# Induced electric field due to a moving wire

1. Dec 27, 2014

### ELB27

1. The problem statement, all variables and given/known data
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]
2. Relevant 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 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.

3. 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, $$\vec{E} = -\frac{\partial}{\partial t}\left(-\frac{\mu_0I\hat{z}}{2\pi}\ln s\right)= \frac{\mu_0I\hat{z}}{2\pi}\left(\frac{\partial\ln s}{\partial s}\frac{\partial s}{\partial y}\frac{dy}{dt}\right)$$ 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 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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Last edited: Dec 27, 2014
2. Dec 28, 2014

### TSny

You want to find the electric field at some fixed point (x, y, z) in space at the specific time, t0, when the wire is coincident with the z-axis.

Thus, you want $\vec{E}(x, y,z, t_0) = -\left[\frac{\partial \vec{A}(x,y,z,t)}{\partial t} \right]_{t=t_0}$.

x, y, and z have fixed values in $\vec{A}(x,y,z,t)$ inside the bracket. Only t is changing.

Can you express the distance $s$ between the fixed point and the wire in terms of the fixed values of x and y, the speed v of the wire, and the time variable t? For convenience, you can assume that the time at which the wire is coincident with the z-axis is t0 = 0.

3. Dec 28, 2014

### ELB27

The position of the wire is $s'=vt\hat{y}$ (If at time $t=0$ it coincided with the z-axis), the position of the arbitrary point is $s=x\hat{x}+y\hat{y}$ (I neglect the z-coordinate as it is irrelevant to the distance between the point and the infinite wire). therefore the distance between them is $|s-s'| = \sqrt{x^2+(y-vt)^2}$. Then, $\vec{E} = -\frac{\partial\vec{A}}{\partial t} = \frac{\mu_0I}{2\pi}\frac{\partial\ln s}{\partial t}\hat{z} = \frac{\mu_0I}{2\pi}\frac{\partial\ln s}{\partial s}\frac{\partial s}{\partial t}\hat{z} = \frac{\mu_0I}{2\pi s}\frac{\partial \sqrt{x^2+(y-vt)^2}}{\partial t}\hat{z} = \frac{\mu_0I}{2\pi s}\frac{(-v)(y-vt)}{\sqrt{x^2+(y-vt)^2}}\hat{z} = -\frac{\mu_0Iv}{2\pi s}\sin\phi\hat{z}$ where, by trigonometry, $\sin\phi = \frac{y-vt}{\sqrt{x^2+(y-vt)^2}}$.

Thanks you very much, I like this argument - more straightforward. (Although I think that it is identical to mine above - here the equations are identical to the case where the arbitrary point moves with velocity $\vec{v}=-v\hat{y}$, right? In either case I should have been more rigorous).

Last edited: Dec 28, 2014
4. Dec 28, 2014

### TSny

That looks very good.

Yes, moving the point in the opposite direction while keeping the wire fixed will yield the same result for $\frac{\partial s}{\partial t}$.