Lorentz force on a wire in earth´s Magnetic field?

1. Oct 5, 2013

leviterande

Hi all!

First, this is neither a part of homework, university or anything like that , it is my own wondering.

I have been searching a lot for how to calculate the Lorentz force acting on a single current carrying wire in Earth´s magnetic field which is 0.6 Gauss. for say a wire of voltage 12v and Current is 3A

Now , I know the formula for Lorentz force is:
F = q [ E + (v * B)] where F = Newtons, q = Columbs, E= Electric-field per meter, V = velocity of electron in wire. B = magnetic field in Tesla.

Just first off please, Where do you take account for the distance between "the current carrying wire" and "the magnetic field". Lorentz force decreases with distance just as everything in electromagnetism, right? but I cant see or understand where do you take that in the calculation. the wire for all I know could be right above a magnet or a mile above the acting magnet...so the force would change dramatically?

So:
1-Where and how do you take account for the distance between the moving charge and the magnetic field in the calculation?

2 What exactly is the "electric field volts per meter" in calculation referring to here in the calculation?

3-What would be the Lorentz force on say ordinary 12v 3A wire 1ft long ( parallel to earth surface and at sea level) in the Earths magnetic field(0.6gauss)? will the force be microscopically small or will it be visible?

I will be so thankfull if I understand these a little better.
Best Reagrds

.

Last edited: Oct 5, 2013
2. Oct 5, 2013

BruceW

for question 1. it is when you calculate the magnetic field that you are taking into account the distance. Remember the magnetic field is defined at each point in space. So if you have a bar magnet, then the magnetic field B due to that magnet will be large near the magnet and will be small very far from the magnet.

Now, you have that the Earth's field is 0.6 Gauss. But clearly, you do not have the position at which this is true. Well, you are supposed to take 0.6 as an approximate value, because the magnetic field due to the earth does not vary much. (for example, if I walk a few meters on the Earth's surface, the magnetic field will only change a tiny bit). So you can think of the magnetic field due to the Earth as being almost constant.

3. Oct 5, 2013

BruceW

I'm not sure what you mean in question 2. and for question 3. what is 1dt? do you mean 1ft?

4. Oct 5, 2013

Staff: Mentor

The strength of the field and hence the value of B decreases with distance, so as long as you're using the value of B at the point where the force is applied, you're OK - the effect of distance is in the value of B.

In practice, the earth's magnetic field is near as no never mind uniform over the distances that you're working with here, so you can just treat B as a constant and be done with it.
(this is somewhat analogous to the way that we use $F=mg$ near the surface of the earth - we know that the value of g is different at different heights, but as long as you stay near the surface, it changes so slowly that we treat it as a constant)

5. Oct 5, 2013

leviterande

HW helper yes sorry I meant 1 ft long.
Really thank you thank you all for your help you so let me see if I can calculate the force as I understand now that I should :
-treat magnetic field as 0.6gauss =0.00006 T
-got an average electron drift velocity in 1mm copper wire at 3A is 0.00028m/s
- electric field "E" is what I don't understand here in the equation F = q [ E + (v * B)] , is it the 12v, I am confused here. Also another variation of Lorentz force calculation is given by a shorter simpler equation=
F = q v * B ??

1-Which equation to follow in my situation?
2-What is the E field I suppose to plug in in the first equation, I am confused, pardon me.

Thanks

6. Oct 5, 2013

Staff: Mentor

The first equation is the more general one that you use if you have both an electrical field and a magnetic field. Set E to zero and you get the second one, and that's what you want to do here (unless you're planning to put your wire near a source of a strong electrical field, such as between the plates of a capacitor).

7. Oct 5, 2013

leviterande

Ah that makes sense
So 1ft 1mm copper wire say carrying 3A at 12v reacting to earths 0.06gauss has an electron drift velocity of around 0.00028m/s

Alright, if I calculate the L force it :

F = q v * B =

3 coulombs * 0.00028m/s * 0.00006Teslas =0,00000005N which seems a tremendously and awfully small force.

1-Am I making a mistake here?
2-hmmm How does the length of the wire itself gets accounted for in the calculation or does it at all?

Thanks

8. Oct 5, 2013

BruceW

you already have the current, so you don't need to calculate the drift velocity.
it is as nugatory says. for this case, the net force on the wire is simply due to qvB, so you can forget about the other part of the equation. So just use E=0. So now, you have that the force is qvB, and the quantities you have are current (I), length (L) and magnetic field (B). So with a little bit of thought about the definition of 'current', how can you write qv in terms of I and L ?

edit: I was writing this before your last reply. yeah, you should not be using the drift velocity. And the charge is not 3 Coulombs, but the current is 3 Amps.

9. Oct 5, 2013

leviterande

Sorry I am a little lost now, I wrote the length of wire just as info, didn't know how or if I needed to plug it in calculation. For 1ft 1mm wire carrying 3A do you mean that
in order to calculate the force I should only multiply 3 * 0.00006?

10. Oct 5, 2013

BruceW

you don't need the drift velocity. You have the equation F=qvB. But it would be difficult to calculate v and q. So you can re-write this equation to include I and L instead of q and v. Have a think about it. Use the definition of current. You can write an 'equivalent' equation.

11. Oct 5, 2013

BruceW

Well, I guess you can use the drift velocity method if you want. so, you have v. And you know that the charge on each electron is 1 Coulomb. so you can calculate the force on each electron. And then multiply that by the number of electrons in your wire.

12. Oct 5, 2013

leviterande

Bruce ok , you are correct, the force should now be F=IL*B which is
1ft= 0.3m
3Amps * 0.3meters * 6E-5 teslas = 5,4E-5N i.e. 0.00005 Newtons= 0.0054 grams
I think this is the correct calculation? The force though seems to be truly very very small?

