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Net Power In a Charged Rotating Cylinder

  1. May 16, 2013 #1
    1. The problem statement, all variables and given/known data
    A long hollow nonconducting cylinder (radius R= 0.060 m, length L= 0.70 m) carries a uniform charge per unit area of σ= 4.0 C/m^2 on its surface. Beginning from rest, an externally applied torque causes the cylinder to rotate at constant acceleration α= 40 rad/s^2
    about the cylinder axis. Find the net power entering the interior volume of the cylinder from the surrounding electromagnetic fields at the instant the angular velocity reaches ω= 200 rad/s.
    (Answer 4.6μW)


    2. Relevant equations
    I=S[itex]_{avg}[/itex]=[itex]\frac{1}{cμ_{0}}[/itex] * [(E[itex]_{max}[/itex])[itex]^{2}[/itex]*sin[itex]^{2}[/itex](kx-ωt)]


    3. The attempt at a solution

    The intensity I of the electromagnetic wave is [itex]\frac{Power}{Area}[/itex] where I assumed the area would be the surface area of the cylinder (2[itex]\pi[/itex]*r*L).

    so to find the net power through the cylinder I thought it would just be I*area.

    From a previous chapter, I determined that E=[itex]\frac{σ}{2ε_{0}}[/itex].

    With angular acceleration α=40rad/s^2 the cylinder would reach the angular velocity
    ω= 200rad/s^2 in t=5 seconds.

    I'm not sure if this is the right approach to the problem. I don't know what to do about x or the angular wave number k in the intensity function.
     
  2. jcsd
  3. May 16, 2013 #2

    rude man

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    Not sure why you're invoking the planar moving e-m wave equation.

    I would think of this as a solenoid once the cylinder is spinning.

    You can imagine N turns in L = 0.7 meter with each turn carrying q/N coulombs with q = 2πRLσ total coulombs.

    Then, as the cylinder spins, the charges represent current in each "winding" and you can solve for B inside the solenoid.

    Once you have B you of course have the total magnetic field energy within the cylinder.
     
  4. May 16, 2013 #3
    I'm sorry, can you explain a bit further? I used the e-m wave equation because that is the topic of the section in which this question was asked. How would I use the total magnetic field to find the power?
     
  5. May 16, 2013 #4

    rude man

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    Energy density = B2/2μ0
    Energy = energy density * volume

    Funny that the question arose in conjunction with e-m waves. Maybe I'm looking at it the wrong way ...

    I guess since these are accelerating charges there will be an electromagnetic wave generated. A bit beyond me I'm afraid. Is this an advanced e-m course?

    EDIT: oh, and it asks for power entering the cylinder. Doing it my way that would be dE/dt where E = Energy. That would account for the need for an accelerating cylinder, so that dE/dt is not zero.
     
    Last edited: May 16, 2013
  6. May 16, 2013 #5
    It's a general physics course. Thanks for the replies, I will keep working on it.
     
  7. May 16, 2013 #6

    TSny

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    Maybe this is a problem on the application of the Poynting vector. As the cylinder spins, a B field is set up inside as rude man o:) said. Since the rate of spin is increasing, B is increasing. The increasing B field will induce an E field. If you can find the E field at the surface of the cylinder, you can construct the Poynting vector and determine the rate of flow of field energy into the interior region of the cylinder.
     
  8. May 16, 2013 #7

    mukundpa

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    I think it wants the rate of increase in energy within the cylinder. Increasing angular velocity increases the B field and changing B field creates E field. Thus we have to calculate the total energy U (due to B and E fields) as a function of time and then dU/dt will the power.
     
  9. May 16, 2013 #8

    TSny

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    Yes, that should give the same result as integrating the Poynting vector over the surface of the cylinder. Since the question asks for the "net power entering the interior volume of the cylinder from the surrounding electromagnetic fields", it made me think of the Poynting vector as the appropriate method. But the question is not really very clear on the method to be used.
     
  10. May 17, 2013 #9

    rude man

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    So maybe del x E = -∂B/∂t in addititon to d/dt B2/2μ0?

    Then E field energy = |E|2/2ε0
    and E power = d/dt of above ...
     
  11. May 17, 2013 #10

    TSny

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    Since B is increasing linearly with time, E will be constant. So, the E-field energy will not be changing with time and there will be no power going into the E field energy at the instant of time of interest.

    EDIT: But del x E = -∂B/∂t is relevant for finding the E field at the surface of the cylinder in order to construct the Poynting vector. You can show that the integral of the Poynting vector over the surface of the cylinder is equal to the rate of change of the total B-field energy inside the cylinder which is also equal to the mechanical input power of whoever/whatever is turning the cylinder.

    EDIT 2: In the last sentence I should have said equal to the mechanical input power done against the electrical force acting on the charged surface of the cylinder by the induced E field. There is also additional mechanical input power necessary to increase the rotational kinetic energy of the cylinder due to its mass.
     
    Last edited: May 17, 2013
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