5.3. A short bar magnet placed with its axis at 30° with a uniform external magnetic field of 0.25 T experiences a torque of magnitude equal to 4.5 \times 10 ^{-2} J . What is the magnitude of magnetic moment of the magnet?

Answers (1)


The angle between axis of bar magnet and external magnetic field, θ = 30°

Magnetic field strength, B = 0.25 T

Torque on the bar magnet, Τ = 4.5 x 10^{-2}J

We know,

Torque experienced by a bar magnet placed in a uniform magnetic field is:

T = m x B = mBsin\theta

m = \frac{T}{Bsin\theta}

\implies m = \frac{4.5\times10^{-2}J}{0.25T\times sin30^{\circ}}

\therefore  m = 0.36 JT^{-1}

Hence, the magnitude of the moment of the Bar magnet is 0.36 JT^{-1}.