Current induced in loop as magnetic monopole passes

In summary, the conversation discusses the induced current in a resistanceless loop of wire with self-inductance L when a magnetic monopole q_m passes through it. The solution involves using equations for curl and divergence of the electric and magnetic fields, as well as applying Stokes theorem and integrating over time. However, there is a discrepancy in the solution as the answer is off by a sign. Both solutions involve the change in magnetic flux being equal to zero, but it is unclear why the provided solution is incorrect.
  • #1
Dazed&Confused
191
3

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.
 
  • #3
It looks like your equation should be:
$$\frac{d\Phi}{dt} = L\frac{dI}{dt}$$
 

FAQ: Current induced in loop as magnetic monopole passes

1. What is a magnetic monopole?

A magnetic monopole is a hypothetical particle that has only one magnetic pole (either north or south). Unlike regular magnets, which have both a north and south pole, a magnetic monopole would have only one pole and would exist as a single, isolated particle.

2. How does a magnetic monopole induce a current in a loop?

When a magnetic monopole passes through a loop of wire, it creates a changing magnetic flux through the loop. This changing magnetic flux induces an electric field, which in turn causes an electric current to flow in the loop.

3. What is the direction of the induced current in the loop?

The direction of the induced current in the loop depends on the direction in which the magnetic monopole is moving. If the monopole is moving towards the loop, the induced current will flow in one direction, and if it is moving away from the loop, the current will flow in the opposite direction.

4. How does the strength of the induced current relate to the magnetic monopole?

The strength of the induced current is directly proportional to the strength of the magnetic monopole. This means that a stronger magnetic monopole will induce a stronger current in the loop.

5. Is the current induced in the loop permanent?

No, the current induced in the loop is not permanent. It will only flow as long as the magnetic monopole is in motion and creating a changing magnetic flux through the loop. Once the monopole stops moving or leaves the vicinity of the loop, the induced current will stop flowing.

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