JEE Main · 2023 · 06 Apr 2023 · Shift 2 · Electromagnetic Induction

The inner coil ($r_1=1$ cm, $N_1=10$) is very small compared with the outer coil ($r_2=1000$ cm, $N_2=200$), so the field of the outer coil is uniform over the inner area and equals $B=\frac{{\mu }_0{N}_{2}I}{2r_2}$. The flux linked with the inner coil is $N_{1}B\pi r_1^2$, so the mutual inductance is $M=\frac{{\mu }_0{N}_1{N}_2\pi r_1^2}{2r_2}$. Substituting, $M=\frac{4\pi \times 10^{-7}\times 10\times 200\times \pi \times (0.01{)}^2}{2\times 10}$, and with ${\pi }^2=10$ this gives $M=4\times 10^{-8}$ H. The required value is $4$.