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: Recall the magnetic moment.
For a single turn loop carrying current $I$ and enclosing area $A$, the magnetic moment is $M = IA$.
Step 2: Relate the wire length to the radius.
The whole wire of length $L$ becomes the circumference, so \[ L = 2\pi r \;\Rightarrow\; r = \frac{L}{2\pi}. \]
Step 3: Write the enclosed area.
\[ A = \pi r^2 = \pi\left(\frac{L}{2\pi}\right)^2 = \frac{\pi L^2}{4\pi^2}. \]
Step 4: Simplify the area.
One factor of $\pi$ cancels: \[ A = \frac{L^2}{4\pi}. \]
Step 5: Insert into the moment.
\[ M = I A = I\cdot\frac{L^2}{4\pi} = \frac{IL^2}{4\pi}. \]
Step 6: Conclude.
So the magnetic moment is $\dfrac{IL^2}{4\pi}$, option (D). \[ \boxed{M = \frac{IL^2}{4\pi}} \]