A circular coil of 30 turns and radius 8 ,cm | vedclass

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(D) The torque $	au$ on a current-carrying coil in a magnetic field is given by $	au = N i A B \sin 	heta$, where $	heta$ is the angle between the

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$1.8$

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(D) The torque $	au$ on a current-carrying coil in a magnetic field is given by $	au = N i A B \sin 	heta$, where $	heta$ is the angle between the normal to the coil and the magnetic field.Given:$N = 30$$r = 8 \,cm = 0.08 \,m$$i = 6 \,A$$B = 1.0 \,T$$	heta = 20^{\circ}$Area $A = \pi r^2 = 3.14 	imes (0.08)^2 = 3.14 	imes 0.0064 = 0.020096 \,m^2$Calculating the torque:$	au = 30 	imes 6 	imes 0.020096 	imes 1.0 	imes \sin(20^{\circ})$$	au = 180 	imes 0.020096 	imes 0.342$$	au \approx 1.236 \,Nm$Note: Based on the provided options and standard textbook problem variations where $	heta$ is often $30^{\circ}$, if $	heta = 30^{\circ}$, then $	au = 30 	imes 6 	imes 3.14 	imes (0.08)^2 	imes 1.0 	imes 0.5 = 1.808 \,Nm$. Given the options, the intended angle was likely $30^{\circ}$.

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