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Infinite current sheet: current suddenly turned on

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djy
#1
Jul29-10, 09:14 PM
P: 33
For an infinite current sheet whose current has always been on, it is well known that a purely magnetic field exists on each side, in all of space.

For an infinite current sheet whose current is initially off, then suddenly turned on everywhere simultaneously, I think I have worked out that a "front" of electromagnetic fields propagates from the sheet at c. Outside the front, there are no fields, of course. Inside the front, there are crossed electric and magnetic fields, both of which are time- and space-invariant.

This result surprised me initially. I thought there would only be a magnetic field inside the front. But EM radiation is constantly arriving from more and more distant radii of the current sheet, contributing to a constant electric field.

Does this sound correct?
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Antiphon
#2
Jul29-10, 09:47 PM
P: 1,781
No.

In order to have a constant electric field the magnetic field would have to increase linearly with time.

The dominant contribution comes from an expanding band, not expanding rings. This is due to the polarization of the current sheet. This means that the time dependent parts of the field fall off with time as 1/r^2 where r is moving out at c.
djy
#3
Jul29-10, 10:08 PM
P: 33
Feynman confirms my thinking (from his Lectures, section 18-4):

"In short, we turn on the current and the magnetic field immediately next to it turns on to a constant value B; then the turning on of B spreads out from the source region. After a certain time, there is a uniform magnetic field everywhere out to some value x, and then zero beyond. Because of the symmetry, it spreads in both the plus and minus x directions.

"The E-field does the same thing. Before t = 0 (when we turn on the current), the field is zero everywhere. Then after the time t, both E and B are uniform out to the distance x = vt, and zero beyond."

Antiphon
#4
Jul30-10, 07:16 PM
P: 1,781
Infinite current sheet: current suddenly turned on

Section 18-4 is quantum mechanics. Rotation for any spin.

If you wait long enough the e fields will vanish.

Please correct your reference to Feynman and I'll look at it.
djy
#5
Jul30-10, 08:38 PM
P: 33
Sorry -- 18-4, volume 2, in the chapter on Maxwell's equations.
Antiphon
#6
Jul30-10, 09:28 PM
P: 1,781
My word.

I can't beleive I'm even typing this, but it seems that Professor Feynman is incorrect.

The electric field will be a function of the rate of acceleration of the charge sheet. This is easily seen in the expression for the vector potential.

Suppose the current sheet ramped up from zero to one amp/sq-meter over the period of one second. Then there would be an outgoing electric field only for one second with diminishing tails as the contribution from distant points arrived.

The magnetic field is given by the curl of the vector potential, no time derivative.

In order to arrive at the magnetostatic solution after a long time the time derivatives must vanish, and they do for any particular fixed observation point.
djy
#7
Jul30-10, 10:12 PM
P: 33
Quote Quote by Antiphon View Post
Suppose the current sheet ramped up from zero to one amp/sq-meter over the period of one second. Then there would be an outgoing electric field only for one second with diminishing tails as the contribution from distant points arrived.
The key is that, even though vector potential drops off with 1/r, the circumference of points contributing to potential at any given time increases with r, thus there is no diminishing tail.

Hopefully a bit later I can post a derivation.
djy
#8
Jul30-10, 10:58 PM
P: 33
I think I got this copied over properly (it is my own derivation). Incidentally, anyone know how to get del to show up? "\del" doesn't seem to work.

Suppose an infinite sheet, in the [itex]y[/itex]-[itex]z[/itex] plane, goes from zero current before [itex]t = 0[/itex] to
[itex]\vec{j} = (0, 0, j)[/itex] current after [itex]t = 0[/itex]. Assume the thickness of the sheet is [itex]T[/itex], with [itex]T[/itex]
negligibly small so that it can be ignored in distance calculations from an observer. The observer is located
at [itex]\vec{x}[/itex]. Without loss of generality, assume [itex]\vec{x} = (x, 0, 0)[/itex]. From symmetry, the fields and potentials
are independent of [itex]y[/itex] and [itex]z[/itex]. Also assume [itex]x > 0[/itex].

