A coil in the shape of an equilateral triangle of side 0.02 m is suspended from a vertex such that it is hanging in a vertical plane between the pole pieces of permanent magnet producing a horizontal magnetic field of  $5\times {10}^{-2}T$. Find the couple acting on the coil when a current 0.1 A is passed through it (the magnetic field is parallel to its plane).

The couple acting on a closed loop is given by  $\tau =NiAB\sin \theta$
Where, N=number of loops, A=area of loop, B=magnetic field and   $\theta$=angle between 
magnetic field and normal to the surface of the loop.
Here,  $N=1,i=0.1A,B=5\times {10}^{-2}T,\theta ={90}^0$(the magnetic field is parallel to its plane)
And  $A=\frac{\sqrt{3}}{4}{a}^2 = \frac{\sqrt{3}}{4}{(0.02)}^2$$=1.732\times {10}^{-4}{m}^2$

$\tau =NiAB\sin \theta$ 
 $\therefore \tau =1\times 0.1\times (1.732\times {10}^{-4})\times (5\times {10}^{-2})\times 1$$=8.66\times {10}^{-7}Nm$ .