Ncert Exercises & Solutions

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 × 10–2 J. What is the magnitude of the magnetic moment of the magnet?

Magnetic field strength, B=0.25T

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 the magnetic moment (M) as:

T = M B \sin θ

\begin{aligned} ∴ M & = \frac{T}{B \sin θ} \\ = \frac{4.5 \times 10^{- 2}}{0.25 \times \sin 30^{\circ}} = 0.36 {JT}^{- 1} \end{aligned}

Hence, the magnetic moment of the magnet is

0.36 J T^{- 1} .

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