The planar concentric rings of metal wire having radii $r_1$ and $r_2$ (with $r_1 \gt r_2$) are placed in…
The planar concentric rings of metal wire having radii $r_1$ and $r_2$ (with $r_1 \gt r_2$) are placed in air. The current I is flowing through the coil of larger radius. The mutual inductance between the coils is given by ($\mu_0=$ permeability of free space)
$\frac{\mu_0 \pi\left(r_1+r_2\right)^2}{2 r_2}$
$\frac{\mu_0 \pi\left(r_1-r_2\right)^2}{2 r_1}$
$\frac{\mu_0 \pi r_1^2}{2 r_2}$
$\frac{\mu_0 \pi r_2^2}{2 r_1}$
Solution
The magnetic field at the centre of a loop is given by
$B=\frac{\mu_0 N I}{2 R}...(i)$
$\therefore \quad$ Magnetic field produced by ring $A, B_A=\frac{\mu_0 I}{2 r_1}$
$\therefore \quad$ Magnetic flux produced in ring B due to $\mathrm{B}_{\mathrm{A}}$,
$\phi_B=B_A A_B$
$\ldots\left(\mathrm{A}_{\mathrm{B}}\right.$ is area)
$\therefore \quad \phi_B=\frac{\pi_0 I}{2 r_1} \times \pi r_2^2=\frac{\mu_0 \pi r_2^2}{2 r_1} I$
...[From(i)]
Mutual Inductance $M=\frac{\phi}{I}$
$\therefore \quad M=\frac{\phi_B}{I}=\frac{\mu_0 \pi r_2^2 I}{2 r_1 I}=\frac{\mu_0 \pi r_2^2}{2 r_1}$