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:
The magnetic moment ($M$) of a current-carrying loop is defined as the product of the current and the area enclosed by the loop.
Key Formula or Approach:
1. Circumference \( L = 2\pi r \).
2. Magnetic Moment \( M = I \cdot A = I \cdot \pi r^2 \).

Step 2: Detailed Explanation:
Given:
Length \( L = 6 \text{ m} \)
Current \( I = 2 \text{ A} \)

Step 1: Find the radius $r$:
\[ 2\pi r = 6 \implies r = \frac{6}{2\pi} = \frac{3}{\pi} \text{ m} \]
Calculate the area $A$:
\[ A = \pi r^2 = \pi \left( \frac{3}{\pi} \right)^2 = \pi \cdot \frac{9}{\pi^2} = \frac{9}{\pi} \text{ m}^2 \]

Step 2: Calculate the magnetic moment $M$:
\[ M = I \cdot A = 2 \times \frac{9}{\pi} = \frac{18}{\pi} \text{ Am}^2 \]

Step 3: Final Answer:
The magnetic moment is $\frac{18}{\pi} \text{ Am}^2$.