A) \[\frac{B\pi {{r}^{2}}\omega }{2R}\]
B) \[\frac{{{(B\pi {{r}^{2}}\omega )}^{2}}}{8R}\]
C) \[\frac{{{(B\pi r\omega )}^{2}}}{2R}\]
D) \[\frac{{{(B\pi r{{\omega }^{2}})}^{2}}}{8R}\]
Correct Answer: B
Solution :
The flux associated with coil of area A and magnetic induction B is \[\phi =BA\cos \theta \] \[=\frac{1}{2}B\pi {{r}^{2}}\cos \omega t\] \[\left( \because A=\frac{1}{2}\pi {{r}^{2}}for\text{ }semi-circle \right)\] \[\therefore \] \[{{e}_{induced}}=-\frac{d\phi }{dt}\] \[=-\frac{d}{dt}\left( \frac{1}{2}B\pi {{r}^{2}}\cos \omega t \right)\] \[=\frac{1}{2}B\pi {{r}^{2}}\omega \sin \omega t\] \[\therefore \]Power \[P=\frac{e_{induced}^{2}}{R}\] \[=\frac{{{B}^{2}}{{\pi }^{2}}{{r}^{4}}{{\omega }^{2}}{{\sin }^{2}}\omega t}{4R}\] Hence, \[{{P}_{mean}}=<P>\] \[=\frac{{{B}^{2}}{{\pi }^{2}}{{r}^{4}}{{\omega }^{2}}}{4R}.\frac{1}{2}\] \[\left( \because <{{\sin }^{2}}\omega t>=\frac{1}{2} \right)\] \[=\frac{{{(B\pi {{r}^{2}}\omega )}^{2}}}{8R}\]
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