A 500-turn armature (a rotating solenoid) with a cross-sectional area of 9 cm2 rotates at 120 rpm in a 4.00 T magnetic field. What is the average electric potential produced when the armature rotates from 0 degree to 90 degree to the field? I can't really understand this question, and what is meant by 'average'.

Expert Answers

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When the armature rotates in the magnetic field, there will be induced electromotive force, or potential, due to the fact that the magnetic flux through the armature will be changing as the armature rotates. According to the Faraday's Law, the changing magnetic flux through a wire loop (or several loops, such as a solenoid), produced electric potential.

The flux through the solenoid can be found as

`Phi = NBAcos(theta)` , where N = 500 is the number of turns, B = 4 T is the magnetic field, A = 9 cm^2 = `9*10^(-4) m^2` is the cross-sectional area of the solenoid, and `theta` is the angle between the axis of the solenoid (or normal to the cross-section of the solenoid) and the magnetic field vector.

As the armature rotates, the angle `theta` will change, and this is what causing the change in the flux and thus the electric potential. This angle will depend on time as

`theta = omegat` , where `omega = 120 ` rpm (revolutions per minute), or `2pi*120 ` radians per minute.

So the flux is the cosine function of time. Then the electromotive force, or potential, is

`epsilon = (dPhi)/(dt)` , which will be the derivative of cosine, or sine function of time.

In this problem, however, we asked not to find the potential at each moment of time, but the average potential during the rotation of the armature between 0 degrees (when the axis is parallel to the field) and 90 degrees (when the axis is perpendicular to the field.) The average electromotive force can be found as

`epsilon = (Delta Phi)/(Delta t) = (Phi(90) - Phi(0))/(Delta t)` : the change of flux divided by the time interval during which the change took place.

`Phi(0) = NBAcos(0) = NBA` : when the cross-section of the solenoid is perpendicular to the field, the flux is maximum.

`Phi(90) = NBAcos(90) = 0` : when the cross-section of the solenoid is parallel to the field, there is no flux.

The time interval during which the solenoid rotates by 90 degrees is the 1/4 of the period of the rotation. Since the frequency is 120 rpm, or 120/60 = 2 revolutions per second, the period is

`T = 1/(f) = 1/2 = 0.5 s`

`Delta t = 1/4T = 1/4*0.5 = 0.125 s`

Now we can plug in everything and calculate the average potential:

`epsilon = (NBA - 0)/(Delta t) = (500*4*9*10^(-4))/(0.125) =14.4 V`

The average potential generated as the armature rotates through 90 degrees in the magnetic field is 14.4 Volts.



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