JEE Main 2023 — Electromagnetic Induction Question with Solution

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

$${\varphi }_{1,2}=N_1{\vec{B}}_2\cdot {\vec{A}}_1=N_1{N}_2\frac{{\mu }_0{I}_2}{2r_2}\cdot \pi r_1^2$$

Since $${\varphi }_{1,2}=M{I}_2$$, we can solve for the mutual inductance $$M$$:

$$M=\frac{N_1{N}_2\frac{{\mu }_0{I}_2}{2r_2}\cdot \pi r_1^2}{I_2}$$

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