PhysicsCBSE Class 12 · Sample paper, session 2025-26 · 3 marks · Short answer

CBSE Class 12 Physics sample paper 2025-26: question 27(I)

From the official CBSE sample question paper for 2025-26, section C, question 27(I), for 3 marks. Try it first; the hints and our step-by-step solution are below.

Q27(I) · 3 marks · Short answerMoving Charges and Magnetism

If a point sized object having charge 1C and mass 1g is projected with velocity of $2 \hat{i}$ m/s from a point (0,2cm,0) in the region of magnetic field $- 0.1 \hat{k}$ T which spreads in the first quadrant.
(A) What will be the shape of the path followed by the given charged particle?
(B) At what point it will cross the X-axis?
(C) What will be the kinetic energy of particle when it will enter in the fourth quadrant?

Figure in the paper: First quadrant with X and Y axes, filled with magnetic field into the page (crosses). Charge q at (0, 2 cm) on the Y-axis with velocity V directed along +X. See it in the official paper.

Source: official sample paper from cbseacademic.nic.in (PDF). See it in the full Sample paper 2025-26.

Hints

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

Hint 1

Find the direction of $q v \times B$ at the starting point.

Hint 2

Radius of the circle is $m v / q B$ ; locate the centre from the force direction.

Hint 3

Does a magnetic force ever change the speed?

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.

$v ⊥ B$ , so the magnetic force provides centripetal force and the path is circular with radius $r = \frac{m v}{q B} = \frac{1 0 ^{- 3} \times 2}{1 \times 0.1} = 0.02 m = 2$ cm. Initial force: $F = q v \times B = 1 (2 \hat{i}) \times (- 0.1 \hat{k}) = + 0.2 \hat{j}$ N.
1 mark
(A), (B) With $B$ along $- \hat{k}$ (into the page) the force at (0, 2 cm) is along $+ y$ , so the charge turns anticlockwise about centre (0, 4 cm): it describes a semicircle of radius 2 cm and leaves the field at (0, 6 cm) on the Y-axis; it does not reach the X-axis. (If the field were along $+ \hat{k}$ , the motion would be clockwise about the origin: a quarter circle meeting the X-axis at (2 cm, 0) and entering the fourth quadrant — the answer given in the scheme.)
1 mark
(C) Magnetic force is always ⟂ velocity, so it does no work and speed is unchanged: $K = \frac{1}{2} m v^{2} = \frac{1}{2} \times 1 0^{- 3} \times 2^{2} = 2 \times 1 0^{- 3}$ J when it leaves the field.
1 mark

Answer: r = 2 cm; with B = −0.1k̂ the path is a semicircle leaving at (0, 6 cm) (scheme, assuming clockwise motion: quarter circle crossing X-axis at (2 cm, 0)); KE = 2×10⁻³ J

Where the marks are

Circular path with radius $m v / q B = 2$ cm (1 mark)
Correct shape/exit point of the path (1 mark)
KE unchanged, $2 \times 1 0^{- 3}$ J, since magnetic force does no work (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 paperSample paper 2025-26: every question, by section.

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CBSE Class 12 Physics sample paper 2025-26: question 27(I)

Hints

Step-by-step solution

Revise the chapter

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