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Physically, what does $$\text{Force}\times d(\text{Current})$$ measure?

For example, if we have a boundary of a system and something flowing into the systems through the boundary. Then what is the physical meaning of $$\text{Force applied to the boundary}\times d(\text{Current passing through the boundary}),$$ where $d$ stands for the differential operator.

Could you give examples, please.

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    $\begingroup$ Do you have any reason to think it has an interesting physical meaning? Where did this combination come from? $\endgroup$
    – Javier
    Commented Mar 18, 2020 at 18:33
  • $\begingroup$ Force times current is generally something like power, so this would give you something like derivative of power? But your force is constant. Constant force and changing current means the system has some kind of inertia, like a flywheel in a water stream? $\endgroup$
    – Stack Exchange Supports Israel
    Commented Mar 18, 2020 at 18:53
  • $\begingroup$ @user253751 why did you assume the force it constant? $\endgroup$
    – Our
    Commented Mar 18, 2020 at 19:23
  • $\begingroup$ @Javier I remember seeing it in electromagnetic theory courses, but I can find it, and that is why I am asking for examples. Note that the cause of the force is the current. $\endgroup$
    – Our
    Commented Mar 18, 2020 at 19:24
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    $\begingroup$ @onurcanbektas OK, just keep it in mind for the future. (You've been around long enough that I thought you might have known, but we understand that people miss things sometimes.) And in general, it's better to write the question title separately, as a summary of the body - copying and pasting is not usually a good way to get a good question title. $\endgroup$
    – David Z
    Commented Mar 18, 2020 at 23:17

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I do not know of an example of Force $\times$ [derivative of current] but there’s an easy example of Force $\times$ current density.

Consider the force on a charge $q$: \begin{align} \vec F = q(\vec E+\vec v\times \vec B)\, . \end{align} The infinitesimal work done over a displacement $d\vec \ell$ is \begin{align} dW = \vec F\cdot d\vec \ell \end{align} and, assuming the force is constant over the small distance $d\vec\ell$, the power is \begin{align} P=\frac{dW}{dt}&= \vec F\cdot \frac{d\vec \ell}{dt}= \vec F\cdot \vec v\\ &= q \vec E\cdot \vec v \end{align} since $\vec v$ is perpendicular to $\vec v\times \vec B$ (i.e. magnetic force does no work). For $N$ charges per volume, this becomes \begin{align} p= \vec E\cdot (N q \vec v)= \vec E\cdot \vec J \end{align} where $\vec J$ is a current density and $p$ is a power density in Watts/meter$^3$ as $N$ is a volume charge density in C/meter$^3$.

Thus suggest that $\vec F\cdot d \vec J$ would be something like $dp/q$, i.e. the power density increment per unit charge.

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