Board Exam Previous Year Questions

Board Exam PYQs — Moving Charges and Magnetism

This chapter is one of the highest-yield topics in the CBSE Class 12 Board paper. In any given year, expect a combined 8–12 marks from Chapter 4 — typically split across the four mark-bands.

How the Board paper is structured

Type	Marks	What is asked
VSA	1	One-line statement, definition, formula, or simple direction question.
SA-I	2	Brief statement of a law or a one-step derivation/numerical.
SA-II	3	Derivation or 2-step numerical, often with a labelled diagram.
LA	5	Full derivation + construction + working (cyclotron, galvanometer, solenoid).

Chapter 4 has appeared every year in at least three of these four buckets. The 20 questions below are organised to match this pattern: 5 questions per mark band.

Most-asked themes (last 5 years)

1. Cyclotron — 5-mark LA (construction + working + numerical) appears in 3 of every 5 years.
2. Moving coil galvanometer — 5-mark LA, almost annually, often with conversion to ammeter/voltmeter.
3. Biot-Savart law / circular loop axis derivation — 3-mark or 5-mark.
4. Ampère's law applied to a solenoid — 3-mark or 5-mark derivation.
5. Force between two parallel wires + SI ampere definition — 2-mark or 3-mark.

Mastering the questions below covers $\sim 95\%$ of the Board pattern.

How to Answer Board Questions for Full Marks

Examiners reward process as much as the final number. Follow this template for any 2-, 3-, or 5-mark question:

• State the law / principle / formula first. One clear sentence. ("By Biot-Savart law, the magnetic field due to a current element $Id\vec{l}$ at a point is…")
• Draw a neat, labelled diagram. Mark current direction, field direction, position vector, angle. A diagram is worth half a mark even if your algebra slips.
• Write the formula in vector form, then take magnitudes. Examiners look for the vector form $\vec{F} = I\vec{L}\times\vec{B}$ before the scalar $F = BIL\sin\theta$.
• Derive step by step — no jumps. Show every algebraic step. Each correct line is a discrete mark.
• Substitute with units written. Don't drop units until the very last line.
• Box the final answer with units. E.g. $\boxed{B = 4 \times 10^{-5}\ \text{T}}$, direction stated.
• State direction explicitly wherever vectors are involved. "Force is directed along $+\hat{i}$" or "into the page" — never leave direction implicit.
• For 5-mark LA: split into clear sub-sections (Construction → Principle → Working → Expression → Conversion/Applications) using mini-headers. CBSE markers love structured answers.

Questions and Answers

VSA -1 Mark Questions

Q1. SI unit of magnetic dipole moment

State the SI unit of magnetic dipole moment.

Answer. The SI unit of magnetic dipole moment is ampere metre-squared ($\text{A m}^2$), equivalently joule per tesla ($\text{J/T}$).

Q2. Lorentz force expression

Write the expression for the Lorentz force experienced by a charge $q$ moving with velocity $\vec{v}$ in the simultaneous presence of an electric field $\vec{E}$ and a magnetic field $\vec{B}$.

Answer. $\boxed{\vec{F} = q(\vec{E} + \vec{v}\times\vec{B})}$.

Q3. Right-hand rule direction

A current flows vertically upward in a straight wire. State the direction of the magnetic field at a point directly to the east of the wire.

Answer. Using the right-hand thumb rule (thumb up along $I$, fingers curl), the field at a point east of the wire points northward (horizontally, parallel to the ground).

Q4. Why radial magnetic field in galvanometer?

In a moving coil galvanometer, why is the magnetic field made radial?

Answer. A radial magnetic field ensures that the plane of the coil always remains parallel to $\vec{B}$, so that the torque on the coil is $\tau = NIAB$ — constant and independent of deflection angle $\theta$. This makes the deflection $\phi \propto I$ (linear scale).

Q5. Why does magnetic force do no work?

Justify in one line: a magnetic force can never do work on a moving charge.

Answer. The magnetic force $\vec{F} = q\vec{v}\times\vec{B}$ is always perpendicular to $\vec{v}$. Since power $= \vec{F}\cdot\vec{v} = 0$, no work is done; the speed (and hence kinetic energy) of the charge remains constant.

SA-I 2 Marks Questions

Q1. State Biot-Savart law

State the Biot-Savart law and write it in vector form. Give the SI unit of the constant $\mu_0/(4\pi)$.

