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Magnetic force between two parallel moving charges

  1. Mar 19, 2008 #1
    Hi everyone,

    I have been trying to find an answer to the following question on several websites on the topic, but no luck untill so far. Maybe there is someone who can help me here:

    question:
    If two charges move parallel to each other (say in vacuum and relative to me), with equal speed, will there be a magnetic force felt by each one from the other?

    All explanations I have found so far say that indeed there is one, and this is often explained by the classical experiment with two parallel currents in wires.

    But, I would say that is a different situation. The two charges in my question do not move relative to each other, and hence they do not see mutual magnetic influence, right?

    In the two wire setup there is relative motion of two types of charges in each wire (the ions and electrons), hence a magnetic field will develop in each wire, causing both wires to attract. Or you could say that the electron current in one wire exerts magnetic force on the ion current in the other and vice versa (I know this sounds strange but hey, it's all relative motion that counts, not?).

    The explanations of the two wire experiment I found seem odd to me, since they almost always explain things by regarding the Lorentz force on the electron movement only. But since there is no relative motion between the electrons (at equal currents and wires) this should not suffice for an explanation.

    Comments are very appreciated, since I am puzzled.

    regards,

    Arjan
     
  2. jcsd
  3. Mar 19, 2008 #2

    clem

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    The force between two moving charges must be determined relativistically.
    The result is a complicate formula in advanced textbooks.
    Even what is meant by "force" in this case needs clarification.
    There will be a magnetic and an electric force.
    It is much more complicated than for two wires.
    Explaining the wire force as between two electrons is used in elementary courses because it seems easy, but it is wrong.
     
  4. Mar 19, 2008 #3

    jtbell

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    In the reference frame in which the two charges are at rest, they exert electrostatic forces on each other. In a reference frame in which they are moving, they exert a combination of electric and magnetic forces on each other. The effects of those forces on the charges' motion are equivalent to each other, and are related by the relativistic Lorentz transformation between the two frames. The electric and magnetic fields produced by the charges in the two frames are also related by the Lorentz transformation, applied to a tensor whose components are the components of the electric and magnetic fields.

    http://farside.ph.utexas.edu/teaching/em/lectures/node121.html

    http://farside.ph.utexas.edu/teaching/em/lectures/node123.html
     
    Last edited: Mar 19, 2008
  5. Mar 21, 2008 #4
    jtbell, does it mean that
    different force situations will be seen by different observors, for instance
    one is at rest, and the other is moving with the two charge?
     
  6. Mar 22, 2008 #5
    zhanghe. excuse me, but nevermind the obscure reference to force. But it is all about relativity. The inertial frame of the observer with respect to charges is involved.

    Say you have this field called F generated by some source like one of your electrons. When you are stationary with respect to the electron, the parts of F that aren't zero valued all look like an electric field.

    But when you are in motion with respect to the electron, some parts of F are measured as the magnetic field.

    It's all really one field. What we measure as the electric and magnetic fields depend on relative motion.
     
  7. Mar 23, 2008 #6
    Thank you, Phrak.
    I think I get to know why the textbook says the electric and magnetic fields are united by the relativity.thanks again.
     
  8. Mar 23, 2008 #7
    anytime, zhanghe
     
  9. Mar 24, 2008 #8

    Philip Wood

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    Start by defining what we mean by a magnetic force in a given frame of reference. It is one felt by a moving 'test' charge, and proportional to the speed of motion. Such forces are generated by other moving charged particles, 'source charges', but not by stationary ones. So there is no problem in saying that electrons moving in parallel paths at the same speed experience magnetic forces: we have a moving 'test charge' and a moving 'source charge'. They are not moving relative to one another, but they ARE moving relative to the frame of reference. That's what matters. [Of course the electrons also experience a repulsive electric-field or 'Coulomb' force. This is greater than the (attractive) magnetic force.]

    If we view the electrons from a frame in which they are stationary (that is, we 'ride along with them'), then, by definition, there is no magnetic force between them. There is, of course, a (repulsive) electrostatic or 'Coulomb' force.

