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A magnetic needle suspended parallel to a magnetic field requires J of work to turn it through 60°. The torque needed to maintain the needle in this position will be :
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- 2√3J
- 3J
- √3J
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3 J 2
Correct Option: B
According to work energy theorem W =Ufinal – Uinitial = MB (cos 0 – cos 60°)
| W = | = √3J ...(i) | |
| 2 |
| τ = M→ × B→ = MBsin60° = | = √3J ...(ii) | |
| 2 |
From equation (i) and (ii)
| τ = | = 3J | |
| 2 |
