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Lorentz Force on a Curved Wire in a Non-Constant Magnetic Field

  1. Mar 15, 2014 #1
    What is the formula to evaluate Lorentz force on a curved current carrying wire in a non-constant magnetic field (given by some known vector field). The standard form of Lorentz force (Fb=BxlI) when B and the wire's length 'l' are constants does not account for this case, nor does the differential form of the equation when the wire can be curved but the magnetic force must be constant (dFb=(Bxdl)I).

    For example, if B is being generated by a simplified dipole and the curve is being generated by some parameterized function, what integral describes the force exerted on the wire? (This is shown below with the wire being a circular loop centered on the dipole. Though this is possibly a trivially symmetrical case that evaluates to zero, its still worth considering.)

    My initial guess is that such an evaluation needs to be done using some sort of line integral or at least a double integral, but the cross product throws the whole thing off for me. Suggestions or known properties?

    (This is in conjunction with an effort to create a quantitative analysis of homopolar motors, two of which with slightly different parameters and properties from the above question are shown below)
     

    Attached Files:

  2. jcsd
  3. Mar 15, 2014 #2

    jtbell

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    Staff: Mentor

    Why do you think ##\vec B## has to be constant? (I assume you mean "constant with respect to position", i.e. "uniform")

    Find some way to express ##d\vec F = \vec B \times d\vec l## as a function of position along the wire, then integrate that. In your examples, it looks like ##d\vec F## is always into or out of the paper so you can call it positive or negative accordingly.

    If the wire forms a closed loop, and the net current "piercing" the loop is zero, and the net flux of ##\vec E## through the loop is constant, then the answer turns out to be very simple. See Ampère's Law in integral form.
     
    Last edited: Mar 15, 2014
  4. Mar 16, 2014 #3
    You've forgotten the current term I in your expression for dF. It would seem that ##d\vec F = I (d\vec B \times d\vec l)## would yield the infinitesimal value of F at every point. In order to solve, can I simply take the double integral of the R.H.S. with respect to each variable? or do I first need to map one function into the other to take the integral? Again, the cross product also adds a layer of complexity in order to sum.

    Your final statement that "If the wire forms a closed loop, and the net current "piercing" the loop is zero, and the net flux of E⃗ through the loop is constant, then the answer turns out to be very simple. See Ampère's Law in integral form." is true, however, I don't see how it applies to Lorentz force as it only utilizes dot products and doesn't explicitly describe forces, only the magnetic fields due to currents or vice versa...
     
  5. Mar 16, 2014 #4

    jtbell

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    Staff: Mentor

    Oops, you're right, I left out the I. It should be ##d\vec F = I(\vec B \times d\vec l)##. It's ##\vec B##, not ##d\vec B##. ##\vec B## is the magnetic field at the point where the infinitesimal wire segment ##d\vec l## is located. You're integrating only along the length of the wire.

    You may be confusing this with problems where you find the ##\vec B## at a point, produced by a current-carrying wire with some shape, by integrating the contributions ##d\vec B## from each segment ##d\vec l## of the "source wire." Here, you already know ##\vec B## at each point, either because you've measured it, or you've calculated it from the properties of the source.

    The current is the same everywhere along the wire that the force is being exerted on, so if the wire is a closed loop, then
    $$\vec F = \int {d\vec F} = I \oint {\vec B \times d\vec l}$$
    You're right, Ampère's Law involves the dot product ##\vec B \cdot d\vec l## instead. Brain fart. :blushing:

    Exactly what you use for ##d\vec l## depends on the symmetry of the problem. If the loop is a rectangular one, you might want to use ##d\vec l = \hat x dx + \hat y dy + \hat z dz## and align the loop along the coordinate axes. For spherical symmetry and a circular loop, look up the equivalent expression for ##d\vec l## in terms of ##\hat r##, ##\hat \theta## and ##\hat \phi##. Whichever one you use, express ##\vec B## using the same coordinate system.
     
    Last edited: Mar 16, 2014
  6. Mar 18, 2014 #5
    I think this will help you.

    [itex]\mathtt{F\ =\ \ BILsin\theta}[/itex]

    http://www.xtremepapers.com/revision/a-level/physics/electromagnetism.php
     
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