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Homework Help: Magnetic forces

  1. Mar 8, 2004 #1
    it is often said that "magnetic forces can do no work".. but a straight wire segment placed in a magnetic field experiences a force of mag. B*I*L (LaPlace' law) which accelerates it and thus does work.. how can this be true?? is it really an electric force that is doing the work here or what?
  2. jcsd
  3. Mar 8, 2004 #2


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    If we look at how the magnetic force is calculated from the Lorentz force law:

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

    Because of the cross product, FB and v must be perpendicular.

    Recall that the general definition of work done by a force is:

    [tex] W_{ab} = \int_a^b \vec{F} \cdot d\vec r [/tex]

    Due to the dot product here, you should see that not work is done by a force that is perpendicular to the displacement. However, from the above Lorentz force discussion, this is exactly what we have. So the magnetic force can do no work.

    Be careful not to confuse this with the magnetic force's ability to accelerate a charged particle by changing its direction (not speed/kinetic energy). Hope that helps.
  4. Mar 9, 2004 #3

    Doc Al

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    Staff: Mentor

    Another excellent question.

    Yes, it is an electric force doing the work!

    Since, as jamesrc reminds us, the magnetic force on a moving charge is always perpendicular to the velocity, it can do no work. If the charges were free, they would simply spiral. Yet a current-carrying wire experiences a force that can do work. What's different there?

    In the wire, the charges are constrained. The magnetic force shifts the charges to one side, until the resulting electric field balances the magnetic force. It is this electric field that pulls the wire.
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