A square loop is carrying a steady current I and the magnitude of its magnetic dipole moment is m. if this square loop is changed to a circular loop and it carries the current, the magnitude of the magnetic dipole moment of circular loop will be :

Options

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

Explanation

To solve the problem, we need to find the magnetic dipole moment of a circular loop when we know the magnetic dipole moment of a square loop carrying the same current.

### Step-by-Step Solution:

1. **Magnetic Dipole Moment of Square Loop**:
The magnetic dipole moment \( m \) of a square loop is given by the formula:
\[ m = n \cdot I \cdot A \]
where:
- \( n \) is the number of turns (which is 1 in this case),
- \( I \) is the current,
- \( A \) is the area of the square loop.

For a square loop with side length \( a \), the area \( A \) is:
\[ A = a^2 \]
Therefore, the magnetic dipole moment for the square loop becomes:
\[ m = I \cdot a^2 \]

2. **Relating the Side Length to the Radius of the Circular Loop**:
When the square loop is transformed into a circular loop, we need to relate the side length \( a \) of the square to the radius \( r \) of the circular loop. The perimeter of the square is equal to the circumference of the circle:
\[ 4a = 2\pi r \]
From this, we can solve for \( r \):
\[ r = \frac{4a}{2\pi} = \frac{2a}{\pi} \]

3. **Area of the Circular Loop**:
The area \( A' \) of the circular loop is given by:
\[ A' = \pi r^2 \]
Substituting the expression for \( r \):
\[ A' = \pi \left(\frac{2a}{\pi}\right)^2 = \pi \cdot \frac{4a^2}{\pi^2} = \frac{4a^2}{\pi} \]

4. **Magnetic Dipole Moment of Circular Loop**:
The magnetic dipole moment \( m' \) of the circular loop is given by:
\[ m' = n \cdot I \cdot A' = 1 \cdot I \cdot \frac{4a^2}{\pi} = \frac{4I a^2}{\pi} \]

5. **Expressing \( m' \) in terms of \( m \)**:
Recall that \( m = I \cdot a^2 \). Therefore:
\[ m' = \frac{4}{\pi} m \]

### Final Answer:
The magnitude of the magnetic dipole moment of the circular loop is:
\[ m' = \frac{4m}{\pi} \]