Question

Two conducting circular loops A and B are placed in the same plane with their centres coinciding as shown in figure. The mutual inductance between them is:

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

Answer 1

The Correct Option is A

The magnetic flux (\(\phi\)) through loop B due to current in loop A is given by:

\[ \phi = M \cdot i = B \cdot A \]

The mutual inductance is:

\[ M = \frac{\mu_0 \pi a^2}{2b} \]

where \(a\) is the radius of loop A, \(b\) is the distance between the loops, and \(\mu_0\) is the permeability of free space.

Answer 2

Step 1: Understand the configuration
Two coplanar circular loops A and B share the same center \( O \). Loop A has radius \( a \), and loop B has radius \( b \) with \( b \gg a \). We are to find the mutual inductance between them.

Step 2: Magnetic field at the center due to the larger loop B
When a current \( I_B \) flows through loop B (radius \( b \)), the magnetic field at its center (and throughout the small region near the center) is nearly uniform and given by: \[ B_B = \frac{\mu_0 I_B}{2b}. \] Since \( b \gg a \), this field can be considered uniform over the area of the smaller loop A.

Step 3: Magnetic flux through loop A
The magnetic flux through loop A due to current \( I_B \) in loop B is: \[ \Phi = B_B \times \text{area of A} = \frac{\mu_0 I_B}{2b} \times \pi a^2. \] By definition of mutual inductance \( M \): \[ \Phi = M I_B \;\; \Longrightarrow \;\; M = \frac{\Phi}{I_B}. \] Substitute the value of \(\Phi\): \[ M = \frac{\mu_0 \pi a^2}{2b}. \]

Step 4: Final result
Thus, the mutual inductance between the two coaxial coplanar circular loops is: \[ M = \frac{\mu_0 \pi a^2}{2b}. \]

Final answer

\( \frac{\mu_0 \pi a^2}{2b} \)