$ , inner radius $r_1$ ; $n_1,n_2$ turns per unit length, $N_1=n_1l$ , $N_2=n_2l$ . It is always the INNER radius that appears. With a core of relative permeability $\mu_r$ , multiply by $\mu_r$ .
$M = \dfrac{\mu_0 N_1 N_2 \pi r_2^2}{2r_1}$	Two concentric coplanar/coaxial circular coils, centres coinciding: outer coil radius $r_1$ with $N_1$ turns, inner coil radius $r_2$ with $N_2$ turns, $r_2 \ll r_1$ . For single turns this is $M = \dfrac{\mu_0 \pi r_2^2}{2r_1}$ .
$M = \dfrac{\mu_0}{4\pi}\cdot\dfrac{2\pi R^2\cdot\pi r^2}{\left(R^2+x^2\right)^{3/2}}$	Small loop of radius $r$ on the axis of a big loop of radius $R$ , centres a distance $x$ apart.
$M = \dfrac{\mu_0 N_1 N_2 A}{l}$	Small loop of area $A$ ( $N_2$ turns) inside a long solenoid of length $l$ with $N_1$ turns, normal to the axis.

Inductance — Self-inductance and Energy

Formula	Symbols, units and conditions
$N\Phi_B = LI \quad\Longrightarrow\quad L = \dfrac{N\Phi_B}{I}$	$L$ is the self-inductance (coefficient of self-induction), SI unit henry (H). It depends only on geometry and on the medium, never on the current.
$\varepsilon = -L\dfrac{dI}{dt}$	Self-induced (back) emf. A graph of $\lvert\varepsilon\rvert$ against $dI/dt$ is a straight line through the origin of slope $L$ .
$L = \mu_0\mu_r n^2 A l = \dfrac{\mu_0\mu_r N^2 A}{l}$	Long solenoid: $n$ turns per unit length, $N=nl$ total turns, cross-section $A$ , length $l$ , core of relative permeability $\mu_r$ ( $\mu_r=1$ for air). Note $L \propto N^2$ and $L \propto 1/l$ .
$U = \dfrac{1}{2}LI^2$	Magnetic energy stored in an inductor, in joule (J). A graph of $U$ against $I$ is an upward parabola through the origin.
$u_B = \dfrac{U}{V} = \dfrac{B^2}{2\mu_0}$	Magnetic energy per unit volume, in $\text{J m}^{-3}$ . Electrostatic analogue: $u_E = \dfrac{1}{2}\varepsilon_0E^2$ .
$\varepsilon_1 = -L_1\dfrac{dI_1}{dt} - M_{12}\dfrac{dI_2}{dt}$	Two nearby coils carrying currents at the same time.

AC Generator

Formula	Symbols, units and conditions
$\Phi_B = NBA\cos\omega t$	Flux linkage with an armature coil of $N$ turns, area $A$ , rotating at angular speed $\omega$ in a uniform field $B$ , taking the coil normal parallel to $\vec{B}$ at $t=0$ .
$\varepsilon = NBA\omega\sin\omega t = \varepsilon_0\sin\omega t$	Instantaneous emf. Peak emf $\varepsilon_0 = NBA\omega$ .
$\varepsilon_0 = NBA\omega = 2\pi f\,NBA$	$f$ = frequency of rotation in revolutions per second (Hz); $\omega = 2\pi f$ . Mains frequency in India is $50\ \text{Hz}$ .
$\varepsilon = \varepsilon_0\sin 2\pi f t,\qquad I = \dfrac{\varepsilon_0}{R}\sin\omega t$	Alternating emf and current; $I_0 = \varepsilon_0/R$ .

Constants, units and dimensions

Quantity Symbol SI unit Dimensions
Magnetic flux $\Phi_B$ weber (Wb) $=\text{T m}^2$ $[\text{M L}^2\text{T}^{-2}\text{A}^{-1}]$
Induced emf $\varepsilon$ volt (V) $[\text{M L}^2\text{T}^{-3}\text{A}^{-1}]$
Mutual inductance $M$ henry (H) $[\text{M L}^2\text{T}^{-2}\text{A}^{-2}]$
Self-inductance $L$ henry (H) $[\text{M L}^2\text{T}^{-2}\text{A}^{-2}]$
Magnetic energy density $u_B$ $\text{J m}^{-3}$ $[\text{M L}^{-1}\text{T}^{-2}]$
Permeability of free space $\mu_0 = 4\pi\times 10^{-7}\ \text{T m A}^{-1}$ ; take $\pi = 3.14$ unless the paper says otherwise.
$1\ \text{gauss} = 10^{-4}\ \text{T}$ ; $1\ \text{Wb m}^{-2} = 1\ \text{T}$ ; $1\ \text{H} = 1\ \text{Wb A}^{-1} = 1\ \text{V s A}^{-1}$ .
Unit conversions the numericals need: $1\ \text{cm}^2 = 10^{-4}\ \text{m}^2$ ; $1\ \text{T cm}^{-1} = 10^{2}\ \text{T m}^{-1}$ ; $1\ \text{A ms}^{-1} = 10^{3}\ \text{A s}^{-1}$ .

Short NotesThe chapter subtopic by subtopic, in the textbook's own order.8 subtopics

The whole chapter in brief

6.1 Introduction — what the chapter says

Oersted, Ampere and others showed that moving charges (currents) produce magnetic fields. Faraday (England) and Henry (USA), around 1830, showed the converse: a CHANGING magnetic field produces an electric current in a closed coil.
The phenomenon of producing an emf, and hence a current, by changing the magnetic flux linked with a circuit is called electromagnetic induction.
Everything in this chapter — generators, transformers, inductors, eddy-current braking — follows from one sentence: an emf appears whenever the magnetic flux linked with a circuit changes.

6.2 The Experiments of Faraday and Henry

Experiment 6.1 — bar magnet and coil. Push the N-pole of a bar magnet towards a coil connected to a galvanometer: the pointer deflects while the magnet moves. Hold it still: no deflection. Pull it away: deflection reverses. Use the S-pole: deflections are opposite to those for the N-pole. Move faster: larger deflection. Move the coil instead of the magnet: identical result. Conclusion — it is the RELATIVE motion between magnet and coil that induces the current.
Experiment 6.2 — two coils, one carrying a steady current. Replace the magnet by coil $C_2$ carrying a steady current from a battery, and move it towards $C_1$ : the galvanometer in $C_1$ deflects; away, the deflection reverses; stationary, nothing. Again, relative motion is what matters.
Experiment 6.3 — no motion at all. Both coils held stationary, $C_2$ connected to a battery through a tapping key K. Pressing K gives a momentary deflection; holding K pressed gives none; releasing K gives a momentary deflection the other way. Inserting an iron rod along the axis increases the deflection dramatically. Conclusion — relative motion is not necessary; a CHANGING current (hence changing flux) is enough.
To make the deflection larger: use a soft-iron core, use a more powerful battery, move the arrangement faster. To show the current without a galvanometer: replace it with a small torch bulb, which glows.

