A small bar magnet experiences a torque of 0. | vedclass

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(D) Given: Torque $	au = 0.016 \ Nm$,angle $	heta = 30^{\circ}$,and magnetic field $B = 0.04 \ T$.The torque on a magnetic dipole is given by $	au

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$8$

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(D) Given: Torque $	au = 0.016 \ Nm$,angle $	heta = 30^{\circ}$,and magnetic field $B = 0.04 \ T$.The torque on a magnetic dipole is given by $	au = mB \sin 	heta$.Substituting the values: $0.016 = m 	imes 0.04 	imes \sin 30^{\circ}$.Since $\sin 30^{\circ} = 0.5$,we have $0.016 = m 	imes 0.04 	imes 0.5 = m 	imes 0.02$.Thus,the magnetic moment $m = \frac{0.016}{0.02} = 0.8 \ Am^2$.For a solenoid,the magnetic moment is $m = NIA$.Given: Area $A = 1 \ cm^2 = 10^{-4} \ m^2$ and number of turns $N = 1000$.Equating the magnetic moments: $0.8 = 1000 	imes I 	imes 10^{-4}$.$0.8 = 0.1 	imes I$.Therefore,the current $I = \frac{0.8}{0.1} = 8 \ A$.

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