Q. 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
A) Zero
B) $IBl$
C) $\frac{\sqrt{3}}{2}Il^{2}B^{2}$
D) $\frac{\sqrt{3}}{4}IBl^{2}$
Explanation
(d) Since $\theta = 90{^\circ}$
Hence $\tau = NIAB = 1 \times I \times \left( \frac{\sqrt{3}}{4}l^{2} \right)B$
$= \frac{\sqrt{3}}{4}l^{2}B\ I$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.