Dazed&Confused
- 190
- 3
Homework Statement
Suppose a magnetic monopole q_m passes through a resistanceless loop of wire with self-inductance L. What current is induced in the loop?
Homework Equations
\nabla \times \textbf{E} = - \mu_0 \textbf{J}_m - \frac{\partial \textbf{B}}{\partial t}
\nabla \cdot \textbf{B} = \mu_0 \rho_m
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
<br /> 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 <br /> -LI = \mu_0 q_m
or
<br /> I = -\frac{\mu_0 q_m}{L}<br />
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 <br /> \oint \textbf{B} \cdot d \textbf{a} = \mu_0 q_m<br />
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.