Answer. The Biot-Savart law states that the magnetic field $d\vec{B}$ at a point P due to a small current element $Id\vec{l}$ at position $\vec{r}$ from the element is

$$d\vec{B} = \frac{\mu_0}{4\pi}\, \frac{I\, d\vec{l}\times\hat{r}}{r^2}.$$

The magnitude is $dB = \frac{\mu_0}{4\pi}\frac{I\,dl\,\sin\theta}{r^2}$, where $\theta$ is the angle between $d\vec{l}$ and $\hat{r}$.

The SI unit of $\mu_0/(4\pi)$ is tesla-metre per ampere ($\text{T m/A}$), with value $10^{-7}\ \text{T m/A}$.

Q2. Ammeter vs voltmeter

Distinguish between an ammeter and a voltmeter. State how each is connected in a circuit.

Answer.

Property	Ammeter	Voltmeter
Function	Measures current	Measures potential difference
Connection	In series with the circuit element	In parallel with the circuit element
Ideal resistance	Zero (or very small)	Infinite (or very large)
Realised by	Galvanometer + small shunt	Galvanometer + large series resistance

Q3. Cyclotron frequency

Derive the expression for cyclotron frequency. Why is this frequency independent of the speed of the particle?

Answer. In a cyclotron, the magnetic force provides centripetal force:

$$qvB = \frac{mv^2}{r} \Rightarrow v = \frac{qBr}{m}.$$

The period is $T = 2\pi r/v = 2\pi m/(qB)$, so frequency

$$\boxed{f_c = \frac{qB}{2\pi m}.}$$

Since $f_c$ depends only on $q$, $B$, and $m$ — not on $v$ or $r$ — it remains constant as the particle speeds up and spirals outward. This is the principle that makes the cyclotron work.

Q4. SI definition of the ampere

State the SI definition of the ampere based on the force between two parallel current-carrying wires.

Answer. Two long, straight, parallel conductors of negligible cross-section, placed 1 metre apart in vacuum, carry equal currents. If the force per unit length between them is $2\times 10^{-7}\ \text{N/m}$, then each conductor carries a current of 1 ampere.

This follows from $F/L = \mu_0 I_1 I_2 /(2\pi d)$ with $I_1 = I_2 = 1$ A, $d = 1$ m, giving $F/L = 2\times 10^{-7}\ \text{N/m}$.

Q5. Charge moving parallel to $\vec{B}$

What is the force on a charged particle moving parallel to a magnetic field? Explain.

Answer. $\vec{F} = q\vec{v}\times\vec{B}$. If $\vec{v}\parallel\vec{B}$, the cross product is zero, so $\vec{F} = 0$.

A charged particle moving along (or against) a magnetic field experiences no magnetic force and continues in a straight line at constant velocity. The magnetic field "ignores" the component of velocity parallel to it.

SA-II 3 Marks Questions

Q1. Derive $B$ at centre of a circular loop

Using Biot-Savart law, derive the expression for the magnetic field at the centre of a circular loop of radius $R$ carrying a current $I$.

Answer. By Biot-Savart, the field due to an element $Id\vec{l}$ at a point distance $r$ from it is

$$dB = \frac{\mu_0}{4\pi}\frac{I\,dl\,\sin\theta}{r^2}.$$

At the centre of a circular loop of radius $R$:

• Every element $d\vec{l}$ is perpendicular to $\hat{r}$, so $\theta = 90°$, $\sin\theta = 1$.
• Distance from element to centre is $r = R$ (constant).
• By symmetry, all contributions add along the axis (perpendicular to the plane of the loop).

Total field:

$$B = \int dB = \frac{\mu_0 I}{4\pi R^2} \int dl = \frac{\mu_0 I}{4\pi R^2}\cdot (2\pi R) = \frac{\mu_0 I}{2R}.$$

$$\boxed{B = \frac{\mu_0 I}{2R}.}$$

Direction: perpendicular to the plane of the loop, given by the right-hand curl rule.

Q2. Force between two parallel wires

Two long straight parallel wires separated by a distance $d$ carry currents $I_1$ and $I_2$ in the same direction. Derive an expression for the force per unit length on each wire and state its nature.

