A) \[IL/4\pi \]
B) \[{{I}^{2}}{{L}^{2/}}4\pi \]
C) \[{{I}^{2}}L/8\pi \]
D) \[I{{L}^{2}}/8\pi \]
Correct Answer: A
Solution :
If r is the radius of circle then \[L=2\pi r\] or \[r=\frac{L}{2\pi }\] \[\therefore \] Area \[=\pi {{r}^{2}}=\pi {{L}^{2}}/4{{\pi }^{2}}\] \[=\frac{{{L}^{2}}}{4\pi }\] Now magnetic moment \[M=NiA=I{{L}^{2}}/4\pi \]You need to login to perform this action.
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