Interesting Simple Magnetic Force Question

Discussion in 'Physics' started by MrAl, Sep 7, 2015.

  1. MrAl

    Thread Starter Well-Known Member

    Jun 17, 2014
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    Hello there,

    Here's a simple magnetic force question.

    We have a uniform magnetic field B pointing into the page, perpendicular to the page.
    We have a wire running across the page with current flowing from left to right, using positive conventional current flow so the current flows from left to right and it is comprised of positive charges (not negative charges like electrons). This wire is perpendicular to B.
    The wire is free to move (the whole wire can drift not just one part of it) and the B field is directed in such a way that it exerts a force on the wire, so the wire gets deflected.

    The question is, in which direction does the wire get deflected?

    A search around the web will show that a technique known as the "Right Hand Rule" can be used. It will also show that on one web site we will end up finding that the wire gets deflected upward, while on another web site we will have to conclude that the wire will get deflected downward. The mix up might come from the fact that one web site states that the index finger points in the direction of current I and middle finger in the direction of B, while another states that the index finger points in the direction of B and the middle finger in the direction of I. Swapping fingers like that means the direction of the force gets reversed, so one way we get "up" and the other way we get "down" where the direction of the force is indicated by the thumb.

    This is pretty amazing that such a simple concept can be so skewed, and probably not presented correctly in the first place.

    So which would you choose, "up" or "down" (or even some other direction), and what theory do you have that backs that claim up? Please feel free to use any math technique you like, if you choose to do so.
    I would like to hear any arguments really, with math or not. Try first to use the Right Hand Rule.

    Also important, how would you suggest orientating the fingers in the Right Hand Rule above?

    Thanks :)
     
    Last edited: Sep 7, 2015
  2. Kermit2

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    I learned that a right hand rule was used for generators and a left hand rule was used for motors. I no longer remember which finger exactly. I believe it was index was B field and middle was current.
     
  3. nsaspook

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    It's not skewed. The 'Right Hand Rule' is a simplification of https://en.wikipedia.org/wiki/Lenz's_law
     
    Last edited: Sep 7, 2015
  4. MrAl

    Thread Starter Well-Known Member

    Jun 17, 2014
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    Hi,

    That's one of the points i was trying to get at and talk about.
    If you do it that way, the Force vector points downward.
    If you do it the other way, the Force vector points upward.

    The cross product which i will get to next leads to even more interesting problems :)
     
  5. MrAl

    Thread Starter Well-Known Member

    Jun 17, 2014
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    Hi,

    Did you meant the Lorentz Force Law?
    That's at play here:
    F=q*v x B
    where 'x' here represents the cross product, and v and B are vectors.
    For a continuous current this boils down into:
    F=I*L x B
    where I is the current and L is the length of the wire, and L and B are vectors.
    That in turn boils down to a multiplication, but i am interested in the direction more than anything else. For L and B perpendicular, the force F is perpendicular to both L and B, and that is satisfied with a direction either up or down, but it can only be one not both.

    There are a couple interesting experiments on the web (not equations, formulas, hand diagrams, hand diagrams of hands with fingers (ha ha ha), etc. which tend to vary). They use actual wires and batteries and large magnets. When the current is turned on, the wire moves one way or the other.

    These experiments show that the correct right hand orientation is with the finger in the direction of conventional current flow and the middle finger in the direction of B, and the force F is then the direction of the thumb.
    Using this rule, the first problem i posed in the first post of this thread would have the wire deflecting UP, and that matches two experiments on video i found. Now we get to the other interesting part...the cross product.

    For two vectors, U and V where:
    U=[1,0,0] (this is a length of wire along the x axis)
    V=[0,0,B]

    the force F is proportional to cross(U,V).

    If we make B=1 for simplicity, the cross product comes out to the vector:
    [0,-1,0]

    which indicates a force F in the NEGATIVE y direction, which is down, not up. The right hand rule says it will be up however, and that seems right because the two experiments both show the same result.

    So what is wrong then? How do we get 'up' for the right hand rule yet 'down' for the cross product?

    Care to take a stab at this ?
     
    Last edited: Sep 7, 2015
  6. BR-549

    Well-Known Member

    Sep 22, 2013
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    The wire does not get deflected. There is no force upon the wire. Only the free electrons on it.

    When you put an object in an electric field, if the object has no net charged, it will not move.

    When you put an object in a magnetic field, if the object does not have an aligned magnetic field, it will not move.

    If you loop the conductor, the external magnetic field will cause current to flow in the wire.

    This current will cause its own magnetic field in the wire. This magnetic field, interacting with the external magnetic field, can cause the wire to move.

    If you had a open conductor with many, many excess free electrons, And put it in a strong magnetic field.............so that you could constitute current flow for a short period of time during polarization, The conductor still would not translate.

    It would rotate. Because one end would move toward and the other side would move away. And the charge flow would oscillate. It would repeat itself.

