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Homework Statement
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}##]
[In griffiths' book, 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]
Homework Equations
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 zaxis (and hence from the wire, since I am interested in the moment it coincides with the zaxis), ##\phi## is the angle ##s## makes with the positive xaxis and ##z## is the usual Cartesian one.
The Attempt at a Solution
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, [tex]\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)[/tex] 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 [tex]\vec{E} = \frac{\mu_0 I}{2\pi s}\left(\frac{dy}{dt}\right)\sin\phi\hat{z}[/tex] 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 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!
EDIT: I attached the figure corresponding to this problem to help visualize this one.
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