A square coil of side 10 cm and 50 turns is placed in a uniform magnetic field of magnitude 0.5 T. The plane of the coil makes an angle of 53^ with the direction of the magnetic field. If the coil experiences a magnetic torque of 0.45 N.m, the current flowing through it is ______ A.

Step-by-step solution

The area of the square coil is A = a^2 = (0.1)^2 = 0.01 m ^2 . The angle given is between the plane of the coil and the magnetic field ( 53^ ). The angle between the area vector (normal to the plane) and the magnetic field is: = 90^ - 53^ = 37^ The magnetic torque acting on the coil is given by: = NIAB Substituting the given values: 0.45 = 50 I 0.01 0.5 (37^ ) Using (37^ ) = 0.6 : 0.45 = 50 I 0.01 0.5 0.6 0.45 = 0.15 I I = 0.45 0.15 = 3 A Answer: 3