PhysicsCBSE Class 12 · Board exam 2025, set 55/1/1 · 5 marks · Long answer

CBSE Class 12 Physics 2025: question 32(b)

Set in the CBSE Class 12 board exam of 2025, paper 55/1/1, 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 torque acting on a rectangular current loop suspended in a uniform magnetic field.
(ii) A charged particle is moving in a circular path with velocity $V$ in a uniform magnetic field $B$ . It is made to pass through a sheet of lead and as a consequence, it looses one half of its kinetic energy without change in its direction. How will (1) the radius of its path (2) its time period of revolution change ?

Source: official question paper from cbse.gov.in (PDF). See it in the full 2025 board paper, set 55/1/1.

Hints

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

Hint 1

For the torque, first find which pairs of opposite sides have forces that cancel completely and which pair forms a couple.

Hint 2

The torque of a couple is one force times the perpendicular distance between the two lines of action. That distance depends on the angle between the normal and $B$ .

Hint 3

For (ii), write $r$ in terms of momentum, $p = 2 m K$ . Then check whether $T = 2 π r / v$ contains $v$ at all.

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) Set-up: Take a rectangular loop PQRS with sides $P Q = R S = b$ and $QR = S P = a$ . It carries current $I$ in a uniform field $B$ . Let $\overset{n}{^}$ be the normal to the loop, making an angle $θ$ with $B$ . The sides $b$ are perpendicular to $B$ . *(Diagram: loop with $B$ in its plane direction, normal $\overset{n}{^}$ at angle $θ$ , forces $F_{1}$ and $F_{2}$ on the arms of length $b$ in opposite directions.)*
½ mark
The forces on the arms of length $a$ are equal, opposite and along the same line (the axis of the loop). They cancel and produce no torque.
½ mark
The forces on the arms of length $b$ are equal and opposite: $F_{1} = F_{2} = I b B$ [since $F = I l \times B$ and $b ⊥ B$ ]. Their lines of action are different, so they form a couple.
½ mark
Each force acts at a distance $\frac{a}{2}$ from the axis, and its lever arm is $\frac{a}{2} sin θ$ . So
$τ = F_{1} \frac{a}{2} sin θ + F_{2} \frac{a}{2} sin θ$
½ mark
$τ = I b B a sin θ = I (ab) B sin θ = I A B sin θ$ , where $A = ab$ is the area of the loop.
½ mark
With $m = I A \overset{n}{^}$ , in vector form: $τ = I A \times B = m \times B$ . For a coil of $N$ turns, $τ = N I A B sin θ$ .
½ mark
(ii)(1) Radius: The magnetic force provides the centripetal force: $q v B = \frac{m v ^{2}}{r}$ , so $r = \frac{m v}{q B} = \frac{2 m K}{q B}$ . Hence $r \propto K$ .
½ mark
When the kinetic energy becomes $\frac{K}{2}$ :
$\frac{r ^{'}}{r} = \frac{K /2}{K} = \frac{1}{2}$ , so $r^{'} = \frac{r}{2}$ . The radius becomes $\frac{1}{2}$ times (about 0.707 times) the original.
½ mark
(ii)(2) Time period: $T = \frac{2 π r}{v} = \frac{2 π m}{q B}$ .
½ mark
$T$ does not depend on the speed or the kinetic energy of the particle. So the time period of revolution remains unchanged.
½ mark

Answer: (i) $τ = I A \times B = m \times B$ , magnitude $I A B sin θ$ ; (ii)(1) radius becomes $r / 2$ ; (2) time period stays the same.

Where the marks are

Shows the forces on the arms of length $a$ cancel and the forces on the arms of length $b$ ( $I b B$ ) form a couple (1 mark)
Writes the torque using the lever arm $\frac{a}{2} sin θ$ for each force (1 mark)
Obtains $τ = I A B sin θ$ , i.e. $τ = m \times B$ (1 mark)
Uses $r = 2 m K / q B$ to get $r^{'} = r / 2$ (1 mark)
Uses $T = 2 π m / q B$ to conclude that $T$ is unchanged (1 mark)

Revise the chapter

Chapter: Moving Charges and Magnetism (notes, formulas, where marks are lost and every past question)
Topic: Torque on a Current Loop and Magnetic Dipole

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 paper2025 board paper, set 55/1/1: 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 2025: question 32(b)

Hints

Step-by-step solution

Revise the chapter

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