Question

A conductor having a cross-sectional area of 0.00161 sq in. carries. a current of 4.5 amp. Calculate the electron velocity, assuming the wire to be (a) copper, (b) aluminum.

Answer 1

A=0.00161inc²

I=4.5 amp

$n_{Al}=18.1\times10^{28} m^{-3}$

$n_{cu}=8.39\times10^{28} m^{-3}$

$I=n_{Al}eAv_d$

$V_d=\frac{I}{n_{Al}eA}$

Put value

$$V_d=\frac{4.5}{18.1\times10^{28}\times1.6\times10^{-19}\times0.00161}=9.65\times10^{-8}m/sec$$

$V_d=\frac{I}{n_{cu}eA}$

Put value

$$V_d=\frac{4.5}{8.39\times10^{28}\times1.6\times10^{-19}\times0.00161}=2.082\times10^{-7}m/sec$$