Question

The drift speed in a copper wire is $7\times {10}^{-5}\frac{m}{s}$for a typical electron current. Calculate the magnitude of the electric field inside the copper wire. The mobility of mobile electrons in copper is $4.5\times {10}^{-3}\left(\frac{m}{s}\right)/\left(\frac{N}{C}\right)$. (Note that though the electric field in the wire is very small, it is adequate to push a sizable electron current through the copper wire.)

Short answer

The magnitude of the electric field inside the copper wire is $1.556\times {10}^{-8}\frac{N}{C}$.

Step-by-step solution

Given information

The drift speed of electron current in a copper wire is,$v_d=7\times {10}^{-5}\frac{m}{s}$.

The mobility of mobile electrons in copper is, $Mobility=4.5\times {10}^{-3}\frac{\left({\displaystyle \frac{m}{s}}\right)}{\left({\displaystyle \frac{N}{C}}\right)}$.

Determine the Drift speed of electrons

Consider a current is supplied to a conducting material then the current travels in form of free electrons. The speed of free electrons inside a conducting material is described as the drift speed of electrons.

The drift speed of the electrons inside a conducting material relies upon the mobility of the free electrons and the electric field in the conducting material

Determine the magnitude of the electric field

The formula for the mobility of mobile electrons in copper is as follows:

$$\begin{gathered} Mobility=\frac{v_d}{E} \\ E=\frac{v_d}{Mobility} \end{gathered}$$

Substitute the values and solve as:

$$\begin{gathered} E=\frac{7\times {10}^{-5}{\displaystyle \frac{m}{s}}}{4.5\times {10}^{-3}{\displaystyle \frac{\left({\displaystyle \frac{m}{s}}\right)}{\left({\displaystyle \frac{N}{C}}\right)}}} \\ E=1.556\times {10}^{-8}\frac{N}{C} \end{gathered}$$

Hence, the magnitude of the electric field inside the copper wire is$1.556\times {10}^{-8}\frac{N}{C}$$1.556\times {10}^{-8}\frac{N}{C}$.