6.3 Magnetic Flux

$\Phi_B = \vec{B}\cdot\vec{A} = BA\cos\theta$ , where $\theta$ is between $\vec{B}$ and the AREA VECTOR (the normal), not the plane.
SI unit weber (Wb) $=\ \text{T m}^2$ ; flux is a SCALAR; dimensions $[\text{M L}^2\text{T}^{-2}\text{A}^{-1}]$ .
Non-uniform field or curved surface: $\Phi_B = \sum \vec{B}_i\cdot d\vec{A}_i$ .
Flux is maximum ( $BA$ ) when the plane of the coil is perpendicular to $\vec{B}$ , and zero when the plane contains $\vec{B}$ .
Flux linkage of an $N$ -turn closely wound coil is $N\Phi_B$ .

6.4 Faraday's Law of Induction

Statement: the magnitude of the induced emf in a circuit is equal to the time rate of change of magnetic flux through the circuit.
$\varepsilon = -\dfrac{d\Phi_B}{dt}$ , and for $N$ turns $\varepsilon = -N\dfrac{d\Phi_B}{dt}$ .
Flux can be changed by changing $B$ , by changing $A$ (stretching, shrinking or moving the loop), or by changing $\theta$ (rotating the coil). Each gives an emf.
Induced current $I = \varepsilon/R$ ; induced charge $q = N\Delta\Phi_B/R$ depends only on the total flux change, not on the time taken.
Increasing $N$ increases the emf but not the charge, if $R$ scales with $N$ .

6.5 Lenz's Law and Conservation of Energy

Statement: the polarity of the induced emf is such that it tends to produce a current which opposes the change in magnetic flux that produced it. It is the physical content of the minus sign in Faraday's law.
N-pole approaching a coil: the near face becomes a north pole (current anticlockwise seen from the magnet) and repels the magnet. N-pole receding: the near face becomes a south pole (current clockwise) and attracts it. Either way, the induced effect opposes the motion.
Energy argument: the opposing force means the external agent must do work to keep the flux changing; that work reappears as electrical energy and finally as Joule heat. If the induced current flowed the other way, the magnet would be pulled in, accelerate on its own and gain kinetic energy for nothing — a perpetual motion machine. Hence Lenz's law is a consequence of conservation of energy.
An open circuit still has an induced emf across its ends; only the current is absent.

Situation	Change in flux	Induced current (as seen by the reader)
N-pole of magnet pushed towards coil	increasing, towards reader	anticlockwise, near face becomes N
N-pole pulled away from coil	decreasing	clockwise, near face becomes S
Current in a nearby straight wire increasing	increasing	so as to oppose, i.e. reverse sense to the wire's field
Current in a nearby straight wire decreasing	decreasing	same sense as the wire's field, to sustain it
Irregular loop becoming circular, field into page	area up, flux into page up	anticlockwise
Circular loop being deformed into a straight wire	area down	clockwise (field into page)
Loop moving entirely inside a uniform field	no change	zero

6.6 Motional Electromotive Force

A rod of length $l$ moved with speed $v$ perpendicular to a uniform field $B$ develops $\varepsilon = Blv$ across its ends. Zero if $\vec{v}$ is parallel to $\vec{B}$ , or if the rod lies along $\vec{v}\times\vec{B}$ .
Two ways to get it, both accepted: (i) from Faraday's law, $\Phi_B=Blx$ so $\lvert\varepsilon\rvert = Bl\,\lvert dx/dt\rvert = Blv$ ; (ii) from the Lorentz force $qvB$ on each free charge, work done $W=qvBl$ over the rod, so $\varepsilon = W/q = Blv$ .
Rotating rod about one end: $\varepsilon = \dfrac{1}{2}B\omega l^2 = B\pi\nu l^2$ . About its mid-point: net emf between the ends is ZERO.
The number of spokes in a rotating wheel is irrelevant — the spoke emfs are in parallel.
For a stationary conductor in a changing field there is no $q\vec{v}\times\vec{B}$ term, so the emf must come from an electric field: a time-varying magnetic field generates an electric field. This is the deeper content of Faraday's discovery.

Energy Consideration: A Quantitative Study

Rod on rails, resistance $R$ : $\varepsilon = Blv$ , $I = \dfrac{Blv}{R}$ , retarding force $F = BIl = \dfrac{B^2l^2v}{R}$ , external power $P = Fv = \dfrac{B^2l^2v^2}{R}$ , Joule heat per second $I^2R = \dfrac{B^2l^2v^2}{R}$ .
The two powers are exactly equal: all the mechanical work done against the magnetic force appears as heat. Nothing is created; Lenz's law is book-keeping for energy.
The force is zero once the loop is wholly inside or wholly outside the field region, because the flux then stops changing.
Rotating rod version: $i = \dfrac{B\omega l^2}{2R}$ , $F = Bil = \dfrac{\pi\nu B^2l^3}{R}$ , $P = Fv$ with $v = 2\pi l\nu$ .

