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Maxwell Magnetic Monopole

  1. Jan 30, 2006 #1


    What is the solution in Gauss' law for a magnetic monopole based upon Maxwell's equations?

    Maxwell's equations:
    [tex]\Phi_E = \oint E \cdot dA = \frac{q_e}{\epsilon_0}[/tex]

    [tex]\Phi_B = \oint B \cdot dA = 0[/tex]

    [tex]\epsilon_e = \oint E \cdot ds = - \frac{d \Phi_B}{dt}[/tex]

    [tex]\epsilon_b = \oint B \cdot ds = \mu_0 \left(I_c + \epsilon_0 \frac{d \Phi_E}{dt} \right)[/tex]

    Gauss' magnetic monopole:
    [tex]\frac{\Phi_E}{\Phi_B} = c^2 \; \; \; q_b = q_e[/tex]
    [tex]\Phi_B = \oint B \cdot dA = \frac{\Phi_E}{c^2} = \mu_0 q_b[/tex]
    [tex]\boxed{\Phi_B = \oint B \cdot dA = \mu_0 q_b}[/tex]

    Is this solution correct?


    Reference:
    http://en.wikipedia.org/wiki/Magnetic_monopole
    http://www.physics.nmt.edu/~raymond/classes/ph13xbook/node172.html
     
    Last edited: Jan 31, 2006
  2. jcsd
  3. Jan 31, 2006 #2

    Meir Achuz

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    Your last equation is correct, but you don't need q_b=q_e for it to hold.
    q_b and q_e are independent unless you get into Dirac's QED argument.
     
  4. Feb 1, 2006 #3


    What is the solution for Faraday's law for the EMF generated by a magnetic monopole current?

    Faraday magnetic monopole:
    [tex]\epsilon_e = \oint E \cdot ds = \frac{I_b}{\epsilon_0 c^2}[/tex]

    Ampere-Maxwell magnetic monopole:
    [tex]\epsilon_e = \oint E \cdot ds = \frac{1}{c^2} \left( \frac{I_b}{\epsilon_0} - \frac{d \Phi_B}{dt} \right)[/tex]

    The negative sign was placed here because of Lenz's law:
    Are these solutions correct?

     
  5. Feb 1, 2006 #4

    Meir Achuz

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    I think the I_b should be -I_b.
     
  6. Feb 4, 2006 #5

    That criteria should hold for both a Faraday and a Ampere-Maxwell magnetic monopole, correct?

    Faraday magnetic monopole:
    [tex]\epsilon_e = \oint E \cdot ds = - \frac{I_b}{\epsilon_0 c^2}[/tex]

    Ampere-Maxwell magnetic monopole:
    [tex]\epsilon_e = \oint E \cdot ds = \frac{1}{c^2} \left( - \frac{I_b}{\epsilon_0} - \frac{d \Phi_B}{dt} \right) = - \frac{1}{c^2} \left( \frac{I_b}{\epsilon_0} + \frac{d \Phi_B}{dt} \right)[/tex]

    Ampere-Maxwell magnetic monopole:
    [tex]\epsilon_e = \oint E \cdot ds = - \frac{1}{c^2} \left( \frac{I_b}{\epsilon_0} + \frac{d \Phi_B}{dt} \right)[/tex]

     
    Last edited: Feb 4, 2006
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