Question

What is the magnetic field \(\vec B\) about a point charge \(Q\) at the origin travelling at velocity \(\vec v\)?

Answer 1

Assuming cylindrical coordinates \((r, \phi, z)\) with the positive \(z\)-axis aligned with \(\vec v\), I get\[\vec B = \frac{\mu_0 Qv}{4\pi r}\left[\frac{1}{z}-\frac{1}{\sqrt{z^2+r^2}} + z\left( \frac{z}{(z^2+r^2)^{\frac{3}{2}}}-\frac{1}{z^2} \right)\right] \hat \phi \]by calculating the displacement current through arbitrary disks coaxial with the \(z\)-axis, which then by Ampère's law with Maxwell's correction gives the circulation of the magnetic field about the boundary of that disk. The expression seems awkwardly complicated, though I've done it twice and haven't managed to get it any simpler. Can anyone verify this? If you need more guidance through my line of thinking, let me know.

Answer 2

Surprisingly, the solution actually looks somewhat correct.
http://www.wolframalpha.com/input/?i=z+%3D+%5Cfrac%7B1%7D%7By%7D%5Cleft%5B%5Cfrac%7B1%7D%7Bx%7D-%5Cfrac%7B1%7D%7B%5Csqrt%7Bx%5E2%2By%5E2%7D%7D+%2B+x%5Cleft%28+%5Cfrac%7Bx%7D%7B%28x%5E2%2By%5E2%29%5E%7B%5Cfrac%7B3%7D%7B2%7D%7D%7D-%5Cfrac%7B1%7D%7Bx%5E2%7D+%5Cright%29%5Cright%5D
Here, I plotted an angular slice of the function (i.e., constant \(\phi\)) setting \(\frac{\mu_0 Qv}{4\pi}=1\) and substituting \(z \rightarrow x\) and \(r \rightarrow y\), so it as if the particle is moving in the positive \(x\) direction along the \(x\) axis. The 3D graph is the magnitude of the magnetic field at every point along this cross-sectional slice. It bulges most about the origin, which is where the particle is, which would make sense, and is oriented along the line of travel. Opinions?

Answer 3

Silly me! I just realized my expression above simplifies greatly, plus a factor of \(-1\), which I fixed below.\[\large \vec B = \frac{\mu_0 Qv r}{4\pi (z^2+r^2)^{\frac{3}{2}}} \hat \phi\]Can anyone confirm this?

Answer 4

Well I haven't done these types of problems in a while so I have only questions, no answers (sorry!) Help me to understand a couple of things since you seem to be more up to date with this kind of problem than I am:
Why is it important to use Maxwell's correction to Amperes Law if you have an actual particle moving, not just a building electric field? Would not the equation work if you just calculated the current based on charge per time? It kinda looks like that's what it simplified to but I haven't worked it out myself.
Also in your plot you've substituted x for z (ok) but you put y in for r. When converting from polar to rectangular, r is a function of x and y, how were you able to simply switch coordinate systems without scale factors?
Anyways you website looks cool. Just thought I'd ask your train of thought on those two things. Thanks!

Answer 5

Oh one more thing, before I clicked on your plot I tried to picture it in my mind and I was imagining a cylindrical-looking vector field about the x axis but yours looks like its two cylinders. Is this result part of what you get from the maxwell correction? Thanks for any help remembering.

Answer 6

Maxwell's correction is just the term involving \(\frac{\partial \Phi_E}{\partial t}\) for electric flux \(\Phi_E\) through the open surface corresponding to the closd loop. This is otherwise known as displacement current.\[\oint\limits_{\text{closed loop}} \vec B \cdot d \vec \ell = \mu_0 \left(I_{\text{penetrating}}+\epsilon_0 \frac{\partial \Phi_E}{\partial t}\right)\]The reason why we look at the displacement current instead of just the current is because the current here is just a particle, and thus localized, so it doesn't really make sense to think of it that way. There are only particular open surfaces that will have a penetrating current in this line of thinking, and looking at the electric field gives us a more general view. Also, since I'm considering a point charge, you'd theoretically have an infinite amount of current in an infinitely small space, which is just a pain to deal with since it's unphysical. (At least this was my line of thinking originally, though now that I look at it, the result is really looking a lot like the Biot Savart law, with the extra factor of \(\frac{r}{\sqrt{r^2+z^2}}\) coming from the \(\sin\theta\) that results from the cross product, though I need to verify this more thoroughly..here I believe that the current would be analogous to the notion of \(Qv\)).

To answer your second question, I only switched them to \(x\) and \(y\) so WolframAlpha would interpret them as two independent variables. What you see on the 3D plot is \(B\) as a function of \(z\) and \(r\)...it's just a cross-sectional slice through the \(z\)-axis, which by itself is rectangular.

