A) \[\frac{{{\mu }_{0}}I}{2R}\omega {{r}^{2}}\sin \omega t\]
B) \[\frac{{{\mu }_{0}}I}{4R}\omega \pi {{r}^{2}}\sin \omega t\]
C) \[\frac{{{\mu }_{0}}I}{2R}\omega \pi {{r}^{2}}\sin \omega t\]
D) \[\frac{{{\mu }_{0}}I}{4R}\omega {{r}^{2}}\sin \omega t\]
Correct Answer: C
Solution :
\[\phi =\overline{B}.\overline{A}=BA\cos wt=\pi {{r}^{2}}b\cos wt\] \[\in =-\frac{d\phi }{dt}\] \[=-\frac{d}{dt}(\pi {{r}^{2}}B\cos wt)\] \[=\pi {{r}^{2}}B\sin wt(w)\] \[=\frac{{{\mu }_{o}}I}{2R}\pi w{{r}^{2}}\sin \,\,wt\left( \therefore B=\frac{{{\mu }_{o}}I}{2R} \right)\]
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