# Induced electric field outside a solenoid

1. Jan 23, 2008

### Ajoo

Hi there.

On my electromagnetism test there was the following question:
A long solenoid with radius R has N turns per unit length and carries a current I = I_0*cos(ωt)

Find the electric field inside and outside the solenoid.

I got the following solutions:

$$\vec{E} = \frac{\mu}{2} N I_0 sin( \omega t ) r \vec{e_\theta}, r<R$$
$$\vec{E} = \frac{\mu}{2} N I_0 sin( \omega t ) \frac{R^2}{r} \vec{e_\theta}, r>R$$

My professor says the electric field outside the solenoid is zero which makes sense because we used the approximation that the magnetic field outside is also zero.

However i still have some doubts in this.
If i look at the expression of Faraday's Law all i can see is that the absence of magnetic field only implies that the electric field is conservative and the field i calculated seems to be so.

And i still don't get how if i draw a circular line around the solenoid there is magnetic flux going through the surface that line supports while there is no electric field outside.

Can someone please throw some lights on me?

Thank you and forgive me for my english :P

2. Jan 24, 2008

### Troels

throw in a factor of $$\omega$$ and I concur

$$\frac{d}{dt}\cos(\omega t) = -\omega\sin(\omega t)$$

(Hint: In future work, *always* check the units of your answer - that can uncover a great many errors. In this case: $$[\mu_0 N I_0 r]=\textrm{s V}/\textrm{m}$$ so you need a factor of $$\textrm{s}^{-1}$$ somewhere)

And with good reason, because your professor is wrong. There doesn't have to be a nonzero magnetic field outside the solenoid for it to setup a electric field faraday field, as much as there doesn't have to be a non-zero current density outside a wire for it to set up a magnetic field.

If the charge density is zero everywhere, then the conservative field has to be zero, and hence, the only possible E-field is a faraday field from the solenoid

The reasoning is exactly correct, and there *is* an electric field outside the solenoid. The analogy between Amperés law and faradays low is complete, so a long soleniod sets up a faraday field in the exact same manner as a fat wire sets up a magnetic field.

Think about this: In a transformer, there is *no* magnetic field outside the iron core; they are designed that way, as it would be a terrible waste of energy if there was. And yet, if you vary the magnetic field in core, you get a induced electric field in the coils wrapped around it, even though the wirering is not inside the iron core. Hence, your professors reasoning is wrong.

Last edited: Jan 24, 2008
3. Jan 24, 2008

### Ajoo

Yes, you're right. I just wrote those expressions out of my head as i recalled them and forgot that omega. I did get them right in the test though :P

Don't understand this part well as I don't know the concept of faraday's field (i study in another language, maybe i just know it by another name).
I'm assuming it means induced electric field by the variation of magnetic field. If yes, what i was trying to say is that this unduced electric field outside the solenoid has to be conservative because if dB/dt = 0 then curl(E) = 0 in that region. If i calculate the curl of the induced electric field for the exterior of the solenoid i get 0 as I expected.

I tried to explain my reasoning to my professor a few days ago and he wasn't convinced by it so I tryed to find examples or similar exercises on my books but i couldn't find any because the exercises on it had no solutions for me to consult.
All i could find was a few examples and articles on the internet but i'm not sure if i can convince him with that. If someone has seen a simillar exercise or something written that i could show him to prove my point i'd be most appreciated.

Thank you

4. Jan 24, 2008

### Troels

No problem

There are two "kinds" of electric fields:

Couloumb-fields as determined by gauss law:

$$\nabla \cdot \vec E = \frac{\rho}{\epsilon_0}$$

$$\nabla \times \vec E = \frac{\partial B}{\partial t}$$

I put "kinds" in quotations, for there is not really much point in distrinquish them, as the affect charges in the same way.

Don't be fooled by the zero curl in the exterior region. If you calculate the divergence of the electric field in the vicinity of a point charge, you will also get zero, and hence by gauss law in differential form you would falsely conclude that E is also zero there! Remember that about the laws in differential forms; they are only non-zero at points, that are inside the source, but that does not imply that the themselves fields are zero at points outside the source.

I have now shown this post to two persons, a PhD in magnetic systems, and a friend of mine who study electrical engineering, they both came to same conclusions as you and I, in less that 30 seconds.

