Board Previous Year Questions

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CBSE Board-pattern questions on Electromagnetic Induction, organised by mark value, with model answers in the phrasing examiners reward — law stated first, formula before substitution, units carried, directions justified by Lenz's law. The chapter is a Board regular: Faraday/Lenz statements, the motional-emf derivations, inductance definitions and the AC generator long answer account for most of the marks.

1-Mark Questions (Definitions & Direct)

Q1. State Faraday's law of electromagnetic induction. Answer: The magnitude of the induced emf in a circuit equals the time rate of change of magnetic flux through it: $\varepsilon = -N\,d\Phi_B/dt$ for an $N$-turn coil.

Q2. State Lenz's law. Answer: 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.

Q3. Write the SI units of magnetic flux and of inductance. Answer: Flux: weber (1 Wb = 1 T m$^2$ = 1 V s). Inductance: henry (1 H = 1 Wb/A = 1 V s/A).

Q4. Define the self-inductance of a coil. Answer: It is the flux linkage per unit current in the same coil, $L = N\Phi_B/I$; equivalently the emf induced per unit rate of change of its own current.

Q5. A coil rotates in a magnetic field. At what orientation is the induced emf maximum? Answer: When the coil's plane is parallel to $B$ (flux through it is zero but changing at the maximum rate).

Q6. What is the power source of a hydroelectric generator? Answer: The potential energy of water falling from a height (a dam), which rotates the turbine/armature.

2-Mark Questions (Short Answer)

Q7. Show that Lenz's law follows from the conservation of energy. Answer: If the induced current aided the change (say, attracted an approaching magnet), the magnet would accelerate, increasing the flux change and the current further — kinetic energy would grow endlessly from a gentle push, a perpetual-motion machine. Energy conservation forbids this, so the induced current must oppose the change. The work done against the opposition is what appears as Joule heat.

Q8. Derive the expression for the motional emf of a rod of length l moving with velocity v perpendicular to a field B. Answer: For the loop of enclosed length $x$, $\Phi_B = Blx$. Then $\varepsilon = -d\Phi_B/dt = -Bl\,dx/dt = Blv$. (Equivalently: the Lorentz force $qvB$ on each free charge does work $qvBl$ across the rod, and emf = work/charge = $Blv$.)

Q9. Define mutual inductance and state two factors on which it depends. Answer: The mutual inductance of a coil pair is the flux linkage in one coil per unit current in the other: $N_1\Phi_1 = MI_2$, with induced emf $\varepsilon_1 = -M\,dI_2/dt$. It depends on the geometry of the coils (turns, areas, lengths, separation/orientation) and the permeability of the medium — not on the currents.

Q10. Why does a spark appear across the switch when a circuit containing a large inductor is suddenly broken? Answer: Breaking the circuit makes $dI/dt$ very large, so the back emf $\varepsilon = -L\,dI/dt$ becomes momentarily enormous — large enough to ionise the air gap and drive the current as a spark.

Q11. The current in a coil falls from 5 A to 0 in 0.1 s, inducing an average emf of 200 V. Find the self-inductance. Answer: $L = |\varepsilon|/(\Delta I/\Delta t) = 200/(5/0.1) = 200/50 = 4$ H.

3-Mark Questions (Derivations & Numericals)

Q12. Derive the emf induced between the ends of a rod of length R rotating with angular speed $\omega$ about one end in a perpendicular field B. Answer: An element at distance r moves with speed $v = \omega r$ and contributes $d\varepsilon = Bv\,dr = B\omega r\,dr$. Integrating from 0 to R: $\varepsilon = B\omega\int_0^R r\,dr = \frac{1}{2}B\omega R^2$. (Check: the rod sweeps area at rate $\frac{1}{2}R^2\omega$, and $\varepsilon = B\,dA/dt$ — same result.)

