The Solar wind does indeed exert a force on the planets, however it turns out that the force is so small that it has no measurable effect.
The force can be calculated using the fact that force is equal to the rate of change of momentum. Suppose the total mass of all the Solar wind particles hitting the Earth per second is $M$, and the average velocity of the particles is $v$, then the force the solar wind exerts on the Earth is simply:
$$ F = Mv $$
Off-hand I don't know what the mass flux and velocity are, but the Wikipedia article on the solar wind reports the pressure, $P$, produced by the wind at the Sun-Earth distance to be 1 to 6 nano-Pascals. The total force on the Earth is this pressure multiplied by the cross sectional area $\pi r^2$. The radius of the Earth is about 6,371,000 metres, so we get:
$$ F = P \times \pi r^2 \approx 130 \,\text{to}\, 800 \,\text{kN} $$
To see why this is negligible, let's compare it with the gravitational force between the Sun and the Earth. This is given by Newton's law of gravity:
$$ F = \frac{GM_\text{Sun}M_\text{Earth}}{r^2} $$
and it works out to be:
$$ F \approx 3.54 \times 10^{22} \,\text{N} $$
so the force from the Solar wind is only about 0.000000000000001% ($10^{-15}\%$) of the gravitational force.
It has nothing to do with pressure in the thermodynamic sense nor with virtual particles. There is an intrinsic magnetic field generated somehow in Earth's core (dynamo discussion could fill volumes) and that field interacts with the magnetic field and charged particles of the solar wind. Since the solar wind is supersonic, there is a bow shock generated. This decelerates and deflects the solar wind around the magnetosphere, which stands off from the Earth. Without this, the solar wind's convective electric field (i.e., basically a $\mathbf{E}_{sw}$ = $-
\mathbf{V}_{sw} \times \mathbf{B}_{sw}$ field due to the motion of charged particles carrying a magnetic field past the Earth) would drag the ionized upper atmosphere off Earth very quickly.
Giving the effect that the earth's magnetic field doesn't protect us from the solar wind at all, it just concentrates it at the poles?
This is wrong, it does protect Earth's atmosphere from the solar wind, as I stated above. The drift velocity induced by the solar wind's convective electric field on newly ionized particles (called pick up ions) is called the ExB-drift, and it ranges in speed from 10s of km/s to 100s of km/s. The escape speed from Earth at the surface is only ~11.2 km/s. Thus, if the ionized upper atmosphere were suddenly exposed to $\mathbf{E}_{sw}$, the ions and electrons would immediately be accelerated up to 10s to 100s of km/s, easily escaping Earth's gravitational field.
Best Answer
Solar wind is neutral overall else the Sun will become globally very strongly charged and we don't see that happening.
It comprises Electrons/Protons and other particles.