Q179 / 200 · MHT CET - Medical 2026 (25 April Shift 1) · Physics
Two conducting circular loops of radii $R_1$ and $R_2$ are placed in same plane with their centers coinciding. If $R_1 > R_2$ and current I is flowing through the circular loop of radius $R_1$, then mutual inductance between them is proportional to
- Correct answerA$\dfrac{R_2^2}{R_1}$
- B$\dfrac{R_2}{R_1}$
- C$\dfrac{R_1^2}{R_2}$
- D$\dfrac{R_1}{R_2}$
Solution
Magnetic field at the center of the larger loop of radius $R_1$ carrying current $I$ is given by
$B = \dfrac{\mu_0 I}{2 R_1}$
Since $R_1 > R_2$, the magnetic field can be considered approximately uniform over the entire area of the smaller loop.
Magnetic flux linked with the smaller loop of radius $R_2$ is
$\Phi = B \times A_2 = \left(\dfrac{\mu_0 I}{2 R_1}\right) (\pi R_2^2)$
The mutual inductance $M$ is given by
$M = \dfrac{\Phi}{I} = \dfrac{\mu_0 \pi R_2^2}{2 R_1}$
Therefore, $M \propto \dfrac{R_2^2}{R_1}$
Answer: $\dfrac{R_2^2}{R_1}$
$B = \dfrac{\mu_0 I}{2 R_1}$
Since $R_1 > R_2$, the magnetic field can be considered approximately uniform over the entire area of the smaller loop.
Magnetic flux linked with the smaller loop of radius $R_2$ is
$\Phi = B \times A_2 = \left(\dfrac{\mu_0 I}{2 R_1}\right) (\pi R_2^2)$
The mutual inductance $M$ is given by
$M = \dfrac{\Phi}{I} = \dfrac{\mu_0 \pi R_2^2}{2 R_1}$
Therefore, $M \propto \dfrac{R_2^2}{R_1}$
Answer: $\dfrac{R_2^2}{R_1}$
Hence, the correct answer is option: 'A'.