Question

Every atom makes one free electron in copper. If $1.1$ ampere current is flowing in the wire of copper having 1mm diameter, then the drift velocity (approx) will be (density of copper = $9\times {10}^{3}kg/m^3$ and atomic weight of copper = $63$)

• A. $0.1mm/s$
• B. $0.2mm/s$
• C. $0.3mm/s$
• D. $0.2cm/s$

Answer 1

The correct option is A $0.1mm/s$

$\text{Step 1: No of copper atoms per unit volume}$

$n=\frac{\rho N_A}{M}$

$\Rightarrow$ $n=\frac{9\times {10}^{3}kg/m^3\times 6.023\times {10}^{23}}{0.063kg}=8.60\times {10}^{28}$ $m^{-3}$

$\text{Step 2: No of electrons per unit volume}$

As one copper atom gives one free electron, thus number of electrons per unit volume is $8.6\times {10}^{28}$ $m^{-3}$

$\text{Step 3: Drift velocity}$

Cross-section area of wire, $A=\frac{\pi d^4}{4}=\frac{3.14\times (0.001m{)}^2}{4}=7.85\times {10}^{-7}$ $m^2$

Drift velocity, $v_d=\frac{I}{Ane}$

$\Rightarrow$ $v_d=\frac{1.1A}{(7.85\times {10}^{-7}{m}^2)(8.6\times {10}^{28}{m}^{-3})(1.6\times {10}^{-19}C)}=0.01\times {10}^{-2}$ $m/s=0.1$ $mm/s$

Hence option(A) is correct.