A small bar magnet placed with its axis at 30 | vedclass

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(D) The torque $	au$ on a bar magnet in an external magnetic field $B$ is given by $	au = MB \sin 	heta$.Given: $	heta = 30^{\circ}$,$	au = 0.018

Vedclass · Accepted Answer

$7.2 	imes 10^{-2} J$

Vedclass · Answer

(D) The torque $	au$ on a bar magnet in an external magnetic field $B$ is given by $	au = MB \sin 	heta$.Given: $	heta = 30^{\circ}$,$	au = 0.018\, Nm$,$B = 0.06\, T$.Substituting the values: $0.018 = M 	imes 0.06 	imes \sin 30^{\circ}$.Since $\sin 30^{\circ} = 0.5$,we have $0.018 = M 	imes 0.06 	imes 0.5 = M 	imes 0.03$.Thus,$M = \frac{0.018}{0.03} = 0.6\, A\cdot m^2$.The potential energy of a magnetic dipole is $U = -MB \cos 	heta$.Stable equilibrium is at $	heta = 0^{\circ}$,so $U_i = -MB \cos 0^{\circ} = -MB$.Unstable equilibrium is at $	heta = 180^{\circ}$,so $U_f = -MB \cos 180^{\circ} = MB$.The work done $W$ is the change in potential energy: $W = U_f - U_i = MB - (-MB) = 2MB$.$W = 2 	imes 0.6 	imes 0.06 = 0.072\, J = 7.2 	imes 10^{-2} J$.

$A$ small bar magnet placed with its axis at $30^{\circ}$ with an external field of $0.06\, T$ experiences a torque of $0.018\, Nm$. The minimum work required to rotate it from its stable to unstable equilibrium position is

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