Question

When the current in a wire is 1A, the drift velocity is $1.2\times {10}^{-4}m{s}^{-1}$. The drift velocity when current becomes 5 A is

• A. $1.2\times {10}^{-4}m{s}^{-1}$
• B. $3.6\times {10}^{-4}m{s}^{-1}$
• C. $6\times {10}^{-4}m{s}^{-1}$
• D. $4.8\times {10}^{-4}m{s}^{-1}$

Answer 1

Answer: (C)

$6\times {10}^{-4}m{s}^{-1}$

Given
Initial current through the wire is $I=1A$
Initial drift velocity is, $v_d=1.2\times {10}^{-4}m{s}^{-1}$
Increased current is, $I^{\prime }=5A$
The current, I through the wire is given by
$I={\mu }_e.e.A.v_d$
where, ${\mu }_e$ is the free electron density, e is the charge on electron, A is the area of cross section of the wire and v_d is the drift velocity of electrons.
Since, the free electron density, the charge on electron and the area of cross section of the wire are constant, hence
$I\propto v_d$.................(1)
Now, current through the wire is increased to 5 A, if the new drift velocity of electrons is $v_d^{\prime }$ then
$I^{\prime }\propto v_d^{\prime }$................(2)

From (1) and (2), we can write

$\frac{v_d^{\prime }}{v_d}=\frac{I^{\prime }}{I}$

$v_d^{\prime }=\frac{I^{\prime }}{I}{v}_d$

$v_d^{\prime }=\frac{5}{1}1.2\times {10}^{-4}$

$v_d^{\prime }=6\times {10}^{-4}m{s}^{-1}$