A rectangular coil of wire of $500$ turns of area $10\times 5c{m}^2$ carries a current of $2A$ in a magnetic field of induction $2\times 10^{-3}T$ . If the plane of the coil is parallel to the field. The torque on the coil is (in$Nm$):

The torque on the rectangular coil due to presence of magnetic field is given, 

$\tau =NIAB\sin \theta$

where number of turns $N=500$,

Current in the coil $I=2A$, 

Area of the coil  $A=(10\times 5)10^{-4}{m}^2$,

Magnetic field $B=2\times 10^{-3}T$,

Angle between area and magnetic field vector is $\theta$ 

As area vector is always normal to plane and given the plane is parallel to field, so the angle between area and field is $90^o$.

So, $\tau =500\times 2\times (10\times 5)\times 10^{-4}\times (2\times 10^{-3})sin90=0.01Nm$