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The meaning of curl in Electrodynamics

  1. Mar 4, 2006 #1
    I have a hard time understanding what the curl really means in Maxwell's equations, for example in a steady-state you have

    [tex]\nabla\times \textbf{E} = 0[/tex]

    and in a time-varying field you have

    [tex]\nabla\times \textbf{E} = -\frac{\partial \textbf{B}}{\partial t}[/tex]

    The meaning of the divergence is like "outflow - inflow". I read that the curl is like the amount of rotation. But what does it means in this situation with the electric field? That electric field lines don't "rotate"/curve like for the magnetic field?
  2. jcsd
  3. Mar 4, 2006 #2


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    It doesn't have to have an actual rotation to have a curl. For example, let's say the electric field points from the left to the right, but it has a change in magnitude as you move vertically, such as:

    -----> -----> -----> -----> ----->
    ----> ----> ----> ----> ---->
    ---> ---> ---> ---> --->
    --> --> --> --> -->
    -> -> -> -> ->

    If you notice, while the E-field vector itself has no "rotation", the whole field itself has variation perpendicular to its direction. Such a field will have a curl.

  4. Mar 4, 2006 #3


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    To make a mechanical analogy, imagine that Zz's vector field diagram represents the flow of a physical fluid like water, flowing faster at the top of the diagram. Place a small paddlewheel in the water. The water pushes harder on the top of the wheel, than on the bottom, so the wheel tends to rotate clockwise. That would be the physical significance of the nonzero curl in this sort of situation.
  5. Mar 4, 2006 #4
    That sounds reasonable.
    While I was thinking about the second equation another question arised.
    When you have Faraday's law in it's differentialform, you only have the time derivatime of B, and not the magnetic flux, so does this mean that the differentialform can ONLY be used if you are dealing with a time-varying B-field, and not with other situations where a flux-change can occure, like with a rotating circuit and ect.?

    And the same question arises for Ampére's generalized law

    [tex] \nabla\times \textbf{B} = \mu_0\left(\textbf{J}+\varepsilon_0\frac{\partial\textbf{E}}{\partial t}\right) [/tex]

    And another question is, why does the differentialform of Ampére's law fail when it comes to a static magnetic field, because isn't [itex] \nabla\times \textbf{B} = \vec{0}[/itex] if B if constant, or am I wrong?
  6. Mar 4, 2006 #5


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    Yes, if B is constant in time then there will be no rotational E-field: curl E=0, and the magnetic flux through a chosen surface is constant. Ofcourse this surface has to be constant in time, if the flux through a surface bounding a loop changes because the loop rotates it will also induce an emf, but not because of an E-field that was created by a changing B-field as in an application of Faraday's law.

    You have to distinguish between a field that is constant in TIME and one that is constant in SPACE. If B is constant in SPACE (homogeneous), then curl B=0. Magnetostatics deal with B-fields constant in TIME, but curl B doesn't have to be zero.
  7. Mar 4, 2006 #6
    1. I know there will an induced emf, but you can't work with that by using the differentialform of Faradays, you need the integral form, because you have a flux-change and not a B-field change.

    2. Yes I was refering to that when the B-field is constant in space. How do you use Ampére's law in differentialform if the field is constant in space? Consider a current carrying wire, and the B-field in a fixed radius from the wire. It's easy to use Ampére's law in integral form, but curl B would be 0 because the field is constant in the fixed radius. Or do I use the differentialform completly wrong?
  8. Mar 5, 2006 #7


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    The integral and differential forms are equivalent. If B is constant in time, the flux is constant in time, because you can't apply Stokes' theorem for time variant loops and surfaces.
    Even in integral form [itex]\oint \vec E \cdot d\vec l=-\partial \Phi_m/\partial t=\epsilon[/itex] the flux change is still zero if B is constant in time. The reason why you can use [itex]\epsilon=-\frac{\partial}{\partial t} \Phi_m[/itex] for a rotating loop in a constant B-field has a different reason. It's essentially the Lorentz-force acting on the charge-carriers that causes the emf, not Faraday's law. So, at this stage, the same equation has different origins. When taking into account relativity you can show they should look the same.

    Again, by Stokes' theorem, the differential and integral forms are identical: [itex]\int(\vec \nabla \times \vec B)\cdot d\vec a=\oint \vec B \cdot d\vec l=\mu_0\int \vec J\cdot d\vec a=\mu_0 I_{pierce}[/itex] (with static fields).
    The integral form is sometimes easier to work with, sometimes they are not.
    Last edited: Mar 5, 2006
  9. Mar 5, 2006 #8
    Then how can you find the B field for example from a current carying wire, by using the differential form of Ampére's law, With the integral form you simply get, that the field-strength from a fixed distance r, is (by symmetry)

    [tex] \oint \vec B \cdot d\vec l=\mu_0 I [/tex]
    [tex] B\oint dl = B2\pi r = \mu_0 I [/tex]
    [tex] B = \frac{\mu_0 I}{2\pi r}[/tex]

    How can you calculate that by using the differential form?
  10. Mar 5, 2006 #9

    Meir Achuz

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    There is a long derivation in advanced texts showing that
    the Biot-Savart law can be derived from Max's Eqs.
  11. Mar 6, 2006 #10


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    In this case you 'd have to solve something like [itex]\vec \nabla \cdot \vec B=0[/itex] and [itex]\vec \nabla \times \vec B=\mu_0 \delta^2(y,z)I \hat x[/itex] under the boundary condition that B->0 as r->infinity
    Well, how do you find the solution to many differential equations? You integrate in which case you do the exact same as in the integral form. That's what I meant to say when I they are equivalent. I`m sure there are other ways to solve these PDE's, but I suspect they are much nastier.
  12. Mar 6, 2006 #11


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    The problem formulated in the post #10 has a solution: One usually choses the gauge [itex] \nabla\cdot\vec{A} =0 [/itex] and then the problem reduces to finding the solution to well-posed problem for the Poisson equation. Usually symmetry helps, in this case, the variables are separated if one choses cylindrical coordinates...

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