A 20 turn square coil has an area 19.0 cm 2 and is rotated with an angular speed

Question

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.

What is the frequency of the sinusoidal voltage induced across its ends? 27.1 Hz

What is the peak voltage induced across the ends of the coil? 1.94 V

What is the average power dissipated in the resistor? 0.0376 W

What is the flux through the coil when the plane of the coil is inclined at 18.0° to the magnetic field?
3.52 mWb

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?

I do not how to do the last question.

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)

The answer of this question is I=V^2/R where V=the peak voltage induced across the ends of the coil=1.94 and R=50 ohm, but I do not why I need to use the the peak voltage induced across the ends of the coil rather than the voltage when it is inclined at 18 deg.

Explanation / Answer

Your solution seems to be perfect, but here we are using peak voltage induced, because question mentions "The axis of the coil is perpendicular to the field at all times." and so , Angle = 90o.

So, V = 1.94 V
And P = v^2/R = 1.94^2/50
P = 0.075 W

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