PiTheory

PiTheory JEE Physics Practice & Revision Pack · Class 12 · Chapter 5 · JEE Main

Magnetism and Matter

Chapter-wise practice and revision for JEE Main: concise notes on the bar magnet as a dipole, the earth's field and dip, and magnetic materials, concept figures, and 40 selected previous-year JEE Main questions covering the major concepts and question patterns, each with a step-by-step solution.

40 selected questions3 topics3 concept figuresPreview · first 3 of 40 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 59 previous-year JEE questions in the PiTheory bank

Materials29
Bar magnet23
Earth's field7

Contents

  1. 1Bar magnet as a magnetic dipoleQ1–Q13
  2. 2Earth's magnetism and dipQ14–Q20
  3. 3Magnetisation and magnetic materialsQ21–Q40
  4. ✓Last-day checklist
01

Topic 1 · Q1–Q13

Bar magnet as a magnetic dipole

23 of the 59 JEE PYQs in the bank test this as their main idea

Look first

SNclosed loops: N→S outside, S→N inside
No monopoles: field lines always close.

Quick notes

$$\tau=mB\sin\theta,\qquad U=-mB\cos\theta,\qquad W_{\theta_1\to\theta_2}=mB(\cos\theta_1-\cos\theta_2),\qquad T=2\pi\sqrt{\frac{I}{mB}}$$

A bar magnet behaves like a current loop: moment $m$ points from S to N inside the magnet. Field lines are closed loops (no monopoles), running N to S outside and S to N inside.

QuantityShort magnet at distance $r$
Field on the axis$\dfrac{\mu_0}{4\pi}\dfrac{2m}{r^3}$
Field on the equator$\dfrac{\mu_0}{4\pi}\dfrac{m}{r^3}$ (opposite to $m$)
Potential on the axis$\dfrac{\mu_0}{4\pi}\dfrac{m}{r^2}$
  • Stable to unstable: $W=2mB$. Perpendicular to field: $U=0$ and torque maximum.
  • Bent magnet: moment $=$ pole strength $\times$ straight distance between ends. Semicircle: $m'=\dfrac{2m}{\pi}$.
  • Oscillations: $m\propto\dfrac{I}{T^2}$; in the earth's field use $B_H$.
  • Net flux through any closed surface (or an infinite plane around a loop) is zero.
Trap. 1 gauss $=10^{-4}$ T.

Practice: first 3 of 13 questions

Q1JEE Main 2024 (4 Apr, Shift 2)Easy

The magnetic moment of a bar magnet is $0.5\ \text{A m}^2$. It is suspended in a uniform magnetic field of $8\times10^{-2}\ \text{T}$. The work done in rotating it from its most stable to its most unstable position is

  1. A$8\times10^{-2}\ \text{J}$
  2. B$4\times10^{-2}\ \text{J}$
  3. CZero
  4. D$16\times10^{-2}\ \text{J}$
Answer and solution

Answer: Option A: $8\times10^{-2}\ \text{J}$

Concept. Energy of a dipole $U=-mB\cos\theta$. Most stable: $\theta=0$ ($U=-mB$); most unstable: $\theta=180^\circ$ ($U=+mB$). Work needed $=2mB$.

Step 1. Apply. $2\times0.5\times8\times10^{-2}=8\times10^{-2}$ J.

Answer. Option A.

Point to note. Stable to unstable: $2mB$.

Q2JEE Main 2026 (24 Jan, Shift 1)Easy

A short bar magnet placed with its axis at $30^\circ$ with an external field of 800 Gauss, experiences a torque of $0.016$ N.m. The work done in moving it from most stable to most unstable position is $\alpha\times10^{-3}$ J. The value of $\alpha$ is ______ .

Numerical answer type

Answer and solution

Answer: 64

Concept. The torque gives the moment ($\tau=mB\sin\theta$); then the work from stable to unstable is $2mB$.

Step 1. Moment. $m=\dfrac{0.016}{0.08\times0.5}=0.4$ A m$^2$ (800 G $=0.08$ T).

Step 2. Work. $2\times0.4\times0.08=0.064$ J.

Answer. $\alpha=64$.

Point to note. $m$ from torque, then $2mB$.

Q3JEE Main 2022 (26 Jun, Shift 2)Easy

A bar magnet having a magnetic moment of $2.0\times10^5\ \text{J T}^{-1}$ is placed along the direction of a uniform magnetic field of magnitude $B = 14\times10^{-5}\ \text{T}$. The work done in rotating the magnet slowly through $60^\circ$ from the direction of field is

  1. A$14\ \text{J}$
  2. B$8\ \text{J}$
  3. C$4\ \text{J}$
  4. D$1.4\ \text{J}$
Answer and solution

Answer: Option A: $14\ \text{J}$

Concept. $W=mB(\cos\theta_1-\cos\theta_2)$.

Step 1. Apply. $2\times10^5\times14\times10^{-5}\times(1-\tfrac12)=14$ J.

Answer. Option A.

Point to note. $mB(1-\cos60^\circ)$.

37 more questions and 2 more topics in the full chapter

  1. 2Earth's magnetism and dip7 questions with solutions
  2. 3Magnetisation and magnetic materials20 questions with solutions
  3. ✓Last-day checklist

The complete Magnetism and Matter 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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