PhysicsCBSE Class 12 · Board exam 2026, set 55/5/1 · 5 marks · Long answer

CBSE Class 12 Physics 2026: question 32(b)

Set in the CBSE Class 12 board exam of 2026, paper 55/5/1, section E, 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 $F$ acting on a conductor of length L and area of cross-section A carrying current I and placed in a magnetic field $B$ .
(ii) A part of a wire carrying 2·0 A current and bent at 90° at two points is placed in a region of uniform magnetic field $B = - (0 \cdot 50 T) \hat{k}$ , as shown in the figure. Calculate the magnitude of the net force acting on the wire.

Figure in the paper: Axes: j up, i to the right, k out toward viewer. A rectangular region of width 50 cm marked with crosses (field into page). A wire carrying 2·0 A enters horizontally from the left at the upper part, goes horizontally, bends 90° downward for a vertical segment of 20 cm, bends 90° again to continue horizontally to the right and exits the region. See it in the official paper.

Source: official question paper from cbse.gov.in (PDF). See it in the full 2026 board paper, set 55/5/1.

Hints

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

Hint 1

Start from the force $q (v \times B)$ on one moving charge, then count how many charges are in the wire.

Hint 2

For part (ii), find the force on each straight segment separately; the horizontal pieces add up to the full width of the field region.

Hint 3

The horizontal and vertical forces are perpendicular, so combine them with Pythagoras.

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) Let the conductor have $n$ free electrons per unit volume, each of charge $- e$ , moving with drift velocity $v_{d}$ . The Lorentz force on one electron is $f = - e (v_{d} \times B)$ .
1 mark
The number of free electrons in the conductor is $n A L$ . The total force is $F = - n A L e (v_{d} \times B)$ .
1 mark
The current density is $j = - n e v_{d}$ , so $F = (j A L) \times B$ . Since $j A = I$ , define $L$ as a vector of length $L$ along the current. Then $j A L$ in the direction of current $= I L$ , giving
$F = I (L \times B), ∣ F ∣ = I L B sin θ$
where $θ$ is the angle between $L$ and $B$ .
1 mark
(ii) The magnetic field is uniform. The net force on the bent wire therefore equals the force on a straight wire joining its entry and exit points in the field. The horizontal parts together span the field width $0.50 m$ along $+ \hat{i}$ . The vertical part is $0.20 m$ along $- \hat{j}$ . So:
$L_{eff} = (0.50 \hat{i} - 0.20 \hat{j}) m$
1 mark
Use $F = I L_{eff} \times B$ :
$F = 2.0 [(0.50 \hat{i} - 0.20 \hat{j}) \times (- 0.50 \hat{k})]$
With $\hat{i} \times \hat{k} = - \hat{j}$ and $\hat{j} \times \hat{k} = \hat{i}$ :
$F = 2.0 (0.25 \hat{j} + 0.10 \hat{i}) = (0.20 \hat{i} + 0.50 \hat{j}) N$
The magnitude is:
$∣ F ∣ = 0.2 0^{2} + 0.5 0^{2} = 0.29 \approx 0.54 N$
(Equivalently, horizontal parts: $F = 2 \times 0.5 \times 0.5 = 0.5$ N along $+ \hat{j}$ ; vertical part: $F = 2 \times 0.2 \times 0.5 = 0.2$ N along $+ \hat{i}$ .)
1 mark

Answer: (i) $F = I (L \times B)$ ; (ii) $F = (0.20 \hat{i} + 0.50 \hat{j})$ N, $∣ F ∣ \approx 0.54$ N

Where the marks are

Writes the Lorentz force on a single drift electron (1 mark)
Multiplies by the number of electrons $n A L$ (1 mark)
Uses $I = n e A v_{d}$ to obtain $F = I L \times B$ (1 mark)
Finds the forces on the horizontal (0.5 N) and vertical (0.2 N) parts, or the effective length vector (1 mark)
Combines them to get the net force $\approx 0.54$ N (1 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 paper2026 board paper, set 55/5/1: every question, by section.

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CBSE Class 12 Physics 2026: question 32(b)

Hints

Step-by-step solution

Revise the chapter

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