Question

A circular current loop of radius \(R\) is placed inside a square loop of side length \(L\) \((L \gg R)\) such that they are co-planar and their centers coincide. The permeability of free space is \(\mu_0\). The mutual inductance between circular loop and square loop is ________.

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

Hint:

Remember:

• Mutual inductance: \[ M = \frac{\Phi}{I} \]
• Magnetic field at center of square loop: \[ B = \frac{2\sqrt{2}\mu_0 I}{\pi L} \]
• Flux: \[ \Phi = BA \]

Answer 1

The Correct Option is D

Concept: Mutual inductance is defined as: \[ M = \frac{\Phi}{I} \] where:

• \(\Phi\) = magnetic flux linked with one loop due to current in the other loop
• \(I\) = current producing the magnetic field

Since \(L \gg R\), magnetic field due to the square loop can be assumed approximately uniform over the circular loop.

Step 1: Find magnetic field at the center of square loop. Magnetic field due to one side of square loop at its center is: \[ B_1 = \frac{\mu_0 I}{4\pi (L/2)} (\sin45^\circ + \sin45^\circ) \] Since: \[ \sin45^\circ = \frac{1}{\sqrt{2}} \] \[ B_1 = \frac{\mu_0 I}{2\pi L} \left(\frac{2}{\sqrt{2}}\right) \] \[ B_1 = \frac{\mu_0 I}{\sqrt{2}\pi L} \] There are four sides of the square, hence total magnetic field: \[ B = 4B_1 \] \[ B = \frac{4\mu_0 I}{\sqrt{2}\pi L} \] \[ B = \frac{2\sqrt{2}\mu_0 I}{\pi L} \]

Step 2: Find magnetic flux through circular loop. Area of circular loop: \[ A = \pi R^2 \] Flux linked: \[ \Phi = BA \] \[ \Phi = \left( \frac{2\sqrt{2}\mu_0 I}{\pi L} \right) (\pi R^2) \] \[ \Phi = \frac{2\sqrt{2}\mu_0 I R^2}{L} \]

Step 3: Calculate mutual inductance. Using: \[ M = \frac{\Phi}{I} \] \[ M = \frac{2\sqrt{2}\mu_0 I R^2/L}{I} \] \[ M = 2\sqrt{2}\frac{\mu_0 R^2}{L} \] Therefore, \[ \boxed{ M = 2\sqrt{2}\frac{\mu_0 R^2}{L} } \]