Question

Two conducting circular loops of radii $R_{1}$ and $R_{2}$ are placed in the same plane with their centres coinciding. If $R_{1} >\, R_{2}$ the mutual inductance $M$ between them will be directly proportional to

• A. $\frac{R_{1}{R_{2}}$
• B. $\frac{R_{2}{R_{1}}$
• C. $\frac{R_{1}^{2}}{R_{2}}$
• D. $\frac{R_{2}^{2}}{R_{1}}$

Answer 1

The Correct Option is D

Let a current $I_{1}$ flows through the outer circular coil of radius $R_{1}$ The magnetic field at the centre of the coil is $B_{1} =\frac{\mu_{0} I_{1}}{2R_{1}}$ As the inner coil of radius $R_{2}$ placed co-axially has small radius $(R_{2} < \, R_{1})$, therefore $B_{1}$ may be taken constant over its cross-sectional area Hence, flux associated with the inner coil is $\phi_{2}=B_{1}\, \pi R_{2}^{2} =\frac{\mu_{0}I_{1}}{2R_{1}} \pi R_{2}^{2}$ As $M=\frac{\phi_{2}}{I_{1}}=\frac{\mu_{0}\pi R_{2}^{2}}{2R_{1}}$ $\therefore M \propto\frac{R_{2}^{2}}{R_{1}}$