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)/(4pi)`
B. `(4M)/(pi)`
C. `(M)/(3pi)`
D. `(5M)/(6pi)`

Answer

To solve the problem of finding the new magnetic moment when a current-carrying square loop is converted into a circular loop, we can follow these steps:

### Step-by-Step Solution:

1. **Determine the Magnetic Moment of the Square Loop:**
- The magnetic moment \( M \) of a current-carrying loop is given by the formula:
\[ M = N \cdot I \cdot A \]
where \( N \) is the number of turns, \( I \) is the current, and \( A \) is the area of the loop.
- For a square loop with side length \( A \):
\[ A = A^2 \quad \text{(Area of the square)} \]
- Therefore, the magnetic moment of the square loop is:
\[ M = 1 \cdot I \cdot A^2 = I \cdot A^2 \]

2. **Convert the Square Loop to a Circular Loop:**
- When the square loop is converted into a circular loop, the perimeter of the square must equal the circumference of the circle.
- The perimeter of the square is:
\[ P_{\text{square}} = 4A \]
- The circumference of the circle is:
\[ C_{\text{circle}} = 2\pi r \]
- Setting these equal gives:
\[ 4A = 2\pi r \]
- Solving for \( r \):
\[ r = \frac{4A}{2\pi} = \frac{2A}{\pi} \]

3. **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 \( r \):
\[ A' = \pi \left(\frac{2A}{\pi}\right)^2 = \pi \cdot \frac{4A^2}{\pi^2} = \frac{4A^2}{\pi} \]

4. **Determine the New Magnetic Moment:**
- The magnetic moment \( M' \) of the circular loop is:
\[ M' = N \cdot I \cdot A' = 1 \cdot I \cdot \frac{4A^2}{\pi} = \frac{4IA^2}{\pi} \]
- Since \( M = IA^2 \), we can express \( M' \) in terms of \( M \):
\[ M' = \frac{4M}{\pi} \]

### Final Answer:
The new magnetic moment \( M' \) when the square loop is converted into a circular loop while maintaining the same current is:
\[ M' = \frac{4M}{\pi} \]