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Homework Help: Reducing Maxwell's equations

  1. May 10, 2007 #1
    I have a question regarding two electric lines running parallel to each other with their current running in opposite direction. Under this set up the electric field points up vertically on the page and the magnetic field points out from the page. I am to show that Maxwell's equations reduce to -dH/dz = epsilon dE/dt and dE/dz = - mu dH/dt. One question I have is dt regards time but what does dz regard. I am guessing that it is the change in displacement that runs along the electric lines, the direction of the current. Is this the correct direction?

    I have expressions for Maxwell's equations that involve integrals. Are these the ones I should start off with? The reason I ask is I found expressions for Maxwell's equations online that are described by differentials. These seem like easier ground to pick up the problem from. In fact I don't see how I can solve the problem from the integral Maxwell equations. Clearly there must be a way but I don't see it.

    This website:


    has Maxwell's equations in integral and differential form. If I start from differential form then it looks like the equation involving the curl of E does not need to change. The equation involving the curl of H however has an extra term in it which I'm guessing I need to prove is zero under the described problem statement. That equation is:

    curl of H = J + dD/dt or curl of B = J + dE/dt expressed in the terms I have used.

    So I guess I need to prove that the curl of H can be expressed as -dH/dz only and that the J term is equal to zero.

    I'm not real sure where to get started with this problem.
  2. jcsd
  3. May 10, 2007 #2
    Maxwell's equations (in differential form) relate the space derivatives of the fields to the time derivatives of the fields. In a general situation, fields are functions of both position and time. the [itex]dz[/itex] that you see stems from the spatial derivatives of the fields. You will have to see how each of these equations has evolved. Let me write two of them:

    [tex]\nabla \times \vec{E} = -\frac{\partial \vec{B}}{\partial t}[/tex]
    [tex]\nabla \times \vec{B} = \mu_{0}\vec{J} + \mu_{0}\epsilon_{0}\frac{\partial \vec{E}}{\partial t}[/tex]

    In matter, the second equation gets modified to

    [tex]\nabla \times \vec{H} = \vec{J} + \frac{\partial \vec{D}}{\partial t}[/tex]

    In each of these equations, the left hand side is a space differential operator (a curl) and the right hand side is a time differential operator--time derivative. Mathematically, this is where the [itex]dz[/itex] and [itex]dt[/itex] come from. Physically, these equations relate a time rate of change of the fields to a space rate of change.

    Now, what is it that you mean by "change in displacement that runs along the electric lines"?

    Can you show us your working--just try it yourself at least once. Try to use the differential Maxwell's equations first.

    Have you been given anything about the material in which these fields exist? If its free space, [itex]H = B/\mu_{0}[/itex] and [itex]D = \epsilon_{0}E[/itex] so using H or B (or D or E) is just a matter of choice.
    Last edited: May 10, 2007
  4. May 11, 2007 #3
    What I meant by change in displacement that runs along the electric lines is along that specific axis of the 3 dimensional space. So there could be a dx and a dy but they don't show up in the equations. Instead there is just a dz. What you're saying I think is dz is just a generalized space rate of change in all 3 dimensions.

    I'll work on the problems today and post a copy of my work later on today. In trying to make the curl of E to become dE/dz should I write out the curl of E into its components so that it looks like this:

    curl E = (dR/dy - dQ/dz)i + (dP/dz -dR/dx)j + (dQ/dx - dP/dy)k

    for E = Pi + Qj + Rk

    I'm thinking that two of those directional scalars (P,Q,R) are zero. Am I wrong? I'm not sure what to call P,Q, and R so I'm probably wrong with the 'directional scalars' name but they relate to the electric field and that only has one direction. Say P is non zero then I'm lost what dP/dz would be and dP/dy. It seems to me they would again be zero. I need a little help understanding this part.

    How to get J to disappear in the curl of H equation is lost on me. I'll see what I can come up with but these are the issues I have with the problem.

    I think it is safe to assume that the problem exists in free space but I don't really know. It's a very shortly described problem.
  5. May 11, 2007 #4
    Okay. This is my work. I started with the Integral form because that actually seemed easier. That's the reverse of what I said before. I was looking at different Maxwell Equations from my Physics book when I arrived at that conclusion. I'm not sure if I'm making a legal step when I set d/dA = d/(dz*dl). The work is on a *.jpg attachment to this post. I'm not sure how to introduce the negative sign to the dH/dz equation. I decided that the epsilon 0 constant must be negative and that is how I introduced the negative sign on one side of the equation.

    Attached Files:

  6. May 11, 2007 #5
    One thing I failed to mention is how I concluded that J = 0. J is defined on the wikipedia website I mentioned earlier as the free current density. I interpretted this to mean the current flowing between the two electric lines. These two electric lines are not necessarily connected so the free current density is 0. Maybe I'm interpretting this wrong. Maybe the free part means electricity flowing through air. Making some assumptions I can guess that it is zero so that the J becomes zero. I think this goes back to the question you had earlier mav.
  7. May 11, 2007 #6
    [tex] \epsilon_0 [/tex] cannot be negative.

    It is a fundamental constant. Recall that you can write it as:

    [tex] \epsilon_0 \mu_0 = \frac{1}{c^2} [/tex]

    So you can conclude that neither [itex] \epsilon_0 [/itex] or [itex] \mu_0 [/itex] can be zero.

    EDIT: Actually you can't reach that conclusion alone. Since both epsilon_0, and mu_0 can be zero, and still satisfy that equation. Just note, that it can't be zero.
    Last edited: May 11, 2007
  8. May 15, 2007 #7
    Hi, sorry for the late reply. I had a look at your work. I think you should start with the differential formulation of the Maxwell's equations (you'll get the same equations, but its more obvious whats happening that way). I am not convinced about the operator


    This is what you probably meant:

    [tex]\frac{d}{dt}\int f(x,t)dx = \int \frac{\partial f(x,t)}{\partial t}dx[/tex]


    [tex]\frac{d}{dx}\int f(x,t)dx = f(x,t)[/tex]

    Or if you write the [itex]da[/itex] explicitly and write the double integral, then you will have to operate the derivative twice. Effectively you would be doing this: [itex]d^{2}()/dzdl[/itex].

    Do you see what I'm saying here?

    Start instead with the differential form of Maxwell's equations (curl and divergence)...work from there.
    Last edited: May 15, 2007
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