A circular loop and a square loop are formed from the same wire and the same current is passed through them . Find the ratio of their dipole moments.

Options

A. (a)`4pi`
B. (b)`4/(pi)`
C. (c)`2/(pi)`
D. (d)`2pi`

Explanation

To solve the problem of finding the ratio of the dipole moments of a circular loop and a square loop formed from the same wire carrying the same current, we can follow these steps:

### Step 1: Define the Length of the Wire
Let the total length of the wire be \( L \).

### Step 2: Calculate the Radius of the Circular Loop
For a circular loop, the circumference is given by:
\[ L = 2\pi r \]
From this, we can express the radius \( r \) of the circular loop as:
\[ r = \frac{L}{2\pi} \]

### Step 3: Calculate the Area of the Circular Loop
The area \( A_1 \) of the circular loop is given by:
\[ A_1 = \pi r^2 \]
Substituting the value of \( r \):
\[ A_1 = \pi \left(\frac{L}{2\pi}\right)^2 = \pi \cdot \frac{L^2}{4\pi^2} = \frac{L^2}{4\pi} \]

### Step 4: Calculate the Dipole Moment of the Circular Loop
The magnetic dipole moment \( m_1 \) of the circular loop is given by:
\[ m_1 = I \cdot A_1 \]
Substituting the area:
\[ m_1 = I \cdot \frac{L^2}{4\pi} = \frac{IL^2}{4\pi} \]

### Step 5: Calculate the Side Length of the Square Loop
For a square loop, the total length of the wire is equal to the perimeter:
\[ L = 4a \]
Thus, the side length \( a \) of the square loop is:
\[ a = \frac{L}{4} \]

### Step 6: Calculate the Area of the Square Loop
The area \( A_2 \) of the square loop is given by:
\[ A_2 = a^2 = \left(\frac{L}{4}\right)^2 = \frac{L^2}{16} \]

### Step 7: Calculate the Dipole Moment of the Square Loop
The magnetic dipole moment \( m_2 \) of the square loop is given by:
\[ m_2 = I \cdot A_2 \]
Substituting the area:
\[ m_2 = I \cdot \frac{L^2}{16} = \frac{IL^2}{16} \]

### Step 8: Find the Ratio of the Dipole Moments
Now, we can find the ratio of the dipole moments \( \frac{m_1}{m_2} \):
\[ \frac{m_1}{m_2} = \frac{\frac{IL^2}{4\pi}}{\frac{IL^2}{16}} = \frac{16}{4\pi} = \frac{4}{\pi} \]

### Conclusion
The ratio of the dipole moments of the circular loop to the square loop is:
\[ \frac{m_1}{m_2} = \frac{4}{\pi} \]