Q144 / 150 · MHT CET 2020 (12 Oct Shift 1) · Physics
Two coaxial coils A and B of radii 'R ' and 'Rz' are placed in the same plane. (R $_{2}>$
$\mathrm{R}_{1}$ ). If a current is passed through coil $\mathrm{B}$, the coefficient of mutual inductance
between the coils is proportional to
- A$\frac{1}{\mathrm{R}_{1} \mathrm{R}_{2}}$
- B$\frac{\mathrm{R}_{2}^{2}}{\mathrm{R}_{1}}$
- C$R_{1} R_{2}$
- Correct answerD$\frac{\mathrm{R}_{1}^{2}}{\mathrm{R}_{2}}$
Solution
Magnetic field at the centre of primary coil
$$
B=\mu_{0} i_{1} / 2 R_{1} .
$$
Considering it to be uniform, magnetic flux passing through secondary coil is
$$
\phi_{2}=B A=\frac{\mu_{0} i_{1}}{2 R_{1}}\left(\pi R_{2}^{2}\right)
$$
Now, $M=\frac{\phi_{2}}{i_{1}}=\frac{\mu_{0} \pi R_{2}^{2}}{2 R_{1}}$
$$
\therefore \quad M \propto \frac{R_{2}^{2}}{R_{1}} .
$$
Hence, the correct answer is option: 'D'.