I would ask your professor if he also thinks that the magnetic field outside a straight wire is also zero, for that would be the only mathematical conclusion of his reasoning with regard to faradays law. Here is how:

Both faradys law and Amperés law has the same basic form:

$$\nabla\times\vec F = \vec a$$

where F and a are some vectors. Hence, if a current density J and a time varying magnetic field B is identical (Or are at least parallel or anti-parallel at every point), the magnetic field and electric field the produce will also be identical. The current density in a straight wire and the magnetic field inside a solenoid is an example of such pair. It would thus be a contradiction to, on basis of amperes law, claim that the magnetic field is nonzero outside a wire, while at the same time claim on basis of faradays law the the electric field outside a solenoid with a varying B-field is zero.

Otherwise, I have only been able to find one explicit mention of this fact. It is from David J. Griffiths's textbook "Introduction to electrodynamics" example 7.8:

Well... there you go; the electric field is induced even though the magnetic field is zero at that location

Last edited: Jan 24, 2008
5. Jan 26, 2008

### trosten

I was wondering about the same question, about the field outside a solenoid with a varying current in its wires.

I'm a bit puzzled by the part that $$\nabla \times \vec E = 0 = \frac{\partial B}{\partial t}$$ outside the solenoid.

I'm guessing this is an affect by the "long" solenoid? Since if you have a time varying B-field inside the solenoid it will sureley varie outside it as well.

What about the outside part that is on top/above the solenoid, there the field will be almost as strong as inside it but it is still outside. This is what they use in thoose induction cookers, the problem I first wondered about.

6. Jan 27, 2008

### Biest

The idea of the "long" solenoid implies that $$B = 0$$ as the magnetic field lines that are unable to "exit" the solenoid. I find this reasoning a bit odd because it is usually only used with steady current and should change with a changing magnetic field. In question 7.12 (pg. 305) in Griffith's "Introduction to Electrodynamics" he clearly says a "long" solenoid and has a B field coming out of it, so that reasoning is defunct. And yes I solved the problem and my answer makes sense according to my prof with magic answer book.

Point your prof to that question, he might finally give in

7. Jan 27, 2008

### Troels

Does he?

It may be that I am terribly bad at understanding written english, but I utterly fail to see *where* Griffiths "clearly says a "long" solenoid and has a B field coming out of it" in the formulation of that particular problem. Or anywhere else in the book for that matter (yes I've read it all)

Care to explain?

8. Jan 27, 2008

### TVP45

9. Jan 27, 2008

### Biest

Sry bad reference on my part. I was looking at the diagram above as a reference, sry error on my part while doing my E&M homework out of that chapter.

Griffith does a have problem where the B field inside a long solenoid induces a changing flux outside the solenoid, which in case an E should be produced. Pg. 309 7.17. I dunno what he means by long in this case because you can supposedly take it out, but by general definition it should fit.

10. Jan 27, 2008

### jambaugh

In line with the advice in post 4 with regard to conceptual traps in the differential form of Maxwell's equations try starting with the integral form.

The line integral $\oint E\cdot d\ell$ around the solenoid must be proportional to the time rate of change of the B flux, $\frac{d}{dt} \int B\cdot dA$ through any area bound by that line integral's path. Since the latter is not zero the E field cannot be zero on all of that path.

11. Jan 27, 2008

### Troels

Fair enough

First of all I think you should dwell a bit on Example 5.9, in which he gives the formal argument on zero field outside an infinitely long solenoid, on basis of amperes law.

It is of course true that it doesn't really make sense to "pull an infinitely long solenoid out of a current loop", but provided that the soleniod is long compared to the radius enclosing loop, the amount of flux inside it is much greater than the one running back on the outside, so that it may safely be neglected in accordance with a truely infinite solenoid.

The essential point of that particular problem is not to concern yourself with the exact field of such a coil, but rather notice that the amount of charge flowing through the resistor is independent of the speed at which you move the solenoid.

12. Jan 28, 2008

### Biest

True, the only problem is see as pointed out earlier that the line integral of E is equal to the change in flux over time. That is given here, since outside the flux will be constant.

I don't understand where the problem with adding the word "infinitely." We physicists are just lazy :P