JEE Main 2016 — Physics Electromagnetism

2016mcqmedium

A conducting metal circular-wire-loop of radius $r$ is placed perpendicular to a magnetic field which varies with time as $B={B}_{0}{e}^{-\frac{t}{\tau }},$ where ${B}_{0} \mathrm{and} \tau$ are constants at time $t=0$ . If the resistance of the loop is $R$ , then the heat generated in the loop after a long time $(t\rightarrow \infty )$ is

Official previous-year question

Held on 10 Apr 2016 · Verified 6 Jul 2026.

Options

A
$\frac{{\pi }^{2}{r}^{4}{B}_{0}^{4}}{2\tau R}$
B
$\frac{{\pi }^{2}{r}^{4}{B}_{0}^{2}}{2\tau R}$
C
$\frac{{\pi }^{2}{r}^{4}{B}_{0}^{2}R}{\tau }$
D
$\frac{{\pi }^{2}{r}^{4}{B}_{0}^{2}}{\tau R}$

Did you get this right?

Solution

Here, $B={B}_{0}{e}^{-\frac{t}{\tau }}$

Area of the circular loop, $A=\pi {r}^{2}$

Flux linked with the loop at any time $t$ ,

$\phi =BA=\pi {r}^{2}{B}_{0}{e}^{-\frac{t}{\tau }}$

Emf induced in the loop, $\epsilon =-\frac{d\phi }{dt}=\pi {r}^{2}{B}_{0}\frac{1}{\tau }{e}^{-\frac{t}{\tau }}$

Net heat generated in the loop

$\int _{0}^{\infty }\frac{{\epsilon }^{2}}{R}dt=\frac{{\pi }^{2}{r}^{4}{B}_{0}^{2}}{{\tau }^{2}R}\int _{0}^{\infty }{e}^{-\frac{2t}{\tau }}dt$

$=\frac{{\pi }^{2}{r}^{4}{B}_{0}^{2}}{{\tau }^{2}R}\times \frac{1}{(-\frac{2}{\tau })}\times {[{e}^{-\frac{2t}{\tau }}]}_{0}^{\infty }$

$=\frac{-{\pi }^{2}{r}^{4}{B}_{0}^{2}}{2{\tau }^{2}R}\times \tau (0-1)=\frac{{\pi }^{2}{r}^{4}{B}_{0}^{2}}{2\tau R}$

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Solution

Here, $B={B}_{0}{e}^{-\frac{t}{\tau }}$

Area of the circular loop, $A=\pi {r}^{2}$

Flux linked with the loop at any time $t$ ,

$\phi =BA=\pi {r}^{2}{B}_{0}{e}^{-\frac{t}{\tau }}$

Emf induced in the loop, $\epsilon =-\frac{d\phi }{dt}=\pi {r}^{2}{B}_{0}\frac{1}{\tau }{e}^{-\frac{t}{\tau }}$

Net heat generated in the loop

$\int _{0}^{\infty }\frac{{\epsilon }^{2}}{R}dt=\frac{{\pi }^{2}{r}^{4}{B}_{0}^{2}}{{\tau }^{2}R}\int _{0}^{\infty }{e}^{-\frac{2t}{\tau }}dt$

$=\frac{{\pi }^{2}{r}^{4}{B}_{0}^{2}}{{\tau }^{2}R}\times \frac{1}{(-\frac{2}{\tau })}\times {[{e}^{-\frac{2t}{\tau }}]}_{0}^{\infty }$

$=\frac{-{\pi }^{2}{r}^{4}{B}_{0}^{2}}{2{\tau }^{2}R}\times \tau (0-1)=\frac{{\pi }^{2}{r}^{4}{B}_{0}^{2}}{2\tau R}$

JEE Main 2016 — Physics Electromagnetism

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JEE Main 2016 — Physics Electromagnetism

Options

Did you get this right?

Solution

Your note

JEE Main Electromagnetism in other years

More JEE Main Electromagnetism questions

Other Physics topics for JEE Main

Mechanics

Modern Physics

Optics

Thermodynamics

Waves & Oscillations

Keep practicing Electromagnetism