Question

Figure-agives the magnitude E(x) of the electric fields that have been set up by a battery along a resistive rod of length 9.00mm(Figure-b). The vertical scale is set by. ${\mathbf{E}}_{\mathbf{s}}\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{00}\mathbf{\times }{\mathbf{10}}^{\mathbf{3}}\mathbf{V}\mathbf{/}\mathbf{m}$The rod consists of three sections of the same material but with different radii. (The schematic diagram of Figure-bdoes not indicate the different radii.) The radius of section 3 is 2.00mm. (a) What is the radius of section 1 and (b) What is the radius of section 2?

Short answer

1. The radius of section 1 is 1.55 mm
2. The radius of section 2 is 1.22 mm

Step-by-step solution

Identification of given data

1. Vertical scale is set by${\mathrm{E}}_{\mathrm{s}}=4.00\times {10}^3\mathrm{V}/\mathrm{m}$
2. Radius of section 3,${\mathrm{r}}_3=2.00\mathrm{mm}$

Understanding the concept of the current density

The current density is the rate of flow of charges per unit time per unit cross-section area of the conductor. First, we have to find the ratios of the current densities of the three sections of the rod. As the three sections of the rod are in series, the current throughout the rod must be the same. By equating the currents through each section, we can find the required radius.

Formulae:

The current density of the wire according to the electric field, $J=\frac{E}{p}$ …(i)

Here, J is current density, E is electric field, and p is resistivity.

The current flowing through the wire, I = JA …(ii)

Here, I is current, J is current density, and A is area of cross-section.

The cross-sectional area of the wire, $A=\pi r^2$ …(iii)

Here, A is area of cross-section, r is radius.

(a) Calculation of the radius of section 1

Since, the vertical scale is set by ${\mathrm{E}}_{\mathrm{s}}=4.00\times {10}^3\mathrm{V}/\mathrm{m}$

The electric fields from the figure are in the ratio:

${\mathrm{E}}_1:{\mathrm{E}}_2:{\mathrm{E}}_3=2.5:\hspace{0.17em}4:1.5$ …(iv)

As the material of the three sections of the rod is same, the resistivity must also be the same, i.e. is constant. This implies that the electric fields are directly proportional to their respective current densities considering equation (i).

Thus, comparing equations (i) to (iv), we can get the ratio of the current densities as follows:

${\mathrm{J}}_1:{\mathrm{J}}_2:{\mathrm{J}}_3=2.5:\hspace{0.17em}4:1.5$

As the three sections of the rod are in series, the current throughout the rod must be same. Thus, the current relations considering equation (ii) can be given as:

role="math" localid="1661412543657" $J_1{A}_1=J_2{A}_2=J_3{A}_3$

…(v)

Thus, the value of the radius for section 1 can be given using the above relation of equation (v) and equation (iii) as:

$\frac{{\mathrm{J}}_1}{{\mathrm{J}}_3}=\frac{\pi r_3^2}{\pi r_1^2}$

$$\begin{gathered} \frac{2.5\mathrm{V}/\mathrm{m}}{1.5\mathrm{V}/\mathrm{m}}=\frac{{\left(2\mathrm{mm}\right)}^2}{{\mathrm{r}}_1^2}\left(\mathrm{substituting}\mathrm{the}\mathrm{above}\mathrm{values}\right) \\ {\mathrm{r}}_1^2=\frac{{\left(2\mathrm{mm}\right)}^2\times 1.5}{2.5} \\ =2.4{\mathrm{mm}}^2 \\ {\mathrm{r}}_1=1.55\mathrm{mm} \end{gathered}$$

Hence, the value of the radius of section 1 is 1.55mm .

(b) Calculation of the radius from section 2

Similarly, the value of the radius for section 2 can be given using the above relation of equation (a) and equation (iii) as follows:

$$\begin{gathered} \frac{{\mathrm{J}}_2}{{\mathrm{J}}_3}=\frac{{\mathrm{\pi r}}_3^2}{{\mathrm{\pi r}}_2^2} \\ \frac{4\mathrm{V}/\mathrm{m}}{1.5\mathrm{V}/\mathrm{m}}=\frac{{\left(2\mathrm{mm}\right)}^2}{{\mathrm{r}}_2^2} \\ {\mathrm{r}}_2^2=\frac{{\left(2\mathrm{mm}\right)}^2\times 1.5}{4} \\ =1.5{\mathrm{mm}}^2 \\ {\mathrm{r}}_2^2=1.22\mathrm{mm} \end{gathered}$$

Hence, the value of the radius of section 2 is 1.22 mm .