A short bar magnet placed with its axis at $30^{\circ}$ 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 the magnetic moment of the magnet?

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Solution

Magnetic field strength, $B = 0.25 T$
Torque on the bar magnet, $τ = 4.5 \times 10^{- 2} J$
The angle between the bar magnet and the external magnetic field, $θ = 30^{\circ}$
Torque is related to magnetic moment $(M)$ as: $τ = M B sin θ$

$∴ M = \frac{τ}{B sin θ} = \frac{4.5 \times 10^{- 2}}{0.25 \times sin 30^{\circ}} = 0.36 J / T$

Hence, the magnetic moment of the magnet is $0.36 J / T$

Final Answer: $M = 0.36 J / T$

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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 4.5×10−2J . What is the magnitude of the magnetic moment of the magnet?

A short bar magnet placed with its axis at $30^{\circ}$ 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 the magnetic moment of the magnet?