Question

A silver wire 2.6 mm in diameter transfers a charge of 420 Cin 80 min. Silver contains 5.8 x 10^{28} free electrons per cubic meter.

What is the current in the wire?
What is the magnitude of the drift velocity of the electronsin the wire?

Answer 1

1) Current in the wire: 0.0875 A

The current in the wire is given by:

`I=(Q)/(t)`

where

Q is the charge passing a given point in the conductor

t is the time elapsed

In this problem, we have

Q = 420 C is the total charge passing through a given point in a time of

t = 80 min = 4800 s

So, the current is

`I=(420 C)/(4800 s)=0.0875 A`

2) Drift velocity of the electrons:
`1.78\cdot 10^(-6) m/s`

The drift velocity of the electrons in the wire is given by:

`u = (I)/(nAq)`

where

I = 0.0875 A is the current

`n=5.8\cdot 10^(28)` is the number of free electrons per cubic meter

A is the cross-sectional area

`q=1.6\cdot 10^(-19) C` is the charge of one electron

The radius of the wire is

`r=(d)/(2)=(2.6 mm)/(2)=1.3 mm=0.0013 m`

So the cross-sectional area is

`A=\pi r^2=\pi (0.0013 m)^2=5.31\cdot 10^(-6) m^2`

So, the drift velocity is

`u = ((0.0875 A))/((5.8\cdot 10^(28))(5.31\cdot 10^(-6))(1.6\cdot 10^(-19)C))=1.78\cdot 10^(-6) m/s`