Verified Solution • 2025–26 Session moderate

QJEEM16Ph3032: JEE Main 2016 Physics - Class 12 JEEM16Ph3032 Solution: A conducting metal circular-wire-loop of radius r is placed…

A conducting metal circular-wire-loop of radius r is placed perpendicular to a magnetic field which varies with time as
B = B0e −tr{^{{{ - t} \over r}}} , where B0 and τ\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 →\to ∞\infty ) is :

Multiple Choice Options

  • A π2r4B042τR{{{\pi ^2}{r^4}B_0^4} \over {2\tau R}}
  • B π2r4B022τR{{{\pi ^2}{r^4}B_0^2} \over {2\tau R}}
  • C π2r4B02Rτ{{{\pi ^2}{r^4}B_0^2R} \over \tau }
  • D π2r4B02τR{{{\pi ^2}{r^4}B_0^2} \over {\tau R}}

Step-by-Step Verified Solution

Given,

B = B0e −tτ^{ - {t \over \tau }}

Area of the circular loop, A = π\pi r2

∴\therefore    Flux ϕ\phi = BA = π\pi r2 B0 e −tτ^{ - {t \over \tau }}

Induced emf in the loop,

ε\varepsilon = −- dϕdt{{d\phi } \over {dt}} = π\pi r2B0 1τ{1 \over \tau } e −tτ^{ - {t \over \tau }}

Heat generated

= ∫0∝i2R dt\int\limits_0^ \propto {{i^2}R\,dt}

= ∫0∝ε2R dt\int\limits_0^ \propto {{{{\varepsilon ^2}} \over R}} \,dt

= 1Rπ2r4B02τ2∫0∝e−2tτ dt{1 \over R}{{{\pi ^2}{r^4}B_0^2} \over {{\tau ^2}}}\int\limits_0^ \propto {{e^{ - {{2t} \over \tau }}}} \,dt

= π2r4B02τ2R×1(−2τ)[e−2tτ]0∝{{{\pi ^2}{r^4}B_0^2} \over {{\tau ^2}R}} \times {1 \over {\left( { - {2 \over \tau }} \right)}}\left[ {{e^{ - {{2t} \over \tau }}}} \right]_0^ \propto

= −π2r4B022τ2R×τ(0−1){{ - {\pi ^2}{r^4}B_0^2} \over {2{\tau ^2}R}} \times \tau \left( {0 - 1} \right)

= π2r4B022τR{{{\pi ^2}{r^4}B_0^2} \over {2\tau R}}

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