PhysicsCBSE Class 12 · Board exam 2023, set 55/5/1 · 5 marks · Long answer

CBSE Class 12 Physics 2023: question 31(a)

Set in the CBSE Class 12 board exam of 2023, paper 55/5/1, section D, question 31(a), for 5 marks. Try it first; the hints and our step-by-step solution are below.

Q31(a) · 5 marks · Long answerCurrent Electricity

(i) Explain how free electrons in a metal at constant temperature attain an average velocity under the action of an electric field. Hence obtain an expression for it. (3)
(ii) Consider two conducting wires A and B of the same diameter but made of different materials joined in series across a battery. The number density of electrons in A is 1.5 times that in B. Find the ratio of drift velocity of electrons in wire A to that in wire B. (2)

Source: official question paper from cbse.gov.in (PDF). See it in the full 2023 board paper, set 55/5/1.

Hints

Open one hint at a time and try again before the next.

Hint 1

Write the velocity of one electron after time t_i since its last collision, then average over all electrons.

Hint 2

In series the current is the same. Use I = neAv_d with equal areas.

Step-by-step solution

Each step shows the marks it earns, as an examiner would award them. Checked by an independent re-solve from the question paper.

(i) Explanation: In the absence of a field, free electrons move randomly and their average velocity is zero. When a field $E$ is applied, each electron experiences a force $F = - e E$ and accelerates between collisions with the ions. Collisions keep randomising its velocity. The net result, averaged over all electrons, is a small constant velocity opposite to $E$ , called the drift velocity. It is constant at a given temperature because the relaxation time $τ$ is fixed.
1 mark
Derivation: Each electron has acceleration $a = - \frac{e E}{m}$ . If electron $i$ had velocity $v_{i}$ just after its last collision and time $t_{i}$ has passed since, its velocity now is $V_{i} = v_{i} - \frac{e E}{m} t_{i}$ .
½ mark
The drift velocity is the average over all electrons: $v_{d} = (V_{i})_{a v g} = (v_{i})_{a v g} - \frac{e E}{m} (t_{i})_{a v g}$ .
½ mark
Because thermal velocities are random, $(v_{i})_{a v g} = 0$ . The average time between collisions is the relaxation time, $(t_{i})_{a v g} = τ$ .
½ mark
Therefore $v_{d} = - \frac{e E}{m} τ$ , with magnitude $v_{d} = \frac{e E τ}{m}$ .
½ mark
(ii) In series, the same current $I$ flows through both wires. With $I = n e A v_{d}$ , this gives $v_{d} = \frac{I}{n e A} = \frac{4 I}{n e π D ^{2}}$ . With the same $I$ and $D$ , $v_{d} \propto \frac{1}{n}$ .
1 mark
$\frac{v _{d A}}{v _{d B}} = \frac{n _{B}}{n _{A}} = \frac{n _{B}}{1.5 n _{B}} = \frac{2}{3}$
1 mark

Answer: (i) $v_{d} = - \frac{e E τ}{m}$ ; (ii) v_A : v_B = 2 : 3

Where the marks are

Explanation of drift under the force −eE with collisions (1 mark)
Velocity of an electron $V_{i} = v_{i} - e E t_{i} / m$ (½ mark)
Averaging over all electrons (½ mark)
(v_i)avg = 0 and (t_i)avg = τ (½ mark)
v_d = eEτ/m (½ mark)
Same I and A, so v_d ∝ 1/n (1 mark)
Ratio = 2/3 (1 mark)

Revise the chapter

Chapter: Current Electricity (notes, formulas, where marks are lost and every past question)
Topic: Drift Velocity, Mobility and Resistivity

Practise this chapterBoard-style questions from Current Electricity, one at a time, marked against the answer key and marking points.Back to the full paper2023 board paper, set 55/5/1: every question, by section.

The question is transcribed from the official paper; figures are described in words. The solution, hints and chapter tag are ours. Spotted an error? Write tosupport@questionbuddy.in.

CBSE Class 12 Physics 2023: question 31(a)

Hints

Step-by-step solution

Revise the chapter

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