Question

An aluminum wire having a cross-sectional area equal to 5.40 10-6 m2 carries a current of 5.50 A. The density of aluminum is 2.70 g/cm3. Assume each aluminum atom supplies one conduction electron per atom. Find the drift speed of the electrons in the wire.

Answer 1

Answer:

Drift speed of electrons will be 1.056x10^-4 m/s

Explanation

Given Data:

A(area)= 5.4 x 10^-6 $m^{2}$

I(current)= 5.5 A

Density= 2.7 $g/cm^{3}$

Calculation:

The equation for drift velocity is:

$v(drift)=I/nqA$

In this case 'q' will be charge of electron which is= 1.6 x 10-19

As each atoms supplies one conduction electron, so number of conduction electrons will be equal to number of atoms.

Hence,

n= no. of conduction electrons/$m^{3}$ = no. of atoms/$m^{3}$

To find 'n' we can use following equation:

$n= (mass/cm^{3} *atoms/mol)/(mass/mol)$

We know atoms/mol is equal to Avogadro`s number i.e 6.02 x 10^23

and molar mass of aluminium is 26.982 g.

Now,

$n=(2.7g/cm^{3} * 6.02*10^{23} )/25.982g$    (putting values in above equation)

$n=6.024*10^{22} electrons/cm^{3}$

$n= 6.024*10^{22} *10^{6} electrons/m^{3}$ (converting electrons/cm3 to electrons/m3)

$n= 6.024*10^{28} electrons/m^{3}$

To find drift velocity, we will use equations mention before:

$v(drift)=I/nqA$

$v(drift)=5.5A/(6.024*10^{28}electrons/m^{3} *1.6*10^{-19}C* 5.4*10^{-6}m^{2} )$

$v(drift)= 1.056*10^{-4} m/s$