91Ó°ÊÓ

The cross-sectional area of the thick wire is,$A_1$.

The cross-sectional area of the thinner wire is,$A_2=\frac{A_1}{3}$ .

The number of electrons per second flowing through the thick wire is,$n_1$.

The number of electrons per second flowing through the thinner wire is,$n_2=n_1$.

The drift speed of the electrons in the thick wire is,$\stackrel{¯}{V_1}=4×{10}^{-5}\frac{m}{s}$.

The drift speed of the electrons in the thinner wire is,$\stackrel{¯}{v_2}$ .

The speed of the electrons moving through a conducting material when a current is supplied is described as the ‘drift speed’ of the electrons.

If the current supplied to the material increases then the value of drift speed of the electrons also increases.

It is given that the thick wire leads into another thinner wire of the same material, then the same current flows through both the wires.

Then, the formula for the current flowing through the thick wire is given by, $$\begin{gathered} i_1=i_2 \\ {n}_1{A}_1{\stackrel{¯}{v}}_1=n_2{A}_2\stackrel{¯}{v_2} \end{gathered}$$

Putting, role="math" localid="1668591417012" $n_1=n_{2}and{A}_2=\frac{A_1}{3}$in the expression and solve as:

$$\begin{gathered} n_1{A}_1{\stackrel{¯}{v}}_1=n_1\frac{A_1}{3}{\stackrel{¯}{v}}_2 \\ {\stackrel{¯}{v}}_2=\frac{3A_1{\stackrel{¯}{v}}_1}{A_1} \\ {\stackrel{¯}{v}}_2=3{\stackrel{¯}{v}}_1 \end{gathered}$$

Putting the value of $$\begin{gathered} v_1 \\ {\stackrel{¯}{v}}_2=3\left(4×{10}^{-5}\frac{m}{s}\right) \\ {\stackrel{¯}{v}}_2=12×{10}^{-5}\frac{m}{s} \end{gathered}$$,

Hence, the drift speed in the thinner wire is $12×{10}^{-5}\frac{m}{s}$.