CBSE 2025, Set 1

Question

(b) (i) Current $I$ $(= 1$ A$)$ is passing through a copper rod ($n = 8\cdot5 	imes 10^{28}$ m$^{-3}$) of varying cross-sections as shown in the figure. The areas of cross-section at points A and B along its length are $1\cdot0 	imes 10^{-7}$ m$^2$ and $2\cdot0 	imes 10^{-7}$ m$^2$ respectively.

![Copper rod with varying cross section](https://www.edvaya.com/images/cbse/physics/2025/set1/q31b-i-copper-rod-varying-cross-section.webp)

Calculate:
(I) the ratio of electric fields at points A and B.
(II) the drift velocity of free electrons at point B.

(ii) Two point charges $q_1$ $(= 16	ext{ }\mu	ext{C})$ and $q_2$ $(= 1	ext{ }\mu	ext{C})$ are placed at points $\vec{r}_1 = (3	ext{ m})\hat{i}$ and $\vec{r}_2 = (4	ext{ m})\hat{j}$. Find the net electric field $\vec{E}$ at point $\vec{r} = (3	ext{ m})\hat{i} + (4	ext{ m})\hat{j}$.

Edvaya · Accepted Answer

The correct answer is Subjective. **(i)(I) Ratio of electric fields at A and B:** Using $\vec{j} = \sigma\vec{E}$ (current density = conductivity × field), and $j = I/A$: $$\frac{E_A}{E_B} = \frac{j_A}{j_B} = \frac{I/A_A}{I/A_B} = \frac{A_B}{A_A} = \frac{2\cdot0 	imes 10^{-7}}{1\cdot0 	imes 10^{-7}} = \boxed{2:1}$$ (The same current flows through both cross-sections — continuity of current.) **(i)(II) Drift velocity at B:** Using $I = neAv_d$: $$v_d = \frac{I}{neA_B} = \frac{1}{8\cdot5 	imes 10^{28} 	imes 1\cdot6 	imes 10^{-19} 	imes 2\cdot0 	imes 10^{-7}}$$ $$= \frac{1}{8\cdot5 	imes 1\cdot6 	imes 2 	imes 10^{28-19-7}} = \frac{1}{27\cdot2 	imes 10^2} = \frac{1}{2720}$$ $$\boxed{v_d \approx 3\cdot68 	imes 10^{-4} 	ext{ m/s}}$$ --- **(ii) Net electric field at $\vec{r} = 3\hat{i} + 4\hat{j}$:** **Position of charges:** - $q_1 = 16	ext{ }\mu	ext{C}$ at $\vec{r}_1 = 3\hat{i}$ - $q_2 = 1	ext{ }\mu	ext{C}$ at $\vec{r}_2 = 4\hat{j}$ - Field point P at $\vec{r} = 3\hat{i} + 4\hat{j}$ **Vector from $q_1$ to P:** $$\vec{r} - \vec{r}_1 = (3\hat{i} + 4\hat{j}) - 3\hat{i} = 4\hat{j}$$, distance = 4 m **Electric field due to $q_1$ at P:** $$\vec{E}_1 = \frac{kq_1}{r_{1P}^2}\hat{r}_{1P} = \frac{9 	imes 10^9 	imes 16 	imes 10^{-6}}{(4)^2}\hat{j} = \frac{144 	imes 10^3}{16}\hat{j} = 9 	imes 10^3\hat{j} 	ext{ N/C}$$ **Vector from $q_2$ to P:** $$\vec{r} - \vec{r}_2 = (3\hat{i} + 4\hat{j}) - 4\hat{j} = 3\hat{i}$$, distance = 3 m **Electric field due to $q_2$ at P:** $$\vec{E}_2 = \frac{kq_2}{r_{2P}^2}\hat{r}_{2P} = \frac{9 	imes 10^9 	imes 1 	imes 10^{-6}}{(3)^2}\hat{i} = \frac{9000}{9}\hat{i} = 10^3\hat{i} 	ext{ N/C}$$ **Net electric field:** $$\boxed{\vec{E}_{net} = \vec{E}_1 + \vec{E}_2 = 10^3\hat{i} + 9 	imes 10^3\hat{j} = ({\hat{i} + 9\hat{j}}) 	imes 10^3 	ext{ N/C}}$$

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