Timeline for Induced electrical field of infinite straight current carrying wire
Current License: CC BY-SA 4.0
22 events
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| Oct 19 at 10:10 | history | edited | LPZ | CC BY-SA 4.0 |
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| Jul 1 at 11:36 | history | edited | LPZ | CC BY-SA 4.0 |
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| Oct 26, 2023 at 19:05 | history | edited | LPZ | CC BY-SA 4.0 |
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| Sep 25, 2023 at 23:00 | comment | added | Ján Lalinský | > "Essentially, it shows that an isolated wire is not physical, and they always come in pairs of opposing currents like in waveguides (this is a form of confinement). " Isolated wire certainly is physical, and they do not always come in pairs. Infinite wire, isolated or not, is unphysical. | |
| Sep 25, 2023 at 22:20 | history | edited | LPZ | CC BY-SA 4.0 |
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| Sep 25, 2023 at 20:26 | history | edited | LPZ | CC BY-SA 4.0 |
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| Aug 24, 2023 at 4:41 | comment | added | LPZ | No, with an oscillating current, the magnetic field will be oscillating as well. Even the OP’s solution shows oscillations, as usual, the electric and magnetic field are merely in quadrature. | |
| Aug 24, 2023 at 4:37 | history | edited | LPZ | CC BY-SA 4.0 |
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| Aug 24, 2023 at 0:03 | comment | added | Ján Lalinský | @LPZ displacement current is vital here: if we assume it vanishes, magnetic field is restricted to be a linear function of time, and OP assumes oscillating current, which will have oscillating magnetic field. | |
| Aug 23, 2023 at 23:23 | history | edited | LPZ | CC BY-SA 4.0 |
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| Aug 23, 2023 at 23:19 | comment | added | LPZ | The OP used the quasistatic approximation, so I just focused on its consistencies. Taking into account the displacement current is not always necessary. | |
| Aug 23, 2023 at 23:17 | history | edited | LPZ | CC BY-SA 4.0 |
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| Aug 23, 2023 at 18:50 | comment | added | Ján Lalinský | The electric field obeys $\nabla \times \mathbf E = - \frac{\partial \mathbf B}{\partial t}$, but this equation is not enough to find the fields, we need the other equation $\nabla \times \mathbf B = \mu_0 \mathbf j + \mu_0\epsilon_0 \frac{\partial \mathbf E}{\partial t}$ too. I think it makes no sense to seek solutions to only $\nabla \times \mathbf E = 0$ here, the magnetic term is important. | |
| Aug 23, 2023 at 18:45 | comment | added | Diger | I see, anyway. I think Jan's answer is more what I was actually looking for, but I wouldn't mind if you elaborate specifically on the last point about the log-divergence and how it arises and cancels with two wires? | |
| Aug 23, 2023 at 18:42 | comment | added | Ján Lalinský | The relevance is that electric field with zero curl is wrong for the assumed situation - the correct field has to have non-zero curl in general (that can be zero only at special position and time). | |
| Aug 23, 2023 at 18:00 | comment | added | LPZ | @Diger It’s rather an “option. The formalism allows you to add a uniform charge distribution (since you are looking at $z$ invariant solutions). If you set $c_1=0$, then there will be no charge. | |
| Aug 23, 2023 at 17:58 | comment | added | LPZ | @JánLalinský yes for the second charge density commentary, it was just a bad copy paste. For your first comment, I don’t see how it’s relevant. I’m just looking at the homogeneous solutions, the discussion is purely mathematical. | |
| Aug 23, 2023 at 17:56 | history | edited | LPZ | CC BY-SA 4.0 |
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| Aug 23, 2023 at 17:18 | comment | added | Diger | Just wondering, if I have a current along the $z$-axis, does it necessarily represent a charge distribution as well? I mean, this is how I interpret this answer, because I was actually only considering a current, not a charge. | |
| Aug 23, 2023 at 17:15 | comment | added | Ján Lalinský | Circulation of $\mathbf E$ has nothing to do with charge distribution on the z axis. For field $\frac{c_2}{2\pi r} \mathbf e_{\varphi}$, circulation is non-zero but does not depend on radius, so curl of such field vanishes except on the z axis. | |
| Aug 23, 2023 at 17:09 | comment | added | Ján Lalinský | $\nabla \times \mathbf E = 0$ does not hold except in static case where $\mathbf B = const.$, or specific instants when $\partial \mathbf B/\partial t = 0$. | |
| Aug 23, 2023 at 16:43 | history | answered | LPZ | CC BY-SA 4.0 |