A circular coil having 200 turns,2.5 times 10 | vedclass

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(D) The work done $W$ in rotating a magnetic dipole in a magnetic field is given by the change in potential energy:$W = U_f - U_i$Potential energy $U

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$5$

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(D) The work done $W$ in rotating a magnetic dipole in a magnetic field is given by the change in potential energy:$W = U_f - U_i$Potential energy $U = -\vec{M} \cdot \vec{B} = -MB \cos 	heta$Initial orientation: $\vec{M}$ is along $\vec{B}$,so $	heta_i = 0^{\circ}$.$U_i = -MB \cos 0^{\circ} = -MB$Final orientation: $\vec{M}$ is perpendicular to $\vec{B}$,so $	heta_f = 90^{\circ}$.$U_f = -MB \cos 90^{\circ} = 0$Work done $W = 0 - (-MB) = MB$Given: $N = 200$,$I = 100 \mu	ext{A} = 100 	imes 10^{-6} 	ext{ A}$,$A = 2.5 	imes 10^{-4} 	ext{ m}^2$,$B = 1 	ext{ T}$.Magnetic moment $M = NIA = 200 	imes (100 	imes 10^{-6}) 	imes (2.5 	imes 10^{-4}) = 5 	imes 10^{-6} 	ext{ A m}^2$.Work $W = MB = (5 	imes 10^{-6}) 	imes 1 = 5 	imes 10^{-6} 	ext{ J} = 5 \mu	ext{J}$.

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