    Now Relativity theory tells us the ('transverse') force between objects will be less in the frame in which they are both moving than when measured in a frame in which they are both stationary. So the repulsive force between the electrons will be less in the laboratory frame than in the ride-along-with frame. But that's just what we observe, only we choose to regard the repulsive force in the laboratory frame as a repulsive Coulomb force added to a weaker attractive magnetic force. In a way we've EXPLAINED the origin of magnetic forces as a frame-modification to Coulomb forces.

    [A word of warning. There is another complication (which doesn't affect the essential argument above.) The ELECTRIC field force (i.e. that part of the force on a testing charge which isn't affected by the testing charge's motion) IS affected by the motion of the source charge. It's greater when the source charge is moving. So the Coulomb force between the electron moving in the laboratory is greater than when seen in the ride-along-with frame. But the increase in repulsive Coulomb force is less than, and outweighed by, the attractive magnetic force which comes into play in the lab frame!]

    Sorry if this seems complicated. I'm afraid it just IS, though the complication is largely due to our insistence on classifying electromagnetic forces as electric or magnetic or both. However, it really is useful so to do - usually.
     
    Last edited: Mar 24, 2008
  10. Mar 24, 2008 #9
    Can i ask a related question,

    If we have two charges that are moving highly relativistically in a straight line, I have been told (and looked in the textbook) that they will repel to a certain distance and stay at that distance (or the force drops to very small), now I am concerned because the book explained this as the magnetic and electric forces canceling, but in the frame of reference of the electrons there is only electric force. So what is actually happening here?

    cheers

    Gareth
     
  11. Mar 24, 2008 #10

    Philip Wood

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    In the charges' frame (my 'ride-along-with-frame') there is, as you say, only the electric repulsion. This will get weaker the further the charges get from each other. Even when the force is so weak that the acceleration is almost zero, they'll still have their recession velocity and will go on receding for ever.

    To find out what happens in the lab frame it's easiest to look at what happens in terms of frame-modified forces (first two sentences of my paragraph 3 above). Thus we can say that the resultant force in the lab frame (in which the charges are moving) will be less than in the ride-with frame. But it won't, for some finite separation of the charges, change from repulsive to attractive, or even drop to zero. Look at it this way: if the force changed direction, that would be a definite recordable event. It couldn't happen in one reference frame (the lab frame) but not in another (the ride-with frame).

    I think there's confusion between this scenario and another. The resultant (repulsive) force in the lab frame is less than in the ride-with frame. But it isn't zero. However, the faster the electrons are travelling (sorry about the double 'l'; I'm English) in their parallel paths, the smaller the resultant force becomes. It approaches zero for charges' speeds approaching the speed of light. When I say 'approaches' I'm not describing a process occurring in time; I'm using it in the mathematical sense.
     
  12. Mar 24, 2008 #11
    I may have gotten confused. I don't have the book I looked at on me, but the one i am referring to is PROBLEMS AND SOLUTIONS ON ELECTROMAGNETISM page 581 the problem is 5019 part b. I will try and post it tomorrow when i get the book to explain where my confusion is
     
  13. May 17, 2008 #12
    Hello! I thought of this problem when I was learning Electromagnetism and read about the scenario of the two wires (the Amper definition). I just want to make sure I got it right. So the magnetic force is just a decrease in the electrostatic force caused by relativistic transformation, but we percieve the electrostatic force as constant and make up a magnetic force so that the results are equivalent? That's how I have interpreted the Philip's explanation. Sorry if I have messed up the terminology :-)
     
  14. May 17, 2008 #13

    Philip Wood

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    Hi. What you say pretty well sums it up. But, as I said, for the two charges moving in parallel paths the analysis is rather tricky. It is extremely easy to show from Special Relativity the factor by which the repulsive force is less when the charges are viewed from the lab frame, in which the charges are moving. But, in terms of electric and magnetic forces, the reduction has to be attributed to a magnetic attractive force AND a somewhat increased repulsive electric field. [Electric field forces on a test charge are, by definition, those which are the same whether or not the particle is moving.]

