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Homework Help: Quick question about Ampere's Law and Lorentz force law

  1. Jan 12, 2012 #1
    Btw I am obtaining all these equations from chapter 5 of Griffiths text.

    Lorentz force law is: Fmag = ∫I(dI x B)

    Does this give the force on a current carrying wire due to an external magnetic field? (i.e. has nothing to do with own wires magnetic field)
    I think the answer is yes since the wire can really exert a force on itself.

    Also I am confused about the dI term. Is this as small segment of current in the wire? (the one we are calculating the force on)

    secondly, Amperes law in integral form is: ∫B . dI = Ienc*μo (the period is the dot product)

    I understand the idea of this integral, but again I am confused by the dI term. I know the LHS of the above equation is just the magnitude of B time the length of the loop enclosing the current. It seems that dI is a segment of the loop enclosing the current, but if dI is a segment of current (as I assumed for Lorentz law) I think I might be missing a piece of this puzzle.
  2. jcsd
  3. Jan 12, 2012 #2

    rude man

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    1. Yes, dF = i(dl x B) is an element of force on a current-carrying element of wire dl. Its own magnetic field does not apply a force to itself; these forces cancel each other. dl is an elemental length of wire, not a current. The current is supplied by "i" in the formula. "i" is a scalar.

    2. In the formula ∫B*dl, B and l are vectors. dl is an element of length around the contour enclosing the current i. For an infinitely long wire and integrating around a circular contour, and no other sources of B present, B is due to just the current-carrying wire, so B is uniform around the contour; B and dl are always in-line so the dot-product is just B*l = B*2∏r where r is the radius of the contour.
  4. Jan 12, 2012 #3

    Indeed this gives the force that on the wire, or put better on the moving charges in the wire. It is straightforward from the Lorentz force law for a charged point particle (with no electric field):


    (Note there is no force if the particle isn't moving)
    But in this case you have a whole lot of point particles moving continuously (or so we pretend) through the wire. So at every point of the wire you could say there is a force given by this law. To calculate the total force on the wire we integrate (sum) them all up to get the expression

    ∫dq(v[itex]\times[\itex]B then you say the infinitesimal bit of charge on the wire is given by the charge density \lambda so dq=[itex]\lambda[\itex]dl where dl is an infinitesimal bit of wire. take the charge densite together with the velocity and you get current. Now comes the tricky part where you made a reading error you don't integrate over dI but dl! you integrate over bits of wire not current. and because the current is actually pointing in the same direction as the wire (it can't very well move in another direction) you take the norm out (which is the current that you measure, 'I', no longer a vector).

    So basically you just made a bit of a misread (read dlength not dcurrent). And a mispel I think if a wire produces a magnetic field to force doesn't act on the wire NO self interaction.

    In ampere's law the dl is something else entirely namely the lengt of the closed loop you're integrating over. This means that dl as a vector is in the [itex]\phi[\itex] direction, but hey so is B so the inner product just yields you the factor in front of what the Biot-Savart law got you. integrate this over the entire loop and you get I_enc \in \mu_0 (actually also an inner product on R) since we all now the length of a circle or radius is just 2[itex]\pi[\itex] r
  5. Jan 12, 2012 #4

    rude man

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    I'm sorry but I could not follow your points you tried to make.
  6. Jan 12, 2012 #5
    Basically, in the first integral, where you were reading dI as in integrating over bits of current you should read dl as in integrating over bits of length of wire.

    In the second integral don't read dI as in bits of current again but this time the integration is over a circle around the wire which is just air (no actual physical loop or something just mathematical). So dl again the length f the loop.
  7. Jan 13, 2012 #6

    rude man

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    I don't think you understood what I said. I looked it over and cannot correlate your complaints with what I wrote. So sorry.
  8. Jan 13, 2012 #7
    Oh sorry I thought Jfuld couldn't follow, I basically said the same you did just a bit more lengthy. No complaints there I meant to say Jfuld, I think, misread the text in Griffiths. He must have read an I where he should have read an l. Then instead of integrating over two different concepts of length, namely length of wire and length of contour, you try to integrate over the same thing twice namely current in a wire which just doesn'treally maken sense. That must be why it seemed like such a puzzle!
  9. Jan 14, 2012 #8

    Right, I did misread. However these comments were are very helpful and I do understand it now thank you.

    lorentz law is sort of like coulombs law, and amperes is sort of like gausses law for magnetostatics.

    Thanks everyone!
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