    Why people insist on using an open circuit to explain the operation of of closed circuit, baffles me.

    Once we have current in the loop, now we can use the right hand rule and determine the N-S poles, and then determine the direction of movement from the external field.

    I use the left hand rule myself, I don’t believe in conventional flow.
     
  7. WBahn

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    Mar 31, 2012
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    Gee, you might tell that to the people that used a current balance as the original standard definition for the ampere.
     
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  8. nsaspook

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    Last edited: Sep 7, 2015
  9. BR-549

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    Sep 22, 2013
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    I believe they have understood this for quite awhile. They used closed loops to measure current.

    Ampere did move conductors with magnetic fields, but only when current is flowing in the conductor.
     
  10. WBahn

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    Once again you are contradicting yourself. First you say that the current carrying wire that the TS described won't be deflected by the magnetic field and now you say that Ampere moved conductors with magnetic fields when current was flowing in it. Which is it?
     
  11. BR-549

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    Pardon me. Your right. I didn’t realize the conductor had current. Please excuse me.

    In that case, the direction that the conductor moves, depends on where the rest of the loop is.

    If the loop lies north, the loop will move away....down deeper into the paper.

    If the loop lays south, it will lift from the paper.

    The field going into the paper is from a south pole.

    The north loop generates a south pole coming out of the paper. It will deflect.

    The south loop generates a north pole coming out of the paper. It will attract.

    Sorry about my previous comments.

    In order to know the direction the wire will move, we have to know what side of the wire is the inside and outside of the current loop.
     
    Last edited: Sep 7, 2015
  12. WBahn

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    How do you reconcile this with the fact that the force on a current imposed on a moving charge in a magnetic field is

    F = qv x B

    Notice that there is nothing there about the details of the loop that the charges that make up this current outside the region of interest.

    What if the loop lies vertically?

    Build this and you will see that you are simply wrong.

    The force on the wire as described (B into the page, I from left to right) will be upward. If you have a current loop in this field then if the current is circulating counter-clockwise the forces on the loop will be compressive trying to reduce the diameter of the loop. If the current is circulating clockwise then the forces on the loop will be tensile trying to expand the diameter of the loop.

    If things were as you described, then I could make a magnet using a coil of current carrying wire and have it translate simply by the current interacting with the magnetic field it creates, which would violate all kinds of conservation laws. And before you say something about currents not being able to interact with the magnetic fields they create, let me point out that an electromagnet most definitely does experience these self-induced forces. In a coil the magnetic field produces will be oriented such that the magnetic forces will be trying to expand the diameter of the coil. These are called hoop stresses and are a major factor in the design of high field magnets.
     
  13. WBahn

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    Mar 31, 2012
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    The easiest way to remember the right hand rule without messing it up is to forget about which finger points in the direction of which vector.

    If you have

    A = B x C

    You want to cross B into C, so start with your fingers (all of them) pointing in the direction of B and then fold ("cross") your fingers so that (all of them) point in the direction of C. Your outstretched thumb will then point in the direction of A. It's that easy.

    In more detail:

    Take your hand with all four fingers outstretched and your thumb pointing outward at a (nominally) right angle. Imagine putting your palm down on a table so that if your fingers point upward your thumb points to the left. This is just to prevent someone from pointing their thumb so that it is "coming out" of their palm.

    Point your fingers in the direction of B.

    Now orient your hand so that as you fold your fingers toward your palm (no more than 180°) they point in the direction of C. You are "crossing" B into C. With your hand oriented this way, your thumb points in the direction of A.
     
  14. BR-549

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    Sep 22, 2013
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    With just the conductor, without the orientation of the loop, all we know is that there is a south magnetic pole on the top of wire and north magnetic pole on bottom of wire. There is no way to know which is the pole from the current of the loop.

    You do agree that if we flip the loop, the direction of movement will change, don't you?

    The loop has established current and field.
     
  15. WBahn

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    Mar 31, 2012
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    No, I don't agree, because it won't. The force on that section of wire is upward (within the plane of the page) regardless of where the wire ends up going.

    Try it!

    Take some flimsy wire and suspend it between two supports so that it is slack. Now run some current through it and hold a permanent magnet near it and note the direction of the force that is exerted on the wire. Route the rest of the current loop however you want (but keep it away from the wire of interest so that they don't interact and complicate things) and note the direction of the force that is exerted on the wire. It won't change.

    The force is the CROSS PRODUCT of the current and the magnetic field. The cross product produces a vector that is perpendicular to EITHER of the vectors being crossed. This, right there, disproves your assertion because if the force is either into or out of the paper then the that would require that the cross product produced a result that was parallel (or anti-parallel) to one of the arguments.
     
  16. MrAl

    Thread Starter Well-Known Member

    Jun 17, 2014
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    Hi,

    Oh ok, so i guess you are trying to do this via the magnetic field of the wire caused by I and the uniform magnetic field B that is present all around the wire and in a given direction.
    So what did you get for the result for the deflection of the wire?
     