6.7 Inductance

Flux linkage is proportional to current: $N\Phi_B \propto I$ . The constant of proportionality is the inductance. It is a scalar, unit henry (H), dimensions $[\text{M L}^2\text{T}^{-2}\text{A}^{-2}]$ , and depends only on geometry and on the medium — never on the current.
Mutual inductance $M$ : $N_1\Phi_1 = MI_2$ and $\varepsilon_1 = -M\dfrac{dI_2}{dt}$ . Reciprocity $M_{12}=M_{21}=M$ always holds, which is what makes the hard direction of a calculation easy.
Two long coaxial solenoids of equal length $l$ : $M = \mu_0 n_1n_2\pi r_1^2 l = \dfrac{\mu_0N_1N_2\pi r_1^2}{l}$ , with $r_1$ the INNER radius.
Concentric coils, outer radius $r_1$ with $N_1$ turns, inner radius $r_2 \ll r_1$ with $N_2$ turns: $M = \dfrac{\mu_0N_1N_2\pi r_2^2}{2r_1}$ .
Self-inductance $L$ : $N\Phi_B = LI$ , $\varepsilon = -L\dfrac{dI}{dt}$ . The self-induced emf is called the BACK emf because it always opposes the change of current, whether growing or decaying.
Long solenoid: $L = \mu_0\mu_r n^2Al = \dfrac{\mu_0\mu_rN^2A}{l}$ . So $L\propto N^2$ , $L\propto A$ , $L\propto 1/l$ , and $L$ jumps when a soft-iron core is inserted.
Energy stored: $U = \dfrac{1}{2}LI^2$ ; energy per unit volume $u_B = \dfrac{B^2}{2\mu_0}$ .
$L$ is the electromagnetic analogue of MASS: it is electrical inertia, resisting any change of current, exactly as $\dfrac{1}{2}LI^2$ mirrors $\dfrac{1}{2}mv^2$ .

Mechanics	Electromagnetism
mass $m$	self-inductance $L$
velocity $v$	current $I$
kinetic energy $\dfrac{1}{2}mv^2$	magnetic energy $\dfrac{1}{2}LI^2$
force opposing acceleration	back emf $-L\,dI/dt$

Self-inductance $L$	Mutual inductance $M$
Single coil; flux of the coil linked with itself	Pair of coils; flux of one linked with the other
$N\Phi = LI$ , $\varepsilon = -L\,dI/dt$	$N_1\Phi_1 = MI_2$ , $\varepsilon_1 = -M\,dI_2/dt$
Depends on turns, area, length, core	Also depends on separation and relative orientation
Solenoid: $\mu_0\mu_r n^2Al$	Coaxial solenoids: $\mu_0n_1n_2\pi r_1^2l$

6.8 AC Generator

Principle: electromagnetic induction. Rotating a coil in a uniform magnetic field changes the flux linked with it and induces an emf. It converts MECHANICAL energy into ELECTRICAL energy.
Parts: armature coil (many turns of insulated copper wire on a soft-iron core, which increases the flux linked); strong field magnets, with the axis of rotation perpendicular to $\vec{B}$ ; two slip rings, which give a sliding contact between the rotating coil and the external circuit; two carbon brushes, which carry the current out to the load.
$\Phi_B = NBA\cos\omega t$ , so $\varepsilon = -\dfrac{d\Phi_B}{dt} = NBA\omega\sin\omega t = \varepsilon_0\sin\omega t$ with $\varepsilon_0 = NBA\omega = 2\pi fNBA$ .
The emf is maximum when the plane of the coil is PARALLEL to $\vec{B}$ (i.e. $\theta=90^{\circ}$ or $270^{\circ}$ ), where the rate of change of flux is greatest, and zero when the coil's plane is perpendicular to $\vec{B}$ , where the flux is maximum.
Direction of the induced current: Fleming's right-hand rule.
Graphs: flux is a cosine curve between $+NBA$ and $-NBA$ ; emf is a sine curve between $+\varepsilon_0$ and $-\varepsilon_0$ , a quarter cycle out of step with the flux.
Commercial machines: hydro-electric, thermal and nuclear generators, up to about $500\ \text{MW}$ . Mains frequency is $50\ \text{Hz}$ in India and $60\ \text{Hz}$ in the USA. In most large generators the coils are held stationary and the electromagnets rotate.

Standard graphs CBSE asks you to plot

Plot	Shape
$\lvert\varepsilon\rvert$ against $dI/dt$ for a coil	straight line through the origin, slope $L$
$\Phi$ against $I$ for an inductor	straight line through the origin, slope $L$
$U$ against $I$ for an inductor	upward parabola through the origin
$\varepsilon$ against $\omega$ for a rod rotating about one end	straight line through the origin
$\varepsilon$ against $l$ for a rod rotating about one end	upward parabola through the origin, since $\varepsilon\propto l^2$
$\Phi$ against $t$ , rectangular loop crossing a field region	trapezium: linear rise, flat top, linear fall
$\varepsilon$ against $t$ , same loop	negative block, zero, positive block
$\varepsilon$ against $t$ , AC generator	sine curve, peak $\varepsilon_0=NBA\omega$

Traps that cost marks

$\theta$ in $BA\cos\theta$ is measured from the NORMAL. If the question says the PLANE makes $60^{\circ}$ with the field, then $\theta = 30^{\circ}$ . Read the wording twice.
A wire passing normally through the centre of a loop gives zero flux for ALL currents — its field lies in the plane of the loop — so however fast the current changes, no emf is induced.
A loop moving entirely inside a uniform, time-independent field has no emf: nothing about $B$ , $A$ or $\theta$ is changing. Only a loop crossing the boundary of the field region has one.
Induced CHARGE $q = N\Delta\Phi/R$ does not depend on the time taken; induced EMF does. Do not confuse the two.
Inductance never depends on the current. Halving the current in a solenoid leaves $L$ unchanged.
$L\propto N^2$ , not $N$ . Doubling the turns at fixed length and area makes $L$ four times; doubling both the turns and the length makes it only twice.
In the coaxial-solenoid formula for $M$ , the radius that appears is the INNER one, whichever solenoid you imagine carrying the current — that is what reciprocity guarantees.
A rod rotating about its mid-point has ZERO potential difference between its ends, even though each half has an emf across it.
For a rod on rails, use the TOTAL circuit resistance in $I=\varepsilon/R$ , but note that when the rails have negligible resistance this is just the rod's own resistance.
Mechanical power supplied equals the Joule heat, not twice it and not half it. Quote $P = \dfrac{B^2l^2v^2}{R}$ once and use it for both.
Emf in an AC generator is maximum when the FLUX IS ZERO (coil plane along $\vec{B}$ ), not when the flux is maximum.
Slip rings, not a split-ring commutator: a commutator would make it a DC generator.
Always write the negative sign in $\varepsilon = -N\,d\Phi/dt$ when the question says "state Faraday's law" — the sign carries a mark as Lenz's law.
Keep units straight: $\text{cm}^2 \to 10^{-4}\ \text{m}^2$ , $\text{T cm}^{-1} \to 10^2\ \text{T m}^{-1}$ , $\text{A ms}^{-1}\to 10^3\ \text{A s}^{-1}$ , $1\ \text{G}=10^{-4}\ \text{T}$ .

