Question

Two wires each of radius of cross-section 𝑟 but of different materials are connected together end to end (i.e. in series). If the densities of charge carries 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. $1:4$
• B. $2:1$
• C. $1:2$
• D. $4:1$

Answer 1

The correct option is D $4:1$

We know, current

$I=neA{v}_d$

($n\to$ Number density of free electrons or density of charge carriers , $v_d\to$ Drift velocity, $A=$ cross section area )

Since, the wires are connected in series, equal current passes through them

$I_1=n_{1}eA{v}_{d_1},I_2=n_{2}eA{v}_{d_2}$

Now, $I_1=I_2=I$

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

(both wires have equal cross section radius and hence equal cross section area 𝐴)

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

But, it is given that $\frac{n_2}{n_1}=4$

$\Rightarrow \frac{v_{d_1}}{v_{d_2}}=4$

Therefore, drift velocity of electrons in the two wires will be in the ratio $4:1$.

Hence, option (c) is correct.