Welcome students! This page presents the complete Class 12 Physics Chapter 6 Question Answer | Electromagnetic Induction | English Medium | ASSEB guide. Every NCERT in-text and exercise question is solved step by step in line with the latest Assam State School Education Board (ASSEB / AHSEC) syllabus. The chapter explores how a changing magnetic field produces an electric current — the principle behind generators, transformers, induction cookers and modern wireless charging.
Chapter Summary
Electromagnetic induction is the phenomenon in which an electromotive force (EMF) is produced in a conductor whenever the magnetic flux linked with it changes. It was discovered independently by Michael Faraday (1831) in England and Joseph Henry in the United States. The two pillars of the chapter are Faraday’s law (which gives the magnitude of the induced EMF) and Lenz’s law (which gives its direction and is rooted in the conservation of energy).
Other key ideas include motional EMF (induction in a conductor moving through a magnetic field), eddy currents (induced loops in solid metal), self-inductance and mutual inductance (the inductive property of a single coil and that between two coils), the energy stored in an inductor, and the working principle of an AC generator.
Key Formulas
| Quantity | Formula |
| Magnetic flux | $\Phi_B = \vec{B}\cdot\vec{A} = BA\cos\theta$ |
| Faraday’s law (single loop) | $\epsilon = -\dfrac{d\Phi_B}{dt}$ |
| Faraday’s law ($N$ turns) | $\epsilon = -N\dfrac{d\Phi_B}{dt}$ |
| Motional EMF | $\epsilon = Bvl$ |
| Self-induced EMF | $\epsilon = -L\dfrac{dI}{dt}$ |
| Inductance of solenoid | $L = \mu_0 n^2 A l$ |
| Mutual induction | $\epsilon_2 = -M\dfrac{dI_1}{dt}$ |
| Energy stored in inductor | $U = \tfrac{1}{2}LI^2$ |
| EMF of AC generator | $\epsilon = NBA\omega\sin(\omega t)$ |
| SI unit of flux | weber (Wb) $= \mathrm{T\,m^2}$ |
| SI unit of inductance | henry (H) $= \mathrm{V\,s\,A^{-1}}$ |
Faraday’s Experimental Setup
Motional EMF in a Sliding Rod
Solenoid (Self-Inductance)
NCERT Exercise Solutions
Q1. Predict the direction of induced current in the situations described by the following figures (a) to (f).
Answer: Using Lenz’s law, the induced current opposes the change in flux that produced it.
| Case | Direction of induced current |
| (a) Bar magnet pushed towards a closed loop | From q to r via the external circuit (so that the face nearer the magnet acts as the same pole, repelling the magnet). |
| (b) Magnet moved away | From p to q and from q to r in the closed loop, so the loop attracts the receding magnet. |
| (c) Tapping key closed in primary | From y to z in the secondary. |
| (d) Rheostat slider moved (current decreasing) | From z to y in the secondary. |
| (e) Steady current in primary, secondary moved | From x to y. |
| (f) No flux change | No induced current. |
Q2. Use Lenz’s law to determine the direction of induced current in the situations described.
Answer:
(a) A wire of irregular shape turning into a circular shape: the area of the loop increases, so the outward flux $\Phi_B$ increases. By Lenz’s law, the induced current must oppose this increase, hence it flows along adcba (i.e., direction such that its own flux opposes B).
(b) A circular loop deforming into a narrow straight wire: the enclosed area decreases, flux decreases, and the induced current flows along a’d’c’b’ to support the decreasing flux.
Q3. A long solenoid with 15 turns per cm has a small loop of area $2.0\,\text{cm}^2$ placed inside it normal to its axis. If the current carried by the solenoid changes steadily from 2.0 A to 4.0 A in 0.1 s, what is the induced EMF in the loop while the current is changing?
Answer:
Number of turns per metre, $n = 15 \times 100 = 1500\ \text{m}^{-1}$.
Loop area $A = 2.0\times 10^{-4}\ \text{m}^2$, $\dfrac{dI}{dt} = \dfrac{4.0-2.0}{0.1} = 20\ \text{A/s}$.
$$\epsilon = \mu_0 n A \frac{dI}{dt} = (4\pi\times 10^{-7})(1500)(2\times 10^{-4})(20)$$
$$\boxed{\epsilon \approx 7.54\times 10^{-6}\ \text{V} = 7.54\ \mu\text{V}}$$
Q4. A rectangular wire loop of sides 8 cm and 2 cm with a small cut is moving out of a region of uniform magnetic field of magnitude 0.3 T directed normal to the loop. What is the EMF developed across the cut if the velocity of the loop is 1 cm/s in a direction normal to the (a) longer side, (b) shorter side of the loop? For how long does the induced voltage last in each case?
Answer: Using $\epsilon = Bvl$.
(a) Motion normal to the longer side ($l = 8\ \text{cm} = 0.08\ \text{m}$):
$$\epsilon = (0.3)(0.01)(0.08) = 2.4\times 10^{-4}\ \text{V} = 0.24\ \text{mV}$$
Time for which EMF lasts $= \dfrac{2\times 10^{-2}}{10^{-2}} = 2\ \text{s}$ (loop must traverse its 2 cm width).
(b) Motion normal to the shorter side ($l = 2\ \text{cm} = 0.02\ \text{m}$):
$$\epsilon = (0.3)(0.01)(0.02) = 0.6\times 10^{-4}\ \text{V} = 0.06\ \text{mV}$$
Time for which EMF lasts $= \dfrac{8\times 10^{-2}}{10^{-2}} = 8\ \text{s}$.
Q5. A 1.0 m long metallic rod is rotated with an angular frequency of 400 rad/s about an axis normal to the rod passing through its one end. The other end of the rod is in contact with a circular metallic ring. A constant magnetic field of 0.5 T parallel to the axis exists everywhere. Calculate the EMF developed between the centre and the ring.
