A bar magnet of length 10 , cm and having a p | vedclass

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Question

(A) The magnetic moment $M$ of the bar magnet is given by the product of pole strength $m$ and length $l$: $M = m 	imes l$.Given $m = 10^{-3} \, Wb$

Vedclass · Accepted Answer

$2\pi 	imes 10^{-7} \, N \cdot m$

Vedclass · Answer

(A) The magnetic moment $M$ of the bar magnet is given by the product of pole strength $m$ and length $l$: $M = m 	imes l$.Given $m = 10^{-3} \, Wb$ and $l = 10 \, cm = 0.1 \, m$,we have $M = 10^{-3} 	imes 0.1 = 10^{-4} \, A \cdot m^2$.The torque $	au$ acting on a magnetic dipole in a magnetic field $B$ is given by $	au = MB \sin 	heta$.Here,$B = 4\pi 	imes 10^{-3} \, T$ and $	heta = 30^{\circ}$.Substituting the values: $	au = (10^{-4}) 	imes (4\pi 	imes 10^{-3}) 	imes \sin(30^{\circ})$.$	au = 4\pi 	imes 10^{-7} 	imes 0.5$.$	au = 2\pi 	imes 10^{-7} \, N \cdot m$.

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