Question

The mutual inductance between two planar concentric rings of radii $ {{r}_{1}} $ and $ {{r}_{2}} $ (with $ {{r}_{1}}>{{r}_{2}} $ ) placed in air is given by

• 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}}} $

Answer 1

Answer: (A)

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}}} $