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NEET · Physics · STD 12 - 5. Magnetism and matter

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

  1. A \(\frac{{\sqrt 3 W}}{2}\)
  2. B \(\;\frac{{2W}}{{\sqrt 3 }}\)
  3. C \(\;\frac{W}{{\sqrt 3 }}\)
  4. D \(\;\sqrt 3 W\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(\;\sqrt 3 W\)

Step-by-step Solution

Detailed explanation

At equilibrium, potential energy of dipole \(U_{i}=-M B_{H}\)
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