A current-carrying rectangular coil is placed | vedclass

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

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The magnetic field is perpendicular to the plane of the coil

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(B) The torque acting on a current-carrying coil in a magnetic field is given by $\vec{	au} = \vec{m} 	imes \vec{B}$,where $\vec{m}$ is the magnetic moment and $\vec{B}$ is the magnetic field.The magnitude of the torque is $	au = mB \sin(	heta)$,where $	heta$ is the angle between the magnetic moment vector $\vec{m}$ and the magnetic field vector $\vec{B}$.The magnetic moment vector $\vec{m}$ is always perpendicular to the plane of the coil.If the magnetic field is perpendicular to the plane of the coil,then the angle between $\vec{m}$ and $\vec{B}$ is $0^o$ or $180^o$.In this case,$	au = mB \sin(0^o) = 0$ or $	au = mB \sin(180^o) = 0$.Therefore,the coil will not tend to rotate when the magnetic field is perpendicular to the plane of the coil.

$A$ current-carrying rectangular coil is placed in a uniform magnetic field. In which orientation will the coil not tend to rotate?

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