Question

# A conducting metal circular-wire-loop of radius r is placed perpendicular to a magnetic field which varies with time as $$B =B_0e^{-t/\tau}$$, where $$B_0$$ 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

A
π2r4B402τR
B
π2r4B202τR
C
π2r4B20Rτ
D
π2r4B20τR

Solution

## The correct option is B $$\cfrac{\pi^2r^4B^2_0}{2\tau R}$$Flux through the loop,$$\phi=\vec B. \vec S$$$$\phi =B_0\pi r^2e^{-t/\tau}$$$$\varepsilon =-\cfrac{d\phi}{dt}=\dfrac{B_0\pi r^2e^{-t/\tau}}{\tau}$$Heat $$\displaystyle Q=\int^{\infty}_0\cfrac{\varepsilon ^2}{R}dt$$$$\displaystyle Q= \int^{\infty}_0 \cfrac{B_o^2\pi^2r^4e^{-2t/\tau}}{R\tau^2}$$$$=\left[ \cfrac{B_o^2 \pi^2 r^4 e^{-2t/\tau}}{2R\tau} \right]_{t=0}^{\infty}$$$$=\cfrac{\pi^2r^4B_o^2}{2\tau R}$$Physics

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