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Homework Help: Magnetic Field Induced by Nonuniform Electric Flux

  1. Apr 6, 2012 #1
    1. The problem statement, all variables and given/known data

    During the design of a new doppler radar unit, you have two circular plates with radius R = 3 cm and a plate separation of d = 5 mm. The magnitude of the electric field between the plates is given as:

    E = (700 V/(m sec2))*(1-r/Rp)*t 2

    where Rp is the radius of the plates, t is the time in sec and r is the distance from the axis of the plates (for r < Rp ).

    a) What is amplitude of the induced magnetic field a distance 1.5 cm from the center axis joining the plates at time t = 2.9 sec?

    2. Relevant equations

    ∫B.ds = [itex]\mu_{0}[/itex][itex]\epsilon_{0}[/itex] d[itex]\Phi[/itex]e/dt

    3. The attempt at a solution

    I did this problem as I worked out previous ones, taking the time derivative of the electric field to find the changing electric flux.

    B=[itex]\mu_{0}[/itex][itex]\epsilon_{0}[/itex] pi*r^2/2*pi*r * dE/dt
    B=[itex]\mu_{0}[/itex][itex]\epsilon_{0}[/itex] (.015/2) * 2*700t(1-.015/.03)
    B = 1.69E-16

    This answer is rejected by the software. I'm at a loss for what I'm missing here. I don't understand why I'm given the separation between the plates, I suspect that may be involved in the solution but I'm not seeing how. Thanks for any help.
  2. jcsd
  3. Apr 6, 2012 #2


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    Gold Member

    Okay, I think I see what you are doing here (it took me awhile to figure out your above result).

    When you tried to calculate the flux, you just multiplied the electric field E by the area, πr2. That would work if the electric field was [STRIKE]constant[/STRIKE] uniform within the entire circle defined by r (it can be time changing, but at a particular point in time it must be uniform across the surface defined within r). But it's not [STRIKE]constant[/STRIKE] uniform for this problem. (Although due to symmetry, the magnetic field strength is [STRIKE]constant[/STRIKE] uniform around the perimeter 2πr, which is why the ∫B·ds = B[2πr] part worked okay.)

    You're going to have to do an integration, one way or the other. You have two choices. You can calculate the flux directly via

    [tex] \Phi_E = \int_S \vec E \cdot \vec{dA}, [/tex]

    and then take the time derivative to find [itex] \frac{\partial \Phi_E}{\partial t} [/itex]. Or, you can take the time derivative first to obtain [itex] \frac{\partial E}{\partial t} [/itex], and then evaluate

    [tex] \frac{ \partial \Phi_E}{\partial t} = \int_S \frac{\partial \vec E}{\partial t} \cdot \vec{dA}. [/tex]

    Either way works fine.

    It may be useful to note that in polar coordinates, [itex] dA = r \ dr \ d\theta [/itex].
    Last edited: Apr 6, 2012
  4. Apr 7, 2012 #3
    Thanks! Haven't had to do a double integral before now.
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