NeetCrackers Physics Practice
Practice Set

Electromagnetic Induction and Alternating Currents — Practice Questions with Answers

30 fully solved questions covering Faraday's law, Lenz's law, self and mutual inductance, and AC circuits.

R
Questions written and solutions verified by Roshan, science educator with 5 years of experience teaching NEET and JEE aspirants. Checked against NCERT and previous years' papers. Last reviewed September 2026.
10MCQs · 1 mark
10Short Answer · 3 marks
10Long Answer · 5 marks
90Total Marks

📑 On This Page

  1. Section A — 10 MCQs (1 mark each)
  2. Section B — 10 Short Answer Questions (3 marks each)
  3. Section C — 10 Long Answer Questions (5 marks each)
Section A

🎯 Multiple Choice Questions

10 questions · 1 mark each · Correct answer highlighted with explanation

Q1 According to Faraday's law, an EMF is induced in a circuit whenever:
Correct Answer: (b). Faraday's law: induced EMF is proportional to the rate of change of magnetic flux, not simply the presence of a field.
Q2 Lenz's law is a consequence of the principle of conservation of:
Correct Answer: (b). The induced current always opposes the change causing it, ensuring work must be done against this opposition, consistent with energy conservation.
Q3 The SI unit of self-inductance is:
Correct Answer: (b) Henry (H). Since $\varepsilon = -L\dfrac{dI}{dt}$, inductance has units of volt-seconds per ampere, called the henry.
Q4 For a purely resistive AC circuit, the current and voltage are:
Correct Answer: (a) In phase. For a resistor, $V=IR$ holds instantaneously, so current and voltage reach their peaks and zeros simultaneously.
Q5 In a purely inductive AC circuit, current:
Correct Answer: (a). In a purely inductive circuit, current lags voltage by a quarter cycle (90°).
Q6 In a purely capacitive AC circuit, current:
Correct Answer: (a). In a purely capacitive circuit, current leads voltage by a quarter cycle (90°) — opposite to the inductive case.
Q7 The RMS value of an alternating current is related to its peak value $I_0$ by:
Correct Answer: (b) $I_{rms} = I_0/\sqrt{2} \approx 0.707I_0$, for a sinusoidal AC waveform.
Q8 A transformer works on the principle of:
Correct Answer: (b). A transformer transfers energy between two coils via mutual induction, using a changing magnetic flux linking both windings.
Q9 At resonance in a series LCR circuit, the impedance is:
Correct Answer: (a). At resonance, inductive and capacitive reactances cancel ($X_L=X_C$), leaving impedance $Z=R$, its minimum possible value.
Q10 Eddy currents are:
Correct Answer: (a). Eddy currents are induced, circulating currents within a bulk conductor's interior due to changing magnetic flux, typically causing undesirable energy loss (heat) in transformer cores and similar devices.
Section B

✍️ Short Answer Questions

10 questions · 3 marks each · Full model answers included

Q1. State Faraday's laws of electromagnetic induction.
3 Marks
Answer

First law: Whenever the magnetic flux linked with a circuit changes, an EMF is induced in the circuit, lasting as long as the flux continues to change.

Second law: The magnitude of the induced EMF is directly proportional to the rate of change of magnetic flux linkage: $\varepsilon = -\dfrac{d\Phi}{dt}$ (for a single-turn coil), or $\varepsilon = -N\dfrac{d\Phi}{dt}$ for $N$ turns.

Q2. State Lenz's law, and explain its significance in terms of energy conservation.
3 Marks
Answer

Lenz's law states that the direction of an induced current is always such that it opposes the change in magnetic flux that produced it.

Significance: This opposing nature ensures energy conservation — if the induced current instead aided the change (rather than opposing it), it would create a self-reinforcing process generating energy from nothing, violating energy conservation. Since the induced current opposes the change, external work must be done to maintain or continue that change, and this external work is precisely what supplies the electrical energy manifested as the induced current.

Q3. Define self-inductance. Derive the expression for self-induced EMF.
3 Marks
Answer

Self-inductance is the property of a coil by which it opposes any change in the current flowing through itself, by inducing an EMF within the coil due to the changing magnetic flux this changing current itself creates.

Since flux linkage $N\Phi \propto I$, we can write $N\Phi = LI$, where $L$ is the self-inductance. Since induced EMF $\varepsilon = -\dfrac{d(N\Phi)}{dt}$:

$\varepsilon = -L\dfrac{dI}{dt}$

Q4. Explain the working principle of an AC generator.
3 Marks
Answer

An AC generator converts mechanical energy into electrical energy, based on electromagnetic induction. A coil is mechanically rotated within a uniform magnetic field (or, equivalently, a magnet is rotated near a stationary coil) — as the coil rotates, the magnetic flux linked with it continuously changes.

