Question

A conductor having a cross-sectional area of 0.00161 Sq. in. carries a current of 4 amp. calculate the electron velocity, assuming the wire to be a copper and aluminum?​

Answer 1

Hence,  the electron velocity for aluminum and copper wire is  8.6 x $10^{-8}$ m/sec and  1.9 x $10^{-7}$ m/sec

Given that;

cross-sectional area = 0.00161sqqThecurrentn

The current carried by the conductor = 4 amp

To find;

The electron velocity, assuming the wire to be of copper and aluminum.

Solution;

As we know,

$n_{cu}$ = 8.39 x $10^{28}$ $m^{-3}$ and $n_{al}$ = 18.1 x $10^{28}$ $m^{-3}$

The current carried in a conductor is given by the equation;

I  = $n_{al}$eA$v_{d}$ where

e = charge of an electron, $v_{d}$ = electron velocity and A = Area of cross-section

$v_{d}$ = $\frac{I}{n_{al} eA}$

Putting the values we get,

$v_{d}$ = $\frac{4}{18.1 X 10^{28} X 1.6 X 10^{-19} X 0.00161 }$ = 8.6 x $10^{-8}$ m/sec (for aluminum)

and,

$v_{d}$ = $\frac{4}{8.39 X 10^{28} X 1.6 X 10^{-19} X 0.00161 }$ = 1.9 x $10^{-7}$ m/sec (for copper)

Hence,  the electron velocity for aluminum and copper wire is  8.6 x $10^{-8}$ m/sec and  1.9 x $10^{-7}$ m/sec

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