An infinite wire carrying a constant current $I$ in the $\hat{z}$ direction is moving in the $$y$$ direction at a constant speed $v$. Find the electric field, in the quasistatic approximation, at the instant the wire coincides with the $$z$$ axis.

Answer: $$-(\mu_0Iv/2\pi s)\sin\phi\hat{z}$$

[This problem is prob.7.51 in Griffiths' Introduction to Electrodynamics 4th edition. There, the answer given is also shown right at the end of the problem and is wrong - it says $\cos\phi$ instead of $\sin\phi$ according to the official errata list].

In order to solve this question, I used two formulas:

Expression for the induced electric field in terms of the vector potential: $$\vec{E} = -\frac{\partial\vec{A}}{\partial t}$$.

Vector potential due to an infinite wire carrying current $$I$: $\vec{A} = -\frac{\mu_0I}{2\pi}\ln s\hat{z}$$.

Here and throughout this post I will use cylindrical coordinates where $s$ is the distance from the z-axis (and hence from the wire, since I am interested in the moment it coincides with the z-axis), $\phi$ is the angle $s$ makes with the positive x-axis and $z$ is the usual Cartesian one.

My reasoning is as follows: Although the problem doesn't make it as clear as I would like, by the answer it is clear that the problem refers to the induced electric field due to the moving magnetic field and that we should disregard the field due to the line charge in the wire. By the two formulas above, $$\vec{E} = -\frac{\partial}{\partial t}\left(-\frac{\mu_0I\hat{z}}{2\pi}\ln s\right)= \frac{\mu_0I\hat{z}}{2\pi}\left(\frac{\partial\ln s}{\partial s}\frac{\partial s}{\partial y}\frac{dy}{dt}\right)$$ Now, $$\frac{\partial\ln s}{\partial s} = \frac{1}{s}$$ and $$\frac{\partial s}{\partial y} = \frac{\partial \sqrt{x^2+y^2}}{\partial y} = \frac{y}{\sqrt{x^2+y^2}} = \sin\phi$$. Thus, the equation becomes $$\vec{E} = \frac{\mu_0 I}{2\pi s}\left(\frac{dy}{dt}\right)\sin\phi\hat{z}$$ Now comes the part that troubles me. According to the problem statement, the wire is moving in the $$+\hat{y}$$ direction and hence, $$\frac{dy}{dt} = v$$. However, this leads to two problems - first, my answer will be off by a sign, and second, I'm troubled by the fact that it is the wire, and hence the magnetic field, that's moving with this speed, and I have never encountered such a case. Moreover, we assume a quasistatic approximation, and thus I (think I) assumed that the wire is stationary right from the start, since the vector potential corresponds to the static magnetic field $$\vec{B} = \frac{\mu_0 I}{2\pi s}\hat{\phi}$$. Thus, my argument goes that instead of the wire moving in the $$+\hat{y}$$ direction, we let the space (or an arbitrary loop in space) move in the $$-\hat{y}$$ direction and hence $$\frac{dy}{dt} = -v$$. This way, the relative velocity remains the same (and according to special relativity, the situation is identical) and the answer turns out correctly. Is my argument valid? Is there anything in this (interesting and new to me) situation of a moving magnetic field that I have overlooked? I hate to be consciously affected by seeing the final answer - I think that I would never have detected any trouble with the former, apparently naive argument...

Any comments and/or corrections will be greatly appreciated!


1 Answer 1


At a fixed time $t$, the wire is the line $(0,vt,z)$ where $z$ can be any value and $s$ can be your instantaneous distance to this line. Specifically, we can consider $\vec{A}(x,y,z,t)=-\frac{\mu_0I}{2\pi}\ln s\hat{z}$ where $s$ is nothing more than a shorthand for $\sqrt{x^2+(y-vt)^2}$.

Thus, taking the quasistatic approximation, $$\vec{E} = -\frac{\partial}{\partial t}\left(-\frac{\mu_0I\hat{z}}{2\pi}\ln s\right)= \frac{\mu_0I\hat{z}}{2\pi}\frac{\partial}{\partial t} ln(\sqrt{x^2+(y-vt)^2}).$$

Just take the partial by treating $x$ and $y$ as constants and evaluate at $t=0$ and the sign works out fine because the very last chain rule gives a factor of $-v$ because the wire moves in the positive $y$ direction.

The quasistatic approximation is handled by the fact that we are just ignoring how or why the charges move in a steady way down the wire, joule losses and what compensates for them, etc. We have a nice vector potential that instantaneously gives a magnetic field around an axis just as if there had always been a steady current there, and we ignore any scalar potential that might help the charge in the wire move along. If you had a wire that moved slowing compared to how long it was, those things would approximately be true very quickly, so we did it by setting up a vector potential that acts like vector potentials act in the quasistaic limit. It doesn't change how we take the partial, and the wire is still moving, giving the position of the wire as $(0,vt,z)$ from the beginning definitely has the wire moving. It's ignoring any electrostatic forces and setting up an $\vec{A}$ that at every moment acts like the charge has always been where it is now that is the quasistatic approximation.

I don't know if you've studied retardation yet, but here are details about quasistatics. A real and finite wire with a steady current that moves slowly has the wire moving so slowly relative to its size that the wire acting due to the parts of the wire it sees is approximately the same as it acting due to what the wire is actually doing. Our vector potential is based on giving a $\vec{B}$ like what you'd get from looking at where the wire is now, as if it had been there and steady all along. In that sense it is a quasistaic type field.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.