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.20

1
Derivation 1
Describe a simple experiment (or activity) to show that the polarity of the emf induced in a coil is always such that it tends to produce a current which opposes the change of magnetic flux that produces it. OR State Lenz's law. Give one example to illustrate this law. "The Lenz's law is a consequence of the principle of conservation of energy." Justify this statement. (3/5, AI 2014)
Asked in: CBSE Class 12 Physics — Long Answer · 2/5, Delhi 2014 · Section E, 5 marks
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Lenz's law states that the direction of the induced emf and the direction of the induced current are always such that they oppose the cause which produces them. Simple experiment/example: When the N-pole of a bar magnet is moved towards a coil connected to a galvanometer, the induced current in the coil flows in the anticlockwise direction (as viewed from the magnet's side), so as to make that face of the coil a north pole, which opposes the motion of the magnet towards the coil by exerting a repulsive force on it. Similarly, when the N-pole of the magnet is moved away from the coil, the induced current flows in the clockwise direction, developing a south pole on that face, which attracts the receding magnet, again opposing its motion (this time, opposing its departure). Justification via conservation of energy: In order to keep changing the magnetic flux linked with the circuit (i.e. to keep moving the magnet against the opposition offered by the induced emf), some work must be done by the external agent (or energy must be spent). This energy spent ultimately appears in the circuit as electrical energy. Suppose, for contradiction, that Lenz's law were not valid, i.e. suppose the induced current flowed in the direction opposite to that predicted by Lenz's law. Then the resulting force on the magnet would make it move faster and faster on its own — the magnet would gain kinetic energy without any equivalent expenditure of energy by an external agent. This would amount to a perpetual motion machine, which violates the law of conservation of energy. Hence Lenz's law must be valid, and it is indeed a direct consequence of the law of conservation of energy.
2
Derivation 2
A rectangular conductor MNPQ, with a movable arm MN (resistance $r$ ), is kept in a uniform magnetic field (into the page), as shown in the figure. Resistances of the arms MQ, QP and PN are negligible, and the arm MN moves with velocity $v$ . Obtain the expression for the (a) current induced in the loop, specifying its direction, and (b) power required to move the arm.
Asked in: CBSE Class 12 Physics — SA · 2023 · Section C, 3 marks
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(a) Let the rod MN of length $l$ move with uniform velocity $\vec{v}$ on two parallel rails, in the presence of a uniform magnetic field $B$ directed into the page. In time $dt$ , the rod moves a distance $dx$ , so the change in area of the loop is $dA = l\,dx$ . Change in flux, $d\phi = B\,dA = Bl\,dx$ , so induced emf $\varepsilon = \left|\dfrac{d\phi}{dt}\right| = Bl\dfrac{dx}{dt} = Blv$ So induced current flows through the circuit: $i = \dfrac{\varepsilon}{R} = \dfrac{Blv}{R}$ , where $R (= r)$ is the resistance of the rails (here, the moving rod MN). By Lenz's law (or Fleming's right-hand rule), this induced current opposes the increase in flux and flows in the sense that produces a retarding force on MN. (b) The retarding (opposing) force exerted on the rod by the induced current is $\vec{F}_m = i(\vec{l}\times\vec{B})$ , i.e. $F_m = Bil = \dfrac{B^2l^2v}{R}$ For the rod to move with uniform velocity, an external force equal and opposite to this must be applied: $F_{ext} = F_m = \dfrac{B^2l^2v}{R}$ Power required to move the arm (mechanical power delivered by the external agent) is $P = F_{ext}\cdot v = \dfrac{B^2l^2v^2}{R}$
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Derivation 3
The figure shows a rectangular conducting frame MNOP of resistance $R$ , placed partly in a perpendicular magnetic field $\vec{B}$ (into the page), with the arm ON (of length $l$ ) lying inside the field. The frame is moved with velocity $\vec{v}$ (directed from O towards P, i.e. pulling the frame out of the field), the perpendicular distance between arms MN and PO being $b$ . Obtain the expression for the (a) force acting on the arm 'ON' and its direction, and (b) power required to move the frame so as to get a steady emf induced between the arms MN and PO.
Asked in: CBSE Class 12 Physics — SA · AI 2019 · Section C, 3 marks
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(a) While the frame is partly inside the magnetic field (with arm ON inside the field, cutting field lines as it's dragged through), a motional emf $\varepsilon = Blv$ is induced, driving a current $i = \dfrac{Blv}{R}$ through the loop. The magnetic force on the current-carrying arm ON is then $F_{ON} = \dfrac{l^2B^2v}{R}$ , directed so as to oppose the motion of the frame (i.e. opposite to $\vec{v}$ ), by Lenz's law. However, once the frame has moved fully inside (or fully outside) the field, there is no further change in flux, so no current flows, and hence $F_{ON} = 0$ in that situation. (b) To keep the frame moving at this steady velocity while ON is inside the field (so that a steady emf is maintained between the arms MN and PO), the external agent must supply a force equal to $F_{ON}$ . Power required to move the frame is $P = F_{ext}\cdot v = \dfrac{B^2l^2v^2}{R}$
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Derivation 4
A metallic rod of length $l$ and resistance $R$ is rotated with a frequency $\nu$ , with one end hinged at the centre and the other end at the circumference of a circular metallic ring of radius $l$ , about an axis passing through the centre and perpendicular to the plane of the ring. A constant and uniform magnetic field $B$ parallel to the axis is present everywhere. (a) Derive the expression for the induced emf and the current in the rod. (b) Due to the presence of the current in the rod and the magnetic field, find the expression for the magnitude and direction of the force acting on this rod. (c) Hence obtain the expression for the power required to rotate the rod.
