A circular conducting coil of radius 1m is… - JEE Main 2021 Physics PYQ

JEE Main 2021 — Physics Electromagnetism

2021integerhard

A circular conducting coil of radius $1m$ is being heated by the change of magnetic field $\vec{B}$ passing perpendicular to the plane in which the coil is laid. The resistance of the coil is $2\mu \Omega$ . The magnetic field is slowly switched off such that its magnitude changes in time as

$B=\frac{4}{\pi }\times {10}^{-3}T(1-\frac{t}{100})$

The energy dissipated by the coil before the magnetic field is switched off completely is $E=______\mathrm{mJ}$ .

Official previous-year question

Held on 25 Jul 2021 · Verified 6 Jul 2026.

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Solution

$\phi =\vec{B}\cdot \vec{S}$

$\phi =\frac{4}{\pi }\times {10}^{-3}(1-\frac{t}{100})\cdot \pi {R}^{2}$

$\phi =4\times {10}^{-3}\times {(1)}^{2}(1-\frac{t}{100})$

$\epsilon =\frac{-\text{d}\phi }{\text{d}t}$

$\epsilon =\frac{-\text{d}}{\text{d}t}(4\times {10}^{-3}(1-\frac{t}{100}))$

$\epsilon =4\times {10}^{-3}(\frac{1}{100})=4\times {10}^{-5}V$

When $B=0$

$1-\frac{t}{100}=0$

$t=100\mathrm{sec}$

Heat $=\frac{{\epsilon }^{2}}{R}t$

Heat $=\frac{{(4\times {10}^{-5})}^{2}}{2\times {10}^{-6}}\times 100J$

Heat $=\frac{16\times {10}^{-10}\times 100}{2\times {10}^{-6}}J$

Heat $=0.08J$

Heat $=80\mathrm{mJ}$

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