Question

A wire of length $6 \text{ m}$ is bent to form a circular loop of single turn. If the current through the loop is $2 \text{ A}$, the magnetic moment of the circular loop (in $\text{Am}^2$ ) is

• A. $18\pi$
• B. $\frac{18}{\pi}$
• C. $\frac{36}{\pi}$
• D. $36\pi$

Hint:

For a fixed length of wire, the magnetic moment is maximum when the wire is bent into a circle because a circle encloses the maximum area for a given perimeter.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
A current-carrying loop acts as a magnetic dipole. The magnetic moment depends on the current and the area enclosed by the loop. Since a straight wire is bent into a circle, the length of the wire becomes the circumference of the circle.
Step 2: Key Formula or Approach:
Circumference: \(L = 2\pi r \implies r = \frac{L}{2\pi}\).
Area of circle: \(A = \pi r^2\).
Magnetic moment: \(M = IA\) (for a single turn, \(N=1\)).
Step 3: Detailed Explanation:
Find the radius of the loop:
\[ L = 6 \text{ m} \] \[ 2\pi r = 6 \implies r = \frac{6}{2\pi} = \frac{3}{\pi} \text{ m} \] Calculate the area of the circular loop:
\[ A = \pi r^2 = \pi \left(\frac{3}{\pi}\right)^2 = \pi \left(\frac{9}{\pi^2}\right) = \frac{9}{\pi} \text{ m}^2 \] Calculate the magnetic moment:
Given current \(I = 2 \text{ A}\).
\[ M = I \times A = 2 \times \left(\frac{9}{\pi}\right) = \frac{18}{\pi} \text{ Am}^2 \] Step 4: Final Answer:
The magnetic moment is \(\frac{18}{\pi}\).