# Why does a larger wire cross section mean less atom-electron collisions when there must be more atoms, too?

Remember that the electron that pops out the end of the wire is not the same one that got pushed in at the far end. It's more like a tube filled with ball-bearings. Push one in at one end and another immediately pops out at the far end. While the speed of the *wave* along the wire (tube) is very high, the speed of the individual electrons (balls) is very low.

Current is given by the equation:

$$I = vAnq$$

Where:

*I*= current (amps, A).*v*= drift velocity (m/s).*A*= cross-sectional area of the conductor (m^{2}).*n*= charge density (m) This is the number of charge carriers that can move per m^{3}. For copper this is 8.5 × 10^{28}electrons/m^{3}.*q*= charge on each charge carrier (coulombs, C) is −1.6 × 10^{−19}C.

Rearranging this we can see that the velocity of the electrons is given by

$$ v = \frac {I}{Anq} $$

For **1 A** in a **2 mm diameter copper** conductor we get a drift velocity of 2.3 x 10^{-5} m/s or **0.023 mm/s**.

If we double the diameter we quadruple the area and get a drift velocity of 0.006 mm/s.

Thus, if the diameter of the wire were larger, it would only make sense that the electrons don't collide as much, therefore creating less resistance due to a larger wire.

This should now make a bit more sense. If each moving electron only has to travel 1/4 the distance in the larger conductor of our examples then it will have 1/4 the collisions.