Answer: EMF developed between the two ends of a rotating rod is
$$\epsilon = \tfrac{1}{2}B\omega L^2 = \tfrac{1}{2}(0.5)(400)(1.0)^2 = 100\ \text{V}$$
Hence the EMF is 100 V.
Q6. A horizontal straight wire 10 m long extending from east to west is falling with a speed of 5.0 m/s, at right angles to the horizontal component of the earth’s magnetic field, $0.30\times 10^{-4}\ \text{Wb/m}^2$. (a) What is the instantaneous value of the EMF induced in the wire? (b) What is the direction of the EMF? (c) Which end of the wire is at the higher electrical potential?
Answer:
(a) $\epsilon = Bvl = (0.30\times 10^{-4})(5.0)(10) = 1.5\times 10^{-3}\ \text{V} = 1.5\ \text{mV}$.
(b) Using $\vec{F} = q\vec{v}\times\vec{B}$, positive charges experience a force from east to west, so the induced EMF acts from east to west.
(c) Hence the western end of the wire is at the higher potential.
Q7. Current in a circuit falls from 5.0 A to 0.0 A in 0.1 s. If an average EMF of 200 V is induced, give an estimate of the self-inductance of the circuit.
Answer:
$$L = \frac{|\epsilon|}{|dI/dt|} = \frac{200}{(5.0-0)/0.1} = \frac{200}{50} = 4.0\ \text{H}$$
Self-inductance $L = \mathbf{4.0\ H}$.
Q8. A pair of adjacent coils has a mutual inductance of 1.5 H. If the current in one coil changes from 0 to 20 A in 0.5 s, what is the change of flux linkage with the other coil?
Answer:
$$\Delta(N\Phi) = M\,\Delta I = 1.5 \times 20 = 30\ \text{Wb-turns}$$
The change in flux linkage with the second coil is 30 Wb-turns.
Q9. A jet plane is travelling towards west at a speed of 1800 km/h. What is the voltage difference developed between the ends of the wing having a span of 25 m, if the earth’s magnetic field at the location has a magnitude of $5\times 10^{-4}\ \text{T}$ and the dip angle is $30^\circ$?
Answer:
Speed: $v = 1800\ \text{km/h} = 500\ \text{m/s}$. The relevant component is the vertical component $B_V = B\sin 30^\circ = 2.5\times 10^{-4}\ \text{T}$.
$$\epsilon = B_V v l = (2.5\times 10^{-4})(500)(25) = 3.125\ \text{V}$$
The voltage developed between the wing tips is about 3.125 V.
Q10. Suppose the loop of Q4 is stationary but the current feeding the electromagnet that produces the magnetic field is gradually reduced so that the field decreases from its initial value of 0.3 T at the rate of $0.02\ \text{T/s}$. If the cut is joined and the loop has a resistance of $1.6\ \Omega$, how much power is dissipated as heat?
Answer:
Area $A = 8\times 2 = 16\ \text{cm}^2 = 16\times 10^{-4}\ \text{m}^2$.
$$\epsilon = A\frac{dB}{dt} = (16\times 10^{-4})(0.02) = 3.2\times 10^{-5}\ \text{V}$$
$$P = \frac{\epsilon^2}{R} = \frac{(3.2\times 10^{-5})^2}{1.6} = 6.4\times 10^{-10}\ \text{W}$$
The dissipated power is about $6.4\times 10^{-10}\ \text{W}$, drawn from the source feeding the electromagnet.
Q11. It is desired to make a 12 cm long solenoid having a self-inductance of 1.0 mH. If the cross-sectional area is $5\ \text{cm}^2$, find the number of turns per unit length and the total number of turns in the solenoid.
Answer: From $L = \mu_0 n^2 A l$:
$$n = \sqrt{\frac{L}{\mu_0 A l}} = \sqrt{\frac{1\times 10^{-3}}{(4\pi\times 10^{-7})(5\times 10^{-4})(0.12)}}$$
$n \approx 3.64\times 10^3\ \text{turns/m}$. Total turns $N = nl \approx 437$.
Q12. A square loop of side 12 cm with its sides parallel to the X- and Y-axes is moved with a velocity of 8 cm/s in the positive X direction in an environment with a magnetic field directed in the +Z direction. The field has a gradient of $10^{-3}\ \text{T/cm}$ along $-X$ and decreases at the rate of $10^{-3}\ \text{T/s}$. The loop has resistance $4.5\ \text{m}\Omega$. Find the direction and magnitude of induced current.
Answer: The two contributions to the EMF are due to motion and to the temporal change of B:
$$\epsilon_{\text{motion}} = \frac{dB}{dx}\,A\,v = (10^{-1}\ \text{T/m})(144\times 10^{-4})(8\times 10^{-2}) = 11.52\times 10^{-5}\ \text{V}$$
$$\epsilon_{\text{time}} = A\frac{dB}{dt} = (144\times 10^{-4})(10^{-3}) = 1.44\times 10^{-5}\ \text{V}$$
Both contributions add (flux increases due to motion towards stronger field; field also decreases with time, so signs are the same when considered carefully): total $\epsilon = 12.96\times 10^{-5}\ \text{V}$. Current $I = \epsilon/R = 12.96\times 10^{-5}/4.5\times 10^{-3} \approx 2.88\times 10^{-2}\ \text{A} = 28.8\ \text{mA}$. The current is induced in such a sense as to oppose the increasing flux, i.e., directed clockwise as viewed from $+Z$.
Q13. It is desired to estimate the magnetic field between the pole pieces of a magnet. A small flat search coil of area $2\ \text{cm}^2$ with 25 closely wound turns is positioned normal to the field. It is quickly snatched out and the total charge through a $0.50\ \Omega$ galvanometer is found to be $7.5\ \text{mC}$. The combined resistance of the coil and galvanometer is $0.50\ \Omega$. Estimate the field.
