A bar magnet of length L and magnetic dipole moment M is bent in the form of an arc as shown in figure. The ne

Question

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:

Accepted Answer

1. **Original Magnetic Moment:** The magnetic dipole moment ($M$) of a bar magnet is the product of its pole strength ($m$) and its length ($L$). 
   $$ M = m 	imes L $$ 
2. **Bent Magnet Geometry:** When the magnet is bent into an arc, the length of the arc remains $L$. Let the radius of the arc be $R$ and the angle subtended at the center be $	heta$ (in radians).
   $$ L = R	heta \implies R = \frac{L}{	heta} $$
3. **New Magnetic Moment ($M'$):** The new magnetic moment is the product of the pole strength ($m$) and the straight-line distance (chord) between the two poles. The chord length is given by $2R \sin(	heta/2)$.
   $$ M' = m 	imes (2R \sin(	heta/2)) $$
4. **Substitution:** Substituting $R = L/	heta$:
   $$ M' = m 	imes 2 \left( \frac{L}{	heta} ight) \sin(	heta/2) = \frac{2(mL)}{	heta} \sin(	heta/2) = \frac{2M}{	heta} \sin(	heta/2) $$
5. **Determining the Angle:** Based on the probable answer $3M/\pi$, we test the angle $	heta = 60^\circ = \pi/3$ radians.
   $$ M' = \frac{2M}{\pi/3} \sin(60^\circ/2) = \frac{6M}{\pi} \sin(30^\circ) $$
   $$ M' = \frac{6M}{\pi} 	imes \frac{1}{2} = \frac{3M}{\pi} $$
   This confirms the magnet is bent into an arc subtending $60^\circ$ at the center.

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