Question

Find the mutual inductance in the arrangement, when a small circular loop of radius R is placed inside a large square loop of side \(L\) (\(L \gg R\)). The loops are coplanar and their centers coincide:

• A. \(M = \frac{\sqrt{2} \mu_0 R^2}{L}\)
• B. \(M = \frac{2 \sqrt{2} \mu_0 R}{L^2}\)
• C. \(M = \frac{2 \sqrt{2} \mu_0 R^2}{L}\)
• D. \(M = \frac{\sqrt{2} \mu_0 R}{L^2}\)

Hint:

 For mutual inductance in coplanar loops:
• Use the magnetic field from the larger loop and calculate flux through the smaller loop.
• Divide flux by the current to obtain mutual inductance.

Answer 1

The Correct Option is C

To find the mutual inductance \( M \) between the small circular loop and the large square loop, we start by considering the magnetic field due to the square loop at the center where the circular loop is located.

1. The magnetic field \( B \) at the center due to one side of the square loop carrying current \( I \) can be derived from Biot-Savart law. For a segment of length \( L \) at a distance \( \frac{L}{2} \) from the center:

B = \frac{\mu_0 I}{2 \pi d} \left( \sin{\theta_1} + \sin{\theta_2} \right)

• Here, \( d = \frac{L}{2} \), and angles \( \theta_1 = \theta_2 = 45^\circ \) for symmetry:

B_{\text{one side}} = \frac{\mu_0 I \sqrt{2}}{2 \pi L}

2. Since there are four sides, the total magnetic field \( B_{\text{total}} \) at the center due to the square loop is given by:

B_{\text{total}} = 4 \times B_{\text{one side}} = \frac{2 \sqrt{2} \mu_0 I}{\pi L}

3. The flux \( \Phi \) through the small circular loop of radius \( R \) is:

\Phi = B_{\text{total}} \times \text{area of circular loop} = \frac{2 \sqrt{2} \mu_0 I}{\pi L} \times \pi R^2

4. Thus, the mutual inductance \( M \), defined by the relation \( \Phi = M I \), is:

M = \frac{2 \sqrt{2} \mu_0 R^2}{L}

This matches with the option: \( M = \frac{2 \sqrt{2} \mu_0 R^2}{L} \).