QJEEM23Phb7c8: JEE Main 2023 Physics - Class 12 JEEM23Phb7c8 Solution: Two concentric circular coils with radii and , and number of…

The magnetic field $B_2$ due to the current $I_2$ in the larger coil with 200 turns is given by:

$B_2 = \frac{N_2 \mu_0 I_2}{2r_2} = \frac{200 \mu_0 I_2}{2 \times 10}$

The magnetic flux $\phi_{1,2}$ through the smaller coil due to this magnetic field is given by:

$\phi_{1,2} = N_1 \vec{B}_2 \cdot \vec{A}_1 = N_1 N_2 \frac{\mu_0 I_2}{2 r_2} \cdot \pi r_1^2$

Since $\phi_{1,2} = MI_2$ , we can solve for the mutual inductance $M$ :

$M = \frac{N_1 N_2 \frac{\mu_0 I_2}{2 r_2} \cdot \pi r_1^2}{I_2}$

Substituting the given values for $r_1$ , $N_1$ , $r_2$ , and $N_2$ :

$M = \frac{10 \times 200 \times 4 \pi \times 10^{-7} \times \pi \times (0.01)^2}{2 \times 10}$

Simplifying the expression, we get:

$M = 4 \times 10^{-8} \mathrm{H}$

So, the mutual inductance between the two concentric coils is $4 \times 10^{-8} \mathrm{H}$ .