Question

A copper wire with a diameter of $1.02mm$ carries a current of $1.7A$. The drift velocity ($v_d$)of electrons in the wire is? (The number density of electrons in copper = $8.5\times {10}^{28}{m}^3$.)

• A. $1.25mm/sec$
• B. $0.15mm/sec$
• C. $1.5mm/sec$
• D. $2.5mm/sec$

Answer 1

The correct option is B

$0.15mm/sec$

Step 1:Given data

The density of electrons in copper $\rho$=$8.5\times {10}^{28}{m}^3$

Diameter of copper wire $d=1.02mm=102\times {10}^{-5}m$

Current $I=1.7A$

Charge of electron $Q=1.6\times {10}^{-19}C$

Let the drift velocity of electrons be $v_d$

Let the cross-sectional area of the wire be $A$

Let the current density be $j$

Step 2: Formula used

$$\begin{gathered} I=j\times A \\ j=v_d\times Q\times \rho \end{gathered}$$

Step 3: Calculating drift velocity

We know, current,

$$\begin{gathered} I=j\times A \\ j=\frac{I}{A}=\frac{I}{{\displaystyle \raisebox{1ex}{$\pi d^2$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}}=\frac{I\times 4}{\pi d^2}=\frac{1.7\times 4}{3.14\times {(102\times {10}^{-5})}^2} \\ j=v_d\times Q\times \rho \end{gathered}$$

Drift velocity,

$$\begin{gathered} v_d=\frac{j}{Q\times \rho } \\ {v}_d=\frac{1.7\times 4}{3.14\times {(51\times {10}^{-5})}^2\times 8.5\times {10}^{28}\times 1.6\times {10}^{-19}} \\ =\frac{1.7\times 4}{3.14\times 8.5\times 51\times 51\times 1.6\times {10}^{-10}\times {10}^{28}\times {10}^{-19}} \\ =\frac{1.7\times 4}{444292.416\times {10}^{-1}} \\ =1.5\times {10}^{-4}m/sec=0.15mm/sec \end{gathered}$$

Drift velocity is equal to $0.15mm/sec$.

Hence, option B is correct.