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Reading through my lecture notes I have written that the electric field $E$ drives a current $I$ around a wire such that $E = \frac{V}{L}$ where $L$ is the length of the wire and $V$ is the potential difference across the wire. Where does this come from?

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  • $\begingroup$ It would be useful if you could provide details from your lecture notes. $\endgroup$ – ZeroTheHero Mar 20 '17 at 14:47
  • $\begingroup$ What sort of details would you like? $\endgroup$ – jaghd Mar 20 '17 at 14:59
  • $\begingroup$ I don't understand what do you actually want to claim. Can you be more specific? $\endgroup$ – Feynman Mar 21 '17 at 1:00
  • $\begingroup$ Please do not post images of texts you want to quote, but type it out instead so it is readable for all users and so that it can be indexed by search engines. For formulae, use MathJax instead. $\endgroup$ – user191954 Nov 7 '18 at 18:05
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  • Step 1 is to find the relation between the resistance $R$, the conductivity $\sigma$ of the material, and the cross-section of your wire.
  • Step 2 is to find the relation between the electric field and the current density $J$. This involves the conductivity $\sigma$.
  • Step 3 is to relate the current density $J$ to the net current $I$ in your wire.

If your text discusses Poynting vectors it should have all the above information as well.

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Notice for constant E the equation is magnitude of potential difference V = EL. You get the form E = V/L. Also J=E/resistivity. Multiply this by are of wire on both sides, and the numerator and denominator of right side by L. Use areaxJ=I the current and area/(Lxresistivity)=1/R. You get the V=IR

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