A rectangular coil of 400 turns and 10^{-2} | vedclass

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(A) The torque (moment of force) acting on a current-carrying coil in a magnetic field is given by $	au = N i A B \sin \alpha$, where $\alpha$ is the

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(A) The torque (moment of force) acting on a current-carrying coil in a magnetic field is given by $	au = N i A B \sin \alpha$, where $\alpha$ is the angle between the normal to the coil and the magnetic field. Given that the plane of the coil makes an angle of $60^{\circ}$ with the magnetic field, the angle $\alpha$ between the normal to the coil and the magnetic field is $\alpha = 90^{\circ} - 60^{\circ} = 30^{\circ}$. Alternatively, using the formula $	au = N i A B \cos 	heta$, where $	heta$ is the angle between the plane of the coil and the magnetic field, we have $	heta = 60^{\circ}$. Substituting the given values: $N = 400$, $i = 0.5 \ A$, $A = 10^{-2} \ m^2$, $B = 1 \ T$, and $	heta = 60^{\circ}$. $	au = 400 	imes 0.5 	imes 10^{-2} 	imes 1 	imes \cos 60^{\circ}$ $	au = 400 	imes 0.5 	imes 10^{-2} 	imes 1 	imes 0.5$ $	au = 200 	imes 10^{-2} = 2 	imes 0.5 = 1 \ Nm$.

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