A 20 turn square coil has an area 190 cm2 and is rotated wit

A 20 turn square coil has an area 19.0 cm² and is rotated with an angular speed of 170.0 rad/s in a uniform 0.30 T magnetic field. The axis of the coil is perpendicular to the field at all times. If the coil forms a closed circuit with a series resistance of 50.0 , what is the instantaneous power being dissipated in the resistor when the plane of the coil makes an angle of 18.0° with the magnetic field?

B = 0.30 T
N = 20
A = 19.0 * 10^-4 m^2
w = 170.0 rad/s
R = 50.0

V = N*B*A*w*sin(90 - 18.0)
V = 20 * 0.30 * 19.0 * 10^-4 * 170.0 * sin(72)

P=V^2/R=0.0679 V*A

But my answer is not correct.

Solution

take angle as 18 degree instead of 72 degree.

It is given that the angle between the plane of coil and magnetic field the angle is 18 degree.

in the formula for V we take the angle between plane of coil and the magnetic field.

the answer should be correct now