Question

Calculate the drift Velocity of elections
in an aluminium wire diameter o.9 mm
carrying current
of 6 assume
that
4.5 x 10^28 electrons /m^3 are available
for conduction.
​

Answer 1

Hope this helps.
I=6A
A= 0.63*10^-6
n= 4.5*10^28
I=neAVd
Vd= I/neA
Putting values
Vd= 6/ 4.5*10^28 x 1.6*10*-19 x 0.63*10-6
Vd= 1.3*10-3 m/s

Answer 2

Answer:

The drift velocity is  $1.3\times10^{-3}\frac{m}{s}$.

Given:

$n=4.5\times10^{28} \frac{electrones}{m^{3} }$

D = 0.0009m

I = 6A

Explanation:

Here,

The drift velocity is denoted by $V_{d}$.

The area is denoted by A.

The current is denoted by I.

The electron's density is denoted by n.

The diameter is denoted by D.

Now,

by the equation,

$A=\frac{\pi D^{2}} {4}$

$A=\frac{\pi 0.0009^{2} }{4}$

$A=0.63\times10^{-6} m^{2}$

Then,

By the equation,

$V_{d}= \frac{I}{NeA} \\V_{d}= \frac{6}{4.5\times10^{28}\times(1.6)\times10^{-19} \times 0.63\times10^{-6} }$

$V_{d} =1.3\times10^{-3}\frac{m}{s}$

So, the drift velocity is $1.3\times10^{-3}\frac{m}{s}$.