Asked in: CBSE Class 12 Physics — Long Answer · AI 2014C · Section E, 5 marks
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(a) Let $\theta$ be the angle traced by a point on the rod (or a free electron in it) in time $t$ , as it rotates with angular speed $\omega = 2\pi\nu$ . Area swept out by the rod in this time is $A = \pi l^2\times \dfrac{\theta}{2\pi} = \dfrac{1}{2}l^2\theta$ Magnetic flux linked, $\phi = B\left(\dfrac{1}{2}l^2\theta\right)\cos 0° = \dfrac{1}{2}Bl^2\theta$ By Faraday's law, induced emf, $\varepsilon = \dfrac{d\phi}{dt} = \dfrac{1}{2}Bl^2\dfrac{d\theta}{dt} = \dfrac{1}{2}Bl^2\omega$ $\therefore$ Induced current, $i = \dfrac{\varepsilon}{R} = \dfrac{Bl^2\omega}{2R}$ (b) Force acting on the rod (carrying current $i$ in field $B$ ), $F = ilB = \dfrac{\pi\nu B^2l^3}{R}$ (using $\omega = 2\pi\nu$ ). This force, by Lenz's law, opposes the rotation of the rod; the external agent must apply a force equal and opposite to this Lorentz force on the rod to keep it rotating at constant $\omega$ . (c) Power required to rotate the rod is $P = Fv = \dfrac{\pi\nu B^2l^3 v}{R}$ , where $v = l\omega = 2\pi l\nu$ is the linear speed of the outer (free) end of the rod.
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Derivation 5
Two concentric circular coils X and Y of radii $r_1$ and $r_2$ ( $r_1 > r_2$ ), having $N_1$ and $N_2$ turns respectively, are placed coaxially with their centres coinciding. Obtain an expression for (i) the mutual inductance of the arrangement, and (ii) the magnetic flux linked with coil Y when a current $I$ flows through coil X.
Asked in: CBSE Class 12 Physics — SA · 2021C · Section C, 3 marks
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Since $r_1 > r_2$ (coil Y lies well within coil X), when a current $I$ flows through the outer coil X, the magnetic field at the centre — and, since $r_2$ is small compared to $r_1$ , essentially uniform over the entire area of the inner coil Y — is $B = \dfrac{\mu_0 N_1 I}{2r_1}$ The magnetic flux linked with coil Y (of area $\pi r_2^2$ and $N_2$ turns) is $N_2\phi_2 = B\times \pi r_2^2\times N_2 = \dfrac{\mu_0 N_1 N_2 \pi r_2^2}{2r_1}I$ (i) $\therefore$ Mutual inductance, $M = \dfrac{N_2\phi_2}{I} = \dfrac{\mu_0 N_1 N_2 \pi r_2^2}{2r_1}$ (ii) Magnetic flux linked with Y is $N_2\phi_2 = MI = \dfrac{\mu_0 N_1 N_2 \pi r_2^2}{2r_1}I$
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Derivation 6
(a) Define mutual inductance and write its S.I. unit. (b) A square loop of side 'a' carrying a current $I_2$ is kept at distance $x$ from an infinitely long straight wire carrying a current $I_1$ , as shown in the figure (the loop is coplanar with the wire, with its near side parallel to the wire at distance $x$ , and its far side at distance $x+a$ ). Obtain the expression for the resultant force acting on the loop.
Asked in: CBSE Class 12 Physics — Long Answer · AI 2019 · Section E, 5 marks
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(a) The phenomenon of inducing an emf in a circuit by changing the magnetic flux in a neighbouring circuit is called mutual induction. The S.I. unit of mutual inductance is the henry (H). (b) Consider two straight, infinitely long, parallel conducting wires carrying currents $I_1$ and $I_2$ , separated by a distance $x$ . The force per unit length between them is $\dfrac{F}{a} = \dfrac{\mu_0 I_1 I_2}{2\pi x}$ The near side of the loop (at distance $x$ , carrying current $I_2$ parallel to $I_1$ ) experiences a force of attraction: $F_1 = \dfrac{\mu_0}{4\pi}\dfrac{2I_1I_2 a}{x}$ The far side of the loop (at distance $x+a$ , carrying current $I_2$ antiparallel to $I_1$ ) experiences a force of repulsion: $F_2 = \dfrac{\mu_0}{4\pi}\dfrac{2I_1I_2 a}{x+a}$ The forces on the two sides of the loop perpendicular to the wire cancel out by symmetry. Net force on the loop, $F_{net} = F_1 - F_2 = \dfrac{\mu_0}{4\pi}\dfrac{2I_1I_2 a^2}{x(x+a)}$ , directed towards the wire (i.e. the loop is attracted towards the wire, since the near-side attraction exceeds the far-side repulsion).
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Derivation 7
Define mutual inductance between a pair of coils. Derive an expression for the mutual inductance of two long coaxial solenoids of the same length, wound one over the other. OR Define the term 'mutual inductance' between two coils. Obtain the expression for the mutual inductance of a pair of long coaxial solenoids, each of length $l$ and radii $r_1$ and $r_2$ ( $r_2 \gg r_1$ ), the total number of turns in the two solenoids being $N_1$ and $N_2$ respectively. (AI 2014)
Asked in: CBSE Class 12 Physics — Long Answer · AI 2017 · Section E, 5 marks
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Mutual inductance: the phenomenon of inducing an emf in a circuit by changing the current or magnetic flux in a neighbouring circuit is called mutual induction, and the constant of proportionality $M$ between the flux linked with one coil and the current in the other is called their mutual inductance. Consider two long coaxial solenoids, each of length $l$ : let $S_1$ (inner, radius $r_1$ ) have $n_1$ turns per unit length, and $S_2$ (outer, radius $r_2$ ) have $n_2$ turns per unit length, with total turns $N_1$ and $N_2$ respectively. When a current $I$ is passed through the outer solenoid $S_2$ , the magnetic field produced inside it is $B_2 = \mu_0 n_2 I$ The flux linked with the inner solenoid $S_1$ is $N_1\phi_1 = B_2(\pi r_1^2)n_1 l = \mu_0 n_1 n_2 \pi r_1^2 lI$ (using the inner solenoid's cross-sectional area, since that is the area common to both) $\therefore M_{12} = \mu_0 n_1 n_2 \pi r_1^2 l$ Similarly, when a current $I$ is passed through the inner solenoid $S_1$ , the flux linked with the outer solenoid $S_2$ gives $M_{21} = \mu_0 n_1 n_2 \pi r_1^2 l$ So $M_{12} = M_{21} = M$ . Hence, the coefficient of mutual induction between the two coaxial solenoids is $M = \mu_0 n_1 n_2 \pi r_1^2 l$ , or equivalently $M = \dfrac{\mu_0 N_1 N_2 \pi r_1^2}{l}$
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Derivation 8
Explain the meaning of the term mutual inductance. Consider two concentric, coplanar circular coils, one of radius $r_1$ and the other of radius $r_2$ ( $r_1 < r_2$ ), placed with their centres coinciding. Obtain the expression for the mutual inductance of the arrangement.
