Case Study Based Question: Motion of Charge in Magnetic Field An electron with speed v_o c moves in a circle of radius r_o in a uniform magnetic field. This electron is able to traverse a circular path as the magnetic force acting on the electron is perpendicular to both v_o and B, as shown in the figure. This force continuously deflects the particle sideways without changing its speed and the particle will move along a circle perpendicular to the field. The time required for one revolution of the electron is T_o. (i) If the speed of the electron is now doubled to 2v_o. The radius of the circle will change to (ii) If v = 2v_o, then the time required for one revolution of the electron (T_o) will change to (iii) A charged particle is projected in a magnetic field B = (2i + 4j)×10² T. The acceleration of the particle is found to be a = (xi + 2j) m/s². Find the value of x. (iv) If the given electron has a velocity not perpendicular to B, then trajectory of the electron is (iv) If this electron of charge (e) is moving parallel to uniform magnetic field with constant velocity v, the force acting on the electron is

PhysicsCBSE Class 12 · Sample paper, session 2024-25 · 4 marks · Case-based

CBSE Class 12 Physics sample paper 2024-25: question 29

From the official CBSE sample question paper for 2024-25, section D, question 29, for 4 marks. Try it first; the hints and our step-by-step solution are below.

Q29 · 4 marks · Case-basedCompetency-basedMoving Charges and Magnetism

Case Study Based Question: Motion of Charge in Magnetic Field
An electron with speed $v_{o} ≪ c$ moves in a circle of radius $r_{o}$ in a uniform magnetic field. This electron is able to traverse a circular path as the magnetic force acting on the electron is perpendicular to both $v_{o}$ and B, as shown in the figure. This force continuously deflects the particle sideways without changing its speed and the particle will move along a circle perpendicular to the field. The time required for one revolution of the electron is $T_{o}$ .

(i)
If the speed of the electron is now doubled to $2 v_{o}$ . The radius of the circle will change to
1. (A) $4 r_{o}$
2. (B) $2 r_{o}$
3. (C) $r_{o}$
4. (D) $r_{o} /2$
1 mark
(ii)
If $v = 2 v_{o}$ , then the time required for one revolution of the electron ( $T_{o}$ ) will change to
1. (A) $4 T_{o}$
2. (B) $2 T_{o}$
3. (C) $T_{o}$
4. (D) $T_{o} /2$
1 mark
(iii)
A charged particle is projected in a magnetic field $B = (2 i + 4 j) \times 1 0^{2}$ T. The acceleration of the particle is found to be $a = (x i + 2 j)$ m/s $^{2}$ . Find the value of x.
1. (A)4 ms $^{- 2}$
2. (B)-4 ms $^{- 2}$
3. (C)-2 ms $^{- 2}$
4. (D)2 ms $^{- 2}$
1 mark
(iv)
If the given electron has a velocity not perpendicular to B, then trajectory of the electron is
1. (A)straight line
2. (B)circular
3. (C)helical
4. (D)zig-zag
1 mark
OR
(iv)
If this electron of charge (e) is moving parallel to uniform magnetic field with constant velocity v, the force acting on the electron is
1. (A)Bev
2. (B)Be/v
3. (C)B/ev
4. (D)Zero
1 mark

Figure in the paper: Electron moving in a circle in a uniform magnetic field B directed into the page (crosses); velocity v_o tangent to circle and magnetic force F directed toward the centre at several positions. See it in the official paper.

Source: official sample paper from cbseacademic.nic.in (PDF). See it in the full Sample paper 2024-25.

Hints

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

Hint 1

Write r and T for circular motion in a magnetic field and see how each depends on v.

Hint 2

The magnetic force is always perpendicular to B; use the dot product.

Hint 3

Split the velocity into components along and perpendicular to B.

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) $r = \frac{m v}{e B} \propto v$ , so doubling v doubles the radius: $r = 2 r_{o}$ . Option (B).
1 mark
(ii) $T = \frac{2 π r}{v} = \frac{2 π m}{e B}$ , which is independent of speed, so T stays $T_{o}$ . Option (C).
1 mark
(iii) The magnetic force, and hence the acceleration, is perpendicular to B, so $a \cdot B = 0$ : $(x) (2) + (2) (4) = 0 \Rightarrow x = - 4 m s^{- 2}$ . Option (B).
1 mark
(iv) The component of v perpendicular to B gives circular motion and the component parallel to B gives uniform motion along B, so the path is a helix. Option (C).
1 mark
(iv) Alternative: $F = e v B sin 0^{\circ} = 0$ when v is parallel to B. Option (D).

Answer: (i) B; (ii) C; (iii) B (x = −4 m s⁻²); (iv) C [alternative (iv): D]

Where the marks are

(i) $2 r_{o}$ , option B (1 mark)
(ii) $T_{o}$ , option C (1 mark)
(iii) x = −4, option B (1 mark)
(iv) helical, option C; or alternative: zero, option D (1 mark)

Revise the chapter

Chapter: Moving Charges and Magnetism (notes, formulas, where marks are lost and every past question)
Topic: Motion of a Charge in a Magnetic Field

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 2024-25: every question, by section.

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CBSE Class 12 Physics sample paper 2024-25: question 29

Hints

Step-by-step solution

Revise the chapter

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