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Energy dissipation in current flow

Physics Asked by Junaid Aftab on January 25, 2021

In section 4.8, Energy dissipation in current flow, of Purcell and Morin’s Electricity in Magnetism, the expression for the power expended by a resistor is derived. The sections includes the following discussion:

The energy thus expended shows up eventually as
heat. In our model of ionic conduction, the way this comes about is quite
clear. The ion acquires some extra kinetic energy, as well as momentum,
between collisions. A collision, or at most a few collisions, redirects its
momentum at random but does not necessarily restore the kinetic energy
to normal. For that to happen the ion has to transfer kinetic energy to the
obstacle that deflects it. Suppose the charge carrier has a considerably
smaller mass than the neutral atom it collides with. The average transfer of kinetic energy is small when a billiard ball collides with a bowling
ball. Therefore the ion (billiard ball) will continue to accumulate extra
energy until its average kinetic energy is so high that its average loss of
energy in a collision equals the amount gained between collisions. In this
way, by first “heating up” the charge carriers themselves, the work done
by the electrical force driving the charge carriers is eventually passed on
to the rest of the medium as random kinetic energy, or heat.

I may be missing out on something elementary but I’m unable to understand the concepts in the paragraph related to energy transfer in the collisions.

One Answer

I think the point made is that the kinetic energy of the electrons (or other charge carrier) will normally be far higher than $kT$. That's because although collisions with the lattice are frequent the electron loses very little energy with each collision.

The point being made is that in a collision between a light object and a heavy one very little of the kinetic energy is transferred to the heavy object. If you bounce a ping pong ball off a battleship the ping pong ball bounces back with almost the same energy it had before the collision. That means the kinetic energy of the electrons keeps growing and growing until it gets big enough for the KE lost in each collision to balance the KE gained from the electric potential. At this point the KE of the electrons is a lot greater than $kT$ so the energy transferred to the lattice is greater than $kT$ and the lattice heats up.

At least, that's what I think the book is saying, though it seems a long winded way to say something really rather obvious.

Answered by John Rennie on January 25, 2021

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