Motion in a magnetic field and relativity

1. Jul 5, 2014

Solibelus

1. The problem statement, all variables and given/known data
We're working in the right-handed Cartesian coordinate system.
Unit system is CGS.
A conducting bar of length L is placed along the x axis. Center of mass at x=0 when t=0.
It's moving with constant velocity V in the +y direction.
There's a uniform magnetic field B, such that:
$$\vec B = Bcosβ \hat y - Bsinβ \hat z$$
a. Find the potential difference within the rod. Find the electric field E.
b. Find E' and B' in the rod's frame of reference, far away from the rod. V=0.99c.

2. Relevant equations
Not given, but I assume I'll need Lorentz force and the relativistic transformation of the fields.

3. The attempt at a solution
Well I think I managed a:
The moving rod's free electrons are affected by the Lorentz force:
$$\vec F_{mag}=qVBsinβ(-\hat x)$$
This moves the electrons to one side of the rod (the +x side) and the "positive charges" to the -x side. This causes a potential difference and an electrical field E inside the conductor.
The force due to this electric field, at least at some time t1, cancels the magnetic component of the Lorentz force:
$$\Sigma F_x = qE + F_{mag} = 0$$
$$\vec E = VBsinβ \hat x$$
$$\Delta V = ∫\vec E.d \vec s = EL$$

Now for b:
I understand that from a we can see that the bar is basically an electric dipole.
Being an electric dipole, it produces an electric field which I don't know how to derive.
If I'm to derive this field, with addition to the given B field, I'll be able to apply Lorentz transformations and get E' and B' outside (and far away from) the bar.
What am I missing here?
I've got a feeling I'm stuck due to something really dumb.

Last edited: Jul 5, 2014
2. Jul 5, 2014

Staff: Mentor

I'm not sure if you have to include the bar in the fields - this would require knowledge of the cross-section. Maybe it is just a transformation of the magnetic field.

You can include the bar, if you assume some cross-section A for it:
With the known charge density everywhere (you can derive this from the electric fields), it is possible to derive the dipole moment. There is a formula for the electric field of a dipole.
This can be added to the other field.

3. Jul 5, 2014

Solibelus

There is no cross section given though. I don't think it'd be correct to just assume it unless it cancels out somewhere during the actual calculations.

I guess I can derive the charge density from Gauss' law, but E's gradient is zero, is it not?
Maybe the E field outside for some reason exists only in the reference frame of the moving observer (sitting on the rod), due to Lorentz's transformations? But if so then why? It is a dipole, it has to have a field in the stationary reference frame.

Something is very wrong with my thinking here :(

4. Jul 6, 2014

Staff: Mentor

Well, the field will depend on the cross-section, so unless you introduce it you won't be able to include the effect of the bar. Which could be fine - I don't know how the problem statement is meant.

And this requires a specific charge density to cancel the external field you see in the frame of the bar.

5. Jul 6, 2014

Solibelus

So I'm guessing that either the cross-section or the density are missing from the original question.
I think I'll just introduce this piece of data and solve using dipole approximation as usual.
Thanks!

6. Jul 7, 2014

rude man

part (a):
There's a standard formula for finding the emf in media moving in a magnetic field, called (sometimes) the "Blv" law. Use the vector formulation of this law to obtain your emf.

I did not get what you did for V.

part (b): later, it's too late here.

7. Jul 7, 2014

Solibelus

That "standard formula" - is it not simply Faraday's law after applying Stoke's theorem?
Does it even apply here, given that I don't have a closed circuit? What's the ∫dl loop here?
If this is not what you had in mind, then I don't know of any "Blv law"... it's just that during the course I took they basically taught us Maxwell's equations in various states, not any specific formulas to remember (heartless university professors!).

EDIT: OK, did some googling on this Blv thingy. I was correct, it's simply emf = -1/c (∂ø/∂t), where ø is the flux, changing in time. This is the result of Faraday's law as I explained above.
I don't think that this applies here, since there's no closed loop. It's not similar to the "rod moving on rails" problem. Am I mistaken?

EDIT 2: my googling result: https://sites.google.com/site/electromagneticinductionmbb1/induced-electromotive-force

For the potential difference, what I did is this: the E and B components of the Lorentz force cancel each other out, because eventually the current there dies out. This allows me to equate the Lorentz force to zero and get the E inside the conductor. It's obvious that there's a charge separation due to the forces.
Now, since I know E inside, I can say that the potential difference between the edges of the rod is basically E*L, L being the length of the rod.
That's what I'd measure with an ideal voltmeter I guess.

Now my original problem was finding E outside the rod, since the rod is a dipole.
If I'm able to do that, it's trivial to apply Lorentz transformations on the constant B and E away from the rod.

Last edited: Jul 7, 2014
8. Jul 7, 2014

rude man

Shocking news for you: Maxwell's equations do not always apply for moving media. In fact, strictly speaking they do not apply ever.

For moving media, the expresssion

∇ x E = -∂B/∂t must be replaced by
∇ x E = -∂B/∂t + ∇ x (v x B).

In your case, B is unchanging in time so ∂B/∂t = 0. The "Blv" law derives from the second term.

So apply Stokes to the second term and for your case you get the circulation of E around a loop including the bar (but with the remaining segments not cutting flux) = emf.

Most physics books approach your problem as follows: form a rectangular loop with the desired medium (in your case the bar) included in the loop, with the other 3 sides of the loop not cutting flux. Then they say "OK, compute the flux rate of change inside the area of the loop which in your case is BLv." However, Maxwell does not speak flux, he speaks only E and B. And B does not change with time. So really you're applying Faraday, not Maxwell, if you go that route, which is perfectly OK, except sometimes it fails.

