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Magnetic Force on current-carrying wire

  1. Aug 28, 2014 #1
    Not necessarily a homework question but this is pretty fundamental. I can't get a decent derivation online.

    If the length of the wire in the uniform magnetic field is [itex]l[/itex] and it moves a distance [itex]\delta s[/itex] in [itex]\delta t[/itex]:

    From the Lorentz Force Law:

    [itex]\vec{F} = q\vec{v}\times\vec{B}[/itex]

    if flux is cut at [itex]\frac{\pi}{2}[/itex]:

    [itex]\vec{F} = q\vec{v}\vec{B}[/itex]

    [itex]\vec{v} = \frac{d\vec{s}}{dt}[/itex], [itex]q = \int I dt[/itex]

    [itex]\vec{F} = \vec{B}\frac{d\vec{s}}{dt} \int I dt[/itex]

    That is about as far as I get, not sure how the length of the conductor [itex]l[/itex] comes into it?
     
    Last edited: Aug 28, 2014
  2. jcsd
  3. Aug 28, 2014 #2

    BvU

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    Let us try to get the question right: The force you want to calculate is the force on a current-carrying wire. (This is what I read in the title). But then the wire moves a distance ##\delta\,\vec s##, so more is happening ?

    "If the flux is cut at ##\pi\over 2##" means what ? The wire is perpendicular to ##\vec B##, ##\delta\,\vec s## is perpendicular to both ?

    Can I see a drawing ?

    You are apparently not happy with this derivation ? Isn't it sufficient to adjust the ##\vec v## vector for the case the wire isn't stationary ?
     
  4. Aug 28, 2014 #3
    Thank you for your response.

    Unfortunately I don't have a diagram, you are correct in your statement that the wire in this example is perpendicular to both the uniform magnetic field and the velocity vector.

    I don't really follow the context of that particular derivation you have referred to. I suppose if I were to describe the problem qualitatively the question I am asking is what is the relation between current and velocity for a wire moving through a uniform magnetic field.

    Maybe I am slightly off track. I find the concept of the induced emf due to the flux linkage relatively easy to grasp. I am now asking what is the relation to current?
     
  5. Aug 28, 2014 #4

    BvU

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    Yes, but everything can be brought back to the Lorentz force: Induced emf is a ##\vec v## of the charge carriers in a conductor, a wire in this case, due to the wire itself moving. Force on a current-carrying wire is a ##\vec v## of the charge carriers moving within the wire.

    My impression was you do not necessarily want to dig into the case of a combination of the two, but I could be wrong, however:
    tells me otherwise.

    Are you comfortable with the q v = q L/t = q/t L = I L steps in the derivation ?

    The effect of the v from moving the wire simply adds (vector wise) to the effect of the movement
    within the wire.

    Suppose the wire is moving downward in their picture, then all I can think of is that the red velocity vector rotates a bit clockwise (with the component along the wire remaining the same magnitude). And with that rotation, the F vector also rotates by the same amount, thus pushing the charge carriers a bit towards the right end of the wire. I can't do much in the sense of physics with that, however: bringing about a current I in the wire has become a little easier. But we don't look at external stuff like a current source or so.

    I think it would be wise to move on to further exercises with wire frames moving through zones with a magnetic field and such. Perhaps the links in the link also provide some insight.
     
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