Question

There is a current of 40 amperes in a wire of ${10}^{-16}{m}^2$ area of cross-section. If the number of free electrons per $m^3$ is ${10}^{29}$, then the drift velocity will be:

• A. $1.25\times {10}^3$ m/s
• B. $2.50\times {10}^{3}m/s$
• C. $2.0\times {10}^{6}m/s$
• D. $25\times {10}^{6}m/s$

Answer 1

The correct option is D $25\times {10}^{6}m/s$

If L is the length of wire so velocity is given by $v=\frac{L}{t}$

Total number of free electrons in the wire, $Q=nqLA$

Current,

$I=\frac{Q}{t}$

$I=\frac{nqLA}{t}$

$I=nqvA$

$v=\frac{I}{nqA}$

Where, n is the number of electron, $n={10}^{29}$

q is the charge of an electron, $q=1.6\times {10}^{-19}C$

A is area, $A={10}^{-16}{m}^2$

I is current, $I=40A$

So, drift velocity,

$v=\frac{40}{{10}^{29}\times 1.6\times {10}^{-19}\times {10}^{-16}}$

$v=25\times {10}^{6}m/s$