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Effective potential energy of a charge in a magnetic field

  1. Dec 13, 2014 #1
    Hi,
    I'm studying the Lagrangian and its applications in electromagnetism. I stumbled across this inconsistency:

    The force of a charge moving through a magnetic field is
    ## F_b = q v \times B ##
    If we define B to be in the ## \hat{z} ## direction, this equation can be written as
    ## F_b = q ( \dot{\rho} \hat{\rho} + \rho \dot{\phi} \hat{\phi} + z \hat{z} ) \times B \hat{z} ##
    ## F_b = q \dot{\rho} B (-\hat{\phi}) + q \rho \dot{\phi} B \hat{\rho} ##
    ## F_b = -q \dot{\rho} B \hat{\phi} + q \rho \dot{\phi} B \hat{\rho} ##

    According to John Taylor (Classical Mechanics) the ## \hat{\phi} ## term is equal to zero and the only force the charge experiences is
    ## F_b = q \rho \dot{\phi} B \hat{\rho} ##

    Why is this?
     
  2. jcsd
  3. Dec 13, 2014 #2
    John who? I just looked up John Taylor and found this pdf...I am reading section 2.5 "Motion of a charge in a uniform magnetic field" and I don't see what you seem to indicate.

    From the formulation shown (in Cartesian coordinates), the charge is going to experience two forces (one for each coordinate) and both are perpendicular to the direction of the magnetic field.

    The same happens in cylindrical coordinates...if a charge moves purely in φ direction, it is going to experience a force that will push it radially out (increasing r); and if a charge moves in the radial direction r, it will experience a force that pushes it in the -φ. If the charge motion also has a z component, this will not produce a force...the magnetic field does not mind such motion as it is not crossing the field in anyway.
     
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