Question

There is a current of 1.344 amp in a copper wire whose area of cross-section normal to the length of the wire is $1m{m}^2$. If the number of free electrons per $c{m}^{3}is8.4\times {10}^{22}$, then the drift velocity would be

• A. $1.0mm/sec$
• B. $1.0m/sec$
• C. $0.1mm/sec$
• D. $0.01mm/sec$

Answer 1

The correct option is A $1.0mm/sec$

Given that,

Current $I=1.344A$

Area of cross-section, $A=1m{m}^2=0.01c{m}^2$

Number of free electrons, $n=8.4\times {10}^{22}$

Charge on electron $e=1.6\times {10}^{-19}$

Drift velocity is given by

$v_d=\frac{i}{nAe}$

$=\frac{1.344}{8.4\times {10}^{22}\times 0.01\times 1.6\times {10}^{-19}}$

$=\frac{1.344}{{10}^3\times 0.1344}$

$={10}^{-2}cm/\sec$

$=1mm/\sec$