Asked in: CBSE Class 12 Physics — Long Answer · 2/5, AI 2016 · Section E, 5 marks
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Mutual inductance of a pair of coils is defined as the emf induced in one of the coils when the rate of change of current in the other coil is unity. When current $I_2$ flows through the outer coil-2 (radius $r_2$ ), the magnetic field produced at its centre is $B_2 = \dfrac{\mu_0 I_2}{2r_2}$ , directed normal to the plane of the coils. Since $r_1 < r_2$ , this field may be taken as uniform over the area of the inner coil-1, so the flux linked with coil-1 is $\phi_{12} = B_2 A_1\cos 0° = \dfrac{\mu_0 I_2}{2r_2}\times \pi r_1^2\times 1$ $\therefore \dfrac{\phi_{12}}{I_2} = \dfrac{\mu_0\pi r_1^2}{2r_2}$ , so the mutual inductance is $M_{12} = \dfrac{\mu_0\pi r_1^2}{2r_2}$
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Derivation 9
(a) Define mutual inductance and write its S.I. units. (b) Derive an expression for the mutual inductance of two long coaxial solenoids of the same length, wound one over the other. (c) In an experiment, two coils $C_1$ and $C_2$ are placed close to each other. Find the expression for the emf induced in coil $C_1$ due to a change in the current through coil $C_2$ .
Asked in: CBSE Class 12 Physics — Long Answer · Delhi 2015 · Section E, 5 marks
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(a) Mutual inductance is the phenomenon of an opposing emf being produced in one coil as a result of a change in current (or flux) in a neighbouring coil. S.I. unit is henry (H). (b) [Same derivation as for two coaxial solenoids of equal length above:] Considering an inner solenoid $S_1$ (radius $r_1$ , $n_1$ turns per unit length) and outer solenoid $S_2$ (radius $r_2$ , $n_2$ turns per unit length), each of length $l$ : $M = \mu_0 n_1 n_2 \pi r_1^2 l = \dfrac{\mu_0 N_1 N_2 \pi r_1^2}{l}$ (c) The mutually induced emf in coil $C_1$ (as capacitor $C_2$ 's current changes) is $\varepsilon = -\dfrac{d\phi}{dt} = -\dfrac{d}{dt}(Mi) = -M\dfrac{di}{dt}$ , where $\dfrac{di}{dt}$ is the rate of change of current in the neighbouring coil $C_2$ .
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Derivation 10
Deduce an expression for the mutual inductance of two long coaxial solenoids but having different radii and different number of turns.
Asked in: CBSE Class 12 Physics — Long Answer · 3/5, AI 2014 · Section E, 5 marks
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Consider two long coaxial solenoids, each of length $l$ . Let $n_1$ be the number of turns per unit length of the inner solenoid $S_1$ (radius $r_1$ ), and $n_2$ the number of turns per unit length of the outer solenoid $S_2$ (radius $r_2$ ). Let $N_1$ and $N_2$ be the total number of turns of $S_1$ and $S_2$ respectively. When a current $I$ is passed through $S_2$ , the field inside it is $B_2 = \mu_0 n_2 I$ ; the flux linked with $S_1$ is $N_1\phi_1 = B_2(\pi r_1^2)n_1 l$ (since flux threads only the common cross-sectional area, that of the narrower, inner solenoid $S_1$ ) $N_1\phi_1 = \mu_0 n_1 n_2 \pi r_1^2 lI \;\Rightarrow\; M_{12} = \mu_0 n_1 n_2 \pi r_1^2 l$ Similarly, passing a current through $S_1$ and finding the flux linked with $S_2$ gives $M_{21} = \mu_0 n_1 n_2 \pi r_1^2 l$ Since $M_{12} = M_{21} = M$ , the mutual inductance of the two coaxial solenoids (of different radii and different turns) is $M = \mu_0 n_1 n_2 \pi r_1^2 l$ , or equivalently $M = \dfrac{\mu_0 N_1 N_2 \pi r_1^2}{l}$ (note: it is the radius $r_1$ of the narrower/inner solenoid that appears here, regardless of which solenoid carries the current, by the reciprocity $M_{12}=M_{21}$ )
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Derivation 11
(a) Define the term 'self-inductance' and write its S.I. unit. (b) Obtain the expression for the mutual inductance of two long coaxial solenoids $S_1$ and $S_2$ wound one over the other, each of length $L$ and radii $r_1$ and $r_2$ , and $n_1$ and $n_2$ number of turns per unit length, when a current $I$ is set up in the outer solenoid $S_2$ .
Asked in: CBSE Class 12 Physics — SA · Delhi 2017 · Section C, 3 marks
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(a) Self-inductance: when the current in a coil is changed, a back emf is induced in the same coil; this is called self-inductance. If $L$ is the self-inductance of the coil, $N\phi = LI \Rightarrow L = \dfrac{N\phi}{I}$ . The S.I. unit of self-inductance is the henry (H). (b) Let $S_1$ (inner solenoid, radius $r_1$ , $n_1$ turns per unit length) and $S_2$ (outer solenoid, radius $r_2$ , $n_2$ turns per unit length) be coaxial, each of length $L$ . When current $I$ flows through the outer solenoid $S_2$ , the field it produces is $B_2 = \mu_0 n_2 I$ The flux linked with the inner solenoid $S_1$ is $N_1\phi_1 = B_2(\pi r_1^2)n_1 L = \mu_0 n_1 n_2 \pi r_1^2 LI$ (using $S_1$ 's cross-sectional area, being the smaller of the two) $\therefore M_{12} = \mu_0 n_1 n_2 \pi r_1^2 L$ By reciprocity, $M_{12} = M_{21} = M = \mu_0 n_1 n_2 \pi r_1^2 L$
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Derivation 12
Define self-inductance of a coil. Obtain the expression for the energy stored in an inductor $L$ connected across a source of emf. OR Define the term self-inductance of a solenoid. Obtain the expression for the magnetic energy stored in an inductor of self-inductance $L$ to build up a current $I$ through it. (AI 2014)
Asked in: CBSE Class 12 Physics — SA · AI 2017 · Section C, 3 marks
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Self-inductance: the phenomenon by which a coil opposes any change in current flowing through it, by inducing a back emf in itself, is called self-inductance. Consider a circuit with an inductor $L$ and resistor $R$ connected to a source of emf. As the current grows, part of the work done by the source is stored in the inductor as magnetic-field energy, and the rest appears as heat in the resistor. If $I$ is the current at time $t$ , the self-induced emf is $\varepsilon = L\dfrac{dI}{dt}$ In a small time $dt$ , work done by the source against this back emf is $dW = \varepsilon I\,dt = L\dfrac{dI}{dt}I\,dt = LI\,dI$ Total work done in building up the current from $0$ to $I_0$ : $W = \displaystyle\int_0^{I_0} LI\,dI = \dfrac{1}{2}LI_0^2$ This entire work done is stored as magnetic potential energy in the inductor: $U = \dfrac{1}{2}LI^2$
