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While applying Ampere's law to derive the magnetic field of a solenoid, why can we consider $\vec{\bf B }$ to be zero just outside the solenoid? For example here it says "Only the upper portion of the path contributed to the sum because the magnetic field is zero outside..". What is the proper justification for this statement?

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For an infinite solenoid, you can argue by symmetry that the $B$-field outside the solenoid has to be parallel to the axis. From this, by varying the size of the loop used in Ampere's law, you can show that the $B$-field outside the solenoid (whatever strength it is) does not vary with distance from the solenoid.

It's pretty easy to show that the $B$-field goes to zero from a solenoid, even an infinite one, as the distance from the solenoid goes to infinity. And so the $B$-field has to be uniformly zero outside the solenoid.

For a finite solenoid, if you are not close to the ends, you can argue that the missing parts of the infinite solenoid shouldn't affect the $B$-field much, and so the field is weak outside the solenoid as compared to inside.

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  • $\begingroup$ I strongly believe this is wrong. Even an infinite solenoid.would.produce.a magnetic field similiar (even the same) like.a wire with.the same current. Or do.you think a current clamp.wouldnt.work for this arrangement. $\endgroup$
    – lalala
    Feb 25 '18 at 14:50
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    $\begingroup$ @lalala the field is constant outside by Ampere's law. It must be $0$ for otherwise a finite current would produce a non-zero field at $\infty$. In particular, given that the magnetic energy density $\sim \vert \vec B\vert^2$, this would in turn imply that a finite current can produce an infinite amount of energy (once the energy density is integrated over the whole space). Indeed one can numerically show that the field becomes arbitrarily small at the waist of an arbitrary long solenoid if the coil windings are tights. $\endgroup$ Feb 25 '18 at 17:38
  • $\begingroup$ @ZeroTheHero That's not a valid argument. It takes infinite energy to get a finite current going in an infinite cylinder (since its inductance is infinite), and the energy integrated over space is infinite anyway (since an infinite solenoid has infinite volume). You have to be more careful with an energy argument here... $\endgroup$
    – Chris
    Feb 25 '18 at 22:08
  • $\begingroup$ @lalala That said, the result is correct. It's an elementary result you should be able to read about in an E&M textbook. $\endgroup$
    – Chris
    Feb 25 '18 at 22:10
  • $\begingroup$ Question is: in your abstraction of the infinite solenoid, is the a current going from minus infinity to plus infinity (then there is a magnetic field outside (due to.Amperes law), with the same strength a straight wire of the same.current. or do you think.of a charged cylinder rotating. To me, an.infinite solenoid is the former. $\endgroup$
    – lalala
    Feb 26 '18 at 3:18
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You are right, there has to be some magnetic field outside the solenoid as well, but that magnetic field is so weak that it is not considered for theoretical purposes.

What actually matters is the Magnetic Flux. Inside a solenoid the magnetic flux is too high (large number of magnetic field lines crossing a small cross-sectional area) whereas, outside the solenoid, the spacing between the field lines increases, i.e., the number of lines crossing per unit area reduces considerably. Thus, in comparison to inside volume of a solenoid, the magnetic field outside the solenoid is relatively zero. You can include that magnetic field in your calculation but that would make the mathematics little more difficult.

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  • $\begingroup$ Yes, I am looking for a proper reasoning to show that the field outside is weak. Does this follow from some kind of symmetry argument? $\endgroup$
    – praveen kr
    Feb 25 '18 at 13:30
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    $\begingroup$ arxiv.org/abs/1610.07876 $\endgroup$ Nov 24 '18 at 12:56

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