(a) At room temperature, what is the strength of the electric field in a

12-gauge copper wire (diameter $\mathbf{2}\mathbf{.}\mathbf{05}\mathbf{}\mathbf{mm}$) that is needed to cause a $\mathbf{4}\mathbf{.}\mathbf{50}\mathbf{-}\mathbf{A}$

current to flow? (b) What field would be needed if the wire were made of silver

instead?

The electric force per unit charge is known as electric.

$E=\rho \frac{I}{A}$

Here, $I$is current in ampere $\left(\mathrm{A}\right)$, $\rho$is resistivity in ohms meter $\left(\mathrm{\Omega }.\mathrm{m}\right)$and is the

area in ${\mathrm{m}}^2$.

(a)

Consider the given resistivity, current and the diameter of the wire.

$$\begin{gathered} \rho =1.72\times {10}^{-8}\Omega .\mathrm{m} \\ I-4.50A \\ d=2.05\times {10}^{-3}\mathrm{m} \end{gathered}$$

Solve for the electric field strength.

$E=\rho \frac{I}{A}$

Substitute the values and solve.

$$\begin{gathered} E=\frac{\left(1.72\times {10}^{-8}\right)\left(4.50\right)}{{\displaystyle \frac{\mathrm{\pi }}{4}}\left(2.05\times {10}^{-8}\right)} \\ =23.45\times {10}^{-3}\mathrm{V}/\mathrm{m} \end{gathered}$$

Hence, strength of the electric field in a $12-$gauge copper wire that is needed to

cause a 4.50-Acurrent to flow is $23.45\times {10}^{-3}\frac{\mathrm{V}}{\mathrm{m}}$.

(b)

Given the value of the resistivity is $\rho =1.47\times {10}^{-8}\Omega .m$.

The strength of the electric field is:

$E=\rho \frac{I}{A}$

Substitute the value and solve.

$$\begin{gathered} E=\frac{\left(1.47\times {10}^{-8}\right)\left(4.50\right)}{{\displaystyle \frac{\mathrm{\pi }}{4}}\left(2.05\times {10}^{-8}\right)} \\ =20.045\times {10}^{-3}\mathrm{V}/\mathrm{m} \end{gathered}$$

Hence, electric field that needed if wire were made of silver instead is $20.045\times {10}^{-3}\frac{\mathrm{V}}{\mathrm{m}}$.