I'm studying electrical engineering and am just learning Ampere's Law.

I'm ok with a simple conductor carrying a current fully enclosed by the integration path: Simple Loop

However, say for example there was a permanent magnet in the vicinity of the conductor so there is more flux present than just that from the current, wouldn't Ampere's law effectively say there was more current present than there actually is?

PM Present

  • $\begingroup$ You now have a magnetic field which is not of the same magnitude around the loop and note that the $\vec B\cdot d\vec l$ would have to take account of the direction of the magnetic field relative to the displacement including the fact that at times they will be in opposite directions. If you did all that around your loop which is concentric with wire I am confident that Ampere's law would still be correct. $\endgroup$
    – Farcher
    Apr 14 '17 at 9:19
  • $\begingroup$ There's no "flux" in case of Amperes law. There's only a line integral $\endgroup$ Apr 15 '17 at 8:51
  • $\begingroup$ Does this answer your question? Ampere's law and external currents $\endgroup$
    – Buraian
    Jan 31 '21 at 22:18

What is called Ampere's Law is the integral form of the 1st Maxwell equation: $$\text{curl} \textbf{H} = \textbf{J} \tag{1}\label{1}$$ Using Stokes's theorem for the vector differential operator $\text{curl}$ Maxwell's 1st differential law is shown to be the same as the integral form, ie., Ampere's law: $$\oint_\mathcal{L} \textbf{H}\cdot \textbf{dl} = \int_\mathcal{A} \textbf{J}\cdot \textbf{da} \tag{2}\label{2}$$ Here $\mathcal{L}$ is any simple closed loop in 3-space that is a boundary to some simple smooth surface $\mathcal{A}$ whose elementary line element $\textbf{dl}$ or surface element $\textbf{da}$ are assigned a vector that is proportional with the loop's tangent or the surface's outward normal, respectively.

The equivalence of the differential and integral forms, $\eqref{1}$ and $\eqref{2}$, is to be understood as being analogous to the fundamental theorem of calculus: differentiation and integration are dual operation of each other where one describes a local the other a global property of the function (field).

In both cases you have to write down all the fields including all the sources. What this means is that you must decompose the current as $\textbf{J}=\textbf{J}_c+\textbf{J}_b$. Here $\textbf{J}_c$ and $\textbf{J}_b$ denote the "true" flowing conduction current density and the "bound" current density within the permanent magnets, respectively. Now if you break it down that way

$$\oint_\mathcal{L} \textbf{H}\cdot \textbf{dl} = \int_\mathcal{A} \textbf{J}_c\cdot \textbf{da} +\int_\mathcal{A} \textbf{J}_b\cdot \textbf{da} \tag{3}\label{3}$$

and you can see that if you select an integrating loop $\mathcal{L}$ bounding a surface $\mathcal{A}$ through which only true current flows, ie., $\textbf{J}_b = 0$ you only have the contribution of the true currents in the integral, $\oint_\mathcal{L} \textbf{H}\cdot \textbf{dl} = \int_\mathcal{A} \textbf{J}_c\cdot \textbf{da}$. This does not mean that $\textbf{H}$ is the same with or without the bounded currents, instead when the magnet is brought in the field changes so that contour integral stay the same if its surface does not cross the permanent magnet that is the source of the bounded currents.

It can be shown that the bounded currents are related to the magnetic polarization of the permanent magnets as $\textbf{J}_b=\text {curl}\textbf{M}$ and $\textbf{B}=\mu _0 (\textbf{H}+\textbf{M})$


Permanent magnets are usually accounted for in Amperes law as:

$$\theta_{PM}=\frac {B_R l_{PM}}{\mu_0\mu_{PM}}$$

$\theta_{PM}$ is the scalar magnetic potential, which is expressed in Ampere (think of ot as equivalent current)

$B_R$ is the remanence flux density of the PM (flux density at zero field, the thing that makes a magnet permanent)

$\mu_0\mu_{PM}$ is the permeability of vacuum multiplied by the relative permeability of the PM.

$l_{PM}$ length of the PM

This is the case if the PM is colinear to the path you want to look at with Amperes law. If the magnet isn't colinear, you can approximately solve your problem by taking $\vec B_{R} \cdot \vec l_{path}$ (dot product) in the upper equation instead of their scalar values.

Important note: when you create a path in Amperes law you may not ommit the magnet length even though it is a source.


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