Question

A small square loop of side 'a' and one turn is placed inside a larger square loop of side b and one turn (\( b \gg a \)). The two loops are coplanar with their centres coinciding. If a current I is passed in the square loop of side 'b', then the coefficient of mutual inductance between the two loops is :

• A. \( \frac{\mu_0}{4\pi} \frac{8\sqrt{2} a^2}{b} \)
• B. \( \frac{\mu_0}{4\pi} \frac{8\sqrt{2} b^2}{a} \)
• C. \( \frac{\mu_0}{4\pi} \frac{8\sqrt{2}}{a} \)
• D. \( \frac{\mu_0}{4\pi} \frac{8\sqrt{2}}{b} \)

Hint:

For any two concentric coplanar loops, the mutual inductance always scales as \( M \propto \frac{(\text{Small Area})}{\text{Large Dimension}} \).

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
Mutual inductance \( M \) is defined by the magnetic flux \( \phi \) through one loop due to current \( I \) in another: \( \phi = MI \).
Since \( b \gg a \), the magnetic field \( B \) produced by the larger loop can be considered uniform over the area of the smaller loop.
Step 2: Key Formula or Approach:
The magnetic field at the center of a square loop of side \( b \) is the sum of fields due to its four sides.
Magnetic field due to a finite wire of length \( L \) at distance \( r \) from its midpoint is \( B = \frac{\mu_0 I}{4\pi r} (\sin \theta_1 + \sin \theta_2) \).
Step 3: Detailed Explanation:
For one side of length \( b \), the distance to the center is \( r = b/2 \). The angles are \( \theta_1 = \theta_2 = 45^\circ \).
\[ B_{\text{side}} = \frac{\mu_0 I}{4\pi (b/2)} (\sin 45^\circ + \sin 45^\circ) = \frac{\mu_0 I}{2\pi b} \left(\frac{2}{\sqrt{2}}\right) = \frac{\mu_0 I}{\sqrt{2}\pi b} \]
Total field at the center due to 4 sides:
\[ B = 4 \times \frac{\mu_0 I}{\sqrt{2}\pi b} = \frac{2\sqrt{2} \mu_0 I}{\pi b} \]
Magnetic flux \( \phi \) through the smaller loop:
\[ \phi = B \cdot \text{Area} = \frac{2\sqrt{2} \mu_0 I}{\pi b} \cdot a^2 \]
Since \( \phi = MI \):
\[ M = \frac{2\sqrt{2} \mu_0 a^2}{\pi b} = \frac{\mu_0}{4\pi} \frac{8\sqrt{2} a^2}{b} \]
Step 4: Final Answer:
The coefficient of mutual inductance is \( \frac{\mu_0}{4\pi} \frac{8\sqrt{2} a^2}{b} \).