A) \[\ell /L\]
B) \[{{\ell }^{2}}/L\]
C) \[L/\ell \]
D) \[{{L}^{2}}/\ell \]
Correct Answer: B
Solution :
[b] \[{{\phi }_{total}}={{B}_{l\arg e}}{{A}_{small}}=\frac{{{\mu }_{0}}}{4\pi }\frac{i}{L/2}(2\sin \,{{45}^{o}})\times {{\ell }^{2}}\] On comparing with \[{{\phi }_{total}}=Mi\], we get \[M\propto \frac{{{\ell }^{2}}}{L}\]You need to login to perform this action.
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