A magnetic dipole of moment 2.5 text{ A m}^2 | vedclass

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(D) The potential energy of a magnetic dipole in a magnetic field is given by $U = -MB \cos 	heta$. Initially, the dipole is in the East-West directi

Vedclass · Accepted Answer

$75$

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(D) The potential energy of a magnetic dipole in a magnetic field is given by $U = -MB \cos 	heta$. Initially, the dipole is in the East-West direction $(	heta_1 = 90^{\circ})$, so $U_i = -MB \cos 90^{\circ} = 0$. Finally, the dipole is in the North-South position $(	heta_2 = 0^{\circ})$, so $U_f = -MB \cos 0^{\circ} = -MB$. By the law of conservation of energy, the loss in potential energy is equal to the gain in kinetic energy: $KE = U_i - U_f = 0 - (-MB) = MB$. Given $M = 2.5 	ext{ A m}^2$ and $B_H = 3 	imes 10^{-5} 	ext{ T}$. $KE = 2.5 	imes 3 	imes 10^{-5} = 7.5 	imes 10^{-5} 	ext{ J}$. Converting to microjoules: $7.5 	imes 10^{-5} 	ext{ J} = 75 	imes 10^{-6} 	ext{ J} = 75 \mu	ext{J}$.

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