How does light travel perpendicular to a sphere? The diagram below shows that the propagation of light is perpendicular to the direction of the E and B fields. This explanation makes sense when looking at this two dimensional representation, shown below. However, with a little imagination the E and B fields can be viewed in three dimensions as spheres created from point sources. Once this is done the clear direction of the A and B fields becomes unclear. My question is: How does light travel perpendicular to a sphere?

 A: *

*Unless you're talking about light ray coming from one vibrating charge only, the so-called point source usually understood to be photons come from small region that can be approximated as spherical boundary.  The EM waves go out incoherently and randomly polarized.  See more on polarization and coherence.

All you observed is rapid switching of wave pulses emitted in all radial directions with random orientation of polarization governed by $\mathbf{k\times E=\omega B}$.
See another question here.

*By rotating polaroid sunglasses (or polarizer), you can see the change in brightness of the blue sky.

Air molecules scatter the unpolarized sunlight and the EM wave re-radiate spherically outwards but only the transverse direction can be transmitted.



*In short antenna, the EM wave is transmitted efficiently along the direction perpendicular to the direction.  Meanwhile, no energy is transmitted along the the axis of symmetry, i.e. the longitudinal direction.

See also the animation in another question here.
The $E$ and $B$ fields produced by an oscillating dipole
$\mathbf{p}=p_0 e^{-i\omega t} \mathbf{e}_z$ at the origin,
in spherical polar coordinates, are given by:
\begin{align}
   \begin{pmatrix} E_r \\ E_{\theta} \\ E_{\phi} \end{pmatrix}  &=
   \begin{pmatrix}
     -\frac{p_0\cos \theta}{2\pi \epsilon_0 r}
     \left( \frac{ik}{r}-\frac{1}{r^2} \right) \\
     -\frac{p_0\sin \theta}{4\pi \epsilon_0 r}
     \left( -k^2-\frac{ik}{r}+\frac{1}{r^2} \right) \\
     0
   \end{pmatrix} e^{i(kr-\omega t)} \\[5pt]
   \begin{pmatrix} B_r \\ B_{\theta} \\ B_{\phi} \end{pmatrix} &=
   \begin{pmatrix}
     0 \\
     0 \\
     \frac{i\omega \mu_0}{4\pi}
     \frac{p_0\sin \theta}{r}
     \left( -ik-\frac{1}{r} \right)
   \end{pmatrix} e^{i(kr-\omega t)}
 \end{align}
The leading terms are
\begin{align}
   \mathbf{E} &=
   \frac{k^2}{4\pi \epsilon_0 r}
   p_0 \sin \theta \, e^{i(kr-\omega t)} \mathbf{e}_{\theta} \\
   \mathbf{B} &=
   \frac{\mu_0 \omega k}{4\pi r}
   p_0 \sin \theta \, e^{i(kr-\omega t)} \mathbf{e}_{\phi}
 \end{align}

Loosely speaking, the anisotropic "spherical" wavefront decays as $\dfrac{1}{r}$ and the amplitude is maximal for $\theta=90^{\circ}$ and zero when $\theta=0^{\circ}$ or $180^{\circ}$ (i.e. the longitudinal direction).  This is due to resolution of vector into the component perpendicular to the direction of propagation, viz., $\mathbf{k}=k\, \mathbf{e}_r$ (i.e. the direction towards the observer).
N.B. For cylindrical (or circular) wave,
$$\psi(r, \phi, z, t) \propto \frac{e^{i(kr-\omega t)}}{\sqrt{r}}$$

*Miscellaneous pictures that may help


