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 $I$?
 A: 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.
A: 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.
