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

Step 1: Understanding the Concept:
Mutual inductance (\(M\)) is a property that quantifies how much magnetic flux is linked between two current-carrying circuits.
By definition, if a current \(I\) flows through one loop (Loop 1), it creates a magnetic field. The flux (\(\Phi_2\)) of this field passing through the second loop (Loop 2) is proportional to the current: \(\Phi_2 = M \cdot I\).
In this problem, it is easier to assume the current flows through the larger square loop and find the flux through the small circular loop.
Because the square loop is much larger than the circle (\(L \gg R\)), we can assume that the magnetic field produced by the square loop is approximately constant and uniform over the small region occupied by the circular loop.
Therefore, we only need to find the magnetic field at the center of the square loop.
Key Formula or Approach:
1. Magnetic field due to a straight wire segment of length \(L\) at its perpendicular distance \(d\): \[ B = \frac{\mu_0 I}{4\pi d} (\sin \theta_1 + \sin \theta_2) \]
2. Flux linkage: \[ \Phi = B_{center} \times \text{Area of small loop} \]
3. Mutual Inductance: \[ M = \frac{\Phi}{I} \]
Step 2: Detailed Explanation:
1. Magnetic Field at the center of the Square Loop:
The square loop has four sides. Due to symmetry, each side contributes equally to the magnetic field at the center.
For one side of length \(L\), the perpendicular distance to the center is \(d = L/2\).
The angles formed from the center to the ends of the side are both \(45^\circ\) (\(\pi/4\) radians).
\[ B_{side} = \frac{\mu_0 I}{4\pi (L/2)} (\sin 45^\circ + \sin 45^\circ) \]
\[ B_{side} = \frac{\mu_0 I}{2\pi L} \left( \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} \right) = \frac{\mu_0 I}{2\pi L} \left( \frac{2}{\sqrt{2}} \right) = \frac{\mu_0 I}{\sqrt{2}\pi L} \]
The total field at the center from all four sides is:
\[ B_{total} = 4 \times B_{side} = 4 \times \frac{\mu_0 I}{\sqrt{2}\pi L} = \frac{2\sqrt{2}\mu_0 I}{\pi L} \]
2. Calculating the Flux (\(\Phi\)):
The flux through the circular loop is the product of the field at the center and the area of the circle (\(\pi R^2\)):
\[ \Phi = B_{total} \times A = \left( \frac{2\sqrt{2}\mu_0 I}{\pi L} \right) \times (\pi R^2) \]
The \(\pi\) terms cancel out:
\[ \Phi = \frac{2\sqrt{2}\mu_0 I R^2}{L} \]
3. Calculating Mutual Inductance (\(M\)):
\[ M = \frac{\Phi}{I} = \frac{2\sqrt{2}\mu_0 R^2}{L} \]
Step 3: Final Answer:
The mutual inductance is \(2\sqrt{2}\frac{\mu_0 R^2}{L}\).