Question

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° is W. Now the torque required to keep the magnet in this new position is
(1) W/√3
(2) √3 W
(3) √3 W/2
(4) 2W/√3

Answer 1

Given:

The energy required to rotate it by 60° is W.

To find:

Now the torque required to keep the magnet in this new position is

Solution:

From given, we have,

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° is W.

The torque acting on a bar magnet kept in a magnetic field,

= m B sin ∅

The work done in the rotation from 0 ° to 60 °,

= m B cos 60°

= √3/2 m B

= √3 W

Option (2) √3 W is correct