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

<p>The magnetic field $B_2$ due to the current $I_2$ in the larger coil with 200 turns is given by:</p> <p>$B_2 = \frac{N_2 \mu_0 I_2}{2r_2} = \frac{200 \mu_0 I_2}{2 \times 10}$</p> <p>The magnetic flux $\phi_{1,2}$ through the smaller coil due to this magnetic field is given by:</p> <p>$$\phi_{1,2} = N_1 \vec{B}_2 \cdot \vec{A}_1 = N_1 N_2 \frac{\mu_0 I_2}{2 r_2} \cdot \pi r_1^2$$</p> <p>Since $\phi_{1,2} = MI_2$, we can solve for the mutual inductance $M$:</p> <p>$M = \frac{N_1 N_2 \frac{\mu_0 I_2}{2 r_2} \cdot \pi r_1^2}{I_2}$</p> <p>Substituting the given values for $r_1$, $N_1$, $r_2$, and $N_2$:</p> <p>$$M = \frac{10 \times 200 \times 4 \pi \times 10^{-7} \times \pi \times (0.01)^2}{2 \times 10}$$</p> <p>Simplifying the expression, we get:</p> <p>$M = 4 \times 10^{-8} \mathrm{H}$</p> <p>So, the mutual inductance between the two concentric coils is $4 \times 10^{-8} \mathrm{H}$.</p>