Question

A 50 turn circular coil of radius 2cm carrying a current of 5.0 A is rotated in a magnetic field of strength 0.20T. In a particular position of the coil. The torque acting is half of the maximum torque. The angle between the magnetic field and the plane of the coil is

Answer 1

According to me the values of B ,n,I,r are of no use as in the question it is clearly written that the value of the torque is Half of the maximum value

Torque=MBsin theta

Sin theta is Max when it's value is 1

Sin 90 = 1

Torque=MB

Half of the max value 1/2

Sin 30 =1/2

Torque''= MB/2

Torque" = Torque/2

Angle= 30 °

I hope it helps(θ‿θ)