Would a gauss rifle based on generated magnetic fields have any kickback?

Simple answer when you think about it:

You are imparting a force to accelerate the slug, so you're going to get an equal and opposite reaction. In a normal rifle, the explosion accelerates the bullet rapidly and you get recoil.

In a gauss rifle, the acceleration will be a bit lower, but for a slightly longer time (the entire length of the barrel), so for the same muzzle velocity you will be able to calculate the recoil in the exact same way.

If the Gauss rifle shoots a projectile with exit speed of $v_1$ and mass $m_1$, then its momentum will be:

$p=m_1v_1$.

Because of momentum conservation law, the rifle will have the same momentum in opposite direction. If the rifles mass is $m_2$, the rifle will start moving in the opposite direction with end speed of:

$v_2 = frac{m_1 v_1}{m_2}$.

But, as the projectile is accelerated for longer time than in a gun, the force acting from rifle on its holder will be lower because $F=frac{dp}{dt}$

About the magnitude. This source claims 2056m/s exit speed, this one around 2382m/s (mach 7).

I have no idea what a personal reilgun's projectile weight should be. Let's take a random pistol round: 7.5g.

Gun length? Let's pick 1m. Gun weight? Let's pick 3kg.

v_gun = - p_projectile / m_gun =
(2382*7.5*(10-3)) / 3 =
5.955m/s

acquired over one meter of travel at let's say constant acceleration.

x = v_0 t + (1/2) a t^2
2 = a t^2
t = sqrt(2/a)
where a = v / t
t = sqrt(2t/v)
t = sqrt(t) sqrt(2/v)
sqrt(t) = sqrt(2/v)
t = 2/v
t ~= 8.4ms

Now the (obviously constant) force is going to be

F = m a
a = 5.955m/s / 8.4*10^-3s = 7089m/ss
F = 3kg * 7089m/ss = 21267N

A typical pro boxer's punch is 5000N. So we are going to need lighter bullets. I wonder what would happen if you shoot this thing in a sandstorm. Or indoors.