A coil is placed in y-z plane making an angle | vedclass

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(C) The torque $	au$ experienced by a current-carrying coil in a magnetic field is given by $	au = |\vec{m} 	imes \vec{B}| = N I A B \sin 	heta$,w

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$200$

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(C) The torque $	au$ experienced by a current-carrying coil in a magnetic field is given by $	au = |\vec{m} 	imes \vec{B}| = N I A B \sin 	heta$,where $	heta$ is the angle between the area vector $\vec{A}$ and the magnetic field $\vec{B}$.The coil is in the $y-z$ plane,so its area vector $\vec{A}$ is along the $x$-axis. However,the problem states the coil makes an angle of $30^{\circ}$ with the $x$-axis. This implies the angle between the normal to the coil (area vector) and the magnetic field (which is along the $x$-axis) is $	heta = 30^{\circ}$.Given values: $N=100$,$I=1 	ext{ A}$,$R=2 	ext{ m}$,$B=\frac{1}{\pi} 	ext{ T}$.Area $A = \pi R^2 = \pi (2)^2 = 4\pi 	ext{ m}^2$.Substituting these into the torque formula:$	au = N I A B \sin 30^{\circ}$$	au = 100 	imes 1 	imes (4\pi) 	imes \frac{1}{\pi} 	imes \sin 30^{\circ}$$	au = 100 	imes 4 	imes \frac{1}{2}$$	au = 200 	ext{ N} \cdot 	ext{m}$.

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