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(C) The torque $	au$ acting on a current-carrying coil placed in a uniform magnetic field $B$ is given by the formula:$	au = n I A B \sin \alpha$whe

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$n I A B$

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(C) The torque $	au$ acting on a current-carrying coil placed in a uniform magnetic field $B$ is given by the formula:$	au = n I A B \sin \alpha$where $\alpha$ is the angle between the normal to the plane of the coil and the magnetic field direction.In the problem,the plane of the coil makes an angle $	heta$ with the magnetic field. Therefore,the angle $\alpha$ between the normal to the coil and the magnetic field is $\alpha = 90^{\circ} - 	heta$.Substituting this into the torque formula:$	au = n I A B \sin(90^{\circ} - 	heta) = n I A B \cos 	heta$However,the question specifically asks for the case where the plane of the magnetic field is horizontal and the plane of the coil is vertical. In this configuration,the normal to the coil is perpendicular to the magnetic field,meaning $\alpha = 90^{\circ}$.Thus,the torque is:$	au = n I A B \sin 90^{\circ} = n I A B$

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