Question

Two wires each of radius of cross-section $r$ but of different materials are connected together end to end (in series). If the densities of charge carriers in the two wires are in the ratio $1:4$, the drift velocity of electrons in the two wires will be in the ratio

• A. $1:2$
• B. $2:1$
• C. $4:1$
• D. $1:4$

Answer 1

The correct option is C $4:1$

If the wires are connected end to end (in series) then current in them will be same.

We know formula of current,

$I=neA{v}_d$

where,
$n\to$ number of free electrons per unit volume
$e=1.6\times {10}^{-19}\text{C}$
$A\to$ Area of cross-section
$v_d\to$ Drift velocity

As current and cross-section is same then

$n_{1}eA{v}_{d_1}=n_{2}eA{v}_{d_2}$

$\Rightarrow \frac{v_{d_1}}{v_{d_2}}=\frac{n_2}{n_1}=\frac{4}{1}$

Hence, option $\left(c\right)$ is correct.

Why this question ? If two wires are connected in series, then current in them will be same Key Formula: If electrons are moving with velocity $v_d,A$ is area of cross-section and $n$ is the number of free electrons per unit volume then, $I=neA{v}_d$