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

Regarding how resistance varies with diameter I've read this simple explanation on Quora:

As electrons move across a wire, they constantly collide with atoms making up a wire. These collisions impede the flow of electrons and are what cause the wire to have resistance.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.

But doesn't a larger wire mean more atoms (I assume the number of atoms increases proportionally with the diameter of the wire?) meaning same amount of collisions ie. same resistance?

I don't question the fact that resistance decreases with diameter, but this explanation doesn't make sense to me.

• The "density" of "collisions" is the same, it's just that "more electrons" are making it through. Many things are in quotes here as its a rather simplistic view. – Eugene Sh. Jul 9 at 14:39
• A larger wire also means there is more total space for a given density of travelling electrons to pass through. – Peter Smith Jul 9 at 14:42
• Keep in mind individual electrons have very little movement, actual propagation is mostly neighbor bumping neighbor; which while not an answer starts to suggest how abstracted from reality the model being contemplated is. – Chris Stratton Jul 9 at 14:46
• Actually, the number of atoms goes up with the square of the diameter. And the spacing between atoms is constant, regardless of the diameter. That's a much too simplistic explanation of the phenomenon of resistance. – Dave Tweed Jul 9 at 15:00
• @DKNguyen I can see why the OP is asking, and I don't think your comment helps to clarify. Suppose we try to move a single electron from one end to the other. Why does it take less energy if the wire has a diameter of 2cm rather than 1cm? Isn't the electron going to make the same number of collisions as it passes through the wire? – Elliot Alderson Jul 9 at 15:35

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 (m2).
• n = charge density (m) This is the number of charge carriers that can move per m3. For copper this is 8.5 × 1028 electrons/m3.
• 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.

No, resistance is a function of cross sectional area and length, the equation is as follows:

$$\ R = \rho\frac{l}{A}\$$

where $$\\rho\$$ is the electrical resistivity of the material.

If the current is the same and we increase the material's cross sectional area, then the resistance must go down (doubling the area of a material is like adding two resistors in parallel) because per a unit area of the material there will be less current. (less current means less collisions per atom if you want to think of it on the quantum scale).

• is there a missing comma after the first word? – jsotola Jul 10 at 1:08