And the reason you see two cylinders is because one is where the magnetic field is going into the page and the other is where the magnetic field is going out of the page. The picture is somewhat misleading since the contour lines look like magnetic field lines of a bar magnet for instance, but you need to remember that these are where the magnetic field is of equal magnitude. It's really just circulating around the \(z\) axis, or \(x\) axis in the graph, and the value of the graph represents the signed magnitude. It's only either pointing into the page or pointing out of the page. Does that make more sense?

Answer 7

@yakeyglee

I'm afraid you solution cannot be correct.
If the charge moves at a classical velocity, you can solve the problem by change of reference frame.

In the reference frame of the particle, there is no magnetic field.
Galilean relativity for (E,B) implies that B stays the same.
So you should find B = 0 in your problem.

Now, if you use Einstein's relativity, then B is not zero in the new frame, but its expression must refer to speed of light, which is not the case in your expression.

Answer 8

Vincent, I follow the Galilean argument about the particle having no magnetic field because it is at rest in its own frame. How is Einstein's result calculated--or more directly, how can one particle "see" its own magnetic field? Thanks for any light you can shed on this for me!

Answer 9

Hi! I could not find the formula I was looking for in an English page.

But here is a page of the French Wikipedia where Galilean and Einsteinian transforms for E and B are stated:

http://fr.wikipedia.org/wiki/Transformations_de_Lorentz_du_champ_%C3%A9lectromagn%C3%A9tique

Point 2. will give you in a frame the new E' and B' in classical mechanics.
Point 3.3 will give you E' and B' in special relativity.

Answer 10

@Vincent-Lyon.Fr The existence of the B field is relative even in the Galilean argument. It is NOT correct to say that since B=0 in one frame that it thus is zero in another. Relativity gives an explanation for the effects of a magnetic field solely in terms of the electric field, leaving B with no real value. This is a Galilean interpretation, which to be self-consistent needs a B field, since it does not incorporate length contraction. The B field in the Galilean view IS dependent on relativity (i.e., notioning of reference frames), though not special relativity. The B field is a geometric interpretation, not a physical thing (think: nothing actually travels in the direction of the B field...not what creates it, nor what it affects). I want to approach this problem from solely a Galilean view, so no special relativity. Assume \(v \ll c\), if that makes you happier.

Answer 11

Look here to see how the magnetic field is just a geometric conceptualization of length contraction: http://en.wikipedia.org/wiki/Relativistic_electromagnetism

Answer 12

I don't know.
These formulae (giving E'x, E'y, E'z, B'x, B'y and B'z) work for an infinite line with linear charge λ travelling at velocity u.

Assuming γ ≈ 1 they allow you to swap from the electric field in frame where charges are at rest to the magnetic field in the frame where they are moving (electric field remains unchanged).

\(E=\Large \frac{\lambda}{2\pi \epsilon_or} \) ; \(B=0\)
to
\(E\:'=\gamma\Large \frac{\lambda}{2\pi \epsilon_or} \) ; \(B\:'=\gamma \:\mu_o\Large \frac{\lambda u}{2\pi r}\)

I'll try to work on your moving point-charge later.

Answer 13

I was too hasty in saying you equation could not be correct, with the speed of light missing in it.

Actually, if γ ≈ 1, then c will not appear here, but c is hidden in \(\mu_o\), so it was here all along.

I think your formula IS Biot & Savart's law for a "point current", although such a current is not physical.

Anyway, congrats for working out the flux of displacement current and deriving expression of B: it is not so easy!

Answer 14

Vincent, thanks for the link. I've brushed up on my relativity and my French all at once!

Answer 15

@Vincent-Lyon.Fr I think it actually is physical if we consider the point charge to be the center of a small sphere of constant charge density and charge \(Q\). The magnetic field would then only hold for the space outside of the sphere. This change of geometry doesn't affect the \(\vec E\) field and since the \(\vec B\) field is only dependent on the change of the \(\vec E\) field by displacement current, the resulting \(\vec B\) fields will be the same. Thus, this analysis is equivalent to the physical movement of a non-point charge.

Answer 16

Hi again!

Your formula \(\ \vec B = \Large \frac{\mu_0 Qv r}{4\pi (z^2+r^2)^{\frac{3}{2}}} \large \hat \phi\) is correct.

I have checked and Biot & Savart law holds for point charges.
You can write:

\(\vec B(\vec r) = \LARGE \frac{\mu_0}{4\pi} \frac{q \vec v \times (\vec r - \vec{r'})}{|\vec r - \vec{r'}|^3}\).

\(\vec r\) is where you measure the field, \(\vec r\:'\) is the position of the charge.

It can actually be derived from the Liénard-Wiechert equations for small velocities and non-accelerated particles.

See: http://en.wikipedia.org/wiki/Li%C3%A9nard%E2%80%93Wiechert_potential

Answer 17

I was not saying that a point-charge is not physical, only a point-current is a weird thing.

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