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Current induced in loop as magnetic monopole passes

  1. Nov 29, 2016 #1
    1. The problem statement, all variables and given/known data
    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?
    2. Relevant 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]

    3. 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
    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]
    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 [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
    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. jcsd
  3. Dec 4, 2016 #2
    Thanks for the thread! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post? The more details the better.
  4. Feb 3, 2018 #3
    It looks like your equation should be:
    $$\frac{d\Phi}{dt} = L\frac{dI}{dt}$$
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