Answer: The induced charge through the circuit is
$$Q = \frac{N\,\Delta\Phi}{R} = \frac{N B A}{R} \quad\Rightarrow\quad B = \frac{QR}{NA}$$
$$B = \frac{(7.5\times 10^{-3})(0.50)}{(25)(2\times 10^{-4})} = 0.75\ \text{T}$$
The magnetic field between the pole pieces is approximately 0.75 T.
Q14. (Rod on rails.) A metal rod of length 15 cm slides on a U-shaped frame in a uniform magnetic field $B = 0.50\ \text{T}$ at a steady speed $v = 12\ \text{cm/s}$. The closed circuit has resistance $9.0\ \text{m}\Omega$. Find (a) induced EMF, (b) current, (c) force needed to push the rod, (d) power dissipated, (e) power supplied by the external agent. Compare with (d).
Answer:
| (a) EMF | $\epsilon = Bvl = (0.5)(0.12)(0.15) = 9\times 10^{-3}\ \text{V} = 9\ \text{mV}$ |
| (b) Current | $I = \epsilon/R = 9\times 10^{-3}/9\times 10^{-3} = 1\ \text{A}$ |
| (c) Force | $F = BIl = (0.5)(1)(0.15) = 0.075\ \text{N}$ |
| (d) Power dissipated | $P = I^2 R = (1)^2(9\times 10^{-3}) = 9\ \text{mW}$ |
| (e) Power by agent | $P_{\text{ext}} = Fv = (0.075)(0.12) = 9\ \text{mW}$ — equal to (d), confirming energy conservation. |
Q15. An air-cored solenoid 30 cm long, area of cross section $25\ \text{cm}^2$, has 500 turns. The current carried by it is 2.5 A which is suddenly switched off in $10^{-3}\ \text{s}$. How much is the average back EMF induced across the ends of the solenoid? Ignore variation of magnetic field near ends.
Answer:
$$L = \mu_0 \frac{N^2}{l}A = (4\pi\times 10^{-7})\frac{(500)^2}{0.30}(25\times 10^{-4}) \approx 2.62\times 10^{-3}\ \text{H}$$
$$|\epsilon| = L\frac{dI}{dt} = (2.62\times 10^{-3})\frac{2.5}{10^{-3}} \approx 6.5\ \text{V}$$
The average back EMF is approximately 6.5 V.
Q16. (a) Obtain an expression for the mutual inductance between a long straight wire and a square loop of side $a$ at a distance $x$ from the wire. (b) For $a = 0.10\ \text{m}, x = 0.20\ \text{m}, v = 10\ \text{m/s}$, current 50 A — find the EMF induced in the loop.
Answer: (a) Flux through the loop is
$$\Phi = \int_x^{x+a} \frac{\mu_0 I}{2\pi r} a\,dr = \frac{\mu_0 I a}{2\pi}\ln\!\left(\frac{x+a}{x}\right)$$
$$M = \frac{\Phi}{I} = \frac{\mu_0 a}{2\pi}\ln\!\left(\frac{x+a}{x}\right)$$
(b) The induced EMF when the loop moves away with velocity $v$ is
$$\epsilon = \frac{\mu_0 I a v}{2\pi}\!\left(\frac{1}{x}-\frac{1}{x+a}\right)$$
$$\epsilon = \frac{(4\pi\times 10^{-7})(50)(0.10)(10)}{2\pi}\!\left(\frac{1}{0.2}-\frac{1}{0.3}\right) \approx 1.7\times 10^{-5}\ \text{V}$$
Q17. A line charge $\lambda$ per unit length is lodged uniformly onto the rim of a wheel of mass M and radius R. The wheel has light non-conducting spokes and is free to rotate without friction about a vertical axis. A uniform magnetic field B exists in a circular region of radius $a$ ($a < R$) coaxial with the wheel. If the field is suddenly switched off, find the angular speed acquired by the wheel.
Answer: By Faraday’s law, the induced electric field along the rim provides a torque on the line charge. Equating impulse of torque to change in angular momentum:
$$I_{\text{wheel}}\,\omega = -\lambda\, 2\pi R\,R \cdot \frac{1}{2\pi R}\!\int E\,dt = -\lambda B \pi a^2\cdot R$$
$$\omega = -\frac{\lambda B \pi a^2 R}{MR^2} = -\frac{\lambda B a^2}{MR}$$
(Treating the rim as a thin ring, $I_{\text{wheel}} = MR^2$.) The wheel begins to rotate about its axis with angular speed $\omega = \dfrac{\pi a^2 \lambda B}{\pi M R}$ in the direction determined by Lenz’s law.
Additional Important Questions (ASSEB)
Q18. Define magnetic flux. State its SI unit.
Answer: Magnetic flux through a surface is the product of the magnetic field and the projection of the area perpendicular to the field, $\Phi_B = \int \vec{B}\cdot d\vec{A}$. Its SI unit is the weber (Wb), where $1\ \text{Wb} = 1\ \text{T}\cdot\text{m}^2$.
Q19. State Faraday’s laws of electromagnetic induction.
Answer:
(i) Whenever the magnetic flux linked with a closed circuit changes, an EMF is induced in the circuit, which lasts only as long as the change persists.
(ii) The magnitude of the induced EMF is equal to the negative of the time rate of change of magnetic flux:
$$\epsilon = -\frac{d\Phi_B}{dt}\quad\text{or}\quad \epsilon = -N\frac{d\Phi_B}{dt}\ (\text{for } N \text{ turns}).$$
Q20. State Lenz’s law and show how it is a consequence of the conservation of energy.
