Homework SourseA 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? 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 W(Incorrect) E = BANsin = BANsin() = 0.30*19*10^-4 *20*170 * sin( 18) = 0.598875 V P =V^2 /R = 0.598875^2 / 50 = 0.00717302531 W(Incorrect)(Incorrect) I think the average power dissipated in the resistor is different than the instantaneous power being dissipated in the resistor. Any idea how to do it?
Solution
Instantaneous power:
V = NBAW*cos()
where = angle between the plane of the coil and magnetic field
So,
V = 20*0.3*(19*10^-4)*170*cos(18 deg)
= 1.84 V
So, current in the circuit, I = V/R
= 1.84/50
= 0.0368 A
So, instantaneous power = I^2*R
= 0.0368^2*50
= 0.0677 W <---------- The answer should be this of the coil i rotated with the magnetic field directed in a particular direction at all times
But if the axis of the coil is perpendicualr to the field at all times:there is no change of flux,
thus the instantaneous emf is 0 and thus instantaneous power = 0 <------ try this
