question_answer

A small square loop of wire of side l is placed inside a large square loop of wire of side L (L > l). The loop are coplanar and their centre coincide. The mutual inductance of the system is proportional to                            [IIT JEE 1998]

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  \[\varphi =B({{l}^{2}})=\frac{{{\mu }_{0}}}{4\pi }\frac{8\pi i{{l}^{2}}}{L}\] \[\therefore \varphi =Mi\Rightarrow M=\frac{\varphi }{i}=\frac{{{\mu }_{0}}}{4\pi }.\frac{8\sqrt{2}{{l}^{2}}}{L}\Rightarrow M\propto \frac{{{l}^{2}}}{L}\]