A) M
B) \[\frac{4}{{{\pi }^{2}}}M\]
C) \[\frac{4}{\pi }M\]
D) \[\frac{\pi }{4}M\]
Correct Answer: D
Solution :
Initially for circular coil \[L=2\pi r\]and \[M=i\times \pi {{r}^{2}}\] \[=i\times \pi {{\left( \frac{L}{2\pi } \right)}^{2}}=\frac{i{{L}^{2}}}{4\pi }\] ..... (i) Finally for square coil \[{M}'=i\times {{\left( \frac{L}{4} \right)}^{2}}=\frac{i{{L}^{2}}}{16}\] ..... (ii) Solving equation (i) and (ii) \[{M}'=\frac{\pi M}{4}\]You need to login to perform this action.
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