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(C) The torque $	au$ acting on a current-carrying coil in a uniform magnetic field $B$ is given by the formula $	au = \vec{M} 	imes \vec{B}$,where

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$\pi r^2 i B$

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(C) The torque $	au$ acting on a current-carrying coil in a uniform magnetic field $B$ is given by the formula $	au = \vec{M} 	imes \vec{B}$,where $\vec{M}$ is the magnetic dipole moment of the coil.The magnitude of the torque is $	au = MB \sin 	heta$,where $	heta$ is the angle between the magnetic moment vector $\vec{M}$ (which is perpendicular to the plane of the coil) and the magnetic field vector $\vec{B}$.Given that the plane of the coil is parallel to the magnetic field,the normal to the plane of the coil makes an angle of $90^\circ$ with the magnetic field. Thus,$	heta = 90^\circ$.The magnetic moment of the circular coil is $M = iA = i(\pi r^2)$.Substituting these values into the torque formula:$	au = (i \pi r^2) B \sin 90^\circ = \pi r^2 i B$.Therefore,the correct option is $(c)$.

$A$ current $i$ flows in a circular coil of radius $r$. If the coil is placed in a uniform magnetic field $B$ with its plane parallel to the field,the magnitude of the torque that acts on the coil is

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