A short bar magnet placed with its axis at 45 | vedclass

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(A) The torque $	au$ experienced by a magnetic dipole in a uniform magnetic field $B$ is given by the formula: $	au = MB \sin 	heta$. Given: Torque

Vedclass · Accepted Answer

$1.8 	imes 10^{-3} \,J \,T^{-1}$

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(A) The torque $	au$ experienced by a magnetic dipole in a uniform magnetic field $B$ is given by the formula: $	au = MB \sin 	heta$. Given: Torque $	au = 3.6 	imes 10^{-5} \,J$ Magnetic field $B = 28.3 	imes 10^{-3} \,T$ Angle $	heta = 45^{\circ}$ Rearranging the formula to solve for the magnetic moment $M$: $M = \frac{	au}{B \sin 	heta}$ Substituting the values: $M = \frac{3.6 	imes 10^{-5}}{28.3 	imes 10^{-3} 	imes \sin 45^{\circ}}$ Since $\sin 45^{\circ} = \frac{1}{\sqrt{2}} \approx 0.707$: $M = \frac{3.6 	imes 10^{-5}}{28.3 	imes 10^{-3} 	imes 0.707} \approx \frac{3.6 	imes 10^{-5}}{20.008 	imes 10^{-3}} \approx 0.1799 	imes 10^{-2} \approx 1.8 	imes 10^{-3} \,J \,T^{-1}$.

$A$ short bar magnet placed with its axis at $45^{\circ}$ with a uniform external magnetic field of $28.3 \times 10^{-3} \,T$ experiences a torque of magnitude equal to $3.6 \times 10^{-5} \,J$. The magnitude of the magnetic moment of the magnet is nearly:

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