Question: $$\nabla \times A = B$$

$A$ is vector magnetic potential, $\mathrm{Wb/m}$

$B$ is magnetic field intensity, $\mathrm{Wb/m^2}$

**Where does one more m come from for $B$?** *Is that from the gradient operator so it is in meter or something?*

Answer: 
It's kind of a funny misconception that the sun is yellow. I mean, astronomically speaking it is indeed a *yellow star*, more precisely [G-type main sequence / yellow dwarf](https://en.wikipedia.org/wiki/G-type_main-sequence_star)... but don't be fooled by the terminology: astronomically speaking, you'll also find that the Earth consists completely of [metal](https://en.wikipedia.org/wiki/Metallicity)!

Actually you should consider **the sun as white**.

The main reason, strangely enough, why we think the sun is yellow is that **we never look at it**. That is, directly enough to judge its colour. When the sun is high in a cloudless sky, it's just too bright to see its colour (and evolution has trained us to not even try, because it would damage the eyes). Only near sunrise or sunset do we actually get to look at the sun, but then it's not so much the colour of the sun but the colour of the *atmosphere* we're noticing – and the atmosphere is, again counter to perception, yellow-orange-red in colour. Well, not quite – the point is that the atmosphere lets red / yellow light through in a straight line whereas bluer frequencies are more [Rayleigh scattered](https://en.wikipedia.org/wiki/Rayleigh_scattering). That's the reason why *the sky* is blue, and also adds to the perception of the sun being yellow: it's yellow-ish in comparison with the surrounding sky colour.

When you see the sun through clouds, you get to see its actual colour more faithfully than usual, both because (as [Mark Bell wrote](https://physics.stackexchange.com/a/627151/3540)) Mie scattering doesn't have the colour-separating effect that Rayleigh scattering does, and because you then see it against a grey / white backdrop instead of against the blue sky.
Sunlight is "white". Blue sky is due to Rayleigh scattering, where the intensity of scattered light depends on the fourth power of the frequency, then this is why the sky is blue, since blue is in the upper bound of the visible spectrum in frequency. This is because the molecules in atmosphere have a size much smaller than the wavelength of light.

The clouds contain droplets of water that are bigger (1-100 $\mu m$) than the particles mentioned before. Since here the size of droplets is bigger than the wavelength of light, we have to use Mie scattering. And in Mie scattering we have that the wavelengths are equally scattered in all directions and that determines the white color.