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By Ampere's Law, using a rectangle as an Amperian loop, I know that it can be derived that the magnetic field is constant throughout all of space (i.e. it does not depend on the distance r from the infinite sheet).

I'm confused as to how this can be possible, as a point close to the current sheet would have lesser distances $R_i$ from every single line of current, and hence the B field produced by each individual current line is larger than for a point further away from the sheet. Since the B fields produced by the individual current lines are larger in magnitude and all have a component in the direction of the net magnetic field, shouldn't the superposed B fields create a larger net B field at a point of lesser r than at a point of greater r? How can the results from Ampere's Law be explained intuitively?

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  • $\begingroup$ Are you sure that thinking of a continuous sheet in terms of individual current "lines" is a valid approach? Suppose, when I double the distance, that I also double the number of lines that I consider. Now we get a constant field again. $\endgroup$ – probably_someone Jan 2 '17 at 1:07
  • $\begingroup$ Can you explain why doubling the distance doubles the number of lines that you would consider? $\endgroup$ – ABCDF Jan 2 '17 at 3:02
  • $\begingroup$ It doesn't, I just make that arbitrary choice. You can divide the continuous charge density into any number of arrays of lines, and I just choose two different ones that can give me the same apparent current density at two different distances. $\endgroup$ – probably_someone Jan 2 '17 at 3:05
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The central question you must ask yourself is: how do you tell how far away you are from a featureless infinite wall? No matter what distance you are from it, the wall looks exactly the same: featureless and infinite. This should intuitively tell you that the field, which depends on "how the wall looks", cannot depend on how far you are away from the wall.

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  • $\begingroup$ Incidentally, this is why thinking of a continuous distribution as discrete "lines" is incorrect; you're adding features to the wall that aren't actually there, so you can tell how far away it is. $\endgroup$ – probably_someone Jan 2 '17 at 1:18
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    $\begingroup$ This is a fallacious argument. If we have an infinite current-carrying wire instead of a sheet, then it would still be true that "matter what distance you are from it," it "looks exactly the same: featureless and infinite." And yet the field of the wire is not constant throughout all of space. $\endgroup$ – user4552 Jan 2 '17 at 5:20
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    $\begingroup$ I may have worded this slightly badly. The reason that the infinite sheet case gives constant current is because the problem looks the same under translations in any direction (which is what I was trying to say in my answer), so we know that it cannot depend on any coordinate. In the case of the wire, it only looks the same if you translate along two axes (i.e. along the direction of the wire and directly away from the wire). Moving in a direction perpendicular to both of those will cause the problem's setup to change, so we don't get the same symmetry. $\endgroup$ – probably_someone Jan 2 '17 at 5:27
  • $\begingroup$ If you continuously move perpendicular to both the along-the-wire dimension and the away-from-the-wire dimension, then you will be moving in circles (orbiting the wire), the wire will continue to look the same, and the field strength will never change. $\endgroup$ – ruakh Jan 2 '17 at 5:33
  • $\begingroup$ Who said I was using polar coordinates? I meant moving in the Cartesian direction perpendicular to those two. $\endgroup$ – probably_someone Jan 2 '17 at 5:33
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The key here is that it is an infinite plane. If you buy that the electric filed is constant outside of an infinite plane of charge, you have to buy this too. I find that the electric field argument is more intuitive though, so think about why that is true first. For any distance away from the plane you say that the electric field should be stronger or weaker, I always have enough plane (and the symmetry about the origin) to exactly cancel the electric field from any point you give me. In short, you give me a distance, I can always match to "cancel that dependence on distance out" because of symmetry in the plane.

Now, let's try to translate this reasoning to magnetic fields. To gain an intuition look at the law of Biot-Savart

$$\mathbf{B}=\frac{\mu_0}{4\pi}\iint\frac{\mathbf{K}\times (\mathbf{r-r'})}{|\mathbf{r-r'}|^3}d^2r'.$$

Since the plane is infinite, it doesn't matter where you place your origin, so put it anywhere. Due to symmetry in the $xy$-plane, it can't depend on the distance from the origin in the $xy$-plane. So the only option is that it depends on the $z$ coordinate. Hence, the equation looks like

$$\mathbf{B}=\frac{\mu_0}{4\pi}\iint\frac{\mathbf{K}\times (\mathbf{z-r'})}{|\mathbf{z-r'}|^3}d^2r'.$$

Now, try to come up with an argument for why the electric field should depend on the position above the plane $z$. For every distance $z$ above the plane you give me, I can always have enough $xy$-plane to make the quantity $\mathbf{B}$ identical to what it would be at some other point $z$.

I suggest simply doing the Biot-Savart integral over all of $\mathbb{R}^2$ very carefully to convince yourself of this. It's one of those things that you should do it at least once in your life with your own hands. It still won't be completely obvious, but you will get a better feel for how it works out to be constant.

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There is a simpler version of this question, which is why the electric field of an infinite sheet of charge is constant throughout all of space. And in fact, the two questions are basically the same. You can transform to a frame moving parallel to the sheet, and then you get a magnetic field as well as an electric one. The electric field $E$ in the sheet's rest frame, at any given point in space, transforms to a combination of fields $E'$ and $B'$ in the new frame. The fields are finite in one frame if and only if they're finite in the other.

So let's consider the simpler question about an infinite sheet of charge. One can calculate the field using Gauss's law, but you asked for an intuitive justification, and I think the intuition is pretty straightforward here, if you just think in terms of field lines. Every charge in the sheet is a source of electric field lines. Because the problem has translational symmetry in the sheet's rest frame, those field lines must be straight and perpendicular to the sheet. Since the field lines are all parallel to one another, their density never changes. Therefore the field is always finite, and is uniform on either side of the sheet.

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I just thought I'd add another argument to InertialObserver's answer in case it helps someone's intuition.

If you think of the plane as composed of infinite parallel wires with current in x, each produces a B-field going around in a circle in the y-z plane. Close to the plane, wires on the plane far away from the origin have almost no y component of field (the only one that survives) - but each wire has a strong field because of the $\frac{1}{r}$ dependenace. But far away from the plane, wires far away on the plane have most of their field in the y-direction. Its these two effects that cancel - a larger component of a weaker field vs. a weaker component of a stronger field.

More formally: Let's say the wires pass a current in the $x$ direction. Because the plane is infinite, any point in the z-direction is equivalent to a point at the origin. So let's say our point of interest is a z-distance $a$ from the origin.

Because these are infinite wires they produce the standard B-field $\frac{\mu_{0}I}{2\pi r}e_{\phi}$, where $e_{\phi}$ is the standard unit vector in polar coordinates, in the y-z plane. If we have a current density J, then for each `wire' $dx$, $dI=Jdx$. If you convert into Cartesian coordinates the y-component integral of all these wires will end up being:

$$ -\frac{\mu_{0}K}{2\pi}\int_{-\infty}^{\infty}\frac{z}{z^2+y^2}dy $$

Just think about the integrand. If $z\to 0$ then the numerator is small. This is because the B-field circles from wire elements far away have not yet started to `bend' into the y-direction. But as $z\to\infty$ the numerator becomes large because for any finite y, the field of the circle is almost all in this direction. Its these two effects that cancel each other out - a larger component of field from further away circles in y against the falling off in distance that you'd expect. Doing the integral will show you the cancellation is exact - and exactly analogous to Gauss's law for parrallel plates as people above have pointed out.

Happy to delete if this is not helpful and clogs up the discussion.

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