Magnetic Effects & Magnetism

JEE Main level, single correct, medium

87 of 276

Question 87 of 276

A conducting metal circular-wire-loop of radius rr is placed perpendicular to a magnetic field which varies with time as B=B0e−tτ,B={B}_{0}{e}^{-\dfrac{t}{\tau }}, where B0andτ{B}_{0}and\tau are constants at time t=0t=0. If the resistance of the loop is RR, then the heat generated in the loop after a long time (t→∞)\left(t\rightarrow \infty \right) is

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Answer: (B) π2r4B022τR\dfrac{{\pi }^{2}{r}^{4}{B}_{0}^{2}}{2\tau R}

Here, B=B0e−tτB={B}_{0}{e}^{-\dfrac{t}{\tau }}
Area of the circular loop, A=πr2A=\pi {r}^{2}
Flux linked with the loop at any time tt ,
ϕ=BA=πr2B0e−tτ\phi =BA=\pi {r}^{2}{B}_{0}{e}^{-\dfrac{t}{\tau }}
Emf induced in the loop, ϵ=−dϕdt=πr2B01τe−tτ\epsilon =-\dfrac{d\phi }{dt}=\pi {r}^{2}{B}_{0}\dfrac{1}{\tau }{e}^{-\dfrac{t}{\tau }}
Net heat generated in the loop
∫0∞ϵ2Rdt=π2r4B02τ2R∫0∞e−2tτdt\int _{0}^{\infty }\dfrac{{\epsilon }^{2}}{R}dt=\dfrac{{\pi }^{2}{r}^{4}{B}_{0}^{2}}{{\tau }^{2}R}\int _{0}^{\infty }{e}^{-\dfrac{2t}{\tau }}dt
=π2r4B02τ2R×1(−2τ)×[e−2tτ]0∞=\dfrac{{\pi }^{2}{r}^{4}{B}_{0}^{2}}{{\tau }^{2}R}\times \dfrac{1}{\left(-\dfrac{2}{\tau }\right)}\times {\left[{e}^{-\dfrac{2t}{\tau }}\right]}_{0}^{\infty }
=−π2r4B022τ2R×τ(0−1)=π2r4B022τR=\dfrac{-{\pi }^{2}{r}^{4}{B}_{0}^{2}}{2{\tau }^{2}R}\times \tau \left(0-1\right)=\dfrac{{\pi }^{2}{r}^{4}{B}_{0}^{2}}{2\tau R}

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