Question

Two wires , each of the radius $r$ 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 of.

• 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 together end to end (in series), then current in them will be same, which is given by, $$i=neA{v}_d$$Where,
$n\to$Number of electrons per unit volume (Number density),
$e\to$Charge of an electron$=1.6\times {10}^{-19}\text{C}$,
$A\to$Area of cross section,
$v_d=$Drift velocity.

As current and area are same then

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

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

Given, charge density, $\frac{n_1}{n_2}=\frac{1}{4}$
So,
$\frac{v_{d_1}}{v_{d_2}}=\frac{4}{1}$

Hence, option (c) is the correct answer.

Why this question? If two wires are connected in series connection,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 number of free electrons per unit volume then, $$i=neA{v}_d$$