Question

A straight wire carrying a current (I) is turned into a circular loop. If the magnitude of the magnetic moment associated with it is 'M', then the length of the wire will be

• A. $\frac{M\pi}{4I}$
• B. $[\frac{4\pi I}{M}]^{\frac{1}{2}}$
• C. $[\frac{4M\pi}{I}]^{\frac{1}{2}}$
• D. $4\pi MI$

Hint:

Physics Tip: If a wire of length $L$ is bent into a polygon of $n$ sides, its area (and thus its magnetic moment) increases as $n$ increases, reaching a maximum when it is bent into a circle.

Answer 1

The Correct Option is C

Concept: Physics (Magnetic Effects of Electric Current) - Magnetic Moment of a Current Loop.

Step 1: State the formula for magnetic moment. The magnetic moment $M$ of a current-carrying loop is the product of the current $I$ and the area $A$ enclosed by the loop. $$ M = I \cdot A = I(\pi R^2) \text{ } $$

Step 2: Relate the wire length to the loop radius. The length $L$ of the wire is equal to the circumference of the circular loop. $$ L = 2\pi R \implies R = \frac{L}{2\pi} \text{ } $$

Step 3: Substitute $R$ into the magnetic moment equation. $$ M = I\pi \left( \frac{L}{2\pi} \right)^2 \text{ } $$ $$ M = \frac{I \cdot L^2}{4\pi} \text{ } $$

Step 4: Solve for $L$. $$ L^2 = \frac{4M\pi}{I} \implies L = \sqrt{\frac{4M\pi}{I}} \text{ } $$ $$ \therefore \text{The length of the wire is } \left[ \frac{4M\pi}{I} \right]^{\frac{1}{2}}. \text{ } $$