Question

A steel wire with magnetic moment \( M \) and length \( l \) is bent into semicircle of radius \( R \). Find new magnetic moment.

• A. \( M \)
• B. \( \frac{2RM}{\pi l} \)
• C. \( \frac{2M}{\pi} \)
• D. \( \frac{2\pi RM}{l} \)

Hint:

When bent into a loop, use area of semicircle to recalculate magnetic moment: \( M = I \cdot A \)

Answer 1

The Correct Option is C

Original magnetic moment: \( M = I \cdot A \)
Straight wire: no area ⇒ magnetic moment zero. After bending into semicircle: \[ l = \pi R \Rightarrow R = \frac{l}{\pi},\quad A = \frac{1}{2} \pi R^2 = \frac{1}{2} \pi \left(\frac{l}{\pi}\right)^2 = \frac{l^2}{2\pi} \] Let original \( M = I \cdot l \), new \( M' = I \cdot A = I \cdot \frac{l^2}{2\pi^2} \) Taking ratio, we get \( M' = \frac{2M}{\pi} \)