Question

A small square loop of wire of side \( \ell \) is placed inside a large square loop of wire of side \( L \) ( \( L = \ell^2 \) ). The loops are coplanar and their centers coincide.The value of the mutual inductance of the system is \( \sqrt{x} \times 10^{-7} \, \text{H} \), where \( x = \) _____.

Answer 1

Correct Answer: 128

Step 1: Magnetic Field (MF) due to the larger square loop at the center 

The magnetic field \( B \) at the center of a square loop is given by:

\[ B = \frac{4\mu_0 I}{4 \pi \left(\frac{L}{2}\right)} \left[ \sin 45^\circ + \sin 45^\circ \right] \]

Simplifying the expression, we get:

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

Step 2: Magnetic Flux (\( \phi \)) in the smaller coil

The magnetic flux \( \phi \) in the smaller coil is the product of the magnetic field \( B \) and the area of the coil \( A = L^2 \), which gives:

\[ \phi = \frac{2 \sqrt{2} \mu_0 I}{\pi L} \cdot L^2. \]

Simplifying, we have:

\[ \phi = 2 \sqrt{2} \frac{\mu_0 I}{\pi} L. \]

Step 3: Mutual Inductance (M)

The mutual inductance \( M \) is given by the ratio of flux \( \phi \) to the current \( I \):

\[ M = \frac{\phi}{I} = 2 \sqrt{2} \frac{\mu_0 I^2}{\pi L}. \]

Substituting the values, we get:

\[ M = 2 \sqrt{2} \times 4 \times 10^{-7} \, \text{H} = \sqrt{128 \times 10^{-7}} \, \text{H}. \]

Therefore, the value of \( x \) is:

\[ x = 128. \]

Answer 2

Flux linkage for inner loop:

\[ \phi = B_{\text{center}} \cdot \ell^2 \] \[ = 4 \times \frac{\mu_0 i}{4\pi} \left( \sin 45^\circ + \sin 45^\circ \right) \ell^2 \] \[ \phi = \frac{2\sqrt{2} \mu_0 i \ell^2}{\pi L} \]

Mutual inductance:

\[ M = \frac{\phi}{i} = \frac{2\sqrt{2} \mu_0 \ell^2}{\pi L} = \frac{2\sqrt{2} \mu_0}{\pi} \]

Calculating:

\[ M = \frac{2\sqrt{2} \times 4\pi}{\pi} \times 10^{-7} \] \[ = 8\sqrt{2} \times 10^{-7} \, \text{H} \] \[ = \sqrt{128} \times 10^{-7} \, \text{H} \]

Thus:

\[ x = 128 \]