A small square loop of `n_(1)` turns and side l is placed at the centre of a large coplanar, circular loop of radius r and `n_(2)` tuns. The mutual inductacne of the combination is

Options

A. `(mu_(0)n_(1)n_(2)pir^(2))/(l)`
B. `(mu_(0)n_(1)n_(2)l^(2))/(2r)`
C. `(mu_(0)n_(2)^(2)l^(2))/(2r)`
D. `(mu_(0)n_(1)^(2)l)/(2r)`

Answer

To find the mutual inductance \( M \) of a small square loop with \( n_1 \) turns and side length \( l \) placed at the center of a large coplanar circular loop with radius \( r \) and \( n_2 \) turns, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Magnetic Field at the Center of the Circular Loop**:
The magnetic field \( B \) at the center of a circular loop carrying a current \( I \) is given by the formula:
\[ B = \frac{\mu_0 I n_2}{2r} \]
where \( \mu_0 \) is the permeability of free space, \( I \) is the current flowing through the circular loop, and \( n_2 \) is the number of turns in the circular loop.

2. **Assume Uniform Magnetic Field**:
Since the radius \( r \) of the circular loop is much greater than the side length \( l \) of the square loop, we can assume that the magnetic field is uniform across the area of the square loop.

3. **Calculate the Area of the Square Loop**:
The area \( A \) of the square loop is given by:
\[ A = l^2 \]

4. **Calculate the Magnetic Flux through the Square Loop**:
The magnetic flux \( \Phi \) through the square loop due to the magnetic field from the circular loop is given by:
\[ \Phi = B \cdot A = \left(\frac{\mu_0 I n_2}{2r}\right) \cdot l^2 \]
Therefore, substituting for \( B \):
\[ \Phi = \frac{\mu_0 I n_2 l^2}{2r} \]

5. **Total Magnetic Flux for \( n_1 \) Turns**:
Since the square loop has \( n_1 \) turns, the total magnetic flux linked with the square loop is:
\[ \Phi_{\text{total}} = n_1 \Phi = n_1 \cdot \frac{\mu_0 I n_2 l^2}{2r} = \frac{\mu_0 n_1 n_2 l^2 I}{2r} \]

6. **Relate Magnetic Flux to Mutual Inductance**:
The mutual inductance \( M \) is defined as the ratio of the total magnetic flux linked with the loop to the current \( I \) flowing through the other loop:
\[ M = \frac{\Phi_{\text{total}}}{I} = \frac{\mu_0 n_1 n_2 l^2}{2r} \]

### Final Result:
Thus, the mutual inductance \( M \) of the combination is:
\[ M = \frac{\mu_0 n_1 n_2 l^2}{2r} \]