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 step by step, we will analyze the magnetic dipole moment of both the square loop and the circular loop.

### Step 1: Understanding the Magnetic Dipole Moment of the Square Loop
The magnetic dipole moment \( m \) of a square loop carrying a current \( I \) can be expressed using the formula:
\[ m = n \cdot I \cdot A \]
where:
- \( n \) is the number of turns (for a single loop, \( n = 1 \)),
- \( I \) is the current,
- \( A \) is the area of the loop.

For a square loop with side length \( A \):
\[ A = A^2 \]
Thus, the magnetic dipole moment becomes:
\[ m = I \cdot A^2 \]
Let's denote this as Equation (1).

### Step 2: Finding the Magnetic Dipole Moment of the Circular Loop
When the square loop is transformed into a circular loop with the same perimeter, we need to find the radius \( r \) of the circular loop. The perimeter of the square loop is:
\[ \text{Perimeter of square} = 4A \]
The circumference of the circular loop is:
\[ \text{Circumference of circle} = 2\pi r \]
Setting these equal gives:
\[ 4A = 2\pi r \]
From this, we can solve for \( r \):
\[ r = \frac{4A}{2\pi} = \frac{2A}{\pi} \]

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

### Step 4: Magnetic Dipole Moment of the Circular Loop
Now, substituting the area \( A' \) into the formula for the magnetic dipole moment of the circular loop \( m' \):
\[ m' = I \cdot A' = I \cdot \frac{4A^2}{\pi} \]

### Step 5: Relating \( m' \) to \( m \)
From Equation (1), we know that \( m = I \cdot A^2 \). Thus, we can express \( m' \) in terms of \( m \):
\[ m' = \frac{4}{\pi} m \]

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

### Final Answer
The magnitude of the magnetic dipole moment of the circular loop will be \( \frac{4m}{\pi} \).

---