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 :
[b] \[\phi =\overset{\to }{\mathop{B}}\,.\overset{\to }{\mathop{A}}\,\]; \[\phi =BA\,\cos \,\omega t\] \[\varepsilon =-\frac{d\phi }{dt}=\omega BA\,\sin \,\omega t\]; \[i=\frac{\omega BA}{R}\,\sin \,\omega t\] \[{{P}_{inst}}={{i}^{2}}R={{\left( \frac{\omega BA}{R} \right)}^{2}}\times R\,{{\sin }^{2}}\,\omega t\] \[{{P}_{avg}}=\frac{\int\limits_{0}^{T}{{{P}_{inst}}\times dt}}{\int\limits_{0}^{T}{dt}}=\frac{{{(\omega \,B\,A)}^{2}}}{R}\frac{\int\limits_{0}^{T}{{{\sin }^{2}}\,\omega \,tdt}}{\int\limits_{0}^{T}{dt}}=\frac{1}{2}\,\frac{{{(\omega \,B\,A)}^{2}}}{R}\] \[\therefore \,\,{{P}_{avg}}=\,\frac{{{(\omega \,B\,\pi \,{{r}^{2}})}^{2}}}{8\,R}\] \[\left[ A=\frac{\pi \,{{r}^{2}}}{2} \right]\]
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