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NEET PHYSICSMAGNETISM AND MATTERMedium

Question

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).

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from MAGNETISM AND MATTER. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

PHYSICSMAGNETISM AND MATTERmagneticneedlesuspendedparallelmagnetic

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