A magnetic needle suspended parallel to a magnetic field requires $\sqrt{3}$ J of work to turn it through $60^

Question

A magnetic needle suspended parallel to a magnetic field requires $\sqrt{3}$ J of work to turn it through $60^\circ$. The torque needed to maintain the needle in this position will be:

Accepted Answer

1. **Work Done Formula:** The work done ($W$) in rotating a magnetic dipole (needle) with magnetic moment $m$ in a uniform magnetic field $B$ from an initial orientation $	heta_1$ to a final orientation $	heta_2$ is given by:
   $$ W = mB(\cos 	heta_1 - \cos 	heta_2) $$
   [Source 211, Eq 5.3 derived]
2. **Calculate $mB$:**
   - Initial angle $	heta_1 = 0^\circ$ (parallel to field).
   - Final angle $	heta_2 = 60^\circ$.
   - Given Work $W = \sqrt{3}$ J.
   $$ \sqrt{3} = mB(\cos 0^\circ - \cos 60^\circ) $$
   $$ \sqrt{3} = mB(1 - 0.5) = \frac{mB}{2} $$
   $$ mB = 2\sqrt{3} $$
3. **Calculate Torque:** The torque ($	au$) required to maintain the needle at an angle $	heta$ is given by:
   $$ 	au = mB \sin 	heta $$
   [Source 211, Eq 5.2]
   - At $	heta = 60^\circ$:
   $$ 	au = (2\sqrt{3}) \sin 60^\circ $$
   $$ 	au = 2\sqrt{3} 	imes \frac{\sqrt{3}}{2} = 3 $$
   - The magnitude of the torque is 3 J (Note: Torque units are typically Nm, but J is used here for magnitude consistency with work options).

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