# Maxwell's Correction to Ampere's Law

I have not yet officially studied Electromagnetism but am trying to teach myself at the moment. I understand Maxwell's equations in the context of Magneto- and Electrostatics: they are equivalent, along with appropriate boundary conditions, to the Biot-Savart and Coulomb's law, respectively. In particular, they give the magnetic field due to a particular steady current distribution and the electric field due to a particular configuration of static point charges.

I am, however, confused about the meaning of Maxwell's equations when all terms are involved. In principle, I understand that a changing magnetic field can induce an electric field and a changing magnetic field can induce a magnetic field.
If both a changing magnetic field and a charge distribution is present will the ${\bf E}$ that we calculate from Gauss' Law be equal to the ${\bf E}$ we calculate from Faraday's Law? (I suspect so because of the freedom coming from the fact that divergence and curl involve derivatives and physically, from the superposition principle, we would expect the total electric field to be the sum of the electric field due to static charges and that produced due to changing magnetic field).

Griffiths, in his text on the subject, rearranges Maxwell's equations so that field sources ($\rho$ and ${\bf J}$) are on the right hand side of the equation while fields are on the left hand side. By doing so, we expect a current to produce a changing electric field and a magnetic field. If this is the case, why when we considered magnetostatics, could we neglect the changing electric field produced by the current?

A couple of other questions:

• In Ampere's Law + Maxwell correction term: ${\bf J}$ produces a magnetic field and a changing electric field. How does the magnetic field produced when the changing electric field term is present compare to the magnetic field produced when the Maxwell's correction term is not present? i.e. is less current required to produce the same magnetic field?

• Potentials ${\bf E} = -\nabla V - d{\bf A}/ dt$, ${\bf B}= \nabla \times {\bf A}$ where ${\bf A}$ is the vector potential and $V$ is the scalar potential. How does the $V$ when there is a changing magnetic field present and so ${\bf E}$ is due to both changing ${\bf B}$ field and static charges compare to the $V$ due to just the same static charges?

Thank you in anticipation of your help.

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Can you be more specific? You spend a lot of time explaining the parts you do understand, but are not very clear in what you need to know... –  Sklivvz Dec 8 '12 at 20:59

0) In your 2nd paragraph, I presume you meant to write "a changing magnetic field can induce an electric field" (Faraday's law.)

1) Your first question concerns, I think, a static charge distribution plus a set of changing magnetic fields (produced, perhaps, by a set of dynamic currents somewhere off-stage). In this case,

• the static charges will produce a conservative (curl-less) electric field.
• the changing magnetic fields will induce a divergence-less (non-conservative) electric field

The total electric field will be the sum of these two. (They obviously cannot be equal, since one has a divergence but no curl, and the other has a curl but no divergence.)

2) While a current is by definition charges in motion, magnetostatics studies cases where the currents do not change with time (and net charge density = 0). In those cases the current density $\boldsymbol{j}$ is constant in time, the resultant fields are magnetic fields that are also unchanging in time, and no changing electric fields are induced. So they're not being neglected, they just don't exist in these cases.

3) (your first bullet) The displacement current (aka Maxwell correction term) is required to maintain the consistency of the equation. The classic demonstration is of a capacitor being charged via wires. Here the current is constant but the charge on the capacitor plates is increasing linearly with time.

Imagine trying to calculate the line integral of the magnetic field around a path surrounding the wire, without the displacement current term.

• If one constructs a surface, bounded by the path, that intersects the wire, you can apply Stoke's theorem and Ampere's law and get a non-zero result.
• However, a second surface, also bounded by the path, that passes between the plates of the capacitor, has no current crossing it and so gives a zero result for the line integral.

The displacement current term removes this inconsistency, since the charging capacitor has a changing electric field between its plates.

• In the static charge case, $V$ is just the normal Coulomb potential.
• For the time-varying case, the potentials are not unique: various possibilities, all related by gauge transformations, give the same fields. For example, the Coulomb gauge (in which $\boldsymbol{\nabla \cdot A} = 0$) gives the same formula for the potential as for the static case. Other gauge choices will give different results. I'm sure your text discusses the gauge conditions.
@ZAC: Ah, I think I understand your question a bit better now. 1) Yes, there's only one $\boldsymbol{E}$ in the equations; it's not different from one to the next. 2) You can decompose that field into "divergence-less" ($\boldsymbol{E_t}$, aka transverse) and "curl-less" ($\boldsymbol{E_l}$ aka longitudinal or irrotational) components, $\boldsymbol{E=E_t+E_l}$; the transverse element is zeroed by the divergence operator of Gauss law, and the irrotational element is zeroed out by the curl in Faraday's law. 3) Yes, it's possible to have exactly the same V function in the two cases. –  Art Brown Dec 9 '12 at 18:20
@ZAC: Re 2) I want to add that I am relying on superposition of sources: a) a set of fixed charges produces $\boldsymbol{E_l}$ (and crucially produces no magnetic fields), and b) a set of time-varying currents (with 0 charge density) somewhere off-stage produces the time varying $\boldsymbol{B}$ fields you specify, and hence $\boldsymbol{E_t}$. By superposition, you can add the two $\boldsymbol{E}$ fields to get the result. Superposition works because the equations are linear in the sources. –  Art Brown Dec 9 '12 at 20:44
I should also add that, at this level, talking about $\mathbf{A}$ is probably going to just create more confusion than usefulness. There is a way to write down maxwell's equations that completely removes any reference to $\mathbf{E}$ and $\mathbf{B}$ and instead talks only about $\mathbf{A}$ (or, if you are not doing special relativity, $\vec A$ and $\phi$), but that is way beyond the level of this question. –  Jerry Schirmer Jan 7 at 23:40