Question

In the circuit shown figure 18.108, two thick copper wires connect a 1.5 V battery to a Nichrome wire. Each thick connecting wire is 17 cm long and has a radius of 9 mm. Copper has $\mathbf{8}\mathbf{.}\mathbf{4}\mathbf{\times }{\mathbf{10}}^{\mathbf{28}}$mobile electrons per cubic meter and electron mobility. The Nichrome wire is 8 cm long and has a radius of 3 mm. Nichrome has $\mathbf{9}\mathbf{\times }{\mathbf{10}}^{\mathbf{28}}$mobile electrons per cubic meter and electron mobility of $\mathbf{7}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{5}}\mathbf{}\mathbf{(}\raisebox{1ex}{${\displaystyle \raisebox{1ex}{$\mathbf{m}$}\!\left/ \!\raisebox{-1ex}{$\mathbf{s}\mathbf{)}$}\right.}$}\!\left/ \!\raisebox{-1ex}{${\displaystyle \raisebox{1ex}{$\mathbf{(}\mathbf{V}$}\!\left/ \!\raisebox{-1ex}{$\mathbf{m}\mathbf{)}$}\right.}$}\right.$.

(a) What is the magnitude of the electric field in the thick copper wire?

(b) What is the magnitude of the electric field in the thin Nichrome wire?

Short answer

The magnitude of the electric field inside copper wire is$8.82\mathrm{V}/\mathrm{m}$.

Step-by-step solution

Identification of the given data

The emf of the battery is $\mathrm{\epsilon }=1.5\mathrm{V}.$

The charge density for copper wire is ${\mathrm{n}}_{\mathrm{c}}=8.4\times {10}^{28}\mathrm{electrons}/{\mathrm{m}}^3$

The length of the copper wire is role="math" localid="1668668353251" $l_{\mathrm{c}}=17\mathrm{cm}$

The radius of copper wire is $r_c=9\mathrm{mm}$

The electron mobility of copper wire is ${\mathrm{\mu }}_{\mathrm{c}}=4.4\times {10}^{-3}\left(\raisebox{1ex}{$\mathrm{m}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{s}$}\right.\right)/\left(\raisebox{1ex}{$\mathrm{V}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{m}$}\right.\right)$

The magnitude of the electric field inside the copper wire is found by dividing the emf of the battery by the length of the copper wire.

Determination of magnitude of the electric field inside copper wire.

The magnitude of the electric field inside copper wire is given as:

$E_c\mathit{=}\frac{V}{l_c}$

Substitute all the values in the above equation.

$$\begin{gathered} E_c=\frac{1.5\mathrm{V}}{\left(17\mathrm{cm}\right)\left({\displaystyle \frac{1\mathrm{m}}{100\mathrm{cm}}}\right)} \\ {E}_c=8.82\raisebox{1ex}{$\mathrm{V}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{m}$}\right. \end{gathered}$$

Therefore, the magnitude of the electric field inside copper wire is $8.82\mathrm{V}/\mathrm{m}$.