Answer: Lenz’s law states that the direction of the induced current is always such that it opposes the cause producing it. When a magnet is pushed towards a coil, the induced current makes the near face of the coil a like pole, so external work has to be done against the repulsion. This work appears as electrical energy in the coil — confirming the principle of energy conservation. If the induced current aided the motion, energy would be created out of nothing, violating the conservation law.
Q21. Derive the expression for motional EMF in a straight conductor moving in a uniform magnetic field.
Answer: Consider a rod PQ of length $l$ moving with velocity $v$ perpendicular to a uniform field $B$. A free electron in the rod experiences a magnetic force $F = evB$ along the length of the rod. This force pushes electrons to one end, leaving a positive charge at the other end. An electric field $E$ builds up till $eE = evB$, i.e., $E = vB$. The potential difference between the ends is
$$\epsilon = E\,l = Bvl$$
This is the motional EMF.
Q22. What are eddy currents? Mention two applications and two undesirable effects.
Answer: Eddy currents are circulating induced currents set up in the body of a bulk conductor when the magnetic flux through it changes. They flow in closed loops (whirlpool-like).
Useful applications: (i) electromagnetic damping in galvanometers and balances; (ii) induction furnace for melting metals; (iii) electric brakes in trains; (iv) speedometers.
Undesirable effects: (i) heating of transformer cores reducing efficiency — minimised by laminating the core; (ii) energy loss in motors and dynamos.
Q23. Define self-inductance and derive the expression for the inductance of a long air-cored solenoid.
Answer: Self-inductance is the property by virtue of which a coil opposes any change in the current through itself, $L = N\Phi/I$. For a solenoid of length $l$, area $A$, and $n$ turns per unit length: $B = \mu_0 n I$ inside. Total flux linkage $= N\Phi = (nl)(BA) = \mu_0 n^2 lAI$, hence
$$L = \frac{N\Phi}{I} = \mu_0 n^2 A l$$
Q24. Define mutual inductance and write its SI unit. On what factors does it depend?
Answer: Mutual inductance of two coils is the flux linkage produced in the secondary per unit current in the primary; $\epsilon_2 = -M\,dI_1/dt$. SI unit: henry (H). It depends on (i) number of turns of both coils, (ii) their geometry and area, (iii) their relative orientation and separation, and (iv) the medium (permeability) between them.
Q25. Show that the energy stored in an inductor is $U = \tfrac{1}{2}LI^2$.
Answer: Work done by the source against the back EMF in time $dt$ is $dW = \epsilon\,I\,dt = LI\,dI$. Total work to raise current from 0 to $I$:
$$W = \int_0^I LI\,dI = \tfrac{1}{2}LI^2$$
This work is stored as magnetic energy: $U = \tfrac{1}{2}LI^2$.
Q26. Describe the construction and working of an AC generator and obtain the expression for the EMF.
Answer: An AC generator consists of (i) a rectangular armature coil of $N$ turns wound on a soft iron core, (ii) a strong field magnet, (iii) two slip rings rigidly attached to the coil, and (iv) two carbon brushes pressing against the slip rings. As the coil rotates with constant angular speed $\omega$, the angle between the area vector and field becomes $\theta = \omega t$. Flux through the coil:
$$\Phi = NBA\cos(\omega t)$$
$$\epsilon = -\frac{d\Phi}{dt} = NBA\omega\sin(\omega t) = \epsilon_0\sin(\omega t)$$
where $\epsilon_0 = NBA\omega$ is the peak EMF. The output is alternating, completing one cycle per revolution.
Q27. Two coils have a mutual inductance of 0.005 H. Calculate the induced EMF in one coil if current in the other changes from 0 to 10 A in 0.1 s.
Answer:
$$|\epsilon| = M\frac{dI}{dt} = (5\times 10^{-3})(100) = 0.5\ \text{V}$$
Q28. The current in a coil of self-inductance 5 mH changes uniformly from 2 A to 6 A in 0.2 s. Find the induced EMF.
Answer: $|\epsilon| = L\,dI/dt = (5\times 10^{-3})(20) = 0.1\ \text{V}.$
Q29. A coil of 100 turns and area $0.05\ \text{m}^2$ rotates at 50 rev/s in a magnetic field of 0.1 T about an axis perpendicular to the field. What is the maximum EMF?
Answer: $\omega = 2\pi(50) = 100\pi\ \text{rad/s}$.
$$\epsilon_0 = NBA\omega = 100\times 0.1\times 0.05\times 100\pi \approx 157\ \text{V}$$
Q30. Why are the cores of transformers laminated?
Answer: Lamination of the iron core breaks the conducting paths into thin, mutually insulated sheets. This greatly increases the resistance of the loops in which eddy currents would flow, sharply reducing eddy-current losses ($P\propto 1/R$) and improving efficiency.
Q31. Explain why an aluminium plate, while being pushed into a region of strong magnetic field, experiences resistance to its motion.
Answer: As the plate enters the field, the flux through it changes, generating eddy currents. By Lenz’s law these currents flow in a direction that opposes the change, producing a retarding force on the plate (electromagnetic damping). The kinetic energy lost is dissipated as heat in the plate.
Q32. A circular coil of radius 8 cm and 20 turns is rotated about its vertical diameter at angular speed 50 rad/s in a uniform horizontal field $B = 3.0\times 10^{-2}\ \text{T}$. Find the maximum and average EMF and the maximum current if resistance of the loop is $10\ \Omega$.
Answer: $A = \pi r^2 = \pi(0.08)^2 \approx 0.0201\ \text{m}^2$.
$$\epsilon_0 = NBA\omega = 20\times 3.0\times 10^{-2}\times 0.0201\times 50 \approx 0.603\ \text{V}$$
Average EMF over a full cycle is zero. Maximum current $I_0 = \epsilon_0/R \approx 0.0603\ \text{A}$.
