How Do Maxwell's Equations Account for Magnetic Monopoles?

[Orion1]

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

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

\Phi_B = \oint B \cdot dA = 0

\epsilon_e = \oint E \cdot ds = - \frac{d \Phi_B}{dt}

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

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

Is this solution correct?

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Reference:
http://en.wikipedia.org/wiki/Magnetic_monopole
http://www.physics.nmt.edu/~raymond/classes/ph13xbook/node172.html

 

[Meir Achuz]

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.

 

[Orion1]

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

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

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

The negative sign was placed here because of Lenz's law:

The polarity of the induced emf is such that it tends to produce a current that will create a magnetic flux to oppose the change in magnetic flux through the loop

Are these solutions correct?

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[Meir Achuz]

I think the I_b should be -I_b.

 

[Orion1]

I think the I_b should be -I_b.

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

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

Ampere-Maxwell magnetic monopole:
\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)

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

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