Question

$1.0\times {10}^{16}$electrons flow through a cross section of a silver wire in $320\mu \mathrm{s}$ with a drift speed of $8.0\times {10}^{-4}\mathrm{m}/\mathrm{s}$. What is the diameter of the wire?

Short answer

The diameter of the wire is$0.93mm$.

Step-by-step solution

Given Information

Number of electrons$=1.0\times {10}^{16}$

Time$=320\mu s$

Drift speed$=8.0\times {10}^{-4}\mathrm{m}/\mathrm{s}$

Explanation

The electron density is

$i_e:=\frac{N}{t}$

The current density is known to be

$i_e=n_{e}A{v}_d.$

The area of a circle is,

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

Write the area as

$A=\frac{i_e}{n_e{v}_d}$

Write the electron current and substitute,

$A=\frac{N}{n_{e}t{v}_d}$

Now, substituting this expression for the area at the one for the diameter,

$D=\sqrt{\frac{4N}{\pi n_{e}t{v}_d}}.$

After substituting the values,

localid="1649322696741" $D=\sqrt{\left(\frac{4\times 1\times {10}^{16}e}{\pi \times 5.8\times {10}^{28}e\times 3.2\times {10}^{-4}\mu s\times 8\times {10}^{-4}m/s}\right)}$

localid="1649322704585" $=9.26\times {10}^{-4}\mathrm{m}.$

Final Answer

Hence, the diameter of the wire is$9.26\times {10}^{-4}\mathrm{m}.$