Question

23. A $0.50$-mm-diameter silver wire carries a $20\mathrm{mA}$current. What are (a) the electric field and (b) the electron drift speed in the wire?

Short answer

(a) The electric field is $1.64\mathrm{mV}/\mathrm{m}$.

(b) The electron drift speed in the wire is $11\mu \mathrm{m}/\mathrm{s}$.

Step-by-step solution

Given information Part (a)

A $0.50$-mm-diameter silver wire carries a $20\mathrm{mA}$ current in the electric field.

Explanation Part (a)

To find the current density, equating the two expressions,
$\sigma E=J=\frac{I}{A}$

$\Rightarrow E=\frac{I}{\sigma A}$

The cross section area is a circle,

$E=\frac{I}{\sigma \pi r^2}$

$=\frac{4I}{\sigma \pi d^2}$

$E=\frac{4\times 0.02A}{\pi \times 6.2\times {10}^7\times 0.{0005}^{2}m}$

$=1.64\mathrm{mV}/\mathrm{m}$

Final Answer Part (a)

The electric field is $1.64\mathrm{mV}/\mathrm{m}$.

Given information Part (b)

A $0.50$-mm-diameter silver wire carries a $20\mathrm{mA}$ current in the electron drift speed in the wire.

Explanation Part (b)

To solve for the drift speed as
$v_d=\frac{4I}{\pi d^{2}e{n}_e}$
In the numerical case,
$v_d=\frac{4\times 0.02A}{\pi \times 0.{0005}^{2}m\times 1.6\times {10}^{-19}\times 5.8\times {10}^{28}}$

$=1.1·{10}^{-5}\mathrm{m}/\mathrm{s}.$

$=11\mu \mathrm{m}/\mathrm{s}$.

Final answer Part (b)

The electron drift speed in the wire is $11\mathrm{\mu m}/\mathrm{s}$.