13. Oct 5, 2013

leviterande

By the way,

-If this wire carried same 3A current but at much higher voltage will the force increase or not?

-Does earths electric field somehow affect the lorentz force on this wire in earths magnetic field?

14. Oct 5, 2013

BruceW

yeah, looks correct to me. That is pretty small. But then, surely we expect it to be small. if you dropped a wire carrying a dc current, would you expect the earth's magnetic field to push it to the side?

15. Oct 5, 2013

BruceW

the voltage doesn't matter, as can be seen in the equation. Also, I have never heard of the earth's electric field, so I would assume it is negligible.

16. Oct 5, 2013

leviterande

Earths electric field is around 100v/m and extends "UPWARD" to a maximum of around 400000v at the upper atmosphere at around 50km. I am not really sure but if the electric field is extending upward which is also parallel to earths magnetic field direction, then this earth electric field should have no importance or affect on the resultant Lorentz force on a wire since the earth E field is not perpendicular to the magnetic field... speculating here, would love to get more view on this if possible

17. Oct 5, 2013

Staff: Mentor

No, the E field does not have to be perpendicular to the B field to matter. You can calculate the effect of the E field on the charged particle at a particular point in space, separately calculate the effect of the B field, and then add the two forces to get the total force. However, the only time that the E and B fields will be acting in the same direction is when the E field is perpendicular to the B field, so in the general case you have to write the force equation out using vectors:
$\widehat{F} = q(\widehat{E} + \widehat{v}\times\widehat{B})$

Here the $\times$ operator is the vector crossproduct, and it just happens to reduce to the non-vector formula you gave above $F=q(E+vB)$ in the case that all the forces are exactly lined up in the same direction. (As a matter of notation, $X$ means the length of the vector $\widehat{X}$ without regard to the direction the vector points).

18. Oct 5, 2013

BruceW

That is pretty interesting, but I'm guessing that electric field only appears in the upper atmosphere, not down here near ground level? I mean, 100 V/m ... that's huge...

19. Oct 5, 2013

leviterande

Not sure if I got you, did you mean 100v/m to be huge? or were you sarcastic? :)

20. Oct 6, 2013

BruceW

haha, no, people often seem to think I'm being sarcastic, when I'm just being serious. Maybe it is the way I say it? Also, I often don't get it when people are being sarcastic. Anyway, yeah 100 V/m seems huge to me. A typical circuit with a DC battery might have a few volts and a few feet of wire. So this 100 V/m electric field would be larger than the electric field caused by the battery. That seems unlikely to me. So, I think it is more likely that this electric field appears in the (very high up) atmosphere. This seems more believable, since up there, matter starts to act quite strangely, due to the super-low pressure.

21. Oct 6, 2013

leviterande

aha yeah I thought too it would be a little strange if you were sarcastic :). However, the E-field according to any physics source is indeed 100v/m starting from earth surface. this voltage is in relation to true earth ground. I agree this is a huge but it exists none the less or do you suggest this is not the case? this E field is vertical to earth surface and reaches 300-400kv at the upper atmosphere. thus, the upper atmosphere and the earth surfaces forms kind of like 2 giant plates of a capacitor. I am wondering if this E field contribute to the Lorentz force I mentioned, because apparently the Lorentz force I am getting is way bigger than the calculated 0.0054
Thanks

22. Oct 6, 2013

BruceW

the Lorentz force you are getting is way bigger than the calculated? what does that mean? Are you actually doing an experiment? Or do you mean the contribution of qE to the Lorentz force is much larger than the contribution due to the magnetic field?

23. Oct 6, 2013

BruceW

ah, whoops. I just realized what I said here. Of course, an electron does not have 1 Coulomb charge! I think I am getting too used to natural units, where the charge of an electron equals 1.

24. Oct 6, 2013

leviterande

Yeah I just noticed that while working with some circuits on my bench. I started to investigate if it was simply some errors from nearby ferromagnetic affecting the cables. but it wasn't. I isolated the pendulum from everything. The force is very clearly a Lorentz force I am seeing since the force direction is always perpendicular to the DC current flow no matter where I put the pendulum in my house and this force is always towards one direction for one polarity. If I flip the polarity the conductor swings the other way, yeah a Lorentz force is the reason and my first thought was that it was the Earths magnetic field of course.. it is not much but it is visible which was what caught my attention.

So I wanted to calculate the force to confirm what I believed which turned out to be 0.0054g. I am not sure how much force is acting in my pendulum tests, but the (12v 1-3A)1ft 1-3mm strait rigid conductors/cables move and are visible so I suspected it must be more than 0.0054grans, maybe around half a gram, I am not sure though, I haven't weighted the force that is seen, so the visible force indeed could be 0.0054grams and is able to move a solid conductor visibly from a short pendulum.

Last edited: Oct 6, 2013
25. Oct 6, 2013

BruceW

after looking around online, I find this too! That seems pretty weird to me... I am trying to think of different situations where it might have an effect, but I can't think of any.

For example, in a metal, it will almost always be mostly neutral, but with some free electrons. These free electrons will all be pushed upward by the background electric field of 100 V/m, and then an opposite internal electric field (caused by the charge density) will at some point create an equilibrium. I guess the electric field of 100 V/m is too small for this to cause any significant effect? Otherwise, any metal object would always give off an electric field, relative to the background. (This would be without us putting any current through the metal or anything).

Another example... if you had some electrolytic fluid in a vertical container, and attached electrodes to the top and bottom, with a wire running between them, then there is a potential between the electrodes. But there is also an opposite potential within the electrolytic fluid itself. So maybe this is the reason that electrolysis would not occur. (if it did occur, then we could recharge batteries for free).