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 :
[d] \[e=-\frac{d\phi }{dt}--A\frac{dB}{dt}=-\pi {{r}^{2}}\frac{d}{dt}({{B}_{0}}{{e}^{-t}})=\pi {{r}^{2}}{{B}_{0}}{{e}^{-t}}\] At\[t=0\],\[e=\pi {{r}^{2}}{{B}_{0}}\], \[P=\frac{{{e}^{2}}}{R}=\frac{{{(\pi {{r}^{2}}{{B}_{0}})}^{2}}}{R}=\frac{{{\pi }^{2}}{{r}^{4}}{{B}_{0}}^{2}}{R}\]You need to login to perform this action.
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