First, let's place our sphere
Let's assume there is a two-layer non-rotating Dyson sphere around the Sun (ignore the scale):
Green is empty air. How far should the inner border of this sphere be from the Sun surface for the man standing on it like shown in the picture to experience normal Earth gravity?
I googled around and found this:
$$g=\frac{GM}{r^2}$$
$g$ – gravitational acceleration on the earth's surface
$G$ – gravitational constant
$M$ – mass of the object (in our case the Sun)
$r$ – distance from the center of the mass of the object
So, for $g=9.8\ \mathrm{m/s^2}$, we have
$$\begin{align} r&=\sqrt{\frac{GM}{g}}\\ &=\sqrt{\frac{6.67\times{10}^{-11}\ \mathrm{m^3\ kg^{-1}\ s^{-2}}\times1.99\times{10}^{30}\ \mathrm{kg}}{9.8\ \mathrm{m\ s^{-2}}}}\\ &=3.68\times{10}^9\ \mathrm m \end{align}$$
So, around 3.68 million km, if we subtract Sun radius we get a little less than 3 million kilometers from the solar surface.
Now the main question
If my calculations are right, our inner shell is very close to the Sun, so it should experience a ton of various effects, like Sun's magnetic field, solar wind, etc. Let's ignore for the moment the actual way our non-rotating sphere manages to stay static and not fall towards the Sun. What effects will we see on the outer side of the inner shell (where the red stick man stands), e.g. Sun's magnetic field effects or maybe some non-Newtonian effects?
Why?
I'm researching this for a sci-fi book that's in the works and I want the sci part to be as accurate as possible, there will be some fictional elements of course—to serve the plot—but I'd like to limit contradictions to existing science as much as possible.