Fyneman exercise II ch1-3

Q.

In copper there is one “conduction” electron for each copper atom. When a current of 10 amperes flows through a piece of No. 10 gauge copper wire, what is the average speed of the conduction electrons? How big is ${v^2\over c^2}$? (Remember that the ratio of “magnetic” effects to “electric” effects is about this small.)

Trial.

$Current = {dQ \over dt} ={Sv\cdot dt \over dt} = Sv$
, where $S$ is cross-sectional area and $v$ is velocity.

Cross-sectional area of No. 10 gauge wire : $S=5.26\times 10^{-6}\ m^2$
(reference : American wire gauge)

$10 = 5.26\times 10^{-6}\times v$

$v=1.90\times 10^6\ m\cdot s^{-2}$

${v^2 \over c^2} = {1.90^2\times 10^{12} \over 9.00 \times 10^{16}} = 4.01\times10^{-5}$