# Writing magnetic monopole analogs for electrical circuits

## Homework Statement

This isn't a 'homework' problem as such. I'm currently in University Physics II (E&M), and I've become really interested in magnetic monopoles. They haven't been discussed in any kind of depth in my course, but I'm trying to figure out how a magnetic monopole (if they exist) would behave in a circuit. I've derived Gauss's Law for Magnetism under the assumption that monopoles exist, and I'm now trying to figure out the equations for magnetic potential, as well as the equations for the behavior of monopoles in capacitors, resistors, and inductors.

## Homework Equations

Gauss's Law for Magnetism (with monopoles)
$$\Phi=\oint \vec B \cdot d \vec A=\mu_0q_m$$

## The Attempt at a Solution

The magnetic vector potential is written in terms of the curl of a vector.

$$\vec B=\nabla x \vec A$$

The curl of a vector is not something that I've encountered yet. I'm currently in differential equations (which comes between calc II and calc III at my school). Is there a way to write magnetic potential in terms of an integral or algebraic expression? I've looked and haven't managed to find anything.

Using the relationship that electrical capacitance is ##C=\frac{q}{V}##, I simply wrote this as ##C_m=\frac{q_m}{V_m}## as a direct analogy for a hypothetical magnetic charge. Then I used the modified Gauss's Law for Magnetism and the relation that ##\Phi=\mu_0 q_m## to modify the magnetic analogy for capacitance as ##C_m=\frac{\Phi_m}{\mu_0 V_m}##

Is this a correct way of looking at it? Or is this way off? In either case, I'm still stuck with a magnetic potential that I don't know how to evaluate.

I've done a fair amount of reading on this, and I keep seeing that magnetic flux is essentially the equivalent of 'magnetic current' in a magnetic monopole circuit. Is this the case? I can't figure out how to mathematically arrive at this conclusion.

In an electrical circuit, ##I=\frac{dq}{dt}##, thus in a magnetic circuit ##I_m=\frac{dq_m}{t}##, and ##q_m=\frac{\Phi_m}{\mu_0}##, therefore ##\frac{dq_m}{dt}=\frac{1}{\mu_0}\frac{d\Phi_m}{dt}##. This leads to a conclusion that $$I_m=\frac{1}{\mu_0}\frac{d\Phi_m}{dt}$$