question_answer

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          [MP PET 2003]

A)            Zero                                         

B)            IBl

C)            \[\frac{\sqrt{3}}{2}I{{l}^{2}}{{B}^{2}}\]                         

D)            \[\frac{\sqrt{3}}{4}IB{{l}^{2}}\]

Correct Answer: D

Solution :

                   Since q = 90° Hence \[\tau =NIAB=1\times I\times \left( \frac{\sqrt{3}}{4}{{l}^{2}} \right)\,B\]        \[=\frac{\sqrt{3}}{4}{{l}^{2}}B\]