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Is there any electric field associated with current? If yes, then charge particle passing through magnetic field due to current should also experience electric force along with magnetic force.

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    $\begingroup$ I would say that the current exists because of the electric field of the other charges. Any charge sets up an electric field - moving in a current or stationary. $\endgroup$
    – Steeven
    Commented Apr 3, 2015 at 16:19
  • $\begingroup$ Could you tell me upon which factors that force depend? $\endgroup$ Commented Apr 3, 2015 at 16:24
  • $\begingroup$ Upon which factors the electric force depends? $$F_{electric}=Eq$$where $E$ is field strenght and $q$ is charge. $\endgroup$
    – Steeven
    Commented Apr 3, 2015 at 16:36
  • $\begingroup$ @Steeven: But there is an extra electric-field-component due to acceleration/current-change $\endgroup$
    – image357
    Commented Apr 3, 2015 at 17:31

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Given a charge density $\varrho$ and a current density $\vec{j}$ the EM-potentials in Lorenz-gauge are given by:

\begin{align} \varphi(\vec{r},t) & = \frac{1}{4\cdot\pi\cdot\epsilon_0} \int \frac{\varrho(\vec{r}\ ', t_r)}{|\vec{r} - \vec{r}\ '|} d^3r' \\ \vec{A}(\vec{r},t) & = \frac{\mu_0}{4\cdot\pi} \int \frac{\vec{j}(\vec{r}\ ', t_r)}{|\vec{r} - \vec{r}\ '|} d^3r' \end{align} with $t_r = t - \frac{|\vec{r} - \vec{r}\ '|}{c} $. The electric field is given by: \begin{align} E & = -\nabla \varphi - \frac{\partial \vec{A}}{\partial t} \\ & = -\nabla \varphi - \frac{\mu_0}{4\cdot\pi} \int \frac{\partial\vec{j}(\vec{r}\ ', t_r)}{\partial t_r} \cdot \frac{1}{|\vec{r} - \vec{r}\ '|} d^3r' \end{align} The first term has nothing to do with the current (it's just the usuall electrostatic term originating form the charges), while the second is zero if the current is stationary. So yes, for a non-stationary (time-varying) current, there may also be an "additional" electric field, that originates from current-change only.

This is like: If charges are moving, there is a magnetic-field. If they have an acceleration $\ne 0$, there is also an extra electric-field-component due to this acceleration.

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See, e.g., http://www.astrophysik.uni-kiel.de/~hhaertel/PUB/voltage_IRL.pdf : a power source (say, a battery) creates surface charges on the conductor, and these surface charges create electric field.

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