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

• A. $2 \sqrt3\, J$
• B. $\sqrt3\, J$
• C. $3\, J$
• D. $\frac{3}{2}\, J$

Answer 1

The Correct Option is C

To solve the problem of determining the torque needed to maintain a magnetic needle in a position turned by 60^\circ, we need to use the following concepts and steps:

1. Understanding Work and Torque in Magnetism:
   • The work done to turn a magnetic needle in a magnetic field is related to the angle through which it is turned.
   • When a magnetic needle is turned in a uniform magnetic field by an angle \theta, the work done (W) is given by:
   • W = mB(1 - \cos \theta)
     where m is the magnetic moment, B is the magnetic field strength, and \theta is the angle in radians.
2. Given Data and Relevant Formulae:
   • The work done is W = \sqrt{3}\, \text{J} to turn the needle by 60^\circ (or \frac{\pi}{3} radians).
   • Using the formula: \sqrt{3} = mB(1 - \cos \frac{\pi}{3})
   • Since \cos 60^\circ = \frac{1}{2}, we have:
   • \sqrt{3} = mB(1 - \frac{1}{2}) = mB \times \frac{1}{2}
   • From this, we find mB = 2\sqrt{3}\, \text{J}
3. Calculating Required Torque:
   • To find the torque \tau needed to maintain the needle at this angle, use:
   • \tau = mB \sin \theta
   • Here, \sin 60^\circ = \frac{\sqrt{3}}{2}, so:
   • \tau = 2\sqrt{3} \times \frac{\sqrt{3}}{2} = 3\, \text{J}
4. Conclusion:
   • The torque required to maintain the magnetic needle in this position is 3 \, J.
   • Thus, the correct answer is 3 J.