Question

$1.0\times {10}^{20}$electrons flow through a cross-section of a $2.0\mathrm{mm}$diameter iron wire in $5.0\mathrm{s}$. What is the electron drift speed?

Short answer

The electron drift speed is$75\mu \mathrm{m}/\mathrm{s}.$

Step-by-step solution

Given Information 

Number of electrons flow through$=1.0\times {10}^{20}$

Diameter$=2.0mm$

Time$=5.0s$

Explanation 

The definition of the electron current includes the electron drift speed, which can be expressed as:

$i_e=n_{e}A{v}_d\Rightarrow v_d=\frac{i_e}{n_{e}A}$

The electron current is just the number of electrons per unit time,

$i_e=\frac{N_e}{t}$

The area of the cross-section, since the latter is a circle, will be

$A=\frac{\pi D^2}{4}$

Combining, we can find

$v_d=\frac{N_e/t}{n_e\pi D^2/4}=\frac{4N_e}{\pi n_{e}t{D}^2}$

Substitute the values,

role="math" localid="1648798324977" $v_d=\left(\frac{4\times {10}^{20}e}{\pi ·8.5·{10}^{28}e\times 5s\times 0.002m{m}^2}\right)$

$=7.5·{10}^{-5}\mathrm{m}/\mathrm{s}.$

Final Answer 

Hence, the electron drift speed is$75\mu m/s.$