  17. BR-549

    Well-Known Member

    Sep 22, 2013
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    Let me try this. Draw a loop, At the bottom of the loop, apply the right hand rule with conventional current going the the right.

    This will cause a south magnetic field out of the loop.

    Now go to the top of the loop. Apply right hand rule with conventional current going to the right.

    This causes a north magnetic pole out of the loop.

    How do we know where that conductor is in the loop. The top or the bottom?

    This is why we need to know the inside from the outside if we only have a small sample of the loop conductor.
     
  18. MrAl

    Thread Starter Well-Known Member

    Jun 17, 2014
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    Hello there,

    I think you need to clarify some of your statements.

    First, the current in the wire constitutes moving charge and since it is bound to the physical wire if and of the charge gets defected then the wire gets defected.

    Second, there is no current induced in the wire from the B field because the B field is constant. There may be a current when the field is first switched on (because then it changes from zero to some maximum) but we dont have to consider that part of the problem because this is a static problem only. We only want to know the direction the wire takes (which way it gets defected). It will stay defected until the current or the field is switched off.

    Third, these problems usually dont deal with an "open circuit" wire they deal with a closed circuit, but not all of the wire in the circuit is included in part of the problem. The only part we need to consider for this problem is the part of the wire that is in the field B, and that can be a short section of the entire length. The short section will move while the rest will only move a little if it is dragged by the short section. An experiment can be set up however such that the short section is made up of a short length and the wire to the left and wire to the right are bent at 90 degree angles to the short section. The two 'sides' are then suspended from two hinges that allow the wire to rotate outward or inward. The current comes in one side, and out the other, but the only part of the wire that experiences a force is the short section that is placed inside the field B.
     
  19. MrAl

    Thread Starter Well-Known Member

    Jun 17, 2014
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    Hi,

    Yes thanks, that helps to clarify the right hand rule problem, which i am now going to call Problem #1. That works with the original wire and field i posted in the first post (wire along x, conventional current left to right along x, B field into the page).

    Problem #2 comes in when we try to do the cross product. As i was saying before, the cross of two vectors U and V where they are:
    U=[1,0,0] (wire length with conventional current flow)
    V=[0,0,1] (mag field B)

    and the cross product of U and V:
    U x V = [0,-1,0]

    shows a force oriented DOWN not UP as the right hand rule shows.

    The reason for this appears to be that when we use the cross product we have to invoke another rule which i do not see mentioned anywhere on the web. That rule, if stated somewhere, would have to state that when we use the cross product B must be aligned with the positive z axis, where the cartesian coordinate system is drawn in standard position, and this is with:
    x running from left to right, negative to the left, positive to the right, and
    y running from bottom to top, bottom negative, top positive, and most importantly,
    z running from inside the page to outside the page (pointing out of the page), negative inside the page and positive outside the page. So if we looked at the z axis we would see the positive end coming out of page.

    Since in this standard position z runs OUT of the page, the original problem had B running INTO the page, and thus B must be made negative, or else the whole setup must be rotated such that B points in the positive direction of z.
    If we make B negative, we get the right cross product (mathematically), and if we rotate the whole system then the current is actually running along the negative x axis (right to left now instead of left to right as in the original problem) and y stays the same. Doing this also corrects the solution.

    So we either have to make B negative:
    U=[1,0,0], B=[0,0,-1]

    or we have to rotate the whole system which makes the direction of the flow of charge negative:
    U][-1,0,0], B=[0,0,1]

    This second way of doing it allows B to be kept positive because now it is pointing in the direction of the positive z axis.

    The result then for either way comes out to [0,1,0] which indicates a deflection in the direction of positive y and this agrees with all the experiments.

    Make sense or not?

    A third problem which i will call Problem #3 is the orientation of the B field relative to the permanent magnets used to produce the magnetic field B. The technique usually shown is that for two magnets close to each other so that they form a small gap (maybe 0.1 inches apart) and oriented with one north pole facing one south pole, the B field is directed from the north pole to the south pole. So an arrow that shows B would have the tail near the north pole and the tip of the arrow near the south pole. This is also important because the direction of the B field must be specified also. It also seems convenient to take this as being along the z axis although this isnt a strict rule.

    Also make sense or not?

    See attachment for an experiment.
     
    Last edited: Sep 7, 2015
  20. BR-549

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    Sep 22, 2013
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    I tried to clarify in post #11. I didn’t catch that there was established current.

    Ok, then the direction of current will decide in and out. Or up and down, or back and forth. As it always does. OR the polarity of the b field.

    If you align the B field, parallel to the field of the current, you will get more efficiency.

    If you’ve ever played with long suspended conductors, like Ampere did, you’ll see what I mean.

    He tried all kinds of angles.

    Today, it’s easy to get a strong static b field. The question will be how much current it takes, for the current field to interact.
     
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