# Magnetic monopole induces current in perfect conducting ring

1. Mar 7, 2006

### xman

I am trying to calculate the current induced in a loop assuming a magnetic monopole exists. The loop is a perfect conductor, which I understand implies the electric field inside must be zero. I picture the problem with the magnetic monopole traveling with a velocity coaxial with the loop. I am given the loop has a self inductance L, so what I've done is take Faraday's law (modified for the existance of a mag. monopole), and the current density I'm associating as follows
$$\vec{\nabla} \wedge \vec{E} =0= -\left\{ \mu_{0} \vec{j}_{m}+\frac{\partial \vec{B}}{\partial t} \right\}$$
Yielding
$$\mu_{0} \vec{j}_{m} = -\frac{\partial \vec{B} }{\partial t}$$
Now relating to the induced emf I have
$$-L\frac{dI}{dt} = -\dot{\phi}_{m} = \int -\frac{\partial \vec{B}}{\partial t} \cdot d\vec{a}$$
From which I make the identification
$$\vec{j}_{m} = \rho_{m} \vec{v}$$
Which gives assuming a planar area loop with the direction of the velocity being parallel to the loop is given by
$$-L \frac{dI}{dt} = \mu_{0} \rho_{m} A_{loop} v \Rightarrow -L\frac{dI}{dt} =\mu_{0} \rho_{m} A_{loop} \frac{dz}{dt}$$
which is where I am stuck. I think I want to integrate over a long time, including the point and past the point the magnetic monopole passes through the loop. My question(s) is does everything I have done so far make sense, and if so where do I go from here?

Last edited: Mar 7, 2006
2. Mar 7, 2006

### Meir Achuz

Your equations are wrong. E is not zero in the metal, because there isinductance. Forget about the fields inside the metal of the loop. "Perfect conductor" in this context just means that there will be no IR in your eventual circuit equation. You can find E at the locus of the loop just as you would find B from a moving electric charge.
Then EMF =2\pi R E, and will be time dependent. Then use EMF=L di/dt.

3. Mar 7, 2006

### xman

Thanks for pointing that out, this problem is throwing me around. Ok so I have
$$\vec{\nabla} \wedge \vec{E} = -\mu_{0} \vec{j} -\frac{\partial \vec{B}}{\partial t} \Rightarrow \oint \vec{E} \cdot d\vec{\ell} = -\int \left(\mu_{0} \vec{j} +\frac{\partial \vec{B} }{\partial t} \right) \cdot d\vec{a}$$
where I've applied Stoke's theorem to the LHS of first equation. So far so good right? Now what I want to do is
$$2\pi a E = -\mu_{0} \rho_{m} \pi a^{2} v+ \int -\frac{\partial \vec{B} } {\partial t} \cdot d\vec{a}$$
where $$a$$ is the radius of the loop, and $$v$$ is the velocity of the monopole. Now make the identification
$$\int -\frac{\partial \vec{B}}{\partial t} \cdot d\vec{a} =- \dot{\phi}_{m} =\varepsilon = -L \dot{I}$$
substitute back
$$2 \pi a E + \mu_{0} \rho_{m} \pi a^{2} v=-L \dot{I}$$
am I on the right path or did I miss step again.

Last edited: Mar 7, 2006
4. Mar 8, 2006

### Meir Achuz

I meant to find E=-gvXr/r^3 at the locus of the ring,
and there is no need for the surface integral.

5. Mar 8, 2006

### xman

I am sorry, you lost me here. Could you explain a little more please? What is 'Xr'? and certainly 'g' isn't gravity right? I think I might need a little hand holding for this part.

6. Mar 12, 2006

### Meir Achuz

Sorry to be slow. I've been away

"I meant to find E=-gvXr/r^3 at the locus of the ring,
and there is no need for the surface integral."
In the equation:
E is the electric field.
g is the strength of the monopole charge (Like q for electric charge.)
X is the vector cross product symbol.
r is the vectpr disstance from the monopole to the ring.
r^3 is the cube of r.
The equation is the same thing you would have for -B at the ring due to an electric charge.

7. Mar 23, 2006

### xman

I have a solution without finding the electric field, here's what I did
$$\vec{\nabla} \wedge \vec{E} = -\mu_{0}\vec{j}-\frac{\partial \vec{B}}{\partial t}$$

converting to integral form the with Stoke's theorem

$$\int_{wire} \vec{E}\cdot d\vec{\ell}=-\mu_{0} \int_{S} \vec{j}\cdot d\vec{a}-\int_{S}\frac{\partial \vec{B}}{\partial t}\cdot d\vec{a}$$

Since perfect conductor then
$$\vec{E} = 0$$
inside the loop and the line integral around the wire is zero. Now make the identification
$$-\mu_{0}\int_{S} \vec{j}\cdot d\vec{a} =-\mu_{0}I_{m}$$
the second term of course is the induced flux, so if we integrate both sides wrt time we have
$$0=-\mu_{0}g-\Delta \phi_{m}\Rightarrow \Delta \phi_{m} = L\,\Delta I\Rightarrow I=\frac{\mu_{0}g}{L}$$

I think this works any ideas

Last edited: Mar 23, 2006