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I Electrictric field due to changing uniform magnetic field

  1. Aug 6, 2017 #1
    How does look like electric field lines due to change of the uniform magnetic field?
    Suppose we have a magnetic field between two infunite plates with surface current $i$ which is lineary change with time. Then [itex]B[/itex]-filel is ([itex]x[/itex] - perpendicular to plates, [itex]z[/itex] and [itex]y[/itex] along plates)
    \begin{equation}
    B_z = \frac{4\pi}{c} i
    \end{equation}
    and from Maxwell equation [itex]curl E = -\frac{1}{c}\frac{\partial B}{\partial t}[/itex] we get:
    \begin{equation}
    \frac{\partial E_y}{\partial x} - \frac{\partial E_x}{\partial y} = -\frac{1}{c}\frac{\partial B_z}{\partial t}
    \end{equation}

    How can I find [itex]E[/itex] -field?
     
  2. jcsd
  3. Aug 6, 2017 #2
    Is this a homework problem? If so you should use the homework template.

    If ##i## is changing linearly with time then how would you write the time dependent form of the current and hence the magnetic field?

    Once you find ##B_{z}## you can find the right hand side of the Maxwell-Faraday equation. Since you have two unkowns, ##E_{y}## and ##E_{x}##, you need a second equation to guarantee a unique solution. This is Gauss's law
    $$\nabla\cdot\mathbf{E}=0$$

    With this it is fairly simple to guess the correct form of ##\mathbf{E}## which satisfies both equations.
     
  4. Aug 7, 2017 #3
    No, it is not a homework. This problem is of interest to me.
    2c8ebdf3202014270e567a917b5812fd.jpg
    I dont think the [itex]\nabla\cdot E = 0[/itex] would be enaught. Let's solve this problem in a different way (usingvector potential). As we know, for a uniform magnetic field, 1) [itex]A_x=0, A_y = xB, A_z = 0[/itex] 2) [itex]A_x=-yB, A_y = 0, A_z = 0[/itex] 3) [itex]A_x=-\frac12 yB, A_y = \frac12 xB, A_z = 0[/itex].
    and for defining electric field we should to use [itex]E = -\frac{1}{c} \frac{\partial A}{\partial t}[/itex].
    Suppose [itex]i = kt[/itex]. And [itex]B = \frac{4\pi}{c} kt[/itex].
    Thus we have three possibilities:
    1) [itex]E_x=0, E_y = x\frac{4\pi}{c} k, E_z = 0[/itex]
    2) [itex]E_x=-y\frac{4\pi}{c} k, E_y = 0, E_z = 0[/itex]
    3) [itex]E_x=-\frac12 y\frac{4\pi}{c} k, E_y = \frac12 x\frac{4\pi}{c} k, E_z = 0[/itex]

    All three fields satisfy equations [itex]\nabla\times E = -\frac{1}{c} \frac{\partial B}{\partial t}[/itex] and [itex]\nabla\cdot E = 0[/itex].
    But which one possibility of three I should to choose as solution? Obviously, I need something else. I need boundary conditions, which is not obvious for me in this case.
     
  5. Aug 8, 2017 #4
    I see. I would choose the third solution because of the symmetry of the problem. If you rotate the system around the z axis the magnetic field looks the same and thus the electric field should also look the same. The first two solutions do not have this property. The third solution has rotational symmetry around the z axis.
     
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