# Maxwell Magnetic Monopole

1. Jan 30, 2006

### 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?

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

Last edited by a moderator: May 2, 2017
2. Jan 31, 2006

### 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.

3. Feb 1, 2006

### Orion1

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

$$\epsilon_e = \oint E \cdot ds = \frac{I_b}{\epsilon_0 c^2}$$

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:
Are these solutions correct?

4. Feb 1, 2006

### Meir Achuz

I think the I_b should be -I_b.

5. Feb 4, 2006

### Orion1

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

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