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Physics · Electromagnetic Induction

JEE Main 2026 — 24 January, Morning Shift — Question 44

A short bar magnet placed with its axis at 30∘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 α×10−3 J\alpha \times 10^{-3} \mathrm{~J}. The value of α\alpha is ____\_\_\_\_ .

Answer: 64

Numerical answer — enter this value.

Step-by-step solution

τ=μBsin⁡θ⇒0.016=μ×B×12\tau=\mu \mathrm{B} \sin \theta \Rightarrow 0.016=\mu \times \mathrm{B} \times \frac{1}{2} ⇒μ=0.032BWext=Uf−Ui=μB−(μB)=2μB=2×0.032B×B=0.064 J\begin{aligned} & \Rightarrow \mu=\frac{0.032}{B} & W_{e x t}=U_{f}-U_{i}=\mu B-(\mu B)=2 \mu B & =2 \times \frac{0.032}{B} \times B & =0.064 \mathrm{~J} \end{aligned}

Answer key and solution verified before publishing.

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Exam
JEE Main 2026
Paper
24 January, Morning Shift
Subject
Physics
Chapter
Electromagnetic Induction
Topic
Motional EMF

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