More Numerical Practice (ASSEB Style)
Q33. The magnetic flux linked with a coil changes from $0.4\ \text{Wb}$ to $0.1\ \text{Wb}$ in $0.6\ \text{s}$. Calculate the average induced EMF.
Answer: Magnitude of induced EMF
$$|\epsilon| = \left|\frac{\Delta\Phi}{\Delta t}\right| = \frac{0.4 – 0.1}{0.6} = \frac{0.3}{0.6} = 0.5\ \text{V}$$
The average EMF induced is 0.5 V.
Q34. A coil of 50 turns and area $4\times 10^{-2}\ \text{m}^2$ is placed in a magnetic field of $0.10\ \text{T}$ such that its plane is perpendicular to the field. The coil is then rotated by $180^\circ$ in $0.20\ \text{s}$. Find the average EMF.
Answer: Initial flux $\Phi_1 = BA = 4\times 10^{-3}\ \text{Wb}$. After $180^\circ$ rotation $\Phi_2 = -4\times 10^{-3}\ \text{Wb}$. Net change $\Delta\Phi = 8\times 10^{-3}\ \text{Wb}$.
$$|\epsilon| = N\frac{\Delta\Phi}{\Delta t} = 50\times \frac{8\times 10^{-3}}{0.20} = 2\ \text{V}$$
Q35. A coil has a self-inductance of $0.5\ \text{H}$. Calculate the energy stored when a current of $4\ \text{A}$ flows through it.
Answer:
$$U = \tfrac{1}{2}LI^2 = \tfrac{1}{2}(0.5)(4)^2 = 4\ \text{J}$$
Q36. A solenoid of length $0.5\ \text{m}$ has a radius of $1\ \text{cm}$ and is made up of $500$ turns. It carries a current of $5\ \text{A}$. What is the magnitude of the magnetic field inside the solenoid?
Answer: $n = N/l = 500/0.5 = 1000\ \text{turns/m}$.
$$B = \mu_0 n I = (4\pi\times 10^{-7})(1000)(5) = 6.28\times 10^{-3}\ \text{T}$$
Hence the field inside the solenoid is about $6.28\ \text{mT}$.
Q37. A circular coil of $200$ turns and mean radius $5\ \text{cm}$ carries a current of $5\ \text{A}$. The coil is placed with its plane perpendicular to a uniform magnetic field of $0.2\ \text{T}$. Find the torque on the coil if the plane of the coil makes an angle of $30^\circ$ with the field.
Answer: Area $A = \pi(0.05)^2 \approx 7.85\times 10^{-3}\ \text{m}^2$. Angle between $\vec{n}$ and $\vec{B}$ is $60^\circ$.
$$\tau = NBIA\sin\theta = 200\times 0.2\times 5\times 7.85\times 10^{-3}\times \sin 60^\circ \approx 1.36\ \text{N\,m}$$
Q38. The mutual inductance of two coils is $0.20\ \text{H}$. The current in the primary coil decreases from $6.0\ \text{A}$ to zero in $0.05\ \text{s}$. Find the EMF induced in the secondary coil.
Answer:
$$|\epsilon_2| = M\frac{dI_1}{dt} = 0.20\times \frac{6.0}{0.05} = 24\ \text{V}$$
Q39. A square loop of side $10\ \text{cm}$ and resistance $0.5\ \Omega$ is placed vertically in the east-west plane. A uniform magnetic field of $0.10\ \text{T}$ is set up across the plane of the loop in the north-east direction. The field decreases to zero in $0.70\ \text{s}$. Find the magnitude of the current during this interval.
Answer: Component of $B$ perpendicular to the loop $= B\cos 45^\circ$.
$$|\epsilon| = \frac{\Delta\Phi}{\Delta t} = \frac{(0.10\cos 45^\circ)(0.01)}{0.70} \approx 1.0\times 10^{-3}\ \text{V}$$
$$I = \frac{\epsilon}{R} = \frac{1.0\times 10^{-3}}{0.5} = 2.0\times 10^{-3}\ \text{A} = 2\ \text{mA}$$
Q40. A copper rod of length $0.5\ \text{m}$ rotates with an angular velocity of $20\ \text{rad/s}$ about an axis through one end and perpendicular to its length, in a magnetic field $B = 0.4\ \text{T}$ parallel to the axis of rotation. Find the EMF induced.
Answer:
$$\epsilon = \tfrac{1}{2}B\omega L^2 = \tfrac{1}{2}(0.4)(20)(0.5)^2 = 1.0\ \text{V}$$
Conceptual Short Answer Questions
Q41. Why is the induced EMF also called the “back EMF” in an inductor?
Answer: When the current through an inductor changes, the self-induced EMF acts in a direction that opposes the change, by Lenz’s law. If the current is increasing, the induced EMF acts opposite to the applied EMF; if it is decreasing, the induced EMF acts to maintain the current. Because of this opposing nature, it is called the back EMF.
Q42. Two identical loops, one of copper and the other of aluminium, are rotated with the same angular velocity in the same magnetic field. In which case will (a) the induced EMF and (b) the induced current be more, and why?
Answer: (a) The induced EMF depends only on $NBA\omega$, which is the same for both loops, so the induced EMF is equal. (b) Induced current is $I = \epsilon/R$. Copper has lower resistivity than aluminium, so the copper loop carries a larger current.
Q43. A bar magnet falls through a long copper pipe held vertically. Will the magnet fall freely? Justify.
Answer: No. As the magnet falls, the flux through different cross-sections of the copper pipe changes, inducing eddy currents in the pipe walls. By Lenz’s law, these currents oppose the motion of the magnet, producing a retarding force. The magnet therefore falls slowly with nearly uniform velocity rather than in free fall.
Q44. Why is the metallic core of a moving coil galvanometer made of soft iron and laminated?
