Timeline for How to evaluate the Lorentz force at a surface where the field is discontinuous?
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 18, 2017 at 12:50 | vote | accept | Cham | ||
| Jul 18, 2017 at 12:47 | comment | added | Cham | 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. | |
| Jul 16, 2017 at 23:53 | comment | added | Cham | @WSAaRV, is there a natural relation between the average field on a current surface and the current there ? | |
| Jul 16, 2017 at 13:57 | comment | added | Cham | 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 ? | |
| Jul 16, 2017 at 13:49 | comment | added | Cham | 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 ! | |
| Jul 16, 2017 at 13:23 | history | answered | Selene Routley | CC BY-SA 3.0 |