Question

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 x 10⁻² J. What is the magnitude of magnetic moment of the magnet?

Answer 1

1. Magnetic field properties  =>  1), field lines do not form closed loops  2), for an uniform magnetic field lines it moves  In parallel   Now coming to the Question   Please refer the attachment below  

Answer 2

★ Given:-

Magnetic field strength, B = 0.25 T

Torque on the bar magnet, $T = 4.5\times 10^{-2}\; J$

Angle between the bar magnet and the external magnetic field, θ = 30°

Torque is related to magnetic moment (M) as:

$T = MBsin\theta$

∴$\: M = \frac{T}{Bsin\theta}$

$=\frac{4.5 \times {10}^{ - 2} }{0.25 \times sin30\degree}$

$= 0.36 J {T}^{ - 1}$

Hence, the magnetic moment of the magnet is $= 0.36 J {T}^{ - 1}$