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

Answer: (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
${\varphi }_2=B_1\pi R_2^2=\frac{{\mu }_0{I}_1}{2R_1}\pi R_2^2$
As $M=\frac{{\varphi }_2}{I_1}=\frac{{\mu }_0\pi R_2^2}{2R_1}$
$\therefore M\propto \frac{R_2^2}{R_1}$