The plane of a rectangular loop of wire with sides 0.05 m and 0.08 m is parallel to a uniform magnetic field of induction  1.5×10−2 T. A current of 10.0 ampere flow through the loop. If the side of length 0.08 m is normal and the side of length 0.05 m is parallel to the lines of induction, then the torque acting on the loop is

Torque τ acting on a current carrying coil of area A placed in a magnetic field of induction B is given by τ=NIBA sinθWhere  I=current in the coil,  θ=angle which the normal to the plane of the coil makes with the lines of induction B. Here,  N=1 ; B=1.5×10−2T A=0.05×0.08=40×10-4 m2I=10.0amp,θ=90°=π/2∴τ=1.5×10-2(10.0)×(1)40×10-4sin(π/2)=6×10-4Nm.