Question

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 will be

• A. $\frac{IL}{4\pi^2}$
• B. $\frac{4\pi}{L^2}$
• C. $\frac{4\pi I}{L^2}$
• D. $\frac{I L^2}{4\pi}$

Hint:

Dimensional analysis provides a quick verification shortcut here. The units of magnetic dipole moment are $\text{Ampere} \cdot \text{meter}^2$ ($\text{A}\cdot\text{m}^2$). Therefore, the variable terms in your final formula must have the form $I \times (\text{Length})^2$, which immediately eliminates options (A), (B), and (C).

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
A straight conducting wire of total length $L$ carries a steady electric current $I$.
This entire wire is reshaped by bending it into a single, closed circular loop. We need to determine the resulting magnetic dipole moment ($M$) of this current loop.

Step 2: Key Formula or Approach:
1. The magnetic dipole moment of a single-turn current-carrying loop enclosing an area $A$ is given by:
$$M = I \cdot A$$ 2. For a circular loop of radius $r$, its enclosed surface area is $A = \pi r^2$.
3. The total perimeter circumference of this circular ring is formed by the original length of the wire: $L = 2\pi r$.

Step 3: Detailed Explanation:
First, express the radius $r$ of the circular loop in terms of the given wire length $L$:
$$L = 2\pi r \implies r = \frac{L}{2\pi}$$ Next, substitute this expression for $r$ into the area equation for a circle to find $A$ in terms of $L$:
$$A = \pi r^2 = \pi \left(\frac{L}{2\pi}\right)^2 = \pi \left(\frac{L^2}{4\pi^2}\right)$$ Canceling out one factor of $\pi$ from the numerator and denominator simplifies the area expression to:
$$A = \frac{L^2}{4\pi}$$ Now, substitute this area formula back into the definition for the magnetic dipole moment ($M = IA$):
$$M = I \left(\frac{L^2}{4\pi}\right) = \frac{I L^2}{4\pi}$$

Step 4: Final Answer:
The magnetic moment of the loop is $\frac{I L^2}{4\pi}$, which corresponds exactly to option (D).