A compass needle whose magnetic moment is 60 | vedclass

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Question

(A) The torque $	au$ experienced by a magnetic needle in a magnetic field $B_H$ is given by $	au = M B_H \sin 	heta$,where $M$ is the magnetic mome

Vedclass · Accepted Answer

$30$

Vedclass · Answer

(A) The torque $	au$ experienced by a magnetic needle in a magnetic field $B_H$ is given by $	au = M B_H \sin 	heta$,where $M$ is the magnetic moment,$B_H$ is the horizontal component of the Earth's magnetic field,and $	heta$ is the angle between the magnetic axis and the magnetic meridian (the angle of declination).Given:$M = 60 \, A \cdot m^2$$B_H = 40 \, \mu Wb/m^2 = 40 	imes 10^{-6} \, T$$	au = 1.2 	imes 10^{-3} \, N \cdot m$Substituting these values into the formula:$1.2 	imes 10^{-3} = 60 	imes (40 	imes 10^{-6}) 	imes \sin 	heta$$1.2 	imes 10^{-3} = 2400 	imes 10^{-6} 	imes \sin 	heta$$1.2 	imes 10^{-3} = 2.4 	imes 10^{-3} 	imes \sin 	heta$$\sin 	heta = \frac{1.2 	imes 10^{-3}}{2.4 	imes 10^{-3}} = 0.5$$	heta = \arcsin(0.5) = 30^o$Therefore,the angle of declination is $30^o$.

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