A circular coil of radius 2 cm and 125 turns carries a current of 1 A. The coil is placed in a uniform magnetic field of magnitude 0.4 T. The axis of the coil makes an angle of 30° with the direction of the magnetic field. The torque acting on the coil is 10-4 N m. The value of is . ( =3.14)
Answer
314
Solution
=NIAB ( = angle between the axis of the coil and B)
A= r2=3.14 (0.02)2=1.256 10-3 m2
=125 1 1.256 10-3 0.4 30
=125 1.256 10-3 0.2
=3.14 10-2 N m =314 10-4 N m
=314
Answer: 314