91影视

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

$V=IR$

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

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

$R=\frac{V}{I}$or$\frac{蚁L}{A}$

Where, $蚁$is resistivity $\mathrm{惟}路\mathrm{m}$, $L$is length in and A is area in $m^2$.

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

Given that, length $\left(L_1\right)$and radii $\left(r_1\right)$ of first wire are 1.2 m and 0.8 mm respectively, length$\left(L_2\right)$ and radii$\left(r_2\right)$ of second wire are and respectively and the current in first section is 2.5 mA .

Consider the current flowing through second section is$I_2$ . As given the both sections are connected in series, the resistance changes with change in radius, and voltage changes with resistance. So current in both sections will be the same.

$I_2=2.5\mathrm{mA}$

Hence, current in the$0.80-mm-$ diameter section is 2.5 mA.

The electric field in$1.60-mm-$ diameter and$0.80-mm-$ diameter sections are:

$$\begin{gathered} E_1=蚁\frac{I}{A} \\ =蚁\frac{I}{{{\mathrm{蟺谤}}_1}^2} \\ =\frac{\left(1.72脳{10}^{-8}\right)\left(0.0025\right)}{蟺{\left(0.0008\right)}^2} \\ =21.4渭\mathrm{V}/\mathrm{m} \\ {E}_2=蚁\frac{I}{A} \\ =蚁\frac{I}{{蟺r_2}^2} \\ =\frac{\left(1.72脳{10}^{-8}\right)\left(0.0025\right)}{蟺{\left(0.0004\right)}^2} \\ =85.5渭\mathrm{V}/\mathrm{m} \end{gathered}$$

Hence, magnitude of$\stackrel{鈫�}{E}$ in the$1.60-mm-$ diameter section is$21.4渭\mathrm{V}/\mathrm{m}$ and the magnitude of$\stackrel{鈫�}{E}$ in the$0.80-mm-$ diameter section is $81.5渭\mathrm{V}/\mathrm{m}$.

By the ohm鈥檚 law

$$\begin{gathered} V_{Ag}=I\left(R_1+R_2\right) \\ =I\left(\frac{蚁L_1}{{{\mathrm{蟺谤}}_1}^2}+\frac{蚁L_1}{{{\mathrm{蟺谤}}_2}^2}\right) \\ =\left(0.025\right)\left\{\frac{\left(1.72脳{10}^{-8}\right)\left(1.2\right)}{\mathrm{蟺}{\left(0.0008\right)}^2}+\frac{\left(1.72脳{10}^{-8}\right)\left(1.8\right)}{\mathrm{蟺}{\left(0.0008\right)}^2}\right\} \\ =0.108\mathrm{mV} \end{gathered}$$

Hence, the potential difference between the ends of the$3.00-m$ length of wire is $0.180\mathrm{mV}$.