00:01
We start with gauss's law to find the electric field, and we're going to assume that it's infinitely long, this line charge.
00:11
So we have an integral that we take of the azimuthal angle around the line charge.
00:17
The length l is along the length of the line charge, and we go ahead and assume that we can integrate out along this length, and we get rid of that length l.
00:30
And from gauss's law we obtained the electric field.
00:34
So here we have obtained the integral from 0 to 2 pi along the azimuthal angle.
00:44
And we get the magnitude.
00:45
This is only the magnitude of the electric field.
00:48
Later i'll specify the vector components.
00:52
We're assuming that the line charge runs along the x -axis, and later we'll be able to see that the vector components are along the y and z directions.
01:03
So the line of charge is typically, if you think about it, is really just charge along the x axis, and really a line of charge is the amount of charge per unit length.
01:21
So if it's per unit length, and we go to another reference frame that's moving relative to us, then there's going to be length contraction for every element of length in the line of charge.
01:33
So this means that the line of charge is going to have a length contraction associated with it.
01:43
Every length element will be length contracted, and this means that the line charge can be written this way.
01:55
So we have gamma times dq over dx in the original reference frame.
02:01
And that means the electric field in the other reference frame will be given by, at least in magnitude, by e prime, equals the original magnitude times gamma.
02:20
And i've written it explicitly in terms of velocity.
02:27
Moving on then, we want to get an expression for the current.
02:32
In the original reference frame where the line charge is at rest, there is no electric current and there is no magnetic field because no charges are moving.
02:41
But in this other reference frame where the line charge is observed to be moving, there will in fact be a current and there will be a magnetic field.
02:49
And so we see that the current is going to be given by charge length times the length per unit time, charge over length, i should say, times the length over time.
03:10
Because after all, current is charge per unit time.
03:13
So if we have charge per length times length per time, then we'll see that the length will cancel and we'll just have charge per unit time.
03:23
So we should have v lambda times v, where v is velocity.
03:29
Now, in this other reference frame, it will observe the length, the line charge to be moving with a velocity negative v.
03:37
And that's why we have a minus sign.
03:39
So it sees a current of negative lambda v times gamma.
03:44
And that's what we get for the electric current, as observed in this reference frame.
03:50
The magnetic field is going to be obtained.
03:54
Using a formula for moving reference frames if you know an electric field, namely negative 1 over c squared with the cross product of velocity of the reference frame with the electric field.
04:08
So in the original reference frame, there was no magnetic field, but there was the electric field.
04:14
If we take the cross product of that electric field with the velocity of the reference frame, then in this reference frame where we will observe a magnetic field, we can get that magnetic field with this formula.
04:28
So now this is where we're going to need to know the vector components of the electric fields that we have calculated.
04:38
So we know that the line charge in either reference frame is along the x -axis.
04:45
And since it is a radially symmetric field, this means that the electric field components need to be along the y and z directions, and they need to be symmetric.
04:55
So this means that the y and z components will be such that the electric field as a vector can be written in this way.
05:07
So we'll have y plus z as vector components divided by the square root of two...