Question

The area of cross section of a conductor is $10c{m}^2$ and a current of $3A$ is flowing through it. The drift velocity of electrons is $4\times {10}^{-4}m{s}^{-1}$. What will be the new cross sectional area if we are increasing the drift velocity to $7\times {10}^{-3}m{s}^{-1}$ keeping the current flow constant?

• A. $0.5714c{m}^2$
• B. $0.3931c{m}^2$
• C. $0.4032c{m}^2$
• D. $0.6618c{m}^2$

Answer 1

The correct option is A $0.5714c{m}^2$

Current $i$ is related to the drift velocity $v_D$ and the area of cross section $A$ as $i=neA{v}_D$ ......(1)
where, $n$ is the number of free electrons per $m^3$ and $e$, the electron charge (both are constant).

Given, $i$ should be kept constant and $v_D$ should be increased to $7\times {10}^{-3}m{s}^{-1}$.

So, $v_{D_1}=4\times {10}^{-4}m{s}^{-1}$ and $v_{D_2}=7\times {10}^{-3}m{s}^{-1}$

Initial area, $A_1=10c{m}^2$ and let the final area be $A_2$.

From (1), we have $\frac{v_{D_1}}{v_{D_2}}=\frac{A_2}{A_1}$ at constant $i$.

Substituting the values, we get $\frac{4\times {10}^{-4}}{7\times {10}^{-3}}=\frac{A_2}{10}$
$⟹A_2=\frac{4\times {10}^{-4}\times 10}{7\times {10}^{-3}}$ = $\frac{4\times {10}^{-3}}{7\times {10}^{-3}}$ = $\frac{4}{7}$
$⟹A_2$ = $0.5714c{m}^2$.