A magnetic needle lying parallel to a magnetic field requires W units of work to turn it through 60°. The torque needed to maintain the needle in its position, will be:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → $\sqrt{3} W$
Total work done (W) to rotate the needle from 0° → 60°
[θ → angle between needle and magnetic field]
[B → Magnetic field]
[m → magnetic moment]
$W=\int\limits_{0°}^{60°}mB\sin θdθ$
$=mB\left[-\cos θ\right]_{0°}^{60°}$
$=mB\left(-\frac{1}{2}+1\right)=\frac{mB}{2}$
and,
To maintain the needle in a fixed position at any angle θ.
$z=mB\sin θ=mB\sin 60°$
$=mB=\frac{\sqrt{3}}{2}$
$=2W\frac{\sqrt{3}}{2}=W\sqrt{3}$