Answer: Soft iron has high permeability, which makes the magnetic field strong and radial. Lamination is used to minimise eddy currents that would otherwise heat the core and waste energy.
Q45. A wheel with $10$ metallic spokes each $0.5\ \text{m}$ long is rotated with a speed of $120\ \text{rev/min}$ in a plane perpendicular to the horizontal component of earth’s magnetic field $H = 0.4\ \text{G}$ at the place. What is the EMF induced between the axle and the rim of the wheel?
Answer: $\omega = 2\pi\times (120/60) = 4\pi\ \text{rad/s}$, $H = 0.4\ \text{G} = 0.4\times 10^{-4}\ \text{T}$.
$$\epsilon = \tfrac{1}{2}H\omega L^2 = \tfrac{1}{2}(0.4\times 10^{-4})(4\pi)(0.5)^2 \approx 6.28\times 10^{-5}\ \text{V}$$
The EMF is the same regardless of the number of spokes since they are in parallel.
Q46. Define the coefficient of coupling between two coils. What is its maximum possible value?
Answer: The coefficient of coupling $k$ is defined as $M = k\sqrt{L_1 L_2}$. It measures the fraction of flux of one coil that links with the other. Its maximum value is $1$ (perfect coupling, as in a well-designed transformer with a common iron core).
Q47. State Fleming’s right-hand rule and explain its use.
Answer: Stretch the thumb, forefinger and middle finger of the right hand mutually perpendicular. If the forefinger points along the magnetic field $\vec{B}$ and the thumb along the velocity $\vec{v}$ of the conductor, then the middle finger gives the direction of induced current. It is used to find the direction of motional EMF or current in a generator.
Q48. Why is mutual inductance between coplanar concentric coils maximum when their planes coincide?
Answer: When the coils share the same axis and plane, all the flux produced by one passes through the other. As one coil rotates, the linked flux $\Phi = BA\cos\theta$ falls — and with it, $M$. Hence $M$ is maximum when the angle between the area vectors of the two coils is zero.
Q49. Distinguish between self-inductance and mutual inductance.
| Self-inductance ($L$) | Mutual inductance ($M$) |
| Property of a single coil. | Property of a pair of coils. |
| Opposes change of current in the same coil. | Causes EMF in the second coil due to change of current in the first. |
| $L = N\Phi/I$ | $M = N_2\Phi_2/I_1$ |
| EMF: $\epsilon = -L\,dI/dt$. | EMF: $\epsilon_2 = -M\,dI_1/dt$. |
Q50. A long solenoid of $1000$ turns per metre carries a current of $4\ \text{A}$. Inside it is placed a small coil of area $8\ \text{cm}^2$ and $20$ turns oriented with its axis along the solenoid axis. The current in the solenoid is reversed in $0.04\ \text{s}$. Find the average EMF induced in the small coil.
Answer: $B$ inside solenoid $= \mu_0 nI = (4\pi\times 10^{-7})(1000)(4) = 5.03\times 10^{-3}\ \text{T}$. On reversal, $\Delta B = 2B$.
$$|\epsilon| = N_2 A\frac{\Delta B}{\Delta t} = 20\times 8\times 10^{-4}\times \frac{2\times 5.03\times 10^{-3}}{0.04} \approx 4.02\times 10^{-3}\ \text{V}$$
Average induced EMF $\approx 4\ \text{mV}$.
Important Derivations (ASSEB Long Answers)
D1. Show that the energy stored per unit volume in a magnetic field is $u_B = \dfrac{B^2}{2\mu_0}$.
Answer: Consider a long solenoid of length $l$ and area $A$ with $n$ turns per unit length carrying current $I$. The energy stored is
$$U = \tfrac{1}{2}LI^2 = \tfrac{1}{2}(\mu_0 n^2 A l)I^2$$
The magnetic field inside is $B = \mu_0 n I$, so $I = B/(\mu_0 n)$. Substituting:
$$U = \tfrac{1}{2}\mu_0 n^2 A l \cdot \frac{B^2}{\mu_0^2 n^2} = \frac{B^2}{2\mu_0}\,(A l)$$
Volume of solenoid is $V = Al$, so energy density is
$$u_B = \frac{U}{V} = \frac{B^2}{2\mu_0}$$
This expression is general — it holds for any magnetic field configuration in vacuum.
D2. Derive the expression for the mutual inductance of two long coaxial solenoids.
Answer: Let the inner solenoid $S_1$ have $N_1$ turns, length $l$ and cross-section $A_1$. The outer solenoid $S_2$ has $N_2$ turns and length $l$. When current $I_1$ flows through $S_1$, the field inside is $B_1 = \mu_0 (N_1/l)I_1$. The flux through one turn of $S_2$ is $\Phi = B_1 A_1$. Total flux linkage with $S_2$ is
$$N_2\Phi = N_2 \cdot \mu_0\frac{N_1}{l}I_1\cdot A_1 = \frac{\mu_0 N_1 N_2 A_1}{l}\,I_1$$
$$M_{21} = \frac{N_2\Phi}{I_1} = \frac{\mu_0 N_1 N_2 A_1}{l}$$
By the reciprocity theorem, $M_{12} = M_{21}$. Thus the mutual inductance depends only on geometry and the medium between the solenoids.
D3. Derive the expression for the EMF induced in a rod of length $L$ rotating with angular speed $\omega$ about one of its ends in a uniform magnetic field $B$ perpendicular to the plane of rotation.
Answer: Consider a small element $dr$ at distance $r$ from the rotation axis. Its linear speed is $v = r\omega$. The motional EMF across this element is
$$d\epsilon = B v\,dr = B r\omega\,dr$$
Integrating from $0$ to $L$:
$$\epsilon = B\omega \int_0^L r\,dr = \tfrac{1}{2}B\omega L^2$$
This is the EMF developed between the centre and the rim of the rotating rod.
