A coil in the shape of an equilateral triangle of side l is suspended between the pole pieces of a permanent magnet such that B vector is in plane of the coil. if due to a current i in the triangle a torque torque acts on it, the side l of the triangle is

A coil is in the shape of an equilateral triangle so each side has length = l m, Magnetic field = B,
when a current (i) is passed through the coil having area A, then the magnetic dipole moment of the coil is given by
M=IA  where A is the area of the coil
 $A=\frac{1}{2}\times base\times height=\frac{1}{2}\times l\times \sin 60°=\frac{\sqrt{3}}{4}{l}^2$ [an equilateral triangle has an angle of 60 degree each]
therefore the magnetic dipole moment of the coil is
$M=iA=\frac{\sqrt{3}}{4}i$l²
The angle between the normal to the plane of the coil and the applied magnetic field is 90 degree.
Now the torque acting on the coil is given by 

 $\tau =MB\sin \theta =\frac{\sqrt{3}}{4}i{l}^{2}B\sin 90=\frac{\sqrt{3}}{4}iB{l}^2$