First slide

Magnetic dipole in a uniform magnetic field

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 $60^{0}$ is W. Now the torque required to keep the magnet in this new position is

Easy

A

$\frac{W}{\sqrt{3}}$
B

$\sqrt{3} W$
C

$\frac{\sqrt{3} W}{2}$
D

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

Solution

At equilibrium, potential energy of dipole, $U_{i} = M B_{H}$ Final potential energy of dipole,
$U_{f} = - M B_{H} \cos 60 ° = - \frac{M B_{H}}{2}$
$W = U_{f} - U_{i} = - \frac{M B_{H}}{2} - (- M B_{H}) = \frac{M B_{H}}{2}$ ………..(2)
$Required torque, τ = M B_{H} \sin 60 °$
$τ = 2 W \times \frac{\sqrt{3}}{2} Using eqn. (2)$
$= \sqrt{3} W$

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