# Loop around an infinite solenoid?

1. Aug 24, 2010

### Curl

Say I have an ideal (long and tightly wound) solenoid, and I put a ring of wire around it (large ring). Then, by Faraday's law, if I put a current through the solenoid there will be an induced current in the wire ring around the solenoid.

But if there are no B-field lines crossing my ring of wire, how can this happen? Usually when you apply Faraday's law, its equivalent to using Lorrentz force law on each infinitesimal piece of wire (which contains free charges) and it turns out to be the same as the change in flux, by Stoke's theorem.

It's just weird that the loop of wire "sees" zero change (if the solenoid is ideal, there are no field lines outside it), so then how can there be an induced current? How does the information about the current in the solenoid travel from the solenoid to the wire loop? If there is zero field outside the solenoid, what carries this information?

Can someone explain this?

2. Aug 24, 2010

### pallidin

Just for clarification, are you suggesting that a magnetic field around a current carrying wire(solenoid or not, "ideal" or not) DOES NOT extend beyond the outside surface boundary of the wire?

3. Aug 24, 2010

### gabbagabbahey

There is zero magnetic field outside the wire, but the electric field will be non-zero whenever the current through the loop is zero. An induced electric field doesn't just exist in places of non-zero magnetic field or places where the field is changing. Only the curl of the induced electric field is limited to the region where the magnetic field is changing ( $\mathbf{\nabla}\times\textbf{E}=-\frac{\partial \textbf{B}}{\partial t}$ ).

4. Aug 24, 2010

### Curl

So the electric field is not limited to the inside of the conductor?

That's what it says in some books, that if the solenoid is long and tightly wound there is zero field outside even if the current is nonzero.

5. Aug 24, 2010

### gabbagabbahey

Correct, only its curl is zero outside the solenoid.

6. Aug 24, 2010

### Integral

Staff Emeritus
Indeed, in an infinitely long solenoid, or a toroidal solenoid, there is no electric field outside the coils it is completely contained within the coils.

You need to lose the model of "lines of force" cutting a wire and inducing a current, though picturesque, it is just wrong.

I had this exact same issue while doing a Rowland's ring experiment (a toroidal solenoid). When I presented this dilemma to the prof, he stood me at the chalk board staring at Faraday's law until I comprehended just what it expresses.

Perhaps you need to examine the concept of flux a little more.

7. Aug 24, 2010

### Bob S

This site has a program for calculating the magnetic field anywhere around (inside and outside) an air-core solenoid of variable length.

http://vizimag.com/calculator.htm

The program uses the formula derived in Smythe, Static and Dynamic Electricity, third edition, pages 290-191. Page 335 derives the mutual inductance of two coplanar loops of different radii, as well as the mutual inductance of a short coil and an infinite solenoid.

Bob S

Last edited: Aug 24, 2010
8. Aug 25, 2010

### gabbagabbahey

I hope you mean magnetic field.

When the current through the solenoid is changing, there will be an electric field outside that falls off with distance from the axis of the solenoid. If the electric field were zero outside, then $\oint\textbf{E}\cdot d\textbf{r}$ for the loop would be aswell, contrary to Faraday's law.

9. Aug 28, 2010

### Integral

Staff Emeritus
I need to post in here that Gabb...y is absolulty correct. I misspoke and should have said magnetic field. Sorry it has been 35yrs since I did that lab. I want you all to know that I reported myself for that post and was able to clear up my confusion with the help of Doc Al and Redbelly.

10. Aug 28, 2010

### pallidin

OK, so, where are we at now?
No magnetic field outside of that coil?

11. Aug 28, 2010

### Staff: Mentor

Yes. All the magnetic flux through the outer loop is due to the field within the solenoid. As gabbagabbahey explained, the changing magnetic field induces a non-coulombic electric field outside of the solenoid (including the location of the outer loop). That induced electric field creates the current in that loop.