Question

The centres of two thin coaxial conducting loops of radius $r$ each are separated by distance $d(d >> r)$. Then, the mutual inductance of the system is (coil lies in vertical plane such that both coils plane facing each other)

• A. $\frac{\mu_0 \pi r^2}{2d}$
• B. $\frac{\mu_0 \pi r^2}{2d^3}$
• C. $\frac{\mu_0 \pi r^4}{2d^3}$
• D. $\frac{\mu_0 \pi r^4}{2d}$

Answer 1

Answer: (C)

$\frac{\mu_0 \pi r^4}{2d^3}$

Answer 2

To find the mutual inductance of two thin coaxial conducting loops of radius $r$ each, separated by a distance $d$ where $d \gg r$:

1. Assume a current in one loop: Let a current $I$ flow through the first loop (Loop 1).

2. Calculate the magnetic field produced by Loop 1 at the location of Loop 2: The magnetic field on the axis of a circular loop of radius $r$ carrying current $I$ at a distance $z$ from its center is given by: $B_z = \frac{\mu_0 I r^2}{2(r^2 + z^2)^{3/2}}$ Since the two loops are separated by a distance $d$ and $d \gg r$, we can approximate the magnetic field at the center of Loop 2 (i.e., at $z=d$) by neglecting $r^2$ in the denominator compared to $d^2$: $B_1 \approx \frac{\mu_0 I r^2}{2(d^2)^{3/2}} = \frac{\mu_0 I r^2}{2d^3}$

3. Calculate the magnetic flux through Loop 2 due to the current in Loop 1: Since $d \gg r$, the magnetic field produced by Loop 1 can be considered approximately uniform over the small area of Loop 2, and its direction is along the axis. The area of Loop 2 is $A_2 = \pi r^2$. The magnetic flux $\Phi_{21}$ through Loop 2 due to the current $I$ in Loop 1 is: $\Phi_{21} = B_1 \times A_2$ $\Phi_{21} = \left(\frac{\mu_0 I r^2}{2d^3}\right) (\pi r^2)$ $\Phi_{21} = \frac{\mu_0 \pi I r^4}{2d^3}$

4. Determine the mutual inductance: The mutual inductance $M$ is defined as the ratio of the magnetic flux through one loop to the current in the other loop: $M = \frac{\Phi_{21}}{I}$ Substituting the expression for $\Phi_{21}$: $M = \frac{\frac{\mu_0 \pi I r^4}{2d^3}}{I}$ $M = \frac{\mu_0 \pi r^4}{2d^3}$

The mutual inductance of the system is $\frac{\mu_0 \pi r^4}{2d^3}$.