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

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 (A) (W/√3) (B) √3 W (C) (√3 W/2) (D) (2W/√3)

Tardigrade Admin · Accepted Answer

W = MB ( cos 0° - cos60°) W = MB (1- (1/2)) = (MB/2) Required torque for this position τ = MB sinθ = MB sin60° = (√3/2) MB = √3 W

Q. 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 $6 0^{\circ}$ is $W$ . Now the torque required to keep the magnet in this new position is

$\frac{W}{3}$

$3 W$

$\frac{3 W}{2}$

$\frac{2 W}{3}$

$W = MB (cos 0^{\circ} - cos 6 0^{\circ})$
$W = MB (1 - \frac{1}{2}) = \frac{MB}{2}$
Required torque for this position
$τ = MB sin θ$
$= MB sin 6 0^{\circ}$
$= \frac{3}{2} MB = 3 W$

Q. 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

3​W​

3​W

23​W​

3​2W​