Question

A wire of length \( L \) meter carrying current \( i \) ampere is bent in the form of a circle. The magnitude of its magnetic moment in SI units is:

• A. \( \frac{iL^2}{4\pi} \)
• B. \( \frac{iL^2}{\pi} \)
• C. \( \frac{iL^2}{2\pi} \)
• D. \( \frac{iL^2}{m} \)

Hint:

When solving problems involving the magnetic moment of a circular loop, remember that the magnetic moment depends on both the current and the area of the loop, and that the area of a circle is \( \pi r^2 \), where the radius is related to the length of the wire.

Answer 1

The Correct Option is A

The magnetic moment \( M \) of a current-carrying loop is given by the formula: \[ M = i \cdot A \] Where:
- \( i \) is the current
- \( A \) is the area of the loop
For a wire bent into a circle, the area \( A \) of the circle is given by: \[ A = \pi r^2 \] Where \( r \) is the radius of the circle. Since the total length of the wire is \( L \), and the wire is bent in the shape of a circle, the circumference of the circle is equal to the length of the wire: \[ 2\pi r = L \] Thus, the radius \( r \) is: \[ r = \frac{L}{2\pi} \] Now, substitute this value of \( r \) into the area formula: \[ A = \pi \left( \frac{L}{2\pi} \right)^2 = \frac{L^2}{4\pi} \] Now, the magnetic moment is: \[ M = i \cdot A = i \cdot \frac{L^2}{4\pi} \] Thus, the magnetic moment is \( \frac{iL^2}{4\pi} \), which corresponds to option (A).