A current-carrying coil experiences a torque | vedclass

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(D) The torque $	au$ experienced by a current-carrying coil in a magnetic field is given by $	au = MB \sin 	heta$,where $	heta$ is the angle betwe

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$	an ^{-1}\left(\frac{4}{3}ight)$

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(D) The torque $	au$ experienced by a current-carrying coil in a magnetic field is given by $	au = MB \sin 	heta$,where $	heta$ is the angle between the magnetic field and the normal to the plane of the coil.Maximum torque $	au_{\max} = MB$ (when $	heta = 90^{\circ}$).Given that $	au = 80 \%$ of $	au_{\max} = 0.8 	au_{\max} = \frac{4}{5} MB$.Equating the two expressions: $MB \sin 	heta = \frac{4}{5} MB$.This simplifies to $\sin 	heta = \frac{4}{5}$.Using the trigonometric identity $	an 	heta = \frac{\sin 	heta}{\cos 	heta} = \frac{\sin 	heta}{\sqrt{1 - \sin^2 	heta}}$,we get $	an 	heta = \frac{4/5}{\sqrt{1 - (4/5)^2}} = \frac{4/5}{\sqrt{9/25}} = \frac{4/5}{3/5} = \frac{4}{3}$.Therefore,$	heta = 	an ^{-1}\left(\frac{4}{3}ight)$.

$A$ current-carrying coil experiences a torque due to a magnetic field. The value of the torque is $80 \%$ of the maximum possible torque. The angle between the magnetic field and the normal to the plane of the coil is

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