Question

The two wires in FIGURE P27.62 are made of the same material. What are the current and the electron drift speed in the $2.0-\mathrm{mm}-$diameter segment of the wire?

Short answer

The current and the electron drift speed in the $2.0-\mathrm{mm}$diameter segment of the wire is $v_{d2}=0.5\frac{\mu m}{\mathrm{s}}$.

Step-by-step solution

Given information

Wire 1 diameter $=2.0mm$

Wire 2 diameter$=1.0mm$

Drift speed $V_d=2.0\times {10}^{-4}m/s$

Current$I=2.0A$

Explanation

The current will be the same,$I=2A$

In this case, the electron current (the number of electrons flowing per unit time) would also be the same because of the same current.

That is, $n_{e1}{A}_1{v}_{d,1}=n_{e2}{A}_2{v}_{d,2}$

Due to the same material on both sides, the densities can be simplified, and we can increase the area thanks to the circular pattern.

Simplify by$\pi$we get,

$D_1^2{v}_{d,1}=D_2^2{v}_{d,2}$

Find the drift speed at the second part as,

$v_{d,2}=\frac{D_1^2}{D_2^2}{v}_{d,1}$

Substitute the values,

$v_{d,2}=\left(\frac{1mm}{2mm}\times 0.2m/s\right)$

$=0.05\mathrm{mm}/\mathrm{s}.$

Final Answer

Hence, the current and the electron drift speed in the$2.0mm$diameter segment of the wire is$0.05mm/s.$