A) \[\frac{{{(B\pi r\omega )}^{2}}}{2R}\]
B) \[\frac{{{(B\pi {{r}^{2}}\omega )}^{2}}}{8R}\]
C) \[\frac{B\pi {{r}^{2}}\omega }{2R}\]
D) \[\frac{{{(B\pi r{{\omega }^{2}})}^{2}}}{8R}\]
Correct Answer: B
Solution :
| \[\phi =\vec{B}.\vec{A};\] \[\phi =BA\,\cos \omega t\] |
| \[\varepsilon =-\frac{d\phi }{dt}=\omega BA\,\sin \,\omega t;\] \[i=\frac{\omega BA}{R}\sin \omega t\] |
| \[{{P}_{inst}}={{i}^{2}}R={{\left( \frac{\omega BA}{R} \right)}^{2}}\times R{{\sin }^{2}}\omega t\] |
| \[{{P}_{avg}}=\frac{\int\limits_{0}^{T}{{{P}_{inst}}\times dt}}{\int\limits_{0}^{T}{dt}}=\frac{{{(\omega BA)}^{2}}}{R}\frac{\int\limits_{0}^{T}{{{\sin }^{2}}\omega t\,dt}}{\int\limits_{0}^{T}{dt}}=\frac{1}{2}\frac{{{(\omega BA)}^{2}}}{R}\] |
| \[\therefore \,\,\,{{P}_{avg}}=\frac{{{(\omega B\pi {{r}^{2}})}^{2}}}{8R}\] \[\left[ A=\frac{\pi {{r}^{2}}}{2} \right]\] |
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