This continuously changing flux induces an alternating EMF in the coil (following Faraday's law), given by $\varepsilon = NBA\omega\sin(\omega t)$ — a sinusoidally varying EMF, which is the defining characteristic of alternating current.

Q5. Define RMS value of an alternating current, and explain why it is a more useful measure than the peak value for practical purposes.
3 Marks
Answer

The RMS (root-mean-square) value of an alternating current is the value of steady direct current that would produce the same amount of heat (same power dissipation) in a given resistance over one complete cycle as the actual alternating current does.

It is more practically useful than peak value because AC power dissipation ($I^2R$) directly depends on RMS current, not peak current — quoting standard household voltage (e.g., 220V AC) as an RMS value directly and meaningfully indicates its actual heating/power delivery capability, unlike the instantaneously varying peak value.

Q6. Explain why an inductor opposes changes in current, but offers no opposition to steady (DC) current.
3 Marks
Answer

An inductor's opposing (back) EMF is given by $\varepsilon = -L\dfrac{dI}{dt}$ — this depends specifically on the rate of change of current, not on the current itself.

For a steady DC current, $I$ is constant, so $\dfrac{dI}{dt} = 0$, meaning the inductor generates no back-EMF and offers no additional opposition beyond its own small wire resistance. Only when current is changing (as during switching transients, or continuously in AC) does the inductor generate a genuine opposing EMF, resisting that specific change.

Q7. What is the impedance of a series LCR circuit? Write its expression.
3 Marks
Answer

Impedance is the total effective opposition an AC circuit offers to current flow, combining resistance and reactance (from inductors and capacitors).

For a series LCR circuit: $Z = \sqrt{R^2 + (X_L-X_C)^2}$, where $X_L = \omega L$ (inductive reactance) and $X_C = \dfrac{1}{\omega C}$ (capacitive reactance) — since inductive and capacitive reactances have opposite phase effects, they partially (or fully, at resonance) cancel within the impedance formula.

Q8. Explain the phenomenon of resonance in a series LCR circuit, and derive the resonant frequency.
3 Marks
Answer

Resonance occurs in a series LCR circuit when the inductive reactance exactly equals the capacitive reactance ($X_L=X_C$), causing these opposing reactive effects to cancel completely, leaving impedance at its minimum value ($Z=R$) and current at its maximum value for a given applied voltage.

Setting $X_L=X_C$: $\omega L = \dfrac{1}{\omega C} \Rightarrow \omega^2 = \dfrac{1}{LC} \Rightarrow \omega_0 = \dfrac{1}{\sqrt{LC}}$

Resonant frequency: $f_0 = \dfrac{1}{2\pi\sqrt{LC}}$

Q9. Explain why the core of a transformer is laminated (made of thin, insulated sheets) rather than a single solid block.
3 Marks
Answer

A solid metal core would allow large eddy currents to circulate within its bulk due to the continuously changing magnetic flux (from the AC input), causing significant energy loss as heat (and reducing transformer efficiency).

By laminating the core — using thin sheets electrically insulated from each other, rather than one solid block — the pathways available for eddy currents to circulate are broken up and significantly restricted, substantially reducing eddy current losses and improving the transformer's overall efficiency.

Q10. Explain the working principle of a step-up transformer, using the turns ratio.
3 Marks
Answer

A step-up transformer increases voltage from the primary (input) coil to the secondary (output) coil. Since the same changing flux links both coils (via a shared iron core), the induced EMF in each coil is proportional to its number of turns: $\dfrac{V_s}{V_p} = \dfrac{N_s}{N_p}$.

By constructing the secondary coil with more turns than the primary ($N_s > N_p$), the output voltage $V_s$ becomes correspondingly larger than the input voltage $V_p$ — this is precisely the working principle of a step-up transformer, achieved purely through the turns ratio, based on mutual induction.

