A bar magnet of magnetic moment 2 text{ A m}^ | vedclass

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Question

(C) The work done $W$ by an external torque to rotate a magnetic dipole in a uniform magnetic field is given by the change in potential energy: $W = U

Vedclass · Accepted Answer

$0.6$

Vedclass · Answer

(C) The work done $W$ by an external torque to rotate a magnetic dipole in a uniform magnetic field is given by the change in potential energy: $W = U_f - U_i = -MB \cos 	heta_f - (-MB \cos 	heta_i) = MB(\cos 	heta_i - \cos 	heta_f)$.Given: Magnetic moment $M = 2 	ext{ A m}^2$, Magnetic field $B = 0.3 	ext{ T}$.Initially, the magnet is aligned with the field, so $	heta_i = 0^\circ$.Finally, the magnet is normal to the field, so $	heta_f = 90^\circ$.Substituting the values:$W = MB(\cos 0^\circ - \cos 90^\circ)$$W = 2 	imes 0.3 	imes (1 - 0)$$W = 0.6 	imes 1 = 0.6 	ext{ J}$.

$A$ bar magnet of magnetic moment $2 \text{ A m}^2$ lies aligned with the direction of a uniform magnetic field of $0.3 \text{ T}$. The amount of work required by an external torque to turn the magnet so as to align its magnetic moment normal to the field direction is (in $\text{ J}$)

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