A) l / L
B) \[{{l}^{2}}/L\]
C) \[L/l\]
D) \[{{L}^{2}}/l\]
Correct Answer: B
Solution :
Magnetic field produced due to large loop \[B=\frac{{{\mu }_{0}}}{4\pi }\frac{8\sqrt{2}i}{L}\] Flux linked with smaller loop \[\varphi =B({{l}^{2}})=\frac{{{\mu }_{0}}}{4\pi }\frac{8\pi i{{l}^{2}}}{L}\] \[\therefore \varphi =Mi\Rightarrow M=\frac{\varphi }{i}=\frac{{{\mu }_{0}}}{4\pi }.\frac{8\sqrt{2}{{l}^{2}}}{L}\Rightarrow M\propto \frac{{{l}^{2}}}{L}\]You need to login to perform this action.
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