A bar magnet,when placed at an angle of 30^o | vedclass

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Question

(A) The torque $	au$ experienced by a magnetic dipole in a magnetic field is given by $	au = MB \sin 	heta$,where $M$ is the magnetic dipole moment

Vedclass · Accepted Answer

$2 	imes 10^{-2} \,A-m$

Vedclass · Answer

(A) The torque $	au$ experienced by a magnetic dipole in a magnetic field is given by $	au = MB \sin 	heta$,where $M$ is the magnetic dipole moment,$B$ is the magnetic field,and $	heta$ is the angle.Since $M = m 	imes L$ (where $m$ is the pole strength and $L$ is the length of the magnet),the formula becomes $	au = (m 	imes L) B \sin 	heta$.Given: $	au = 25 	imes 10^{-6} \,N-m$,$B = 5 	imes 10^{-2} \,T$,$L = 5 \,cm = 5 	imes 10^{-2} \,m$,and $	heta = 30^o$.Substituting the values: $25 	imes 10^{-6} = (m 	imes 5 	imes 10^{-2}) 	imes (5 	imes 10^{-2}) 	imes \sin 30^o$.Since $\sin 30^o = 0.5$,we have $25 	imes 10^{-6} = m 	imes 25 	imes 10^{-4} 	imes 0.5$.$25 	imes 10^{-6} = m 	imes 12.5 	imes 10^{-4}$.$m = \frac{25 	imes 10^{-6}}{12.5 	imes 10^{-4}} = 2 	imes 10^{-2} \,A-m$.

$A$ bar magnet,when placed at an angle of $30^o$ to the direction of a magnetic field of induction $5 \times 10^{-2} \,T$,experiences a torque of $25 \times 10^{-6} \,N-m$. If the length of the magnet is $5 \,cm$,what is its pole strength?

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