Question

A copper wire has a diameter of 2.05 mm and carries a current of 5 A due solely to electrons. Given that the free (capable to move) electron concentration in copper is 1023 electrons per cm3, find the average velocity of the electrons in the wire. Give the velocity in mm/s and in miles/hour.

Answer 1

To solve this problem we will apply the concepts related to the average velocity in an electron, defined as the value of the current on the product between the charge, the number of electrons, the constancy pi and the squared radius. Our values are given as,

`D = 2.05 mm`

`I = 5A`

`e = -1.6 x 10^(-19) C`

`Q = 10^(23) e/cm^3 = 10^(29)e/m^3`

The radius would b,

`R = (D)/(2) = (2.05*10^(-3)m)/(2)`

`R = 1.025*10^(-3)m`

Now the average velocity

`V = (I)/((Q e \pi R^2))`

Here,

I = Current

Q = Electron concentration

e = Charge of electron

R = Radius

`V = (5)/((10^(29))(-1.6*10^(-19))(\pi)(1.025*10^(-3))^2)}`

`V = 9.46*10^(-5)m/s`

Converting to the values required we have that,

`V = 9.46*10^(-5) m/s ((1000mm)/(1m))`

`V = 9.46*10^(-2)mm/s`

And,

`V = 9.46*10^(-5)m/s ((3600s)/(1hour))((0.000621371miles)/(1m))`

`V = 2.116*10^(-4)mph`