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B Qualitative understanding of Maxwell's addition

  1. Aug 26, 2016 #1
    Can somebody please give me a qualitative understanding of how a moving electric field creates a magnetic field and vice versa per maxwells equations? Thanks!
     
    Last edited: Aug 27, 2016
  2. jcsd
  3. Aug 26, 2016 #2
    Perhaps this can help?
     
  4. Aug 26, 2016 #3

    jtbell

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    Staff: Mentor

    Did you perhaps mean to write "magnetic field" instead of the second "electric field"?
     
  5. Aug 27, 2016 #4
    Yea, let me edit that.
     
  6. Aug 27, 2016 #5

    jtbell

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    Staff: Mentor

    OK, I'm going to assume you've seen simple applications of Ampere's law without the Maxwell "addition", for finding e.g. the magnetic field produced by straight wire. We define an "Amperian loop", integrate B around it, and relate that to the current "enclosed" by the loop: $$\oint {\vec B \cdot d \vec l} = \mu_0 I$$ More precisely, to get the "enclosed" current, we define a surface whose boundary is the loop, and integrate the current density over that surface: $$\oint {\vec B \cdot d \vec l} = \mu_0 \int {\vec J \cdot d \vec a}$$ In a steady-state situation (magnetostatics) it doesn't matter what shape the surface is, so long as it doesn't have any "holes" in it. We get the same current through it regardless of the shape, and the same B field around the boundary.

    ampere1.gif

    Now suppose we put a capacitor in the circuit.

    ampere2.gif

    No current actually passes through the surface any more! Instead we have an E field whose magnitude changes as the current dumps charge on one plate and removes it from the other. If we use only the current through the surface to calculate the B field, we get different results depending on the shape of the surface, i.e. whether the current actually "pierces" the surface or not. No good!

    We can restore the consistency if we assume that the time-varying flux of E through the surface also contributes to the calculation. Define a new "displacement current" $$I_\textrm{d} = \varepsilon_0 \frac {\partial}{\partial t} \int {\vec E \cdot d \vec a}$$ and add it to the normal "conduction current": $$\oint {\vec B \cdot d \vec l} = \mu_0 (I + I_d)$$ Substituting the definitions of the I's we get the full Ampere-Maxwell law in integral form: $$\oint {\vec B \cdot d \vec l} = \mu_0 \int {\vec J \cdot d \vec a} + \mu_0 \varepsilon_0 \frac {\partial}{\partial t} \int {\vec E \cdot d \vec a}$$ Or in differential form: $$\vec \nabla \times \vec B = \mu_0 \vec J + \mu_0 \varepsilon_0 \frac {\partial \vec E}{\partial t}$$
     
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