Physics · Electromagnetic Induction

JEE Main 2024 — 6 April, Shift 2 — Question 55

A coil having 100 turns, area of $5 \times 10^{-3} \mathrm{~m}^{2}$ , carrying current of 1 mA is placed in uniform magnetic field of 0.20 T such a way that plane of coil is perpendicular to the magnetic field. The work done in turning the coil through $90^{\circ}$ is $\qquad$ $\mu \mathrm{J}$ .

Answer: 100

Numerical answer — enter this value.

Step-by-step solution

$\mathrm{W}=\Delta \mathrm{U}=\mathrm{U}_{\mathrm{f}}-\mathrm{U}_{\mathrm{i}}$ $\mathrm{W}=(-\vec{\mu} \cdot \overrightarrow{\mathrm{B}})_{\mathrm{f}}-(-\vec{\mu} \cdot \overrightarrow{\mathrm{B}})_{\mathrm{i}}$

$=0+(\vec{\mu} . \vec{B})_{i}$ $=\left(100 \times 5 \times 10^{-3} \times 1 \times 10^{-3}\right) \times 0.2 \mathrm{~J}$

$=1 \times 10^{-4} \mathrm{~J}=100 \mu \mathrm{~J}$

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Exam: JEE Main 2024
Paper: 6 April, Shift 2
Subject: Physics
Chapter: Electromagnetic Induction
Topic: Magnetic Flux, Faraday's Law and Lenz's Law

In a coil, the current changes form -2 A to +2 A in 0.2 s and induces an emf of 0.1 V . The selfinductance of the coil is :

Answer: 100

Step-by-step solution

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