91Ó°ÊÓ

According to Ohm’s law the current flowing through conductor is directly proportional to the voltage across the two points.

V = IR

Where, I is current in ampere(A) , R is resistance in ohms $\left(\mathrm{Î\copyright }\right)$ and V is the potential difference volt (V) .

The ratio of V to I for a particular conductor is called its resistance R:

$R=\frac{V}{I}$ or $\frac{\mathrm{ÒÏ}L}{A}$

Where, $\mathrm{ÒÏ}$is resistivity $\mathrm{Î\copyright }â‹\dots m$, L is length in m and A is area in $m^2$.

The equation of electric field is $E=\mathrm{ÒÏ}J$, where $J=\frac{I}{A}$.

Given that, the length of copper wire is 1.20 m, the length of silver wire is 0.80 m, radius of both wire is 0.3 mm and the voltage across the wire is 9.0 V.

The current flowing the both copper and silver wire is:

$$\begin{gathered} \begin{array}{r}I=\frac{V}{R_{Ag}+R_{CU}}\\ =\frac{9}{\left\{\frac{\left(1.47×{10}^{−8}\right)\left(1.2\right)}{\mathrm{Ï€}(0.0003{)}^2}\right\}+\left\{\frac{\left(1.72×{10}^{−8}\right)\left(0.8\right)}{\mathrm{Ï€}(0.0003{)}^2}\right\}}\left(⊗R=\frac{\mathrm{ÒÏ}L}{A}\right)\end{array} \\ \begin{array}{r}=\frac{9}{0.062+0.049}\\ =81.0\mathrm{A}\end{array} \end{gathered}$$

Hence, current in the copper section is 81.0 A and the current in the silver section is 81.0 A.

The electric field in copper and silver section are:

$$\begin{gathered} E_{Cu}=\mathrm{ÒÏ}\frac{I}{A} \\ =\mathrm{ÒÏ}\frac{1}{\mathrm{Ï€}{r}^2} \\ =\frac{\left(1.72×{10}^{−8}\right)\left(81.0\right)}{\mathrm{Ï€}(0.0003{)}^2} \\ =4.92\mathrm{V}/\mathrm{m} \\ {\mathrm{E}}_{\mathrm{Ag}}=\mathrm{ÒÏ}\frac{1}{\mathrm{A}} \\ =\mathrm{ÒÏ}\frac{1}{{\mathrm{Ï€°ù}}^2} \\ =\frac{\left(1.47×{10}^{−8}\right)\left(81.0\right)}{\mathrm{Ï€}(0.0003{)}^2} \\ =4.21\mathrm{V}/\mathrm{m} \end{gathered}$$

Hence, magnitude of$\stackrel{→}{E}$in the copper is 4.92 V/m and the magnitude of$\stackrel{→}{E}$in the silver is 4.21 V/m.

By the ohm’s law

$$\begin{gathered} V_{Ag}=I{R}_{Ag} \\ =\left(81.0\right)\frac{\left(1.47×{10}^{−8}\right)\left(1.2\right)}{\mathrm{π}(0.0003{)}^2} \\ =5.02\mathrm{V} \end{gathered}$$

Hence, the potential difference between the ends of the silver section of wire is 5.02 V.