Two concentric circular coils with radii 1 cm and 1000 cm and number of turns 10 and 200 respectively are placed coaxially with centers coinciding. The mutual inductance of this arrangement will be ______ $$\times 10^{-8}$$ H.
(Take, $$\pi^2 = 10$$)

Let the inner coil have radius $$r_1 = 1\ \text{cm} = 10^{-2}\ \text{m}$$ and turns $$N_1 = 10$$. Let the outer coil have radius $$r_2 = 1000\ \text{cm} = 10\ \text{m}$$ and turns $$N_2 = 200$$.

The magnetic field ($$B_2$$) produced at the center by a current $$I_2$$ flowing in the outer coil is:  $$B_2 = \frac{\mu_0 N_2 I_2}{2r_2}$$

The total magnetic flux ($$\phi_1$$) linked through the inner coil due to this field is:  $$\phi_1 = N_1 B_2 A_1 = N_1 \left(\frac{\mu_0 N_2 I_2}{2r_2}\right) (\pi r_1^2)$$

Since $$\phi_1 = M I_2$$, the mutual inductance $$M$$ is:  $$M = \frac{\mu_0 N_1 N_2 \pi r_1^2}{2r_2}$$

$$M = \frac{(4\pi \times 10^{-7}) \times 10 \times 200 \times \pi \times (10^{-2})^2}{2 \times 10}$$ $$\implies$$ $$M = \frac{4\pi^2 \times 10^{-7} \times 2000 \times 10^{-4}}{20}$$

$$M = \frac{4(10) \times 10^{-7} \times 2000 \times 10^{-4}}{20}$$ $$\implies$$ $$M = \frac{80000 \times 10^{-11}}{20} = 4000 \times 10^{-11} = 4 \times 10^{-8}\ \text{H}$$