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Induced electric field outside a solenoid 
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#1
Jan2308, 07:34 PM

P: 17

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: [tex] \vec{E} = \frac{\mu}{2} N I_0 sin( \omega t ) r \vec{e_\theta}, r<R [/tex] [tex] \vec{E} = \frac{\mu}{2} N I_0 sin( \omega t ) \frac{R^2}{r} \vec{e_\theta}, r>R [/tex] 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
Jan2408, 03:42 AM

P: 125

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


#3
Jan2408, 12:23 PM

P: 17

Thank you for replying.
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. 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
Jan2408, 01:42 PM

P: 125

Induced electric field outside a solenoid
Couloumbfields as determined by gauss law: [tex] \nabla \cdot \vec E = \frac{\rho}{\epsilon_0} [/tex] And faraday fields as determined by faradays law: [tex] \nabla \times \vec E = \frac{\partial B}{\partial t} [/tex] 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 nonzero at points, that are inside the source, but that does not imply that the themselves fields are zero at points outside the source. 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: [tex] \nabla\times\vec F = \vec a [/tex] 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 antiparallel 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 Bfield 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: 


#5
Jan2608, 02:56 AM

P: 46

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 [tex] \nabla \times \vec E = 0 = \frac{\partial B}{\partial t} [/tex] outside the solenoid. I'm guessing this is an affect by the "long" solenoid? Since if you have a time varying Bfield 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
Jan2708, 02:42 AM

P: 67

The idea of the "long" solenoid implies that [tex] B = 0 [/tex] 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
Jan2708, 07:07 PM

P: 125

Care to explain? 


#8
Jan2708, 07:16 PM

P: 1,127




#9
Jan2708, 07:56 PM

P: 67

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
Jan2708, 08:13 PM

Sci Advisor
PF Gold
P: 1,776

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 [itex]\oint E\cdot d\ell [/itex] around the solenoid must be proportional to the time rate of change of the B flux, [itex]\frac{d}{dt} \int B\cdot dA[/itex] 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
Jan2708, 09:49 PM

P: 125

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
Jan2808, 10:18 PM

P: 67

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


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