Back to Directory
NEET PHYSICSMedium

A bar magnet is hung by a thin cotton thread in a uniform horizontal magnetic field and is in the equilibrium state. The energy required to rotate it by 60∘60^\circ is WW. Now the torque required to keep the magnet in this new position is:

A

W3\frac{W}{\sqrt{3}}

B

3W\sqrt{3}W

C

3W2\frac{\sqrt{3}W}{2}

D

2W3\frac{2W}{\sqrt{3}}

Step-by-Step Solution

  1. Work Done (Energy): The work done (WW) in rotating a magnetic dipole with magnetic moment mm in a uniform magnetic field BB from an initial angle θ1\theta_1 to a final angle θ2\theta_2 is given by: W=mB(cos⁡θ1−cos⁡θ2)W = mB(\cos \theta_1 - \cos \theta_2) Given that the magnet starts from equilibrium, θ1=0∘\theta_1 = 0^\circ. It is rotated by 60∘60^\circ, so θ2=60∘\theta_2 = 60^\circ. W=mB(cos⁡0∘−cos⁡60∘)=mB(1−0.5)=mB2W = mB(\cos 0^\circ - \cos 60^\circ) = mB(1 - 0.5) = \frac{mB}{2} From this, we find: mB=2WmB = 2W.
  2. Torque: The torque (τ\tau) required to keep the magnet at an angle θ\theta is given by: τ=mBsin⁡θ\tau = mB \sin \theta At θ=60∘\theta = 60^\circ: τ=mBsin⁡60∘\tau = mB \sin 60^\circ
  3. Substitution: Substitute the value of mBmB derived above: τ=(2W)(32)=3W\tau = (2W) \left( \frac{\sqrt{3}}{2} \right) = \sqrt{3}W
Practice Mode Available

Master this Topic on Sushrut

Join thousands of students and practice with AI-generated mock tests.

Get Started