question_answer

A circular coil of 20 turns and radius 10 cm is placed in uniform magnetic field of \[0.10\text{ }T\]normal to the plane of the coil. If the current in coil is 5 A, then the torque acting on the coil will be

A)  \[31.4\text{ }Nm\]      

B)  \[3.14\text{ }Nm\]

C)  \[0.314\text{ }Nm\]     

D)  zero

Correct Answer: D

Solution :

Torque \[(\tau )\] 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   of coil is parallel to the field applied, hence \[\theta ={{0}^{o}}\] and \[\sin \,\,{{0}^{o}}=0\] \[\therefore \] \[\tau =0\]