A) \[\frac{{{\mu }_{0}}\pi r_{2}^{2}}{2{{r}_{1}}}\]
B) \[\frac{{{\mu }_{0}}\pi r_{1}^{2}}{2{{r}_{2}}}\]
C) \[\frac{{{\mu }_{0}}\pi ({{r}_{1}}+{{r}_{2}})}{2{{r}_{1}}}\]
D) \[\frac{{{\mu }_{0}}\pi {{({{r}_{1}}+{{r}_{2}})}^{2}}}{2{{r}_{2}}}\]
Correct Answer: A
Solution :
Magnetic field due to the larger coil at its centre is \[B=\frac{{{\mu }_{0}}I}{2{{r}_{1}}}\] where\[I\]is the current in the larger coil. Flux through the inner coil is \[\phi =B\times \pi r_{2}^{2}=\frac{{{\mu }_{0}}I}{2{{r}_{1}}}\times \pi r_{2}^{2}\] But \[\phi =ML\] Therefore, \[M=\frac{{{\mu }_{0}}\pi r_{2}^{2}}{2{{r}_{1}}}\]
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