A current I is flowing in a conductor of length L when it is bent in the form of a circular loop its magnetic moment

Options

A. `(IL)/(4pi)`
B. `(4pi)/(IL^(2))`
C. `4piIL^(2)`
D. `(IL^(2))/(4pi)`

Explanation

To find the magnetic moment of a circular loop formed by bending a conductor of length \( L \) through which a current \( I \) is flowing, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Problem**:
We have a straight conductor of length \( L \) carrying a current \( I \). When this conductor is bent into the shape of a circular loop, we need to find its magnetic moment.

2. **Perimeter of the Circular Loop**:
When the conductor is bent into a circular shape, the length of the conductor becomes the circumference of the circle. Therefore, we have:
\[ L = 2\pi r \]
where \( r \) is the radius of the circular loop.

3. **Finding the Radius**:
We can rearrange the equation to solve for \( r \):
\[ r = \frac{L}{2\pi} \]

4. **Area of the Circular Loop**:
The area \( A \) of the circular loop can be calculated using the formula for the area of a circle:
\[ A = \pi r^2 \]
Substituting the expression for \( r \):
\[ A = \pi \left(\frac{L}{2\pi}\right)^2 = \pi \cdot \frac{L^2}{4\pi^2} = \frac{L^2}{4\pi} \]

5. **Magnetic Moment Formula**:
The magnetic moment \( M \) of a loop is given by the formula:
\[ M = N \cdot I \cdot A \]
where \( N \) is the number of turns. Since we have only one loop, \( N = 1 \):
\[ M = I \cdot A = I \cdot \frac{L^2}{4\pi} \]

6. **Final Expression for Magnetic Moment**:
Therefore, the magnetic moment of the circular loop is:
\[ M = \frac{I L^2}{4\pi} \]

7. **Conclusion**:
The correct answer is:
\[ \text{Magnetic Moment } M = \frac{I L^2}{4\pi} \]