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I'm wondering how an external electric field influences an electric current in a conducting wire and how could we measure this effect.

Without an external electric field, the conducting medium resistance, the current passing through it, and the potential difference applied to its extremities can be related by the formula

\begin{equation} U=R\,i\,. \end{equation}

If an external electric field is applied and it opposes the movement of the charge carriers this will be perceived as an increase in the medium resistance, right? Conversely, if its orientation is favorable to the movement of the charge carriers it will be perceived as a decrease in the resistance. If $U'$ is the electric potential of such an external electric field, then we have that

\begin{eqnarray} U+U' &=& (R+R')\,i\, \\ U_{\mathrm{eff}} &=& R_{_{\mathrm{eff}}}\,i, \end{eqnarray}

which leads us to

\begin{eqnarray} U' = (R_{\mathrm{eff}} - R_{\mathrm{nom}})i\,, \end{eqnarray}

where $R_{\mathrm{nom}}$ is the nominal resistance, measured without the external electric field. So, with two measurements, one with and one without the external electric field, it is possible to determine the electric potential and, consequently, the external electric field amplitude and orientation. Is this correct?

Thank you very much.

NeonGabu
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1 Answers1

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No, it won't be perceived as an increased resistance. The resistance of a medium is due to the nature and geometry of the medium itself (and it is linked to irreversible processes that transform energy into heat), and it has nothing to do with the electric field it is immersed in.

By the superposition principle, the effect of the external electric field is to provide an additional potential difference, so as to have

$$ U + U' = R i' $$

which means that changes is the current flowing through the medium, not its resistance.

lr1985
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