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NEET PHYSICSMedium

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

A

232\sqrt{3} J

B

3 J

C

3\sqrt{3} J

D

32\frac{3}{2} J

Step-by-Step Solution

  1. Work Done Formula: The work done (WW) in rotating a magnetic dipole (needle) with magnetic moment mm in a uniform magnetic field BB from an initial orientation θ1\theta_1 to a final orientation θ2\theta_2 is given by: W=mB(cos⁡θ1−cos⁡θ2)W = mB(\cos \theta_1 - \cos \theta_2) [Source 211, Eq 5.3 derived]
  2. Calculate mBmB:
  • Initial angle θ1=0∘\theta_1 = 0^\circ (parallel to field).
  • Final angle θ2=60∘\theta_2 = 60^\circ.
  • Given Work W=3W = \sqrt{3} J. 3=mB(cos⁡0∘−cos⁡60∘)\sqrt{3} = mB(\cos 0^\circ - \cos 60^\circ) 3=mB(1−0.5)=mB2\sqrt{3} = mB(1 - 0.5) = \frac{mB}{2} mB=23mB = 2\sqrt{3}
  1. Calculate Torque: The torque (τ\tau) required to maintain the needle at an angle θ\theta is given by: τ=mBsin⁡θ\tau = mB \sin \theta [Source 211, Eq 5.2]
  • At θ=60∘\theta = 60^\circ: τ=(23)sin⁡60∘\tau = (2\sqrt{3}) \sin 60^\circ τ=23×32=3\tau = 2\sqrt{3} \times \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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