Question

Two conducting circular loops of radii Rl and R2 are placed in the same plane with their centres coinciding. If R1 > R2 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

Answer: (D)

$\frac{R_{2}^{2}}{R_{1}}$

Answer 2

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 $B_{1}$ maybe taken constant over its cross sectional area.

Hence flux associated with the inner coil is

$\varphi_{2} = B_{1}\pi R_{2}^{2} = \frac{\mu_{0}I_{1}}{2R_{1}}\pi r_{2}^{2}$

As $M = \frac{\varphi_{2}}{I_{1}} = \frac{\mu_{0}\pi R_{2}^{2}}{2R_{1}}$

$\therefore M \propto \frac{R_{2}^{2}}{R_{1}}$