# Induced electric field due to a moving wire

• ELB27
In summary: I think this way of looking at it makes it seem more like the wire is moving and the point is stationary - that's the perspective you want for this problem.

## 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 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.

## 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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• IMG_20141227_125305.jpg
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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.

ELB27
TSny said:
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.[...]
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).

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ELB27 said:
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}}##.

That looks very good.

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).

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

ELB27
I am not the author of this figure.OP 2017-07-03: After some thinking and reading, I came to the following conclusion:
The answer given by Griffiths is correct, and the argument that I present in the original post is not.
Indeed, the current carrying wire is moving with a constant speed ##v## in the ##+\hat{y}## direction. This means that the wire itself is moving, and not the space surrounding it. In other words, the wire is not stationary in any way. The field that we are calculating is the field due to the magnetic field generated by the current in the wire, and not the field due to the moving wire itself. This is a very important point, since in the latter case, the quasistatic approximation would not hold, and we would have to use the full expression of the electric field due to a moving line charge. In the case of interest, however, the quasistatic approximation holds and thus, the vector potential that we are using corresponds to the static magnetic field generated by the current in the wire. This is the reason why the wire is assumed to be stationary in the original problem statement.
Thus, the electric field is indeed \vec{E} = \frac{\mu_0 I}{2\pi s}\left(\frac{dy}{dt}\right)\sin\phi\hat{z} as I had calculated. In this case, ##\frac{dy}{dt} = v##, since the wire is moving in the ##+\hat{y}## direction and thus, the relative velocity between the wire and the space is ##+v##. This means that the electric field is in the ##+\hat{z}## direction and is thus the same as the one given by Griffiths.
To summarize, the vector potential corresponds to the static magnetic field generated by the current in the wire, and the electric field is the one that arises due to the change in the magnetic field as the wire moves. The current in the wire remains constant but the magnetic field changes, and this is why there is an induced electric field.

## 1. What is an induced electric field due to a moving wire?

An induced electric field is a type of electromagnetic field that is created when a wire carrying an electric current moves through a magnetic field. This movement causes a change in the magnetic flux, which in turn creates the induced electric field.

## 2. How is the direction of the induced electric field determined?

The direction of the induced electric field is determined by two factors: the direction of the wire's movement and the direction of the magnetic field. The induced electric field will always be perpendicular to both of these directions.

## 3. What is the relationship between the strength of the induced electric field and the speed of the moving wire?

The strength of the induced electric field is directly proportional to the speed of the moving wire. This means that the faster the wire moves, the stronger the induced electric field will be.

## 4. How does the distance between the wire and the magnetic field affect the induced electric field?

The distance between the wire and the magnetic field does not directly affect the strength of the induced electric field. However, the strength of the magnetic field itself can affect the induced electric field. A stronger magnetic field will result in a stronger induced electric field.

## 5. Can the induced electric field be used for practical applications?

Yes, the induced electric field has many practical applications. It is used in devices such as generators and transformers to convert mechanical energy into electrical energy. It is also used in technologies such as magnetic levitation and induction cooktops.