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What is the electric field outside a cylindrical solenoid when inside is turned on a magnetic field? The question is related to the question aharonov-bohm-effect-electricity-generation

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  • $\begingroup$ Note that experimental tests of the Aharanov-Bohm effect have used the vector potential generated by electrically neutral ferromagnets (an iron whisker, a shielded toroid), to which the answers below about resistive solenoids don't apply. $\endgroup$ – rob Jul 28 '14 at 19:14
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For what it's worth, it is stated in http://arxiv.org/abs/1407.4826 and references therein in the context of the Aharonov-Bohm effect that even a constant-current solenoid has outside electric fields: "always there is an electric field outside stationary resistive conductor carrying constant current. In such ohmic conductor there are quasistatic surface charges that generate not only the electric field inside the wire driving the current, but also a static electric field outside it...These fields are well-known in electrical engineering." Sorry, I have not checked that, but it sounds plausible.

EDIT (07/25/2014) Seems there is a confirmation here: http://www.astrophysik.uni-kiel.de/~hhaertel/PUB/voltage_IRL.pdf , see, especially, Fig.4 therein.

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  • $\begingroup$ Not sure why this was accepted. One must consider the relative strength of this effect to that of Faraday's law in the specific geometry. That is, this is not the all-encompassing answer; this is an effect that may or may not be important relative to the EMF induced by Faraday's law. Note that the OP asked about a change in magnetic field, so this doesn't even apply. $\endgroup$ – user3814483 Jul 26 '14 at 4:08
  • $\begingroup$ @user3814483: Of course, only the original poster can explain why my answer was accepted, but I cannot agree that the answer "doesn't even apply". English is not my native language, but the OP is about the electric field when the magnetic field "is turned on", not "is being turned on". No "change" in OP. I believe my interpretation of the question is at least reasonable, the more so as it is the relative phase (between the cases where the magnetic field is turned on and turned off) that is important for the Aharonov-Bohm effect (which is explicitly mentioned in the question). $\endgroup$ – akhmeteli Jul 26 '14 at 8:32
  • $\begingroup$ I interpret "when inside is turned on a magnetic field" to mean a change in magnetic field. My point is, knowing nothing about what the poster knows, it should be pointed out that the AB effect could be minute and irrelevant if you do indeed interpret the question as I do. $\endgroup$ – user3814483 Jul 26 '14 at 17:37
  • $\begingroup$ @user3814483: I offered some arguments to explain why I interpreted the question the way I did. You offered no arguments to support your interpretation, so why should I interpret the question the way you do? I insist that my answer is fully applicable to the question. I could agree that my answer "is not the all-encompassing answer" as, for example, I did not estimate the magnitude of the effect that I referred to, but I hope you'll agree that answers do not have to be "all or nothing". $\endgroup$ – akhmeteli Jul 26 '14 at 18:08
  • $\begingroup$ You shouldn't and you clearly didn't. I'm justifying my earlier comment about the relevance of the answer. $\endgroup$ – user3814483 Jul 26 '14 at 18:57
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The magnetic field inside of a cylindrical solenoid of radius $R$ is given by

$$ \textbf{B} = \mu_0 n I \hat{z}, $$

where n is the turn density in turns/m and I is the current. The field everywhere outside the solenoid is zero. Let's assume that the current $I(t)$ is linearly increasing, so

$$ I(t) = at . $$

For a cylinder of radius $r \ge R$, the flux through its center is

$$ \Phi(t) = \int_{S} \textbf{B} \cdot d\textbf{A} = (\mu_0 n at)(\pi R^2). $$

We can apply Faraday's law to find the electric field,

$$ \int_{\partial S} \textbf{E} \cdot d\textbf{r} = -\frac{d}{dt} \int_S \textbf{B} \cdot d\textbf{A}, $$

and use the cylindrical symmetry to assume that $\textbf{E}$ is constant along the circular boundary. Thus,

$$ \textbf{E}(r, t) = - \hat{\phi} \frac{1}{2 \pi r} \frac{d}{dt} (\mu_0 \pi R^2 n a t) = - \frac{\mu_0 na R^2}{2r} \hat{\phi}. $$

In this problem, all distances are measured from the center of the solenoid ($r = 0$). Unlike the magnetic field, the electric field is everywhere nonzero. For the more general problem, where $I(t)$ is an arbitrary function, the solution is

$$ \textbf{E}(r, t) = - \frac{\mu_0 n R^2}{2r} \frac{dI}{dt} \hat{\phi}. $$

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  • $\begingroup$ I think there's an inconsistency in your equations - the capital $R$ should be a lowercase $r$ (not a constant). If you're evaluating the flux at a particular $r$ then the line integral for $E$ needs to be evaluated there as well... $\endgroup$ – user3814483 Jul 25 '14 at 5:02
  • $\begingroup$ The other issue is the flux. JUST outside the solenoid, the flux is $(\mu_0 n a t)(\pi R^2)$. Inside, replace R with r. Outside (beyond R) the flux will actually start falling off... but definitely won't increase. $\endgroup$ – user3814483 Jul 25 '14 at 5:08
  • $\begingroup$ Thanks @user3814483, fixed the mistakes. Complete oversight on my part. Should have paid more attention to how I defined r and R. Thanks! $\endgroup$ – Ultima Jul 25 '14 at 5:09

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