# Electron Velocity vs Field Strength?

1. Apr 21, 2008

### HarryWertM

I have two cathode ray tubes which generate the same amount of beam, i.e., one tube fires off x electrons per second to the screen at velocity m; the other fires off x electrons per second at velocity n. Say m=2n.

1. Is magnetic field FROM BEAMS ALONE the same??

2. One twice the other? I.e. linear field vs. velocity relationship?
2a.If yes, up to relativistic speeds?

-Harry Wertmuller

2. Apr 21, 2008

### Gokul43201

Staff Emeritus

3. Apr 21, 2008

### HarryWertM

This wasn't from any textbook I ever read. Trying to re-learn physics from the web. Not that I've read many. Thunk up the Q all by myself. If its in some text and you remember what text; course; or even what college [where I might look at their catalog], please advise.

4. Apr 21, 2008

### lbrits

The magnetic field strengths are proportional to the ordinary velocity, for all velocities. Magnetic fields due to currents hold their relationship regardless of the velocities of the charge carriers, although that relies on the fact that the current is uniform. Cheers.

5. Apr 21, 2008

### HarryWertM

"Proportional to the ordinary velocity" and "regardless of the velocities" seem to me to contradict.

"Relies...... on.... current is uniform". Intuitively to me, all currents are non-uniform at the subatomic level - electrons jumping thru atomic shells and electrons flying hru CRT vacuum seem both equally non-uniform. So are you saying the two CRTs yield the same field strength around their respective beams?

6. Apr 21, 2008

### HarryWertM

Perhaps I should amplify why I pose this question. The wikipedia article on "Relativistic Electromagnetism" made me foggy, but here is my understanding: The magnetic force is a by-product of relativistic length contraction. Thus, the when a beam travels faster, it will generate a stronger field [because more length contraction], even for same current strength [number of electrons per second]. However, this increase is non-linear, increasing markedly near c. So a ten percent velocity increase from 90 to 99 percent of c will yield much more than a ten percent increase in magnetic force.

BUT various discussions [including lbrits above] seem to contradict my wikipedia understanding. So not sure of anything. [I generally trust wikipedia.]

7. Apr 21, 2008

### lbrits

I wouldn't trust Wikipedia on everything.

The magnetic field strength is proportional to the velocity, and this proportionality persists for all velocities. That is what I meant.

Yes, currents are essentially non-uniform on a quantum level, but at the level at which asking the question "what is the strength of the magnetic field" is a well defined question, uniform currents is a fairly good approximation.

The "relativistic electromagnetism" viewpoint, while correct, is not really all that correct. It is heavily biased towards one particular reference frame, and if all reference frames are to be on equal footing...

Mmm...

The magnetic field of a relativistic moving charge is (Griffiths)
$$\mathbf{B} = \frac{\mu_0}{4\pi} \frac{ q v (1- \beta^2) \sin\theta } { (1-\beta^2 \sin^2\theta)^{3/2} } \frac{\hat\phi}{R^2}$$
assuming the particle is moving in the z-direction. There is indeed a factor of $$\gamma$$ from transforming from the rest-frame of the particle into the frame of the observer. My bad.

8. Apr 22, 2008

### HarryWertM

Oh no, oh no, oh no. Not math. My question was framed to obviate the need for math.

A more fully defined question:

We have four CRT beams. Each generates the same current [electrons per second]. One beam velocity is set to fifty percent of c. The second at fifty five percent of c. Third ninety percent c. Fourth, ninety nine percent c. So we have two sets of two beams; a ten percent difference in beam velocity in each set. Do we get:

1. No difference in any mag field?

2. A ten percent difference in each set of beams? [As I understand lbrits was starting to argue for this case.]

3. A much greater mag field difference in the pair of CRTs running near light speed. This is the Wikipedia result, I believe.

My intuition is equal on all three cases and my math is dead. Incidentally, case 3. implies that particle beams like LHC would create enormous mag fields around themselves. Is so?

9. Apr 22, 2008

### lbrits

Intuition can only take you so far. You'll need math if you want the right answer. In any case, the effect is relativistic. If you calculate the electric field of the charges in their rest frame, you get the familiar coulomb result. Now, transform the electric/magnetic field into the moving frame, and you pick up factors of gamma.

I don't think the magnetic fields in the LHC are that enormous, because it also depends on luminosity. Yes, relativistic effects are great, but like I said, the devil is in the details...

I believe there's also a self-focusing or "pinching" effect due to the magnetic field generated.

10. Apr 26, 2008

### HarryWertM

Yes, physics requires math.

Physics Forums user "dropout"
[see physicsforums/special general relativity/OK FINE Magnetism is explained....]
apparently accepts the idea that relativity "explains away" magnetism. I do not. My doubt is at least partially detailed in this Physics Forum thread.

User "dropout" cites a university lecture from South Africa which is identical in substance to the relativistic explanation of magnetism presented in the Wikipedia article "Relativistic Electromagnetism". The same explanation was also presented in a university lecture I attended at UCBerkeley. The explanation apparently originated with EMPurcell.

The problem is much greater than mere trouble for my intuition. The Purcell explanation distinctly conflicts with classical theory, and apparently no actual laboratory experiment has been done to prove/disprove Purcell's relativistic theory. I have tried to google "two electron beams". Found only irrelevant subjects.

In his attempts to answer my question PF user "lbrits" quoted an equation from a text by Griffiths. Not having Griffiths, and not knowing the exact meaning of the variables in lbrits equation, I have no idea whether Griffiths is expounding a theory much different from Purcell or not.

A proposal for a real experiment:

A length of glass tube replaces the stainless steel vacuum shroud somewhere in the middle of a very modest electron beam machine. The machine must be capable of generating very steady electron beams at selected velocities and intensities. I suggest velocities of 49.5% c and 99% c. Call this the "generator" beam. A "measurement" beam is set up near the glass tube section of the generator beam. The measurement beam has low current; low velocity; and is exactly parrallel to the generator beam. Deflection will measure the magnetic and electric fields created by the generator beam.

The surprising classical prediction is: no change in magnetic field force for the different generator beams. I derive this by Ampere's Law
[as in hyperphysics..../hbase/magnetic/magcur.html#c3 ]
which depends only on current intensity, not velocity. [[ Obviously, when Ampere, Maxwell, et al were writing equations, electrons were hardly observed and electron beams in high vacuum hardly available. "Electric current" existed only in conductors. But you would think modern texts would need updating, even without Relativity. ]]

Purcell's relativistic theory yields an even more surprising result: no magnetic force at all arising from the generator beam, regardless of velocity. This is because Purcell/Witwatersrand/UCBerkeley explain the mag field force arising only from the difference in velocity in positive and negative charge carriers. Without both charges, Purcell predicts only a change in Coulomb replusion between the electron beams. This repulsion varies non-linearly with generator beam velocity, being dependent on the term $$\sqrt{}$$(1-(v sq/ c sq)). [[ I believe relativity theorists call this term 'gamma'? ]]. Again, you would think texts would mention this.

A subject ripe for lab work? Does anyone quarrel with my "results" for theories above?

Wouldn't cost millions like some experiments. JPL used to have a giant five or ten foot wide CRT [and I do mean CRT] before these days of projectors. Drag that out of storage and play with it.

-Harry Wertmuller