- #1

Dazed&Confused

- 191

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## Homework Statement

Suppose a magnetic monopole [itex]q_m[/itex] passes through a resistanceless loop of wire with self-inductance [itex]L[/itex]. What current is induced in the loop?

## Homework Equations

[tex] \nabla \times \textbf{E} = - \mu_0 \textbf{J}_m - \frac{\partial \textbf{B}}{\partial t}[/tex]

[tex] \nabla \cdot \textbf{B} = \mu_0 \rho_m [/tex]

## The Attempt at a Solution

For a resistanceless loop [itex]\textbf{E}=0[/itex]. Thus integrating [itex] \nabla \times \textbf{E}[/itex], applying Stokes theorem, and then integrating over all time we have

[tex]

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 [/tex]

where [itex]\Phi[/itex] is the flux due to the magnetic field.

No we have that [itex] d \Phi / dt = -L dI /dt [/itex] so that [itex] \Delta \Phi = - LI[/itex]. Thus we have [tex]

-LI = \mu_0 q_m[/tex]

or

[tex]

I = -\frac{\mu_0 q_m}{L}

[/tex]

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 [itex]-LI[/itex] by saying that [itex]-LdI/dt[/itex] was equal to the loop integral of [itex]\textbf{E}[/itex] and that the change in magentic flux was in fact zero as [tex]

\oint \textbf{B} \cdot d \textbf{a} = \mu_0 q_m

[/tex]

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.