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I know that there are well defined equations explaining the magnetic flux density in the solenoid.

However what about magnetic field outside the solenoid?

How is the magnetic flux density related with the current?

UPDATE : Sorry may be the original post is misleading. I have updated some of the terms. I was wanted to ask about the magnetic flux density

To be more precise:

Consider the case where we have one solenoid placed at coordinate $(0,0,0)$ and AC current $I$ is passing through it.

At the same time at point $(x,y,z)$ we observed the magnetic flux density $B$

How can we write $B$ in terms of $d=\sqrt{(x^2+y^2+z^2)}$ and $I$?

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    $\begingroup$ All magnetic field lines are closed, so obviously the total flux on the outside of a solenoid is exactly the same as on the inside, the field is just weaker because the area is much larger. For a short solenoid this amounts to rather strong fields on the outside, for a long solenoid they are only strong at the ends of the coil. Such a solenoid basically has the same field as a long cylindrical permanent magnet. Of you want almost zero field on the outside, then you have to make the solenoid into a skinny torus. That shape has very low flux leakage. $\endgroup$ – CuriousOne May 14 '15 at 20:04
  • $\begingroup$ I am surprised on the VtC. I can clearly see both the effort and the concept. Thus, I vote for leave open. $\endgroup$ – peterh says reinstate Monica Aug 18 '16 at 21:14
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Approximately zero for a solenoid of infinite length. As far as the magnetic field goes, nothing changes from the situation of a direct current passing through the solenoid. The magnetic flux is homogenous inside, and the magnetic flux outside is approximately zero (it's the same magnetic field as inside the solenoid but spread out in all the space around it (to infinity), so you have nearly zero magnetic flux).

What changes inside AND outside is that the changing current causes a changing magnetic flux inside the solenoid (also outside but they are negligible for the reasons stated before). That causes an induced azimuthal electric field (its direction with respect to the current has to do with the rate of change of $I$ which is $dI/dt$).
That electric field is induced inside and outside of the solenoid.

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  • $\begingroup$ My intuition says, there should be radiation (even for an infinite solenoid). The solenoid is basically a magnetic antenna. $\endgroup$ – Sebastian Riese May 14 '15 at 19:44
  • $\begingroup$ I really do not have an opinion on radiation.I know very well the things i know,but i have not yet (self)studied radiation! $\endgroup$ – TheQuantumMan May 14 '15 at 19:51
  • $\begingroup$ If you mean that the induced quantities will induce other quantities due to the time dependence,then i agree with you.Magnetic and electric fields will induce one another i think(although not sure) $\endgroup$ – TheQuantumMan May 14 '15 at 19:52
  • $\begingroup$ The mystery is simply resolved by all the flux leaking at the ends. The skinnier the solenoid, the smaller the field around it except directly at its ends. If you want zero flux to leak, then you have to bend the solenoid into a torus, which is an often used magnetic configuration for high Q, low loss, low radiation RF coils. This works very well with and without a ferromagnetic core. The largest air core toroidal inductors I have ever seen were used for the resonator of an inductive plasma heating device, I believe. $\endgroup$ – CuriousOne May 14 '15 at 20:08
  • $\begingroup$ it also leaks in a torus.The zero magnetic field outside it is again an approximation. $\endgroup$ – TheQuantumMan May 14 '15 at 20:11
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If we are talking about the ideal solenoid made of parallel loops of current, then the field outside will be close to zero, as in @TheQuantumMan's answer. However, if the solenoid is made from a helical coil of wire, the magnetic field outside the solenoid will be largely similar to that of a straight wire. To see this, imagine a circle around the solenoid that lies perpedicular to the solenoid axis. There is a current $I$ passing through the surface defined by this circle, so the line integral around the circle is non-zero. This is the same argument for the magnetic field of a straight wire, so you get the same field. $$B(r) = \frac{\mu_0 I}{2\pi r}$$ where $r$ is the perpendicular distance from the wire/solenoid.

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