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Derivation 13
State the basic principle behind the working of an ac generator. Briefly describe its working and obtain the expression for the instantaneous value of the emf induced.
Asked in: CBSE Class 12 Physics — SA · 2023 · Section C, 3 marks
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Principle: An ac generator (dynamo) is a device that converts mechanical energy into electrical energy. It works on the principle of electromagnetic induction — whenever the magnetic flux linked with a coil changes, an emf is induced in it. Working: The armature coil (of $N$ turns, area $A$ ) is rotated with a constant angular speed $\omega$ inside a uniform magnetic field $B$ , kept fixed. As the coil rotates, the angle between the area vector and $B$ keeps changing, so the flux linked with the coil changes, inducing an alternating emf, which is picked off via slip rings and brushes connected to the external circuit. If $\theta = \omega t$ is the angle between $\vec{B}$ and the area vector of the coil at time $t$ , the flux linked is $\phi = NBA\cos\omega t$ By Faraday's law, induced emf, $\varepsilon = -\dfrac{d\phi}{dt} = NBA\omega\sin\omega t = \varepsilon_0\sin\omega t$ , where $\varepsilon_0 = NBA\omega$ is the peak (maximum) value of the induced emf.
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Derivation 14
(a) Write the principle of working of an ac generator. Derive the expression for the induced emf generated in it. (b) Write the function of slip rings in an ac generator.
Asked in: CBSE Class 12 Physics — Long Answer · 2020C · Section E, 5 marks
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(a) Principle: an ac generator works on the principle of electromagnetic induction — when a coil is rotated in a uniform magnetic field, an emf is induced in it. If the coil (of $N$ turns, area $A$ ) rotates with angular speed $\omega$ in field $B$ , the flux linked at time $t$ is $\phi = NBA\cos\omega t$ Induced emf, $\varepsilon = -\dfrac{d\phi}{dt} = NBA\omega\sin\omega t = \varepsilon_0\sin\omega t = \varepsilon_0\sin 2\pi f t$ , where $\varepsilon_0 = NBA\omega = 2\pi f NBA$ , and $f$ is the frequency of rotation of the coil. (b) Slip rings are used to provide a movable (sliding) contact between the rotating armature coil and the external circuit containing the load.
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Derivation 15
(a) Draw a schematic diagram for an ac generator. Explain its working and obtain the expression for the instantaneous value of the emf in terms of the magnetic field $B$ , number of turns $N$ of the coil of area $A$ rotating with angular frequency $\omega$ . Show how an alternating emf is generated by a loop of wire rotating in a magnetic field. (b) A circular coil of radius 10 cm and 20 turns is rotated about its vertical diameter with angular speed $50\text{ rad s}^{-1}$ in a uniform horizontal magnetic field of $3.0\times 10^{-2}$ T. (i) Calculate the maximum and average emf induced in the coil. (ii) If the coil forms a closed loop of resistance $10\,\Omega$ , calculate the maximum current in the coil and the average power loss due to Joule heating.
Asked in: CBSE Class 12 Physics — Long Answer · AI 2019 · Section E, 5 marks
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(a) The ac generator consists of an armature coil rotated by some external mechanical means in a uniform magnetic field between the poles of a strong field magnet. The rotating armature is connected to the external circuit via slip rings and brushes. As the coil rotates, the flux linked with it changes as $\phi = BA\cos\omega t$ , so the induced emf is $\varepsilon = -\dfrac{d\phi}{dt} = BA\omega\sin\omega t$ ; for a coil of $N$ turns, $\varepsilon = NBA\omega\sin\omega t = \varepsilon_0\sin\omega t$ . This varies sinusoidally with time, so the emf generated is alternating. (b) Given $r = 10\text{ cm} = 0.1\text{ m}$ , so $A = \pi r^2 = 3.14\times (0.1)^2 = 0.0314\text{ m}^2$ ; $N = 20$ ; $\omega = 50\text{ rad s}^{-1}$ ; $B = 3.0\times 10^{-2}\text{ T}$ (i) Maximum induced emf, $\varepsilon_{max} = NAB\omega = 20\times 0.0314\times 3.0\times 10^{-2}\times 50 = 0.942\text{ V}$ Average induced emf (over the range of the cycle, taken as the mean of the maximum and minimum values), $\varepsilon_{avg} = \dfrac{\varepsilon_{max} + \varepsilon_{min}}{2} = \dfrac{0.942 + 0}{2} = 0.471\text{ V}$ (ii) Given $R = 10\,\Omega$ . Maximum current in the coil, $I_0 = \dfrac{\varepsilon_0}{R} = \dfrac{0.942}{10} = 0.0942\text{ A}$ Average current, $I_{avg} = \dfrac{\varepsilon_{avg}}{R} = \dfrac{0.471}{10} = 0.0471\text{ A}$ Average power loss due to Joule heating, $P = I_{avg}^2 R = (0.0471)^2\times 10 = 0.022\text{ W}$
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Derivation 16
State the principle of an ac generator and explain its working with the help of a labelled diagram. Obtain the expression for the emf induced in a coil having $N$ turns, each of cross-sectional area $A$ , rotating with a constant angular speed $\omega$ in a magnetic field $\vec{B}$ , directed perpendicular to the axis of rotation. OR Draw a labelled diagram of an ac generator. Obtain the expression for the emf induced in the rotating coil, for $N$ turns each of cross-sectional area $A$ , in the presence of a magnetic field $\vec{B}$ . (3/5, AI 2017)
Asked in: CBSE Class 12 Physics — Long Answer · 3/5, 2018 · Section E, 5 marks
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Principle: An ac generator works on the principle of electromagnetic induction — it converts mechanical energy into electrical energy. Construction: The ac generator consists of (i) an armature coil of a large number of turns of insulated copper wire, wound over a soft iron core (the soft iron core increases the magnetic flux linked with the coil), (ii) strong field magnets that provide the magnetic field, with the armature coil rotated between their pole faces (axis of rotation perpendicular to the field lines), (iii) slip rings, which provide a movable contact for the armature coil with the external circuit, and (iv) carbon brushes, which pass the current from the armature coil on to the external circuit containing the load. Working: As the armature coil is rotated in the magnetic field, the orientation of the coil relative to the field, and hence the magnetic flux linked with it, keeps changing. This changing flux induces an emf in the coil (Faraday's law). If $\theta = \omega t$ is the angle turned through at time $t$ , the flux linked with the coil is $\phi = NBA\cos\omega t$ Induced emf, $\varepsilon = -\dfrac{d\phi}{dt} = NBA\omega\sin\omega t = \varepsilon_0\sin\omega t$ , where $\varepsilon_0 = NBA\omega$ The direction of the induced current is given by Fleming's right-hand rule.
