If a magnetic monopole travels through a wire loop, which way does the current flow in the circuit? I am confused about the sign convention for the electric field as described in the wikipedia article "Magnetic_monopole" and I would appreciate some help clarifying the topic. Take a positive (north) magnetic monopole traveling at some velocity, v, as I've illustrated in the figure below. The wikipedia article has the electric field rotating clockwise about the velocity vector and identifies this as conventional current. If the monopole travels through a wire loop (let's say copper) where the circuit terminates in a capacitor, where do the electrons end up gathering: Terminal 1 or Terminal 2 of the capacitor? Here, Terminal 1 is on the clockwise end of the wire while Terminal 2 is on the counter clockwise end of the wire. Terminal 2 is my choice but I'm unsure.
 
 A: From the wikipedia page on magnetic monopoles we have Gauss's law for magnetism, $\text{A}\cdot \text{m}$ convention:
$$\vec{\nabla}\cdot\vec B=\mu_0\rho_m$$
For a magnetic monopole, $\rho_m(\vec {r_f})=q_m\delta^{(3)}\left(\vec {r_f}-\vec{r_s}\right)$ where $q_m$ is the magnetic monopole moment, $\vec{r_s}$ is the location of the monopole, and $\vec{r_f}$ is the location of the field point. The $\vec B$ field is then
$$\vec B\left(\vec{r_f}\right)=\frac{\mu_0}{4\pi}\frac{q_m\hat {r}}{r^2}$$
Where $\vec r=\vec{r_f}-\vec{r_s}$ as may be verified by checking that $\vec{\nabla_f}\cdot\vec B=0$ for $\vec{r_f}\ne\vec{r_s}$ and that the flux out of a sphere of radius $b$
$$\oint_{\lVert\vec{r_f}-\vec{r_s}\rVert=b}\vec B\left(\vec{r_f}\right)\cdot d^2\vec {A_f}=\mu_0q_m$$
Then we put the monopole on the $z$-axis so that $\vec{r_s}=\langle0,0,z\rangle$ and center our loop at the origin in the $xy$-plane so that $\vec{r_f}=\langle r\cos\phi,r\sin\phi,0\rangle$. Then
$$d\vec{r_f}=\langle\cos\phi,\sin\phi,0\rangle dr+\langle-r\sin\phi,r\cos\phi,0\rangle d\phi$$
So
$$d^2\vec{A_f}=\pm\langle\cos\phi,\sin\phi,0\rangle dr\times\langle-r\sin\phi,r\cos\phi,0\rangle d\phi=\pm\langle0,0,r\rangle drd\phi=\langle0,0,r\rangle drd\phi$$
Because we want the positive sense of flux to be in the $+z$-direction (up). Then
$$\vec B\left(\vec{r_f}\right)=\frac{\mu_0q_m}{4\pi}\frac{\langle r\cos\phi,r\sin\phi,-z\rangle}{\left(r^2+z^2\right)^{3/2}}$$
Then we may calculate the flux of $\vec B$ through the loop of radius $a$ as
$$\begin{align}\Phi_B&=\int_0^{2\pi}\int_0^a\frac{\mu_0q_m}{4\pi}\frac{\langle r\cos\phi,r\sin\phi,-z\rangle}{\left(r^2+z^2\right)^{3/2}}\cdot\langle0,0,r\rangle drd\phi\\
&=-\frac{\mu_0q_mz}{4\pi}(2\pi)\left[-\left(r^2+z^2\right)^{-1/2}\right]_0^a\\
&=\frac{\mu_0q_m}2\left[\frac z{\sqrt{a^2+z^2}}-\text{sgn}z\right]\end{align}$$
Ignoring the retarded time,
$$\frac{d\Phi_B}{dt}=\frac{\mu_0q_m}2\left[\frac{a^2}{\left(a^2+z^2\right)^{3/2}}-2\delta{(z)}\right]\frac{dz}{dt}$$
Since we considered the positive sense of flux to be up, the positive sense of $d\vec{r_f}$ will be counterclockwise as viewed from above the loop. Then from Faraday's law with magnetic monopoles,
$$\begin{align}\oint_{r=a}\vec E\left(\vec{r_f}\right)\cdot d^2\vec{r_f}&=-\frac{d\Phi_B}{dt}-\mu_0I_m\\
&=-\frac{\mu_0q_m}2\left[\frac{a^2}{\left(a^2+z^2\right)^{3/2}}-2\delta{(z)}\right]\frac{dz}{dt}-\mu_0q_m\delta(z)\frac{dz}{dt}\\
&=-\frac{\mu_0q_ma^2}{2\left(a^2+z^2\right)^{3/2}}\frac{dz}{dt}\end{align}$$
The scary Dirac $\delta$ was canceled out by the current. It really had to be because any surface with the same boundary could have been used in Faraday's law. When the monopole is right next to the surface the flux is half the total flux of the monopole in one sense, and then half the flux but in the opposite sense after it crosses the surface. It can be seen that the flux increases at all times except when it crosses the surface and the rate of change of flux is undefined. It's proportional to the solid angle of the loop as seen from the monopole.  
So if the magnetic monopole was coming up from below at constant speed $\frac{dz}{dt}=v_0$ the counterclockwise line integral of $\vec E$ is negative, so $\vec E$ is clockwise as seen from above the loop.
$$\vec E=\frac{\mu_0q_mav_0}{4\pi\left(a^2+z^2\right)^{3/2}}\langle\sin\phi,-\cos\phi,0\rangle$$
A: 
Your electric field should be anti-clockwise; your illustration has it clockwise.   The diagram from the Wikipedia article is reproduced below; the magnetic monopole diagram is the lower figure, in blue and red.
You use the right hand rule, with your thumb pointing in the direction of travel, and the fingers curl in the direction of the E field.
This is anti-clockwise from the point of view of the observer being approached by the particle, which is the usual convention.
A: To model a magnetic monopole, imagine instead a very long, very thin solenoid with enormously tight winding density and high current. (I vaguely recall this is a "Dirac string.") Due to the usual symmetry considerations, the magnetic field outside the solenoid is zero over most of its length; field lines exit the solenoid at its north end, and enter at its south end. By making the solenoid thinner and more tightly wound, you can make these magnetic field "source regions" arbitrarily small.
You know what would happen if you passed the north end of this magnetic string through a conducting loop: you'd get a current in the loop attempting to oppose the change in the magnetic flux carried by the solenoid (though the information about the solenoid reaches the loop via the solenoid's fringing field). That's also what would happen if you passed a real magnetic monopole through the loop.