We need to integrate over the entire sheet to find the potential [itex]\vec{A}(t, \vec{x})[/itex]. (Note that [itex]\phi = 0[/itex]
because there is never any net charge density anywhere). For a given point in spacetime [itex](t, \vec{x})[/itex], we
simply need find the portion of the sheet whose fields caused by the current being on have had time to
reach [itex]\vec{x}[/itex]---i.e., the portion of [itex]y[/itex] and [itex]z[/itex] on the sheet for which [itex]\sqrt{x^2 + y^2 + z^2} \le ct.[/itex]

From symmetry, it is obvious that this region is circular and we can say [itex]r = \sqrt{y^2 + z^2}[/itex], the radius
of the circle. Thus, we want to integrate over the region,
[itex] r \le \sqrt{(ct)^2 - x^2}. [/itex]
Then we have,
[itex]
\begin{align*}
\vec{A}(t, \vec{x}) \text{ (if } ct \ge x)&= \frac{\mu_0}{4 \pi} \int_0^{\sqrt{(ct)^2 - x^2}} \frac{2 \pi r T \vec{j}}{\sqrt{x^2 + r^2}} \: dr \\
&= \frac{\mu_0 T \vec{j}}{2} \int_0^{\sqrt{(ct)^2 - x^2}} \frac{r}{\sqrt{x^2 + r^2}} \: dr \\
&= \frac{\mu_0 T \vec{j}}{2} \left [ \sqrt{x^2 + r^2} \right ]_{r=0}^{r=\sqrt{(ct)^2 - x^2}} \\
&= \frac{\mu_0 T \vec{j}}{2} (ct - x). \\
\vec{A}(t, \vec{x}) \text{ (if } ct < x) &= 0.
\end{align*}
[/itex]
Then, the non-zero field components are,
[itex]
E_z = -\frac{\partial A_z}{\partial t} =
\everymath{\displaystyle}
\begin{cases}
0 & ct < x \\
\frac{-\mu_0 c T j}{2} & ct \ge x
\end{cases}.
\everymath{}
[/itex]
And [itex]\vec{B} = \text{curl} \vec{A}[/itex] implies
[itex]
\everymath{\displaystyle}
B_y = -\frac{\partial A_z}{\partial x} =
\begin{cases}
0 & ct < x \\
\frac{\mu_0 T j}{2} & ct \ge x
\end{cases}.
\everymath{}
[/itex]
So, after the current is switched on, the ``front'' of the electromagnetic field reaches an observer
after [itex]x/c[/itex] seconds, after which there are perpendicular [itex]\vec{E}[/itex] and [itex]\vec{B}[/itex] fields, constant
in both space and time. In fact, it is a null electromagnetic field, since [itex](B_y)^2 = (E_z)^2/c^2.[/itex]
Antiphon
#9
Jul31-10, 08:51 AM
P: 1,781
Ok, you've convinced me.

For any finite plane the electric field would vanish once the current was visible over the entire plane.
djy
#10
Jul31-10, 03:54 PM
P: 33
Quote Quote by Antiphon View Post
Ok, you've convinced me.

For any finite plane the electric field would vanish once the current was visible over the entire plane.
If you have a finite plane, then the current will continuously pile up as more and more net charge at the ends.

I think a wire loop, in which the current is suddenly turned on, is a better example. Once the entire current loop is within the past light cone of an observer, there should be a pure magnetic field identical to if the current had always been on.
Born2bwire
#11
Jul31-10, 08:57 PM
Sci Advisor
PF Gold
Born2bwire's Avatar
P: 1,758
Hah, neat. Infinity always messes things up with you.
chickpea
#12
Mar29-12, 04:00 PM
P: 1
This is also my result, I just did this problem and was confused enough by the E-field result to google "infinite sheet of current that turns on". Thank you for your explanation.


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