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Derivation 17
(a) Draw a labelled diagram of an ac generator. Derive the expression for the instantaneous value of the emf induced in the coil. (b) A circular coil of cross-sectional area $200\text{ cm}^2$ and 20 turns is rotated about the vertical diameter with angular speed of $50\text{ rad s}^{-1}$ , in a uniform magnetic field of magnitude $3.0\times 10^{-2}$ T. Calculate the maximum value of the current in the coil.
Asked in: CBSE Class 12 Physics — Long Answer · Delhi 2017 · Section E, 5 marks
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(a) [As derived above:] With the armature coil (area $A$ , $N$ turns) rotating at angular speed $\omega$ in field $B$ , the flux linked is $\phi = NBA\cos\omega t$ , so the instantaneous induced emf is $\varepsilon = -\dfrac{d\phi}{dt} = NBA\omega\sin\omega t$ (b) Here, $A = 200\text{ cm}^2 = 2\times 10^{-2}\text{ m}^2$ , $N = 20$ , $\omega = 50\text{ rad s}^{-1}$ , $B = 3\times 10^{-2}\text{ T}$ Maximum emf induced in the coil, $\varepsilon_0 = NBA\omega = 20\times 3\times 10^{-2}\times 2\times 10^{-2}\times 50 = 0.6\text{ V}$ If $R$ is the resistance of the coil, the maximum value of the current is $I_{max} = \dfrac{\varepsilon_0}{R} = \dfrac{0.6\text{ V}}{R}$
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Derivation 18
A rectangular coil of area $A$ , having number of turns $N$ , is rotated at 'f' revolutions per second in a uniform magnetic field $B$ , the field being perpendicular to the coil. Prove that the maximum emf induced in the coil is $2\pi f NBA$ .
Asked in: CBSE Class 12 Physics — Long Answer · 3/5, AI 2016 · Section E, 5 marks
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As derived above, when the coil rotates with angular speed $\omega$ in the field $B$ , the induced emf is $\varepsilon = NBA\omega\sin\omega t$ , so its maximum (peak) value is $\varepsilon_0 = NBA\omega$ Since the coil makes $f$ revolutions per second, its angular speed is $\omega = 2\pi f$ $\therefore \varepsilon_0 = NBA(2\pi f) = 2\pi f NBA$ , as required.
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Derivation 19
(a) Draw a labelled diagram of an ac generator and state its working principle. (b) How is the magnetic flux linked with the armature coil changed in a generator? (c) Derive the expression for the maximum value of the induced emf and state the rule that gives the direction of the induced emf. (d) Show the variation of the emf generated versus time, as the armature is rotated with respect to the direction of the magnetic field.
Asked in: CBSE Class 12 Physics — Long Answer · Delhi 2014C · Section E, 5 marks
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(a) The ac generator has an armature coil (many turns of copper wire on a soft iron core), field magnets, slip rings and carbon brushes. Principle: electromagnetic induction — mechanical energy is converted into electrical energy by rotating the coil in a magnetic field. (b) In a generator, the armature coil is mechanically rotated while the field magnets are kept fixed. As the coil rotates, the orientation of its plane relative to the (fixed) field direction keeps changing, so the magnetic flux linked with it changes continuously. (c) If the coil ( $N$ turns, area $A$ ) rotates with angular speed $\omega$ , the flux is $\phi = NBA\cos\omega t$ , so the induced emf is $\varepsilon = -\dfrac{d\phi}{dt} = NBA\omega\sin\omega t$ The maximum value of the induced emf is $\varepsilon_0 = NBA\omega$ . The direction of the induced current is given by Fleming's right-hand rule. (d) Taking the coil's plane to be initially perpendicular to the field (i.e. flux maximum, emf zero at $t=0$ ), the emf $\varepsilon = \varepsilon_0\sin\omega t$ traces a sine curve against time: zero at $t=0$ , rising to peak $\varepsilon_0$ at a quarter period, back to zero at half a period, down to $-\varepsilon_0$ at three-quarters of a period, and back to zero after a full period $T = 2\pi/\omega$ .
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Derivation 20
(a) State the principle of an ac generator. (b) Explain, with the help of a well-labelled diagram, its working, and obtain an expression for the emf generated in the coil. (c) Is it possible to generate emf without rotating the coil? Explain.
Asked in: CBSE Class 12 Physics — Long Answer · 2020-21 · Section E, 5 marks · CBSE sample question
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(a) Principle: an ac generator is based on the principle of electromagnetic induction. It converts mechanical energy into electrical energy. (b) The generator has an armature coil ABCD (many turns of insulated copper wire, wound over a soft iron core to boost the flux), rotated between the poles of strong field magnets; slip rings R1, R2 give a moving contact for the coil, and carbon brushes B1, B2 carry the current to the external load $R_L$ . As the armature is rotated in the magnetic field, its orientation relative to the field keeps changing, so the flux linked with it, $\phi = BA\cos\omega t$ , changes with time. By Faraday's law, the induced emf is $\varepsilon = -N\dfrac{d\phi}{dt} = NBA\omega\sin\omega t$ , and the induced current is $i = \dfrac{\varepsilon}{R} = \dfrac{NBA\omega}{R}\sin\omega t$ The direction of the induced current is given by Fleming's right-hand rule. (c) It is not possible to generate an emf without rotating the coil (or otherwise changing the relative orientation between the coil and the field). Simply keeping the coil stationary in a magnetic field — however strong — produces no change in the flux linked with it, and by Faraday's law, no emf can be induced without a changing flux.

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