Exploring the Interaction of Magnetic and Electric Fields

[david316]

Hello,

If you have an appropriately oriented conductive ring in a constantly changing magnetic field, current will flow in the ring. There will also be a magnetic field associated with the current in the ring. I understand (maybe ... ) that the current is due to the electric field which is present when a magnetic field is changing.

Assuming there is a point in time when the external magnetic field begins to change, does the magnetic field that is "induced" as a result of the current in the ring interact with the changing magnetic field and result in some sort of dynamics with the magnetic and electric fields or does every thing happen simultaneously (like a electromagnetic wave i.e. you can't have a oscillating electric field without a oscillating magnetic field).

Thanks a lot for any clarity.

 

[vanhees71]

The electromagnetic field, on a classical (i.e. non-quantum) level, does not interact with itself. The sources of the field are the charge and current distributions. The field interacts with the electric charges, including the contribution of these charges themselves to the field.

 

[david316]

Thank you. I understand that in classic physics the electromagnetic field doesn't interact with itself but in the scenario aren't there two "separate" magnetic fields? The changing magnetic field and the "induced" magnet field. Or should this just be considered one single electromagnetic field?

Also, when you drop a magnet in a hollow metal tube, it slows its descent. I assumed that this was because the induced magnetic field is interacting with the magnets magnetic field. Is this not the case?

 

[rude man]

The changing magnetic field induces an electric field; this is true whether or not you have any conducting ring.
In the case of your ring the electric field produces a current. This current produces its own magnetic field so as to oppose the external changing field. Faraday's law, $emf = - d\phi/dt$ refers to the vector sum of the two mag fields, not just the external one. Computing the induced flux is prohibitively difficult (for you) which is why, whenever you tackle such a problem the assumption is always that the induced B field is << external B field.

If the external field is produced by a coil of wire there is a further complication, which is the effect on the external field by the induced field. That's called mutual inductance.

Your last statement "you can't have a oscillating electric field without a oscillating magnetic field" is certainly true.

 

[david316]

"whenever you tackle such a problem the assumption is always that the induced B field is << external B field." Thanks. This will help me sleep at night!

 

[david316]

In the case of your ring the electric field produces a current. This current produces its own magnetic field so as to oppose the external changing field. Faraday's law, $emf = - d\phi/dt$ refers to the vector sum of the two mag fields, not just the external one.

The way this is loosely taught essentially goes like this:

-Changing magnetic field = electric field produced
-Electric field results in current in ring
-Current in ring results in magnetic field
-Magnetic field forms opposing external field
-Total field at the end of this process is vector summation of the two fields.

When this process occurs, is it correct to think about it as a dynamic process (I kind of assume it isn't given that in electric magnetic waves one field doesn't induce the other ...)? For example as the induced magnetic field "forms" does this contribute to $emf = - d\phi/dt$ . When the induced field is fully formed $d\ phi/dt$ will only be the changing external field so its only during this "forming" period you would see some dynamics if said dynamics exist.

 

[rude man]

The way this is loosely taught essentially goes like this:

-Changing magnetic field = electric field produced
-Electric field results in current in ring
-Current in ring results in magnetic field
-Magnetic field forms opposing external field
-Total field at the end of this process is vector summation of the two fields.

Fine.

When this process occurs, is it correct to think about it as a dynamic process (I kind of assume it isn't given that in electric magnetic waves one field doesn't induce the other ...)? For example as the induced magnetic field "forms" does this contribute to $emf = - d\phi/dt$ . When the induced field is fully formed $d\ phi/dt$ will only be the changing external field so its only during this "forming" period you would see some dynamics if said dynamics exist.

Yes you're absolutely correct.

Let's assume the exernal B flux $\phi$ is a ramp: $\phi = \dot \phi t$ with $\dot \phi$ constant.
Then what happens is the current starts to rise from zero exponentially until it reaches a constant value: $i = \dot \phi /R$ with R the coil's resistance.

So now you ask: how fast?
Answer: the actual current $i = \frac {\dot \phi} {R}[ (1 - exp(-Rt/L)]$
So, what is L? L is the inductance of the coil, i.e. self-flux divided by current.
The above equation is the solution to $L~ di/dt + iR = \dot \phi$. The L term accounts for the self-generated flux.
As I said in most problems we ignore the L term and just say $i = \dot \phi/R$. That's because actually calculating L for a single-turn coil is prohibitively difficult.

(You could also solve that ODE with a sinusoidal extrnal flux instead of a ramp, or any other function of your choice).

But now you know the rest of the story.
(Well, everythng except if the self-induced flux feeds back to the coil generating the external flux. That's more of an EE problem than a physics problem anyway).

 

[vanhees71]

Thank you. I understand that in classic physics the electromagnetic field doesn't interact with itself but in the scenario aren't there two "separate" magnetic fields? The changing magnetic field and the "induced" magnet field. Or should this just be considered one single electromagnetic field?

Also, when you drop a magnet in a hollow metal tube, it slows its descent. I assumed that this was because the induced magnetic field is interacting with the magnets magnetic field. Is this not the case?

It's a bit misleading to say that one field induces another field. You can indeed give integral relations between the fields to this effect, but they are not very useful from a physics point of view. There's one and only one electromagnetic field with electric and magnetic components, dependent on the reference frame you measure the electromagnetic field in.

The physical solution for the components are all expressed in terms of retarded integrals over the only true sources, which are the charge and current densities of the electric charge.

 

[david316]

Let's assume the exernal B flux $\phi$ is a ramp: $\phi = \dot \phi t$ with $\dot \phi$ constant.
Then what happens is the current starts to rise from zero exponentially until it reaches a constant value: $i = \dot \phi /R$ with R the coil's resistance.

So now you ask: how fast?
Answer: the actual current $i = \frac {\dot \phi} {R}[ (1 - exp(-Rt/L)]$
So, what is L? L is the inductance of the coil, i.e. self-flux divided by current.
The above equation is the solution to $L~ di/dt + iR = \dot \phi$. The L term accounts for the self-generated flux.

Thank you very much. This is helpful. Another question... does the derived ODE make the following assumption?

"whenever you tackle such a problem the assumption is always that the induced B field is << external B field."

I assume it does since the RHS of the ODE only includes the derivative of the external field?

 

[rude man]

Thank you very much. This is helpful. Another question... does the derived ODE make the following assumption?

"whenever you tackle such a problem the assumption is always that the induced B field is << external B field."

I assume it does since the RHS of the ODE only includes the derivative of the external field?

No. The L di/dt term represents the self-induced B field. The equation is complete except for the possibility of the self-induced field causing an emf to be generated back into the external coil.

L di/dt = $\dot \phi_{22}$ where $\phi_{22}$ is the self-induced flux ("flux on coil 2 due to coil 2"). The external flux we can designate as $\phi_{21}$ ("flux on coil 2 due to coil 1".) Coil 1 produces the external flux, coil 2 is your ring.

 

[david316]

Gotcha. Thanks!