• # question_answer A circular coil of 20 turns and radius 10 cm is placed in uniform magnetic field of 0.10 T normal to the plane of the coil. If the current in coil is 5A, then the torque acting on the coil will be: A)  $31.4$ N-m                  B)  $3.14$ N-m. C)  $0.314$ N-m                D)  zero

Torque (T) acting on a loop placed in a magnetic field B is given by  $\tau \,=nBIA\,\sin \theta$where, A is area of loop, I the current through it, n the number of turns, and $\theta$ the angle which axis of loop makes with magnetic field B. Since, magnetic field  is parallel to the axis of the coil hence $\theta =0{}^\circ$ and s$\sin {{0}^{0}}=0$$\tau \,=0$