A) \[\frac{{{\mu }_{0}}iv}{2\pi x}\left( \frac{l+b}{b} \right)\]
B) \[\frac{{{\mu }_{0}}{{i}^{2}}v}{4{{\pi }^{2}}x}\log \left( \frac{b}{l} \right)\]
C) \[\frac{{{\mu }_{0}}ilbv}{2\pi x(l+x)}\]
D) \[\frac{{{\mu }_{0}}ilbv}{2\pi }\log \left( \frac{x+l}{x} \right)\]
Correct Answer: C
Solution :
We can show the situation as since, loop is moving away from the wire, so the direction of current in the loop will be as shown in the figure. Net magnetic field on the loop due to wire, \[B=\frac{{{\mu }_{0}}i}{2\pi }\,\left( \frac{1}{x}-\frac{1}{x+l} \right)=\frac{{{\mu }_{0}}il}{2\pi x(l+x)}\] So, the magnitude of the emf in the loop\[e=vBb=\frac{{{\mu }_{0}}i/bv}{2\pi x\,(l+x)}\]
You need to login to perform this action.
You will be redirected in
3 sec
