Current induced in loop as magnetic monopole passes

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SUMMARY

The discussion centers on the induced current in a resistanceless loop of wire with self-inductance L when a magnetic monopole q_m passes through it. The derived equation for the induced current I is I = -μ₀q_m/L, where μ₀ is the permeability of free space. The participants clarify the relationship between the induced electromotive force (EMF), magnetic flux, and the self-inductance of the loop. A discrepancy in the sign of the induced current is noted, emphasizing the importance of accurately applying Faraday's law of induction and the concept of magnetic flux.

PREREQUISITES
  • Understanding of electromagnetic theory, specifically Faraday's law of induction.
  • Familiarity with the concepts of magnetic monopoles and their theoretical implications.
  • Knowledge of self-inductance and its role in electrical circuits.
  • Proficiency in vector calculus, particularly Stokes' theorem and curl operations.
NEXT STEPS
  • Study the implications of magnetic monopoles in theoretical physics.
  • Learn about the applications of Faraday's law in electrical engineering.
  • Explore the mathematical derivation of self-inductance in various circuit configurations.
  • Investigate the relationship between magnetic flux and induced EMF in more complex systems.
USEFUL FOR

This discussion is beneficial for physics students, electrical engineers, and researchers interested in electromagnetic theory and the behavior of induced currents in circuits involving magnetic monopoles.

Dazed&Confused
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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.
 
It looks like your equation should be:
$$\frac{d\Phi}{dt} = L\frac{dI}{dt}$$
 

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