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

Vedclass · Accepted Answer

$	au = niAB \cos 	heta$

Vedclass · Answer

(A) The torque $	au$ on a current-carrying coil in a magnetic field is given by $	au = \vec{M} 	imes \vec{B}$,where $\vec{M}$ is the magnetic dipole moment and $\vec{B}$ is the magnetic field.The magnitude of the torque is $	au = MB \sin \alpha$,where $\alpha$ 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 makes an angle $	heta$ with the magnetic field $B$,the angle $\alpha$ between the normal to the coil (the direction of $\vec{M}$) and the magnetic field $\vec{B}$ is $\alpha = 90^\circ - 	heta$.Substituting this into the torque formula: $	au = MB \sin(90^\circ - 	heta) = MB \cos 	heta$.Since the magnetic moment $M = niA$,we have $	au = niAB \cos 	heta$.

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