A) 2 : p
B) p : 2
C) p : 4
D) 4 : p
Correct Answer: C
Solution :
Suppose length of each wire is l. \[r\propto \frac{1}{B}\] \[{{A}_{cirde}}=\pi {{r}^{2}}=\pi {{\left( \frac{l}{2\pi } \right)}^{2}}=\frac{{{l}^{2}}}{4\pi }\] \[\because \] Magnetic moment \[M=iA\] \[\Rightarrow \frac{{{M}_{square}}}{{{M}_{cirde}}}=\frac{{{A}_{square}}}{{{A}_{cirde}}}\] \[=\frac{{{l}^{2}}/16}{{{l}^{2}}/4\pi }\]\[=\frac{\pi }{4}\]You need to login to perform this action.
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