A) zero
B) \[\frac{{{\mu }_{0}}I(b-a)}{24ab}\]
C) \[\frac{{{\mu }_{0}}I}{4\pi }\left[ \frac{b-a}{ab} \right]\]
D) \[\frac{{{\mu }_{0}}I}{4\pi }\left[ 2(b\_a)+\frac{\pi }{3}(a+b) \right]\]
Correct Answer: B
Solution :
\[B=\frac{1}{12}\]of\[\frac{{{\mu }_{0}}I}{2}\left( \frac{1}{a}-\frac{1}{b} \right)\] \[=\frac{{{\mu }_{0}}I(b-a)}{24ab}\].You need to login to perform this action.
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