A) \[\frac{{{\mu }_{0}}I}{\pi R}\left( \tan \frac{\pi }{n} \right)\]
B) \[\frac{{{\mu }_{0}}I}{4\pi R}\left( \tan \frac{\pi }{n} \right)\]
C) \[\frac{{{\mu }_{0}}I}{2\pi R}\left( \tan \frac{\pi }{n} \right)\]
D) \[\frac{{{\mu }_{0}}I}{2\pi R}\left( \cos \frac{\pi }{n} \right)\]
Correct Answer: C
Solution :
[c] \[B=\frac{{{\mu }_{0}}I}{4\pi r}\left[ 2\,\sin \frac{\pi }{n} \right]\] \[\cos \frac{\pi }{n}=\frac{r}{R}\] But or \[r=R\cos \frac{\pi }{n}\] \[\therefore \,\,\,B=\frac{{{\mu }_{0}}I}{4\pi \,R\,\cos \frac{\pi }{n}}\left[ 2\,\sin \frac{\pi }{n} \right]=\frac{{{\mu }_{0}}I}{2\pi R}\left[ \tan \,\frac{\pi }{n} \right]\]You need to login to perform this action.
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