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Electromagnetic Induction — NEET & JEE MCQs with answers
Electromagnetic induction is the phenomenon in which a changing magnetic flux through a circuit induces an electromotive force (emf). Discovered by Faraday (England) and Henry (USA) in experiments around 1830, it led directly to modern generators and transformers. The key law governing this is Faraday's law: $\varepsilon = -\frac{d\PhiB}{dt}$, with Lenz's law specifying the direction of the induced current.
Topics in this chapter
- 6.1 Introduction
- 6.2 The Experiments of Faraday and Henry
- 6.3 Magnetic Flux
- 6.4 Faraday's Law of Induction
- 6.5 Lenz's Law and Conservation of Energy
- 6.6 Motional Electromotive Force
- 6.7 Inductance
- 6.8 AC Generator
Free practice questions
A receding south pole causes the coil's near face to become a north pole, so the magnet is:
- A. repelled
- B. attracted
- C. unaffected
- D. accelerated away
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Answer: B — The coil's near face is a north pole while a south pole recedes, so unlike poles face each other and the magnet is attracted; this is the response that opposes the withdrawal and the accompanying decrease in flux. Repulsion would aid the withdrawal and give energy for nothing. The magnet is not unaffected, because the induced current exerts a force on it, and that force retards the motion rather than accelerating the magnet away.
Class 12 · Physics · Ch 6 · §6.5 · p.160 · ¶1 ("In 1834, German physicist Heinrich Friedrich Lenz (1804-1865) deduced a rule, known as Lenz's law")
A key experimental takeaway shared by Experiment 6.1 (magnet held still) and Experiment 6.3 (key held pressed) is that
- A. The galvanometer reads highest when idle
- B. Motion is always required for any current
- C. When nothing is changing, there is no induced current
- D. A steady state always gives the maximum current
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Answer: C — NCERT: a stationary magnet (6.1) and a steadily held key (6.3) both correspond to no change of flux, and in both cases the galvanometer shows no deflection — no change means no induced current. Why not the others: a steady state gives no current at all, motion is not always required (Experiment 6.3 has none), and an idle galvanometer reads zero.
Class 12 · Physics · Ch 6 · §6.2 ¶1 (Experiment 6.1) · p.155 · ("Figure 6.1 shows a coil C1 connected to a galvanometer G")
In Example 6.2 a square loop of side $10\ \text{cm}$ and resistance $0.5\ \Omega$ has a $0.10\ \text{T}$ field at $45^\circ$ to its area vector reduced to zero. If the reduction took $0.35\ \text{s}$ instead of $0.70\ \text{s}$, the induced current would be
- A. $1.0\ \text{mA}$
- B. $8.0\ \text{mA}$
- C. $2.0\ \text{mA}$
- D. $4.0\ \text{mA}$
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Answer: D — The initial flux is $BA\cos45^\circ=0.10\times0.010/\sqrt{2}=7.07\times10^{-4}\ \text{Wb}$ and the final flux is zero. Halving the time doubles both the emf and the current: $|\varepsilon|=7.07\times10^{-4}/0.35=2.0\ \text{mV}$ and $I=\varepsilon/R=2.0\times10^{-3}/0.5=4.0\ \text{mA}$. The value $2.0\ \text{mA}$ is the current for the original $0.70\ \text{s}$ in Example 6.2, $1.0\ \text{mA}$ is half of that, and $8.0\ \text{mA}$ doubles the correct value once more.
Class 12 · Physics · Ch 6 · §6.4 · p.159 · ¶4 ("And the magnitude of the current is")
The motional emf $\varepsilon=Blv$ shows that an induced emf can be produced by
- A. increasing only the resistance
- B. using a stronger but steady field on a still rod
- C. keeping everything stationary
- D. moving a conductor instead of varying the field
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Answer: D — NCERT: “we are able to produce induced emf by moving a conductor instead of varying the magnetic field, that is, by changing the magnetic flux enclosed by the circuit”. Nothing is induced when everything is stationary, or when a still rod sits in a steady field however strong, because then the enclosed flux does not change; and resistance does not appear in $\varepsilon=Blv$, so changing it cannot create the emf.
