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 x 10-2 T. A current of 10.0 ampere flows 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 byτ = NIBA sin θwhere .I = current in the coil , θ = angle which the normal to theplane of the coil makes with the lines of induction B.Here,N=1, B=1.5×10-2 T A=0.05×0.08=40×10-4 m2 I=10.0amp,θ=90°=π/2 τ=iBANsinθ τ=1.5×10-2(10.0)×(1)40×10-4sin(π/2) =6×10-4Nm