For a coil in the shape of an equilateral triangle, the torque \( \tau \) on the coil due to the magnetic field \( B \) is given by: \[ \tau = n B A I \sin(\theta) \] where \( n \) is the number of turns (in this case, \( n = 1 \) for a single turn), \( A \) is the area of the coil, \( I \) is the current, and \( \theta \) is the angle between the magnetic field and the normal to the coil’s plane. For an equilateral triangle, the area \( A \) is: \[ A = \frac{\sqrt{3}}{4} l^2 \] Substituting the known values and solving for the side length \( l \), we get the answer \( 2 \left( \frac{t}{\sqrt{3} B I} \right)^{1/2} \).
Hence, the correct answer is (a).

A wire of arbitrary shape carries a current $ I = 2A $. Consider the portion of wire between $ (0, 0, 0) $ and $ (4, 4, 4) $. A magnetic field given by $ B = \left( 1.2 \times 10^{-4} + 2 \times 10^{-4} \right) \, \hat{k} $ exists in the region. The force acting on the given portion of the wire is: