Induced electric field outside a solenoid

[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

 

[Troels]

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:

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

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)

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.

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 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.

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

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.

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.

 

[Ajoo]

Thank you for replying.

throw in a factor of 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: so you need a factor of somewhere)

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

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

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.

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.

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

 

[Troels]

Thank you for replying.

No problem

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.

There are two "kinds" of electric fields:

Couloumb-fields as determined by gauss law:

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

And faraday fields as determined by faradays law:

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

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 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.

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:

A line carge $$\lambda$$ is glued onto the rim of a wheel of radius b. In the central region out to a radius of a there is a uniform magnetic field pointing up, (parallel to the axis of the wheel -Troels). Now someone turn the field off. What happens? The changing magnetic field will induce an electric field, curling around the axis of the wheel. This electric field exerts a force on the rarges at the rim and the wheel starts to turn(...) A final word on this example: It's the electric field that did the rotating. To convince you of this, I deliberately set things up so the magnetic field is zero at the location of the charge (on the rim)

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

 

[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.

 

[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

 

[Troels]

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.

Does he?

A long solenoid of radius a is driven by an alternating current, so that the field inside is sinusoidal: $$\textbf{B}(t)=B_0\cos(\omega t)\,\textbf{\hat z}$$. A circular loop of wire of radius a/2 and resistance R is placed inside the solenoid, and coaxial with it. Find current induced in the loop as a function of time. (emph. added)

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?

 

[TVP45]

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

I think Serway's Principles of Physics might help you. You can find it in Google books.

 

[Biest]

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?

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 don't know what he means by long in this case because you can supposedly take it out, but by general definition it should fit.

 

[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.

 

[Troels]

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.

Fair enough

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 don't know what he means by long in this case because you can supposedly take it out, but by general definition it should fit.

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 truly 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.

 

[Biest]

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 truly 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.

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