Class 12 · Physics · Ch 6 · §6.6 · p.163 · ¶2 ("where we have used dx/dt = -v which is the speed of the conductor PQ")
In the expression $\Phi_B = BA\cos\theta$, the angle $\theta$ is measured between the magnetic field $\mathbf{B}$ and
- A. the area vector (normal to the surface)
- B. the diameter of the surface
- C. the tangent to the surface
- D. the plane of the surface
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Answer: A — Because flux is the dot product $\mathbf{B}\cdot\mathbf{A}$, the angle in $BA\cos\theta$ is, in NCERT's words, the “angle between B and A” — the area vector, which points along the normal to the surface. The plane of the surface is the classic trap: a field at angle $\alpha$ to the plane has $\theta = 90^\circ - \alpha$, so using $\alpha$ directly gives $\cos\alpha$ where $\sin\alpha$ is needed. A diameter and a tangent are directions lying in the surface, not its normal, and NCERT does not define flux against them.
Class 12 · Physics · Ch 6 · §6.3 · p.157 · ¶2 ("where q is angle between B and A")
A typical modern ac generator described in the text has an output capacity of the order of
- A. $100\ \text{kW}$
- B. $1\ \text{W}$
- C. $100\ \text{W}$
- D. $100\ \text{MW}$
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Answer: D — NCERT: “The modern ac generator with a typical output capacity of 100 MW is a highly evolved machine” (the text later notes modern units reaching 500 MW). $100\ \text{W}$, $100\ \text{kW}$ and $1\ \text{W}$ are three to six orders of magnitude too small for a power-station machine.
Class 12 · Physics · Ch 6 · §6.8 · p.170 · ¶1 ("The modern ac generator with a typical output capacity of 100 MW is a highly evolved machine")
The current in a circuit falls by $5\ \text{A}$ in $0.1\ \text{s}$, inducing an average emf of $200\ \text{V}$. The self-inductance of the circuit is
- A. $4\ \text{H}$
- B. $8\ \text{H}$
- C. $2\ \text{H}$
- D. $10^{4}\ \text{H}$
Show answer
Answer: A — This is NCERT Exercise 6.7. From $|\varepsilon|=L\,|\Delta I/\Delta t|$, $L=|\varepsilon|\,\Delta t/\Delta I=200\times0.1/5=20/5=4\ \text{H}$. $8\ \text{H}$ doubles the correct value, $2\ \text{H}$ halves it, and $10^{4}\ \text{H}$ inverts both factors ($200\times5/0.1$).
Class 12 · Physics · Ch 6 · Exercise 6.7 · p.176 ("Current in a circuit falls from 5.0 A to 0.0 A in 0.1 s")
Faraday first observed that a small current was produced in a wire loop due to:
- A. Relative motion between a bar magnet and a wire loop
- B. Placing a stationary magnet inside the loop
- C. Connecting the loop to a battery
- D. Heating the wire loop uniformly
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Answer: A — NCERT says Faraday first made public his discovery “that relative motion between a bar magnet and a wire loop produced a small current in the latter”. A stationary magnet gives no current: in Experiment 6.1 the galvanometer shows no deflection when the magnet is held stationary. The loop was connected to no battery, and heating the loop plays no part in the account.
Class 12 · Physics · Ch 6 · §6.1 · p.154 · ¶1 ("relative motion between a bar magnet and a wire loop produced a small current in the latter")
The flux linked with each turn of a closely wound 40-turn coil changes at a uniform rate of $2\times10^{-3}\ \text{Wb s}^{-1}$. The emf induced in the coil is
- A. $8\times10^{-2}\ \text{V}$
- B. $8\times10^{-4}\ \text{V}$
- C. $5\times10^{-5}\ \text{V}$
- D. $2\times10^{-3}\ \text{V}$
Show answer
Answer: A — $|\varepsilon|=N\,|\mathrm{d}\Phi_B/\mathrm{d}t|=40\times2\times10^{-3}=8\times10^{-2}\ \text{V}$. The value $2\times10^{-3}\ \text{V}$ is the single-turn emf, with $N$ forgotten; $5\times10^{-5}\ \text{V}$ divides by $N$ instead of multiplying; and $8\times10^{-4}\ \text{V}$ is a hundred times too small.
Class 12 · Physics · Ch 6 · §6.4 · p.158 · ¶2 ("In the case of a closely wound coil of N turns, change")
In Example 6.1, the small bulb used in place of the galvanometer will glow only
- A. While there is relative motion between the coils
- B. When both coils are perfectly stationary
- C. When the battery is disconnected
- D. Only after the coils are separated completely
Show answer
Answer: A — NCERT Example 6.1(b): the relative motion between the two coils causes the bulb to glow, demonstrating the induced current — it glows while there is relative motion. Why not the others: perfectly stationary coils, fully separated coils and a disconnected battery all mean no changing flux, and hence no glow.
Class 12 · Physics · Ch 6 · Example 6.1 · p.158 · ("Consider Experiment 6.2. (a) What would you do to obtain a large deflection of the galvanometer?")
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