A) \[\frac{B_{0}^{2}\pi {{r}^{2}}}{R}\]
B) \[\frac{{{B}_{0}}10{{r}^{3}}}{R}\]
C) \[\frac{B_{0}^{2}{{\pi }^{2}}{{r}^{4}}R}{5}\]
D) \[\frac{B_{0}^{2}{{\pi }^{2}}{{r}^{4}}}{R}\]
Correct Answer: D
Solution :
\[P=\frac{{{e}^{2}}}{R};\ \ e=-\frac{d}{dt}(BA)=A\frac{d}{dt}({{B}_{o}}{{e}^{-t}})=A{{B}_{o}}{{e}^{-t}}\] \[\Rightarrow P=\frac{1}{R}{{(A{{B}_{o}}{{e}^{-t}})}^{2}}=\frac{{{A}^{2}}B_{o}^{2}{{e}^{-2t}}}{R}\] At the time of starting t = 0 so \[P=\frac{{{A}^{2}}B_{o}^{2}}{R}\] \[\Rightarrow P=\frac{{{(\pi {{r}^{2}})}^{2}}B_{o}^{2}}{R}=\frac{B_{o}^{2}{{\pi }^{2}}{{r}^{4}}}{R}\]
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