The square loop ABCD,carrying a current i,is | vedclass

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Question

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

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$90^o + 	heta$

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(C) The torque on a current-carrying loop in a magnetic field is given by $\vec{	au} = \vec{m} 	imes \vec{B}$,where $\vec{m}$ is the magnetic dipole moment vector perpendicular to the plane of the loop.The torque is zero when the magnetic moment $\vec{m}$ is parallel or anti-parallel to the magnetic field $\vec{B}$. This occurs when the plane of the loop is perpendicular to the magnetic field.In the given figure,the plane of the loop makes an angle $	heta$ with the magnetic field $B$. The angle between the normal to the loop (direction of $\vec{m}$) and the magnetic field $\vec{B}$ is $(90^o + 	heta)$.The loop will rotate until the plane becomes perpendicular to the magnetic field,which corresponds to the position where the torque becomes zero. The total angle of rotation required is $(90^o + 	heta)$.

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