Question

Suppose that a wire leads into another, thinner wire of the same material that has only a third the cross-sectional area. In the steady state, the number of electrons per second flowing through the thick wire must be equal to the number of electrons per second flowing through the thin wire. If the drift speed$\overline{{\mathbf{V}}_{\mathbf{1}}}$in the thick wire is $\mathbf{4}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{5}}\frac{\mathbf{m}}{\mathbf{s}}$, what is the drift speed ${\overline{\mathbf{V}}}_{\mathbf{2}}$in the thinner wire?

Short answer

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

Step-by-step solution

Given information

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,$\overline{V_1}=4\times {10}^{-5}\frac{m}{s}$.

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

Determine the concept of the Drift Speed

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.

The drift speed in the thinner wire

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, $$\begin{gathered} i_1=i_2 \\ {n}_1{A}_1{\overline{v}}_1=n_2{A}_2\overline{v_2} \end{gathered}$$

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

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

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

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