Q13. Derive the self-inductance of a long solenoid and state how it changes with a soft-iron core. Answer: Interior field $B = \mu_0 nI$; flux linkage $N\Phi_B = (nl)(\mu_0 nI)(A) = \mu_0 n^2 Al\,I$. Hence $L = \mu_0 n^2 Al$. With a core of relative permeability $\mu_r$: $L = \mu_r\mu_0 n^2 Al$ — increased $\mu_r$-fold (hugely, for soft iron).

Q14. Obtain the mutual inductance of two long coaxial solenoids of length l, inner radius $r_1$, with turn densities $n_1, n_2$. Answer: Current $I_2$ in the outer gives $B_2 = \mu_0 n_2 I_2$ threading the inner solenoid's $n_1 l$ turns of area $\pi r_1^2$: $N_1\Phi_1 = (n_1 l)(\pi r_1^2)(\mu_0 n_2 I_2)$, so $M = \mu_0 n_1 n_2 \pi r_1^2 l$. The same value results with current in the inner solenoid ($M_{12} = M_{21}$); only the shared inner area carries mutual flux.

Q15. A 500-turn coil of radius 10 cm and resistance 2 $\Omega$, with its plane perpendicular to the earth's horizontal field $3\times10^{-5}$ T, is flipped through 180 degrees in 0.25 s. Estimate the average emf and current. Answer: Per-turn flux change $= 2BA = 2 \times 3\times10^{-5} \times \pi\times10^{-2} = 1.88\times10^{-6}$ Wb. $\varepsilon = N\Delta\Phi/\Delta t = 500 \times 1.88\times10^{-6}/0.25 = 3.77\times10^{-3}$ V; $I = \varepsilon/R = 1.89\times10^{-3}$ A.

5-Mark Questions (Long Answer)

Q16. With a labelled diagram, describe the construction and working of an AC generator and derive the expression for the instantaneous emf. Answer:

1. Principle: electromagnetic induction — rotating a coil in a magnetic field changes the flux $\Phi_B = BA\cos\theta$ through it, inducing an emf.
2. Construction (describe/draw): an armature coil of $N$ turns and area $A$ on a rotor shaft, placed between the poles of a magnet with the rotation axis perpendicular to $B$; the coil ends join two slip rings on the shaft, on which stationary brushes press to connect the external circuit.
3. Working: rotated at constant angular speed $\omega$ (by water, steam, etc.), the angle at time t is $\theta = \omega t$, so $\Phi_B = BA\cos\omega t$ per turn.
4. Derivation: $\varepsilon = -N\,d\Phi_B/dt = -NBA\,\frac{d}{dt}\cos\omega t = NBA\omega\sin\omega t = \varepsilon_0\sin\omega t$, with peak $\varepsilon_0 = NBA\omega$.
5. The emf (and current) alternates in polarity each half rotation — alternating current; emf is maximum when the coil plane is parallel to $B$ and zero when perpendicular.

Q17. (a) Show that the energy stored in an inductor carrying current I is $\frac{1}{2}LI^2$. (b) Hence obtain the magnetic energy density of a solenoid. (c) A 4 H inductor carries 2 A; find the stored energy. Answer:

1. (a) To grow the current, the source works against the back emf at rate $dW/dt = |\varepsilon|I = LI\,dI/dt$. Integrating, $W = \int_0^I LI'\,dI' = \frac{1}{2}LI^2$ — stored as magnetic energy (the analogue of $\frac{1}{2}mv^2$, with L as electrical inertia).
2. (b) For a solenoid, $L = \mu_0 n^2 Al$ and $I = B/\mu_0 n$, so $U_B = \frac{1}{2}\mu_0 n^2 Al \cdot \frac{B^2}{\mu_0^2 n^2} = \frac{B^2}{2\mu_0}\,Al$. Per unit volume: $u_B = B^2/2\mu_0$ (general, like $u_E = \frac{1}{2}\varepsilon_0 E^2$).
3. (c) $W = \frac{1}{2} \times 4 \times 2^2 = 8$ J.