A triangular loop of side l carries a current | vedclass

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(D) The torque $	au$ on a current-carrying loop in a magnetic field is given by $	au = NIAB \sin 	heta$, where $	heta$ is the angle between the no

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$\frac{\sqrt{3}}{4} I B l^2$

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(D) The torque $	au$ on a current-carrying loop in a magnetic field is given by $	au = NIAB \sin 	heta$, where $	heta$ is the angle between the normal to the loop and the magnetic field $B$.Since the plane of the loop is parallel to the magnetic field $B$, the angle between the normal to the loop and the magnetic field is $	heta = 90^\circ$.The area $A$ of an equilateral triangle of side $l$ is $A = \frac{\sqrt{3}}{4} l^2$.Substituting these values into the torque formula:$	au = 1 	imes I 	imes \left( \frac{\sqrt{3}}{4} l^2 ight) 	imes B 	imes \sin(90^\circ)$$	au = \frac{\sqrt{3}}{4} I B l^2$Thus, the correct option is $(d)$.

$A$ triangular loop of side $l$ carries a current $I$. It is placed in a magnetic field $B$ such that the plane of the loop is in the direction of $B$. The torque on the loop is

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