Current induced in loop as magnetic monopole passes

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1. Nov 29, 2016

Dazed&Confused

1. The problem statement, all variables and given/known data
Suppose a magnetic monopole $q_m$ passes through a resistanceless loop of wire with self-inductance $L$. What current is induced in the loop?
2. Relevant equations
$$\nabla \times \textbf{E} = - \mu_0 \textbf{J}_m - \frac{\partial \textbf{B}}{\partial t}$$
$$\nabla \cdot \textbf{B} = \mu_0 \rho_m$$

3. The attempt at a solution
For a resistanceless loop $\textbf{E}=0$. Thus integrating $\nabla \times \textbf{E}$, applying Stokes theorem, and then integrating over all time we have
$$0=\int \oint \textbf{E} \cdot d\textbf{l} dt = -\mu_0 \int \textbf{J}_m \cdot d \textbf{a} dt - \int \int \textbf{B} \cdot d \textbf{a} dt = -\mu_0 q_m + \Delta \Phi$$
where $\Phi$ is the flux due to the magnetic field.
No we have that $d \Phi / dt = -L dI /dt$ so that $\Delta \Phi = - LI$. Thus we have $$-LI = \mu_0 q_m$$
or
$$I = -\frac{\mu_0 q_m}{L}$$

It just so happens that I have the solution. My answer is off by a sign. The method was outwardly similar, except there the left hand side of my equation was found equal to be $-LI$ by saying that $-LdI/dt$ was equal to the loop integral of $\textbf{E}$ and that the change in magentic flux was in fact zero as $$\oint \textbf{B} \cdot d \textbf{a} = \mu_0 q_m$$
and that when the charge is far away (on either side) the flux through a flat surface will be zero so that the change is also zero. I cannot see why mine is incorrect assuming a resistanceless wire, but on the other hand I see that a zero change in magnetic flux must also be true.

2. Dec 4, 2016