How Do Maxwell's Equations Account for Magnetic Monopoles?

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Homework Help Overview

The discussion revolves around the application of Maxwell's equations in the context of magnetic monopoles. Participants are exploring how these equations can be modified or interpreted to account for the existence of magnetic monopoles, particularly focusing on Gauss' law and Faraday's law.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants are examining the validity of equations related to magnetic monopoles derived from Maxwell's equations. There are questions about the independence of electric and magnetic charge and the implications of Lenz's law on induced electromotive force (emf).

Discussion Status

Some participants have provided corrections and alternative perspectives on the equations presented, particularly regarding the signs of terms and the relationship between electric and magnetic charges. The discussion is ongoing, with various interpretations being explored without a clear consensus.

Contextual Notes

There are references to specific conditions under which the equations hold, as well as the potential implications of Dirac's argument in quantum electrodynamics (QED). The discussion also highlights the complexity of integrating magnetic monopoles into classical electromagnetic theory.

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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?

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


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:
Lenz's Law said:
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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I think the I_b should be -I_b.
 
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:
[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]

[/Color]
 
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