A short bar magnet placed with its axis at 30 | vedclass

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(B) Magnetic field strength,$B = 0.25 \; T$.Torque on the bar magnet,$	au = 4.5 	imes 10^{-2} \; J$.Angle between the bar magnet and the external ma

Vedclass · Accepted Answer

$0.36$

Vedclass · Answer

(B) Magnetic field strength,$B = 0.25 \; T$.Torque on the bar magnet,$	au = 4.5 	imes 10^{-2} \; J$.Angle between the bar magnet and the external magnetic field,$	heta = 30^{\circ}$.Torque is related to the magnetic moment $(M)$ as:$	au = M B \sin 	heta$Rearranging for $M$:$M = \frac{	au}{B \sin 	heta}$Substituting the values:$M = \frac{4.5 	imes 10^{-2}}{0.25 	imes \sin 30^{\circ}}$Since $\sin 30^{\circ} = 0.5$:$M = \frac{4.5 	imes 10^{-2}}{0.25 	imes 0.5} = \frac{4.5 	imes 10^{-2}}{0.125} = 0.36 \; J \; T^{-1}$.Hence,the magnetic moment of the magnet is $0.36 \; J \; T^{-1}$.

$A$ 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 magnetic moment (in $J \; T^{-1}$) of the magnet?

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