PiTheory

PiTheory JEE Physics Practice & Revision Pack · Class 12 · Chapter 6 · JEE Main

Electromagnetic Induction

Chapter-wise practice and revision for JEE Main: concise notes on Faraday's and Lenz's laws, motional emf, self and mutual inductance, LR circuits and energy, and AC generators and eddy currents, concept figures, and 50 selected previous-year JEE Main questions covering the major concepts and question patterns, each with a step-by-step solution.

50 selected questions5 topics4 concept figuresPreview · first 3 of 50 questions

How to study this chapter

  1. Look at the topic's figures, then read its notes (5 to 10 minutes). They give the major equations and methods the questions use.
  2. Solve the topic's questions with a timer: 2 minutes for Easy, 3 to 4 for Medium and Hard.
  3. Check the solution. Read the Point to note line even if you were right: it is the idea you will reuse.
  4. Mark each question Got it or Revise. Come back to the Revise list after two days.

Where the marks come from

Main idea of each of the 121 previous-year JEE questions in the PiTheory bank

Faraday & Lenz33
Motional emf27
Inductance26
LR & energy22
Generator13

Contents

  1. 1Faraday's law and Lenz's lawQ1–Q13
  2. 2Motional emfQ14–Q24
  3. 3Self and mutual inductanceQ25–Q35
  4. 4LR circuits and magnetic energyQ36–Q43
  5. 5AC generator and eddy currentsQ44–Q50
  6. ✓Last-day checklist
01

Topic 1 · Q1–Q13

Faraday's law and Lenz's law

33 of the 121 JEE PYQs in the bank test this as their main idea

Quick notes

$$\phi=\vec B\cdot\vec A=BA\cos\theta,\qquad \varepsilon=-N\frac{d\phi}{dt},\qquad q=\frac{N\Delta\phi}{R}$$

Lenz's law (the minus sign): the induced current opposes the change in flux that causes it. This is energy conservation.

What changesEmf
Field, area fixed$NA\dfrac{dB}{dt}$
Area (radius $r$), field fixed$B\cdot2\pi r\dfrac{dr}{dt}$
Angle (rotation)$NBA\omega\sin\omega t$
  • Charge depends only on the total flux change, not the rate.
  • Heat $=\int\dfrac{\varepsilon^2}{R}dt$; pulling a loop out uniformly: work $=$ heat.
  • Only $\vec B$ along the normal counts (a loop in the $xy$ plane sees only $B_z$).
  • Coil resistance $R=\dfrac{\rho\,N\cdot\text{perimeter}}{\pi r_w^2}$.
Trap. "Plane at $\theta$ to $B$" means the normal is at $90^\circ-\theta$.

Practice: first 3 of 13 questions

Q1JEE Main 2024 (31 Jan, Shift 2)Easy

The magnetic flux $\phi$ (in weber) linked with a closed circuit of resistance $8\ \Omega$ varies with time (in seconds) as $\phi = 5t^2 - 36t + 1$. The induced current in the circuit at $t = 2\ \text{s}$ is ______ A.

Numerical answer type

Answer and solution

Answer: 2

Concept. Faraday's law: $\varepsilon=-\dfrac{d\phi}{dt}$; current $I=\dfrac{|\varepsilon|}{R}$.

Step 1. Rate. $\dfrac{d\phi}{dt}=10t-36=-16$ Wb/s at $t=2$.

Step 2. Current. $\dfrac{16}{8}=2$ A.

Answer. $2$.

Point to note. $|d\phi/dt|/R$.

Q2JEE Main 2022 (26 Jun, Shift 1)Easy

The magnetic flux through a coil perpendicular to its plane is varying according to the relation $\phi = (5t^3 + 4t^2 + 2t - 5)$ weber. If the resistance of the coil is $5\ \text{ohm}$, then the induced current through the coil at $t = 2\ \text{s}$ will be

  1. A$15.6\ \text{A}$
  2. B$16.6\ \text{A}$
  3. C$17.6\ \text{A}$
  4. D$18.6\ \text{A}$
Answer and solution

Answer: Option A: $15.6\ \text{A}$

Concept. Same: differentiate the flux, divide by $R$.

Step 1. Rate. $15t^2+8t+2=60+16+2=78$ at $t=2$.

Step 2. Current. $\dfrac{78}{5}=15.6$ A.

Answer. Option A.

Point to note. $|d\phi/dt|/R$.

Q3JEE Main 2026 (5 Apr, Shift 2)Medium

A circular loop of radius $20$ cm and resistance $2\ \Omega$ is placed in a time varying magnetic field $\vec{B} = (2t^2 + 2t + 3)$ T. At $t = 0$, for the plane of the loop being perpendicular to the magnetic field and, the induced current in the loop at $t = 3$ s is $\dfrac{\alpha}{50}$ A. The value of $\alpha$ is ______. (Take $\pi = 22/7$)

Numerical answer type

Answer and solution

Answer: 44

Concept. Area fixed, field changing: $\varepsilon=A\dfrac{dB}{dt}$.

Step 1. Rate. $\dfrac{dB}{dt}=4t+2=14$ T/s at $t=3$.

Step 2. Emf. $\dfrac{22}{7}\times0.04\times14=1.76$ V.

Step 3. Current. $\dfrac{1.76}{2}=0.88=\dfrac{44}{50}$ A.

Answer. $\alpha=44$.

Point to note. $A\,dB/dt$.

47 more questions and 4 more topics in the full chapter

  1. 2Motional emf11 questions with solutions
  2. 3Self and mutual inductance11 questions with solutions
  3. 4LR circuits and magnetic energy8 questions with solutions
  4. 5AC generator and eddy currents7 questions with solutions
  5. ✓Last-day checklist

The complete Electromagnetic Induction chapter is in the Class 12 Pack and the Complete Pack, as a PDF on WhatsApp. Want to see a whole chapter first? Laws of Motion and Current Electricity are free in full.

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