Magnetic moment of a current carrying square loop be `M`. If it is converted in form of circle and same current is passed through it then find the new magnetic moment.

Options

A. `(m)/(pi)`
B. `(3m)/(pi)`
C. `(2m)/(pi)`
D. `(4m)/(pi)`

Answer

To solve the problem, we need to find the new magnetic moment of a circular loop formed from a square loop with the same current flowing through it.

### Step-by-Step Solution:

1. **Identify the magnetic moment of the square loop**:
The magnetic moment \( M \) of a square loop carrying current \( I \) is given by the formula:
\[ M = I \times A \]
where \( A \) is the area of the square loop. If the side of the square is \( A \), then the area \( A \) is:
\[ A = a^2 \]
Thus, the magnetic moment of the square loop becomes:
\[ M = I \times a^2 \]

2. **Calculate the perimeter of the square loop**:
The perimeter \( P \) of the square loop is:
\[ P = 4a \]

3. **Relate the perimeter of the square to the circumference of the circle**:
When the square loop is converted into a circular loop, the perimeter of the square becomes the circumference of the circle. Therefore, we have:
\[ 4a = 2\pi r \]
where \( r \) is the radius of the circular loop.

4. **Solve for the radius \( r \)**:
Rearranging the equation gives:
\[ r = \frac{4a}{2\pi} = \frac{2a}{\pi} \]

5. **Calculate the area of the circular loop**:
The area \( A' \) of the circular loop is given by:
\[ A' = \pi r^2 \]
Substituting the value of \( r \):
\[ A' = \pi \left(\frac{2a}{\pi}\right)^2 = \pi \times \frac{4a^2}{\pi^2} = \frac{4a^2}{\pi} \]

6. **Calculate the new magnetic moment \( M' \)**:
The magnetic moment \( M' \) of the circular loop is:
\[ M' = I \times A' = I \times \frac{4a^2}{\pi} \]

7. **Express \( M' \) in terms of \( M \)**:
Since \( M = I \times a^2 \), we can express \( M' \) as:
\[ M' = \frac{4}{\pi} \times (I \times a^2) = \frac{4}{\pi} M \]

### Final Answer:
Thus, the new magnetic moment \( M' \) of the circular loop is:
\[ M' = \frac{4M}{\pi} \]