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This is a question from Young and Freedman's University Physics. The book's answer says that the electric forces are zero because the wire is electrically neutral, with as much positive charge as there is negative charge.

But why is this true? If we have current flowing, that means there is a potential difference driving the current, so how are adjacent turns or opposite sides of the same turn of the coil both neutral? I mean if the wire is connected to a battery as the source, then that means there is an electric field driving the current so how can the electric forces be zero?

enter image description here

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  • $\begingroup$ Textbooks often neglect wire resistance, so no potential difference is required. $\endgroup$
    – John Doty
    Commented Jun 24 at 15:02
  • $\begingroup$ @JohnDoty But without any potential difference from the terminal how is there a current flow? $\endgroup$ Commented Jun 24 at 16:43
  • $\begingroup$ Ohm's law, $E=IR$. If $R$ is zero, $E$ is zero, even if $I$ isn't zero. Works on the whiteboard. Even works in the lab with superconductors. $\endgroup$
    – John Doty
    Commented Jun 24 at 18:45
  • $\begingroup$ The current is driven by a battery, say, and upon turn on to an ideal battery $V_{bat}$ you have the equation $V_{bat} = Ldi/dt$ implying that $i=V_{bat}t/L$ a linearly increasing function. In a real battery there is also an internal resistance so it is $V_{bat} = Ldi/dt+r_{bat}i$ exponentially limiting the current in time to max out at $I_{max}=V_{bat}/r_{bat}$. Alternatively it could be magnetically linked via a flux of another coil. $\endgroup$
    – hyportnex
    Commented Jun 24 at 18:46

2 Answers 2

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If you do account for resistance then there will be charge accumulated on the windings to create the $E$-field associated with the (presumably small) voltage drop between the terminals, like in this picture: (where we assume symmetric $+V/2$ and $-V/2$ supply voltages on the terminals)

enter image description here

To create the linear voltage drop between the ends of the coil, the precise amount of charge on each turn has a complicated (non-linear) dependence on position, but that was not the question here. Without going into precise details we can conclude that the neighboring turns will electrically repel each other (except those in the middle) but we expect this effect to be negligible compared with to the magnetic force.

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  • $\begingroup$ So if I understood your answer correctly, there will be a charge difference between neighboring turns but with small resistance, the effects will be negligible? I guess the assumption in the book is that the resistance is zero, so by Ohm's law, there is no potential difference in the coil hence no electric force. But I'm confused if we make this assumption, how is there a current at all? Or if the coil is connected to a terminal then the terminal gives a potential difference at the two ends so how does it make sense to not have any voltage drop in the wire? $\endgroup$ Commented Jun 24 at 16:49
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    $\begingroup$ I think all commenters now agree that neglecting the potential difference is an approximation. If you want to explain why current is flowing you should not make that approximation. The approximation is fine if you are only interested in the $B$-field. But if (as in this book) there is an explicit question about the $E$-field then it isn't, and the only correct answer then is: "small but not zero". $\endgroup$ Commented Jun 24 at 16:55
  • $\begingroup$ Thanks for the clarification. But how do you assume symmetric supply voltages on the solenoid? Is this because the whole system is neutral, so there will be a symmetric charge distribution on the left and right of the solenoid? $\endgroup$ Commented Jun 24 at 17:26
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    $\begingroup$ We don't see the "whole system", only the drawing of the solenoid that goes with the question. We therefore assume that the environment is not having any important effect (or else it would be unfair not to mention that in questions about the fields!) In particular: 1) No big charged objects outside the solenoid that strongly influence the $E$-field, or else our conclusions might change. 2) The two terminals simply go to the poles of a battery or some other source in a more or less symmetric way (or else the charge distribution would be less symmetric, but conclusions wouldn't change much). $\endgroup$ Commented Jun 24 at 17:51
  • $\begingroup$ Resistance is needed for the surface charge to appear when current is constant. But if current changes in time, even on a perfect inductor with zero ohmic resistance, there will be non-zero potential difference, and thus surface charges. $\endgroup$ Commented Jun 25 at 0:39
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I'll do the one dimension wire (which can be coiled or not). The length is parameterized by $x$, a Cartesian coordinate.

As stated, it has a current:

$$ j(x) = j_0 $$

and that means there is an electric field in the while, because:

$$ j = \sigma E $$,

so

$$ E(x) = j(x)/\sigma \equiv E_0 $$

Now let's apply a Maxwell equation to find the charge:

$$ \nabla E = \rho/\epsilon_0 $$

which becomes:

$$ \frac{dE(x)}{dx} = \rho(x)\epsilon_0 $$

$$ \frac{d}{dx}E_0 = 0$$

$$ \rho(x) = 0 $$

You could also look at the continuity equation which involves $\dot{j}(x, t)$, but that is zero.

Now if you wrote down some Heaviside step functions to bracket $E(x)$ with 0 field, then the derivative would leave some delta functions of charge in each end, which is the battery.

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  • $\begingroup$ What's your point? Your oversimplified example finds a field $E_0$ that is not $0$, so do you mean that OP is right and the book mentioned in the question is wrong? (OP would probably like to see a clear confirmation...) $\endgroup$ Commented Jun 24 at 15:40
  • $\begingroup$ I mean: we could also conclude that the book is right and OP is wrong since you find no charge on the wire, but that is only true (surprising as it is) in this simplified example, not in OP's solenoid! $\endgroup$ Commented Jun 24 at 16:43

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