D4. From Faraday’s law, derive the expression for motional EMF in a sliding rod and show that the power dissipated equals the mechanical power supplied.
Answer: A rod PQ of length $l$ slides with velocity $v$ on parallel rails closed by a resistor $R$ in a uniform field $B$ perpendicular to the plane. After time $t$ the area enclosed is $A = lx$, where $x = vt$. Flux $\Phi = Blx$, so
$$\epsilon = -\frac{d\Phi}{dt} = -Bl\frac{dx}{dt} = -Blv$$
Magnitude of EMF $|\epsilon| = Blv$. Current $I = Blv/R$. Force on the rod due to this current is $F = BIl = B^2l^2v/R$. Mechanical power delivered:
$$P_{\text{mech}} = Fv = \frac{B^2 l^2 v^2}{R}$$
Electrical power dissipated:
$$P_{\text{elec}} = I^2 R = \left(\frac{Blv}{R}\right)^2 R = \frac{B^2 l^2 v^2}{R}$$
The two are equal, confirming energy conservation.
Multiple Choice Questions (1 mark)
MCQ-1. The SI unit of magnetic flux is:
(a) tesla (b) weber (c) henry (d) gauss
Answer: (b) weber.
MCQ-2. Lenz’s law gives:
(a) magnitude of induced EMF (b) direction of induced current (c) flux change (d) self-inductance
Answer: (b) direction of induced current.
MCQ-3. A coil with $N$ turns has self-inductance proportional to:
(a) $N$ (b) $N^2$ (c) $1/N$ (d) independent of $N$
Answer: (b) $N^2$.
MCQ-4. Eddy currents are minimised in transformer cores by:
(a) using copper cores (b) lamination (c) heating the core (d) using a single thick block
Answer: (b) lamination.
MCQ-5. The dimensional formula of self-inductance $L$ is:
(a) $\mathrm{ML^2T^{-2}A^{-2}}$ (b) $\mathrm{ML^2T^{-1}A^{-1}}$ (c) $\mathrm{ML^{-2}T^{2}A^{2}}$ (d) $\mathrm{MLT^{-2}A^{-2}}$
Answer: (a) $\mathrm{ML^2T^{-2}A^{-2}}$.
MCQ-6. Two coils with self-inductances $L_1$ and $L_2$ have a maximum possible mutual inductance equal to:
(a) $L_1 + L_2$ (b) $L_1 L_2$ (c) $\sqrt{L_1 L_2}$ (d) $\dfrac{L_1+L_2}{2}$
Answer: (c) $\sqrt{L_1 L_2}$.
MCQ-7. A square loop of side $a$ moves with velocity $v$ perpendicular to its plane in a uniform field $B$. The induced EMF is:
(a) $Bav$ (b) $Bv$ (c) $Ba^2 v$ (d) zero
Answer: (d) zero — flux through the loop does not change.
MCQ-8. The magnetic flux through a coil is $\Phi = 5t^2 + 3t + 2\ \text{Wb}$. The EMF at $t = 2\ \text{s}$ is:
(a) $-23\ \text{V}$ (b) $23\ \text{V}$ (c) $-13\ \text{V}$ (d) $13\ \text{V}$
Answer: (a) $-23\ \text{V}$. ($d\Phi/dt = 10t + 3 = 23$, so EMF $= -23\ \text{V}$.)
MCQ-9. Two pure inductors of self-inductance $L$ each connected in series (without mutual inductance) gives total inductance:
(a) $L/2$ (b) $L$ (c) $2L$ (d) $4L$
Answer: (c) $2L$.
MCQ-10. A bar magnet falls vertically through a hollow conducting tube. The acceleration of the magnet is:
(a) $g$ (b) less than $g$ (c) more than $g$ (d) zero
Answer: (b) less than $g$ — eddy currents oppose the fall.
MCQ-11. If the number of turns of a coil is doubled and area is halved, the self-inductance becomes:
(a) same (b) doubled (c) halved (d) four times
Answer: (b) doubled. ($L \propto N^2 A$ so factor $= 4\times \tfrac{1}{2} = 2$.)
MCQ-12. An aircraft of wingspan $40\ \text{m}$ flies horizontally at $1000\ \text{km/h}$ where the vertical component of earth’s magnetic field is $1.6\times 10^{-5}\ \text{T}$. The voltage induced between wing tips is:
(a) $0.18\ \text{V}$ (b) $0.36\ \text{V}$ (c) $0.018\ \text{V}$ (d) $1.8\ \text{V}$
Answer: (a) $0.18\ \text{V}$. ($v = 277.8\ \text{m/s}; \epsilon = Bvl = 1.6\times 10^{-5}\times 277.8\times 40 \approx 0.18\ \text{V}$.)
HOTS / Application Questions
H1. A train moves on rails with velocity $v$ in a region where earth’s vertical magnetic component is $B_V$. The two rails are separated by a distance $l$. Will a galvanometer connected between the rails register any current? Explain.
Answer: A motional EMF $\epsilon = B_V v l$ is generated between the rails. However, since both rails are at rest and only the train moves on them, the EMF does not appear in a stationary galvanometer connected across the rails because no flux change occurs in that loop. If the galvanometer is mounted on the moving train (a moving conductor between the rails), then a current can be observed.
H2. A rectangular loop is partially inside a uniform magnetic field. As it is pulled out, both the EMF and the work done against the magnetic force are non-zero. Where does this work go?
Answer: The work done by the external agent against the magnetic braking force is converted entirely into electrical energy in the loop, which is then dissipated as Joule heat ($I^2R$) in the loop’s resistance. Energy conservation is maintained.
H3. Why is it dangerous to disconnect a circuit carrying a large current through an inductor abruptly?
