# What is the magnetic flux density "outside" the solenoid when AC current is passing through it?

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 $I$?

• 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. May 14, 2015 at 20:04
• I am surprised on the VtC. I can clearly see both the effort and the concept. Thus, I vote for leave open. Aug 18, 2016 at 21:14

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 $dI/dt$).
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.