    Two current-carrying wires (each consisting of a stationary line of + charges and a moving line of - charges) don't experience mutual ELECTRIC field forces, only magnetic field forces. That makes life easier, but this time the relativistic analysis is more complicated (than for two point charges), but not difficult if you keep a clear head.

    There is, in my opinion, a good treatment of these things in Robert Resnick's "Introduction to Special Relativity'. I'm afraid it's probably out of print and expensive secondhand.

    Best wishes.
     
  15. May 17, 2008 #14

    pam

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    The force exerted on a particle q' traveling at velocity v' a distance r from a particle q traveling at constant velocity v is given by
    [tex]\frac{qq'[{\bf r}+{\bf v'}\times({\bf v}\times{\bf r)}]}
    {\gamma_v^2[{\bf r}^2-({\bf v}\times{\bf r})^2]^{\frac{3}{2}}}[/tex].
     
    Last edited: May 17, 2008
  16. May 17, 2008 #15
    If the two charges are at rest relatively to each other, then it seems natural to expect them to experience only static electric force. Magnetic force appears with relative velocity between the charge and the ids element, which, in this case, is the other moving charge. The charge, thus, is at rest respectively to this element of current.

    Wire case may be explained differently foer the existence of protons at rest in the laboratory reference frame is part of the cause of the force between the wires.

    best regards

    DaTario
     
  17. May 18, 2008 #16
    Oh thank you very much, I'm looking forward to relativity classes I will have in 2 years. I know
    just the basics of special relativity from secondary school and I'm not familiar with the necessary calculus to do it properly. But it's nice to know where the catch is, thanks!
     
  18. May 18, 2008 #17
    Magnetic force on a moving point charge is centripetal force.

    Point charge movement within a curved path (windings in an inductor) increases magnetic energy per current, because the curved path increases inductance.

    U[tex]_{M}[/tex] = LI[tex]^{2}[/tex]

    Ion migration is a function of volts, not bends versus straight wire. The increased magnetism at windings is not due to higher electron velocity.

    v(drift) is a function of charge * volts / friction
     
    Last edited: May 18, 2008
  19. May 21, 2008 #18

    Philip Wood

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    Being at rest relative to each other doesn't mean no magnetic forces. In the particles' 'own' frame, i.e. that in which they are both at rest, there are, I agree, no magnetic forces. In another frame (the 'lab' frame) they will both have the same velocity as each other. If the particles' separation is transverse to that velocity (as in the cases discussed above), then there IS a magnetic force between them in the lab frame. How do we know this? The key thing to remember is that the magnetic force on a charge is DEFINED as that part of the Lorentz force which is proportional to the charge's velocity. The relativistic analysis and the classical electromagnetic analysis of the case of the moving charges both predict that there IS just such a non-zero force term in the lab frame for the particles moving in parallel paths with the same velocities, even though there is no RELATIVE velocity between the particles.
     
  20. May 21, 2008 #19
    In high frequency circuit design, all bends are 90 degrees. Curves increase inductance. The inductance of the curve increases the time the wire requires to reach full current. During this time, electromagnetic energy (voltage's energy) is converting to magnetic energy in the wire, instead of being delivered to the load.

    I do not know why bending a (superconductive) wire in an area outside of a voltage field instantly increases magnetic energy (velocity?) per electron (inductance) in the wire. Any curvature in electron path determines location of the center of a magnetic field.

    It is hard for me to understand current flow among atoms. Any non-zero quantized mass movement requires infinite energy. For an electron to travel from one molecule to the next, a chemical bond must exist between the two atoms. For superconductivity to occur, giant molecules the length of the wire in a superconductive magnet must exist. How does an atom get a superconductive energy level (free electrons are not superconductive)?
     
  21. Oct 28, 2009 #20
    So is the net force repulsive or attractive?

    In the comoving frame, there's no B field but in the lab frame there is. E2-B2 is conserved, so by going to the comoving frame, the
    electric field is responsible for the force. In the lab frame, B exists, so E must be greater.
    Is that right?

    It's been a while since I took E & M.
     
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