If you're interested in a case where Faraday does not work I am appending a pdf file for you.

File size:
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9. Jul 7, 2014

rude man

The circulation of E around a closed loop = - dø/dt is independent of any medium. It applies even in thin air. In fact, we wouldn't be communicating now if it didn't since it's the basis of e-m propagation.

So you always can form a closed loop. In your case, the loop would include the bar with the other 3 sides of the rectangular loop stationary & in thin air (in the B field). This IS the rail problem you cited but with infinite resistance along the other 3 sides. Then you can apply Faraday as the rate of change of flux within the loop, with all the emf applied across the bar only.

Faraday determines the emf around any loop but does not by itself describe where along the loop the emf drops occur. You have to take as given that the emf drop is restricted to the moving element(s). That's why you really need the BLv law applied to every part of your loop..

Are you sticking with your result for the emf in part (a)? As I said, I didn't get that answer.

10. Jul 7, 2014

Solibelus

I'm not that easily shocked...
OK, so I think I understand your reasoning. Let me try again!
Now according to what you're saying,
∇ x E = -∂B/∂t + ∇ x (v x B)
B is constant and I wish to work in CGS (in hope that I'm doing it correctly).
So, applying stokes theorem:
$$\oint \vec E.d \vec l = \oint (-\vec {v \over {c}} \times -Bsinβ \hat z).d \vec l$$
Now, let's say that my loop is including the bar.
$$\vec E = \vec {v \over {c}} \times Bsinβ \hat z = -{v \over c}Bsinβ \hat x$$

(This is the same result I got beforeת using Lorentz force)

Where is my mistake?

Now, for V:
$$\vec E = - \vec \nabla {V}$$
$$-\int E . dx = V$$
$$\int {1 \over c} vBsinβ dx = V$$
$$V = {1 \over c} vBLsinβ$$

Then again, E here is the field inside the bar, is it not?

11. Jul 7, 2014

rude man

First, an apology. You got the emf right, I didn't. I took a second cross product instead of the dot product. FYI the emf = (B x L)*v, the * indicating the dot-product, but this is in SI units. I haven't used cgs in over 50 years!

However: the field inside the bar is zero. There are two E fields. One is electrostatic due to charge pileup at the ends of the bar, the other is due to the induced emf. The two exactly cancel. Now, there IS an E field OUTSIDE the bar, and ∫E dl = emf from one end of the bar to the other.

Now, as to part (b), since the B field is stationary w/r/t the observer & the bar is moving at v = 0.99c w/r/t the observer, I know of no reason to invoke the Lorentz transformation. Seems to me the B field remains unchanged but there is rapid cutting of flux lines by the bar so the induced emf would be very large.

However, I could be wrong on this. I've just never encountered relativistic corrections to e-m fields.

As for the E field, note that the problem asks for the E field far away from the moving bar but in the coordinate system of the bar. Consider a trick question!

12. Jul 7, 2014

rude man

BTW also, say the bar is stationary & you're asked for the E field. You're right in assuming a dipole. What do the charges q and -q have to be in order for the electrostatic force on one of them to be equal & opposite to the induced force? (You have a dipole with moment qL.)

13. Jul 7, 2014

Solibelus

Alright, I see, so I must have made a mistake there with the signs, because it does make sense why I have two identical results for E. The signs just need to be opposite I guess...

Now for Q, if I understood you correctly, I just need to use Coulomb's law?
If so, I get this:
$$Q = {v \over c} L^2 B sinβ$$

For (b), I do need to transform because I want to be sitting on the rod. They simply asked to do that, that's all.

14. Jul 7, 2014

rude man

Right! (Assuming your cgs units are right).

If you're sitting on the rod and looking at the E field produced by the rod, what meaning is there to v = 0.99c?

15. Jul 7, 2014

Solibelus

According to the transformations, E should grow by factor γ, but in addition it will have the 0.99γBsinβ component.
B will of course also depend on that 0.99...

Gotta say that I haven't plugged in any numbers to actually get a feeling for the significance here.

16. Jul 7, 2014

rude man

OK, let's try this: your bar is at rest on Earth and you here on Earth are looking at the E field produced by the bar.

As is well known, some galaxies are receding from Earth at close to v = 0.99c. Do you care?

17. Jul 9, 2014

Solibelus

Sorry for being away, had an exam yesterday and some family events.

I do care, because if the galaxy is receding from earth at close to v=0.99c, length and time differs whether I'm measuring from earth or from that galaxy. Now, I'm standing on Earth and we have that receding galaxy.

If the bar is at rest w/r/t me, then the E field is as calculated in that system in (a). I would still care about the galaxy, because maybe it's an interesting galaxy and worth observing while taking into account length-contraction and time-dilation, but the bar isn't in the galaxy's system so I wouldn't care about that specific bar.

If the bar is at rest w/r/t the galaxy, then the E field is obviously different, and I would indeed care to calculate the new E and of course the new B of the bar, just as they asked me to do in (b).

Did I get what you mean? If not, then please elaborate more! :)

18. Jul 9, 2014

rude man

I agree completely. So, what is the E field due to the bar's dipole moment, in the reference system of the bar?

19. Jul 9, 2014

Solibelus

Thanks for the help so far! I'm afraid I'll have to leave this hanging until Sunday since I've got another test in a completely different subject.
However my plan now that I know Q is to approximate E using dipole expansion in the reference frame where the bar is moving and then apply Lorentz transformations for fields on E and B.

20. Jul 9, 2014

rude man

I guess we'll have to agree to disagree, at least on the E field.

Good luck on your other test!