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 it is transformed from a square loop that carries a steady current \( I \) and has a magnetic dipole moment \( m \).

### Step-by-Step Solution:

1. **Understanding the Magnetic Dipole Moment**:
The magnetic dipole moment \( m \) for a current-carrying loop is given by the formula:
\[ m = n \cdot I \cdot A \]
where \( n \) is the number of turns (which is 1 for a single loop), \( I \) is the current, and \( A \) is the area of the loop.

2. **Area of the Square Loop**:
For a square loop with side length \( a \), the area \( A \) is:
\[ A_{\text{square}} = a^2 \]
Therefore, the magnetic dipole moment of the square loop is:
\[ m_{\text{square}} = I \cdot a^2 \]
Given that this is equal to \( m \), we have:
\[ m = I \cdot a^2 \]

3. **Circumference of the Square Loop**:
The perimeter of the square loop is:
\[ P_{\text{square}} = 4a \]

4. **Finding the Radius of the Circular Loop**:
When the square loop is transformed into a circular loop, the circumference of the circular loop is equal to the perimeter of the square loop:
\[ 2\pi r = 4a \]
Solving for \( r \):
\[ r = \frac{4a}{2\pi} = \frac{2a}{\pi} \]

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

6. **Magnetic Dipole Moment of the Circular Loop**:
Now, substituting the area of the circular loop into the magnetic dipole moment formula:
\[ m_{\text{circle}} = I \cdot A_{\text{circle}} = I \cdot \frac{4a^2}{\pi} \]
Since \( m = I \cdot a^2 \), we can express \( m_{\text{circle}} \) in terms of \( m \):
\[ m_{\text{circle}} = \frac{4}{\pi} \cdot m \]

### Final Result:
Thus, the magnitude of the magnetic dipole moment of the circular loop is:
\[ m_{\text{circle}} = \frac{4m}{\pi} \]