Answer: When the switch is opened suddenly, $dI/dt$ becomes very large. The induced back EMF $\epsilon = -L\,dI/dt$ may become enormous — large enough to break down the air gap and cause a spark or even an electric shock. To avoid this, induction-circuit switches are usually fitted with capacitors or freewheeling diodes that absorb the energy.
H4. A coil with $N$ turns is rotated in a uniform magnetic field. Sketch how the induced EMF varies with time and indicate the maximum value.
Answer: The induced EMF is sinusoidal: $\epsilon = \epsilon_0\sin(\omega t)$ with $\epsilon_0 = NBA\omega$. The maximum (peak) value occurs each time the coil’s plane is parallel to $\vec{B}$, while the EMF is zero when the plane is perpendicular to $\vec{B}$.
H5. Suggest two practical applications each of (i) self-induction, (ii) mutual induction, (iii) eddy currents.
Answer:
(i) Self-induction: ignition coil of an automobile; choke coil in fluorescent lamps.
(ii) Mutual induction: transformers; induction coil; metal detectors.
(iii) Eddy currents: induction motor; electromagnetic braking; induction furnace; analogue speedometer.
Glossary
| Term | Meaning |
| Magnetic flux | $\Phi_B = BA\cos\theta$ — measure of magnetic field passing through an area. |
| Faraday’s law | Induced EMF equals the negative time-rate of change of flux. |
| Lenz’s law | Induced current opposes the change producing it (energy conservation). |
| Motional EMF | EMF $\epsilon = Bvl$ produced in a conductor moving across a field. |
| Eddy current | Loop currents induced in bulk conductors due to changing flux. |
| Self-inductance ($L$) | Property of a coil opposing change of its own current; SI unit henry. |
| Mutual inductance ($M$) | Coupling between two coils; flux linked in one per unit current in the other. |
| Inductor energy | Magnetic energy stored, $U = \tfrac{1}{2}LI^2$. |
| AC generator | Device converting mechanical energy to alternating electrical energy via a rotating coil in a magnetic field. |
| Henry (H) | Unit of inductance: $1\ \text{H} = 1\ \text{V}\cdot\text{s/A}$. |
| Weber (Wb) | Unit of magnetic flux: $1\ \text{Wb} = 1\ \text{T}\cdot\text{m}^2$. |
Faraday’s Original Experiments
Experiment 1 (Magnet and Coil): When a bar magnet is moved towards a stationary closed coil connected to a galvanometer, the galvanometer needle deflects in one direction. When the magnet is pulled away, the deflection is in the opposite direction. The deflection lasts only as long as the relative motion exists, and the magnitude depends on the speed of motion. This shows that an EMF is induced when the magnetic flux through the coil changes.
Experiment 2 (Two Coils): Two coils are placed close to each other. The primary is connected to a battery through a key, and the secondary to a galvanometer. When the key is pressed (current rising in the primary), the galvanometer in the secondary deflects briefly. When the key is released (current dying), it deflects the other way. While a steady current flows through the primary, the galvanometer reads zero. Thus, an induced EMF appears in the secondary only when the primary current is changing — establishing the concept of mutual induction.
Experiment 3 (Single Coil with Changing Current): Faraday demonstrated that an EMF can also be induced in a coil due to a change in its own current. This is the principle of self-induction.
Together these experiments led Faraday to the law: “The induced EMF in a closed circuit is equal to the negative time-rate of change of magnetic flux through the circuit.”
Quick Revision Sheet
Use this sheet on the day before the examination for a final brush-up of Class 12 Physics Chapter 6 Electromagnetic Induction.
| Topic | Key Point |
| Faraday’s Law | $\epsilon = -N\,d\Phi_B/dt$. Induced EMF lasts only while flux changes. |
| Lenz’s Law | Induced current opposes change of flux; based on conservation of energy. |
| Motional EMF | $\epsilon = Bvl$ for rod moving perpendicular to $B$ with speed $v$. |
| Rotating Rod EMF | $\epsilon = \tfrac{1}{2}B\omega L^2$ between centre and rim. |
| Self-Induction | $\epsilon = -L\,dI/dt$. Solenoid: $L = \mu_0 n^2 A l$. |
| Mutual Induction | $\epsilon_2 = -M\,dI_1/dt$. For coaxial solenoids $M = \mu_0 n_1 n_2 A_1 l$. |
| Energy in Inductor | $U = \tfrac{1}{2}LI^2$; energy density $u_B = B^2/(2\mu_0)$. |
| AC Generator | $\epsilon = NBA\omega\sin(\omega t)$; peak $\epsilon_0 = NBA\omega$. |
| Eddy Currents | Loop currents in solid conductors; reduced by lamination. |
| Coupling Coefficient | $M = k\sqrt{L_1 L_2}$, $0\le k\le 1$. |
Examination Tips
1. While solving numerical problems, always start by writing the relevant formula and clearly label every quantity. This wins easy marks even if a numerical mistake creeps in later.
2. The minus sign in Faraday’s law is crucial — it represents Lenz’s law. Do not omit it in derivations.
3. When applying Fleming’s right-hand rule to find the direction of induced current in a moving conductor, remember: thumb $=$ velocity, forefinger $=$ field, middle finger $=$ induced current.
4. If the question involves a rotating coil, check whether the EMF asked is the instantaneous, peak, or average value over a cycle. The average over a complete cycle is zero, but the average over a half cycle is $2\epsilon_0/\pi$.
5. For mutual inductance problems involving a square loop and a long wire, do not forget to integrate $\mu_0 I/(2\pi r)$ over the appropriate range — the answer is logarithmic in nature.
This complete ASSEB Class 12 Physics Chapter 6 Question Answer | Electromagnetic Induction | English Medium guide covers every NCERT exercise question, derivations, MCQs, HOTS and additional ASSEB-style questions to help you score full marks. Practice each derivation and numerical until you can reproduce it without looking — that is the key to success in your HS Final examination.