Question

A steel wire of length $l$ has a magnetic moment $M.$ It is then bent into a semi-circular arc. The new magnetic moment is

• A. M/l
• B. M
• C. 2$M/\pi$
• D. $M\times X$

Answer 1

Answer: (C)

2$M/\pi$

$\text{Solution}:$

Here, pole strength,

$m=\frac{M}{l}$

If r is radius of semicircle, then

$\pi r=l$

$\therefore r=\frac{1}{\pi }$

New magnetic moment

$M^{,}=m\times 2r=\frac{M}{l}\frac{2l}{\pi }=\frac{2M}{\pi }$

$\text{HenceC is the correct option}$