Question

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

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

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
Mutual inductance is the ratio of the magnetic flux linked with one circuit to the current in the other circuit.
Because \(L \gg R\), we can assume the magnetic field produced by the large square loop is practically uniform over the entire small area of the inner circular loop.
Step 2: Key Formula or Approach:
The magnetic field at the center of a square loop of side \(L\) carrying current \(I\) is derived using the Biot-Savart law: \(B = \frac{\mu_0 I}{4\pi d} (\sin \theta_1 + \sin \theta_2)\) for each of the 4 sides.
Magnetic flux \(\Phi = B \times A_{circle}\).
Mutual inductance \(M = \frac{\Phi}{I}\).
Step 3: Detailed Explanation:
For the square loop, the perpendicular distance from the center to any side is \(d = L/2\).
The angles made by the ends of a side at the center are \(\theta_1 = 45^\circ\) and \(\theta_2 = 45^\circ\).
The magnetic field due to one side of the square at the center is:
\[ B_{side} = \frac{\mu_0 I}{4\pi (L/2)} (\sin 45^\circ + \sin 45^\circ) = \frac{\mu_0 I}{2\pi L} \left(\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}\right) = \frac{\mu_0 I}{2\pi L} \left(\frac{2}{\sqrt{2}}\right) = \frac{\sqrt{2} \mu_0 I}{\pi L} \] Since there are 4 identical sides, the total magnetic field at the center is:
\[ B_{total} = 4 \times B_{side} = 4 \left( \frac{\sqrt{2} \mu_0 I}{\pi L} \right) = \frac{4\sqrt{2} \mu_0 I}{\pi L} \] Wait, let's re-evaluate this carefully.
\(B_{side} = \frac{\mu_0 I}{4\pi (L/2)} (\sin 45^\circ + \sin 45^\circ) = \frac{2\mu_0 I}{4\pi L} (\sqrt{2}) = \frac{\sqrt{2}\mu_0 I}{2\pi L}\).
Total field \(B_{total} = 4 \times \frac{\sqrt{2}\mu_0 I}{2\pi L} = \frac{2\sqrt{2}\mu_0 I}{\pi L}\).
(Correction applied: The factor of 2 was initially misplaced).
Now, calculate the magnetic flux through the small circular loop of area \(A = \pi R^2\):
\[ \Phi = B_{total} \times A = \left( \frac{2\sqrt{2} \mu_0 I}{\pi L} \right) (\pi R^2) = \frac{2\sqrt{2} \mu_0 R^2 I}{L} \] Finally, the mutual inductance \(M\) is the flux per unit current:
\[ M = \frac{\Phi}{I} = \frac{2\sqrt{2} \mu_0 R^2}{L} \] Step 4: Final Answer:
The mutual inductance is \(2\sqrt{2} \mu_0 R^2 / L\).