91Ó°ÊÓ

The electric force per unit charge is known as electric.

$E=\mathrm{ÒÏ}\frac{I}{A}$

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

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

(a)

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

$$\begin{gathered} \mathrm{ÒÏ}=1.72×{10}^{-8}\mathrm{Î\copyright }.\mathrm{m} \\ I-4.50A \\ d=2.05×{10}^{-3}\mathrm{m} \end{gathered}$$

Solve for the electric field strength.

$E=\mathrm{ÒÏ}\frac{I}{A}$

Substitute the values and solve.

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

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

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

(b)

Given the value of the resistivity is $\mathrm{ÒÏ}=1.47×{10}^{-8}\mathrm{Î\copyright }.m$.

The strength of the electric field is:

$E=\mathrm{ÒÏ}\frac{I}{A}$

Substitute the value and solve.

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

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