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A bar magnet of length 'l' and magnetic dipole moment 'M' is bent in the form of an arc as shown in figure. The new magnetic dipole moment will be
A
$\frac{M}{2}$
B
M
C
$\frac{3}{\pi}M$
D
$\frac{2}{\pi}M$
Detailed Solution
Let the pole strength be m; then $M = ml$.
After bending into an arc subtending $60^\circ$: $r\left(\frac{\pi}{3}\right) = l \Rightarrow r = \frac{3l}{\pi}$
Distance between the poles $= 2r\sin\frac{60^\circ}{2} = r$
$M' = m\times2\left(\frac{3l}{\pi}\right)\left(\frac{1}{2}\right) = \frac{3ml}{\pi} = \frac{3M}{\pi}$
After bending into an arc subtending $60^\circ$: $r\left(\frac{\pi}{3}\right) = l \Rightarrow r = \frac{3l}{\pi}$
Distance between the poles $= 2r\sin\frac{60^\circ}{2} = r$
$M' = m\times2\left(\frac{3l}{\pi}\right)\left(\frac{1}{2}\right) = \frac{3ml}{\pi} = \frac{3M}{\pi}$