Question

A small square loop of wire of side l is placed inside a large square loop of wire L(L>>l). As shown in figure, both loops are coplanar and their centers coincide at point O. The mutual inductance of the system is:

• A. \(\frac{2\sqrt2 \mu_0L^2}{\pi l}\)
• B. \(\frac{\mu_0L^2}{2\sqrt2 \pi L}\)
• C. \(\frac{2\sqrt 2 \mu_0 l^2}{\pi L}\)
• D. \(\frac{\mu_0L^2}{2\sqrt2\pi l}\)

Answer 1

The Correct Option is C

To find the mutual inductance between the two coplanar square loops, we need to understand the concept of magnetic flux linkage and how it applies to this configuration.

Mutual inductance M is defined between two coils as:

M = \frac{\Phi}{I}

where \Phi is the magnetic flux through one coil due to the current I in the other coil.

Step 1: Magnetic Field Due to the Larger Loop

The magnetic field B at the center of a square loop of side L carrying a current I is given by:

B = \frac{2\sqrt{2}\mu_0 I}{\pi L}

where \mu_0 is the permeability of free space.

Step 2: Flux Through the Smaller Loop

The flux \Phi through the smaller loop of side l is:

\Phi = B \times \text{Area of smaller loop} = \left(\frac{2\sqrt{2}\mu_0 I}{\pi L}\right) \times (l^2)

Step 3: Mutual Inductance Calculation

Substitute \Phi in the formula for mutual inductance:

M = \frac{\Phi}{I} = \frac{2\sqrt{2}\mu_0 l^2}{\pi L}

Therefore, the mutual inductance of the system is:

\frac{2\sqrt 2 \mu_0 l^2}{\pi L}

This matches the correct answer: \(\frac{2\sqrt 2 \mu_0 l^2}{\pi L}\) .