Question

Two wires of the same material but of different radii carry the same current i. if the ratio of their radii is 4:2, then the corresponding ratio of their mean drift velocities will be.

Answer 1

Step 1: Given

Ratio of radii $\frac{r_1}{r_2}=\frac{4}{2}$

Here, radius of first wire is $r_1$ , radius of second wire is $r_2$.

$I_1=I_2$

Here, current flowing in first wire is $I_1$ and current flowing in second wire is $I_2$.

Step 2: Calculating mean drift velocity

$V_{d1}=\frac{1}{n{\mathrm{\pi r}}_1^2{\mathrm{q}}_1}$

Here, drift velocity for the first wire is $V_{d1}$, number of electrons is $n$, radius of first wire is $r_1$, and charge flowing in the wire is $q_1$.

$V_{d2}=\frac{1}{n{\mathrm{\pi r}}_2^2{\mathrm{q}}_2}$

Here, drift velocity for the second wire is $V_{d2}$, number of electrons is $n$, radius of second wire is $r_2$, and charge flowing in the wire is $q_2$.

Since same current is flowing. So, $q_1=q_2$.

Step 3: Calculating ratio of mean drift velocity

Divide $V_{d1}$ to $V_{d2}$.

$$\begin{gathered} \frac{V_{d1}}{V_{d2}}=\frac{1}{n{\mathrm{\pi r}}_1^2{\mathrm{q}}_1}\times \frac{n{\mathrm{\pi r}}_2^2{\mathrm{q}}_2}{1} \\ =\frac{r_2^2}{r_1^2}\times \frac{q_1}{q_1} \\ =\frac{2^2}{4^2} \\ =\frac{1}{4} \end{gathered}$$

Hence, the required ratio of drift velocity is $\frac{1}{4}$.