A circular coil of radius R and a current I,w | vedclass

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(A) The potential energy of a magnetic dipole in a magnetic field is given by $U = -\vec{M} \cdot \vec{B} = -MB \cos 	heta$,where $	heta$ is the ang

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$\pi R^2 BI$

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(A) The potential energy of a magnetic dipole in a magnetic field is given by $U = -\vec{M} \cdot \vec{B} = -MB \cos 	heta$,where $	heta$ is the angle between the magnetic moment $\vec{M}$ and the magnetic field $\vec{B}$.Initially,the plane of the coil lies along the magnetic field,meaning the area vector (and thus the magnetic moment $\vec{M}$) is perpendicular to the magnetic field. Thus,$	heta_i = 90^{\circ}$.However,the problem states the coil rotates through $90^{\circ}$ from its initial position. If the plane was along the field,the initial angle between $\vec{M}$ and $\vec{B}$ was $90^{\circ}$. After rotating $90^{\circ}$,the plane becomes perpendicular to the field,so the angle between $\vec{M}$ and $\vec{B}$ becomes $0^{\circ}$.Loss in potential energy = Gain in kinetic energy.$KE = U_i - U_f = (-MB \cos 90^{\circ}) - (-MB \cos 0^{\circ}) = 0 - (-MB) = MB$.Since the magnetic moment $M = I 	imes A = I(\pi R^2)$,the kinetic energy is $KE = \pi R^2 IB$.

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