Question

Find the average speed of the electrons in a 1.0 cm diameter, copper power line, when it carries a current of 20 A.

Answer 1

Average speed
`v_d=(i)/(neA)=(20)/(8.5* 10^(28)* 1.6* 10^(-19)* 7.8* 10^(-5))=1.87* 10^(-5)m/sec`

Step-by-step explanation:

We have given current through power i = 20 A

Diameter d = 1 cm = 0.01 m

So radius r = 0.005 m

So area
`A=\pi r^2=3.14* 0.005^2=7.8* 10^(-5)m^2`

Charge on electron e =
`1.6* 10^(-19)C`

We know that current is given by
`i=neAv_d`, here n is nuber density of free electron, e is charge on electron, A is area and
`v_d` is average speed

We know that for copper n =
`8.5* 10^(28)per\ m^3`

So average speed
`v_d=(i)/(neA)=(20)/(8.5* 10^(28)* 1.6* 10^(-19)* 7.8* 10^(-5))=1.87* 10^(-5)m/sec`