Section C

📘 Long Answer / Numerical Questions

10 questions · 5 marks each · Detailed step-by-step solutions

Q1. A coil of 200 turns has a magnetic flux of $5\times10^{-3}$ Wb linked with it. If this flux is reduced to zero in 0.1 s, find the induced EMF.
5 Marks
Answer

$\varepsilon = N\dfrac{\Delta\Phi}{\Delta t} = 200\times\dfrac{5\times10^{-3}-0}{0.1}$

$\varepsilon = 200\times\dfrac{5\times10^{-3}}{0.1} = 200\times0.05$

$\varepsilon = \mathbf{10\text{ V}}$

Q2. A step-down transformer has a primary coil of 1000 turns and a secondary coil of 100 turns. If the primary voltage is 220 V, find the secondary voltage. If the secondary current is 5 A, find the primary current (assuming an ideal, 100% efficient transformer).
5 Marks
Answer

Secondary voltage: $\dfrac{V_s}{V_p} = \dfrac{N_s}{N_p} \Rightarrow V_s = 220\times\dfrac{100}{1000} = 22\text{ V}$

Primary current (for an ideal transformer, power in = power out): $V_pI_p = V_sI_s \Rightarrow I_p = \dfrac{V_sI_s}{V_p} = \dfrac{22\times5}{220}$

$I_p = \dfrac{110}{220} = \mathbf{0.5\text{ A}}$

Q3. An AC voltage source has a peak value of 311 V. Find its RMS value (this is a standard household AC supply).
5 Marks
Answer

$V_{rms} = \dfrac{V_0}{\sqrt{2}} = \dfrac{311}{1.414}$

$V_{rms} \approx \mathbf{220\text{ V}}$

This confirms the standard 220 V household AC supply corresponds to a peak voltage of about 311 V.

Q4. A series LCR circuit has $R=30\,\Omega$, $X_L=50\,\Omega$, and $X_C=10\,\Omega$. Find (a) the impedance, and (b) the current, if the applied RMS voltage is 200 V.
5 Marks
Answer

(a) Impedance: $Z = \sqrt{R^2+(X_L-X_C)^2} = \sqrt{30^2+(50-10)^2} = \sqrt{900+1600} = \sqrt{2500} = 50\,\Omega$

(b) Current: $I = \dfrac{V}{Z} = \dfrac{200}{50} = \mathbf{4\text{ A}}$

Q5. Find the resonant frequency of a series LCR circuit with $L=0.5$ H and $C=8\,\mu\text{F}$.
5 Marks
Answer

$f_0 = \dfrac{1}{2\pi\sqrt{LC}}$

$LC = 0.5\times8\times10^{-6} = 4\times10^{-6}$

$\sqrt{LC} = \sqrt{4\times10^{-6}} = 2\times10^{-3}$

$f_0 = \dfrac{1}{2\pi\times2\times10^{-3}} = \dfrac{1}{4\pi\times10^{-3}} \approx \dfrac{1}{0.01257}$

$f_0 \approx \mathbf{79.6\text{ Hz}}$

Q6. A conducting rod of length 0.5 m moves with a velocity of 4 m/s perpendicular to a magnetic field of 0.8 T, with velocity, rod, and field all mutually perpendicular. Find the EMF induced across the ends of the rod.
5 Marks
Answer

Motional EMF: $\varepsilon = Blv$

$\varepsilon = 0.8\times0.5\times4$

$\varepsilon = \mathbf{1.6\text{ V}}$

Q7. A coil has a self-inductance of 2 H. If the current through it changes from 5 A to 1 A in 0.2 s, find the magnitude of the induced EMF.
5 Marks
Answer

$\varepsilon = L\dfrac{\Delta I}{\Delta t} = 2\times\dfrac{|1-5|}{0.2} = 2\times\dfrac{4}{0.2}$

$\varepsilon = 2\times20 = \mathbf{40\text{ V}}$

Q8. A pure inductor of inductance 0.2 H is connected to an AC source of frequency 50 Hz. Find the inductive reactance.
5 Marks
Answer

$X_L = \omega L = 2\pi f L$

$X_L = 2\pi\times50\times0.2$

$X_L = 2\pi\times10 = 20\pi$

$X_L \approx \mathbf{62.8\,\Omega}$

Q9. A pure capacitor of capacitance 10 μF is connected to an AC source of frequency 60 Hz. Find the capacitive reactance.
5 Marks
Answer

$X_C = \dfrac{1}{\omega C} = \dfrac{1}{2\pi fC}$

$X_C = \dfrac{1}{2\pi\times60\times10\times10^{-6}}$

$X_C = \dfrac{1}{2\pi\times6\times10^{-4}} = \dfrac{1}{3.77\times10^{-3}}$

$X_C \approx \mathbf{265.3\,\Omega}$

Q10. A circular coil of area 0.02 m² and 100 turns rotates with an angular speed of 50 rad/s in a magnetic field of 0.4 T. Find the peak EMF generated.
5 Marks
Answer

Peak EMF for an AC generator: $\varepsilon_0 = NBA\omega$

$\varepsilon_0 = 100\times0.4\times0.02\times50$

$\varepsilon_0 = 100\times0.4\times1$

$\varepsilon_0 = \mathbf{40\text{ V}}$

📚 Continue Studying