I'll take a simple case as an example. You have a constant and uniform magnetic field inside an ideal infinitely long solenoid, with currents circulating all around the thickless coils (so there's a surface current with a field discontinuity there). \begin{equation}\tag{1} \vec{B}_{\text{inside}} = \mu_0 \, n \, I \: \vec{z}, \end{equation} where $\vec{z}$ is the unit vector oriented along the solenoid's main axis. The field is 0 outside the solenoid (ideal case) : $\vec{B}_{\text{outside}} = 0$.

The magnetic field exerts a Lorentz force density $\vec{f} = \vec{J}_{\text{sol}} \times \vec{B}$ on the currents that are creating that same field, so there is magnetic pressure acting on the solenoid. That pressure should be proportional to the field's energy density.

Now, how do you evaluate the magnetic field, at the solenoid's surface, that should act on the current density $\vec{J}_{\text{sol}}$ ? $\vec{B}_{\text{inside}}$ given above ? $\vec{B}_{\text{outside}} = 0$ ? The average defined as this : \begin{equation}\tag{2} \vec{B}_{\text{average}} = \frac{\vec{B}_{\text{inside}} + \vec{B}_{\text{outside}}}{2} = \frac{1}{2} \, \mu_0 \, n \, I \: \vec{z} \quad ? \end{equation} or what else ? If it's the average (2), how can you justify it ?


It is indeed the average, as you suspect. To understand why, witness that the current is spread over a nonzero thickness. Begin with an azimuthal directed current density $J(r)$ (sketch this) and you get two equations:

$$\frac{\mathrm{d} \,B(r)}{\mathrm{d}\,r} = -\mu_0\,J(r)\quad\text{(}\vec{B}\text{ axially directed)}$$ $$\frac{\mathrm{d} \,F(r)}{\mathrm{d}\,r} = r\,\theta\,L\,J(r)\,B(r)\quad\text{(Lorentz force on sector of azimuthal subtense}\,\theta,\,\text{length}L)$$

Eliminate $J$ to find that $B$ varies with $F$ according to $\frac{\mathrm{d} \,F}{\mathrm{d}\,B} = -r/\mu_0$; this simple equation shows you very simply that the force on an infinitely thin current sheet is calculated with the average of the magnetic fields inside and outside the coil.

The same principle works in calculating the force exerted by a plane wave on a perfect conductor when the former is reflected or absorbed by the latter. You get the same answer whether you use half the magnetic field at the on the outer surface acting on the current sheet, or whether you do the full calculation with exponentially dwindling fields with depth owing to the skin effect.

  • $\begingroup$ I just came to the same conclusion, using an integral instead of a differential equation. So the solution is to "dilute" the current density, instead of considering a mathematical thickless idealization, that has nothing to do with reality. This is interesting ! $\endgroup$ – Cham Jul 16 '17 at 13:49
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    $\begingroup$ I diluted the surface current density on a thin layer of thickness $2 \varepsilon$, replacing the Dirac $\delta(r - R)$ by an uniform finite distribution $\delta_{\varepsilon}(r - R)$ over the interval $R - \varepsilon < r < R + \varepsilon$, and then applying the limit $\varepsilon \rightarrow 0$ at the end of the calculattion. This does the job, but feels a but tricky, since I could have distributed the whole current entirely inside the solenoid, or outside. That gives different results. So the procedure feels ad hoc to me. How to justify centering the distr. around the boundary ? $\endgroup$ – Cham Jul 16 '17 at 13:57
  • $\begingroup$ @WSAaRV, is there a natural relation between the average field on a current surface and the current there ? $\endgroup$ – Cham Jul 16 '17 at 23:53
  • $\begingroup$ Ok, I found the relation I was looking in my previous comment : \begin{equation}\vec{J}_s = \frac{1}{\mu_0} \: \vec{n} \times \Delta \vec{B}_s \, \delta(r - R),\end{equation}where $\vec{n}$ is the normal to the discontinuity surface and $\Delta \vec{B}_s \equiv \vec{B}_{\text{ext}} - \vec{B}_{\text{int}}$ is the field discontinuity. $\delta$ is a Dirac distribution. $\endgroup$ – Cham Jul 18 '17 at 12:47

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