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

CBSE Class 12 Physics 2023: question 32(b)

Set in the CBSE Class 12 board exam of 2023, paper 55/1/1, section D, question 32(b), for 5 marks. Try it first; the hints and our step-by-step solution are below.

Q32(b) · 5 marks · Long answerMoving Charges and Magnetism

(i) Derive an expression for the force acting on a current carrying straight conductor kept in a magnetic field. State the rule which is used to find the direction of this force. Give the condition under which this force is (1) maximum, and (2) minimum.
(ii) Two long parallel straight wires A and B are $2 \cdot 5$ cm apart in air. They carry $5 \cdot 0$ A and $2 \cdot 5$ A currents respectively in opposite directions. Calculate the magnitude of the force exerted by wire A on a 10 cm length of wire B.

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

Hints

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

Hint 1

Start from the force on one moving charge and multiply by the number of carriers in the rod.

Hint 2

Express $n q v_{d} A$ in terms of the current I.

Hint 3

For (ii), use the force per unit length between parallel wires and convert 10 cm and 2.5 cm to metres.

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) Consider a straight rod of length l, cross-section A, with n mobile charge carriers (charge q) per unit volume; total carriers $= n l A$ .
½ mark
Each carrier moving with drift velocity $v_{d}$ in field $B$ experiences $q (v_{d} \times B)$ , so total force $F = n l A q (v_{d} \times B)$
½ mark
Current density $j = n q v_{d}$ , so $F = (j \times B) A l$
½ mark
Since $j A = I$ and taking $l$ along the current: $F = I (l \times B)$ , magnitude $F = I l B sin θ$
½ mark
Fleming's left-hand rule: stretch the forefinger, middle finger and thumb of the left hand mutually perpendicular; if the forefinger points along the magnetic field and the middle finger along the current, the thumb gives the direction of the force on the conductor.
½ mark
(1) Maximum force when $θ = 9 0^{\circ}$ (conductor perpendicular to $B$ ): $F_{ma x} = I l B$ .
½ mark
(2) Minimum (zero) force when $θ = 0^{\circ}$ or $18 0^{\circ}$ (conductor parallel or antiparallel to $B$ ): $F = 0$ .
½ mark
(ii) Force on length l of wire B: $F = \frac{μ _{0} I _{1} I _{2} l}{2 π d}$
½ mark
$F = \frac{2 \times 1 0 ^{- 7} \times 5.0 \times 2.5 \times 0.10}{2.5 \times 1 0 ^{- 2}}$
½ mark
$F = 1.0 \times 1 0^{- 5}$ N (repulsive, since currents are opposite).
½ mark

Answer: (i) $F = I (l \times B)$ ; max at θ = 90° ( $I l B$ ), zero at θ = 0° or 180°; (ii) $F = 1 0^{- 5}$ N (repulsive)

Where the marks are

Number of carriers $n l A$ (½ mark)
Total force $n l A q (v_{d} \times B)$ (½ mark)
Introduces $j = n q v_{d}$ (½ mark)
$F = I (l \times B)$ (½ mark)
Statement of Fleming's left-hand rule (½ mark)
Maximum at θ = 90° (½ mark)
Minimum at θ = 0° or 180° (½ mark)
Formula $μ_{0} I_{1} I_{2} l /2 π d$ (½ mark)
Substitution (½ mark)
Answer $1 0^{- 5}$ N (½ mark)

Revise the chapter

Chapter: Moving Charges and Magnetism (notes, formulas, where marks are lost and every past question)

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

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CBSE Class 12 Physics 2023: question 32(b)

Hints

Step-by-step solution

Revise the chapter

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