Answer. Wire 1 produces at the location of wire 2 a field

$$B_1 = \frac{\mu_0 I_1}{2\pi d}\quad\text{(directed into the page if I}_1\text{ flows up and we look from the right)}.$$

Wire 2 carries current $I_2$ in this field. The force per unit length on wire 2 is

$$\frac{F}{L} = I_2 B_1 = \frac{\mu_0 I_1 I_2}{2\pi d}.$$

By Newton's third law, wire 1 feels an equal and opposite force per unit length.

• Same direction currents $\Rightarrow$ attraction.
• Opposite direction currents $\Rightarrow$ repulsion.

$$\boxed{\frac{F}{L} = \frac{\mu_0 I_1 I_2}{2\pi d}.}$$

Q3. Torque on a current loop

Derive an expression for the torque on a rectangular current loop placed in a uniform magnetic field, with the loop's plane making angle $\theta$ with $\vec{B}$.

Answer. Consider a rectangular loop of sides $a$ and $b$ carrying current $I$, placed in field $\vec{B}$. Let the area vector $\vec{A}$ make angle $\theta$ with $\vec{B}$.

The two sides of length $b$ that are perpendicular to $\vec{B}$ each feel a force $F = BIb$, equal and opposite, forming a couple. The perpendicular distance between these forces is $a\sin\theta$.

Torque:

$$\tau = F\cdot(a\sin\theta) = BIb\cdot a\sin\theta = BIA\sin\theta,$$

where $A = ab$ is the loop area. For $N$ turns,

$$\boxed{\tau = NIAB\sin\theta = mB\sin\theta},$$

with magnetic moment $m = NIA$. In vector form: $\vec{\tau} = \vec{m}\times\vec{B}$.

Q4. Convert galvanometer to voltmeter — numerical

A galvanometer of resistance $G = 50\ \Omega$ gives full-scale deflection at a current $I_g = 4$ mA. Convert it into a voltmeter of range 0–20 V. What series resistance is required?

Answer. For a voltmeter, a high series resistance $R$ is connected with the galvanometer. The total resistance must allow current $I_g$ when the voltmeter reads $V$:

$$V = I_g (R + G) \Rightarrow R = \frac{V}{I_g} - G.$$

Substituting:

$$R = \frac{20}{4\times 10^{-3}} - 50 = 5000 - 50 = 4950\ \Omega.$$

$$\boxed{R = 4950\ \Omega,\ \text{in series with the galvanometer.}}$$

Q5. Compare ammeter and voltmeter (table)

Compare an ammeter and a voltmeter in five key aspects.

Answer.

#	Aspect	Ammeter	Voltmeter
1	What it measures	Current ($I$)	Potential difference ($V$)
2	Connection in circuit	In series	In parallel
3	Resistance (ideal)	Zero	Infinite
4	How constructed	Galvanometer + low shunt resistance $S$ in parallel	Galvanometer + high resistance $R$ in series
5	Range modification	Reduce $S$ to extend range	Increase $R$ to extend range

Both are derived from a moving coil galvanometer, but their internal resistances are tuned to opposite extremes so as to minimally disturb the quantity they measure.

LA 5 Marks Questions

Q1. Magnetic field of a solenoid via Ampère's law

Using Ampère's circuital law, derive the magnetic field inside a long straight current-carrying solenoid. State the assumptions made.

Answer.

Setup. A solenoid is a long cylindrical coil with $n$ turns per unit length, carrying current $I$. For a long, tightly-wound solenoid:

• The field inside is uniform, axial, and strong.
• The field outside is negligible.

Amperian loop. Choose a rectangular loop $abcda$ with:

• $ab$ of length $L$ inside the solenoid, parallel to the axis.
• $cd$ of length $L$ outside, parallel to $ab$.
• $bc$ and $da$ perpendicular segments.

Line integral.

• Along $ab$: $\int \vec{B}\cdot d\vec{l} = BL$ (field uniform, parallel to path).
• Along $cd$: $\int \vec{B}\cdot d\vec{l} = 0$ (field outside is zero).
• Along $bc$ and $da$: $\int \vec{B}\cdot d\vec{l} = 0$ (field perpendicular to path, or zero outside).

So $\oint \vec{B}\cdot d\vec{l} = BL$.

Enclosed current. Number of turns enclosed = $nL$, each carrying current $I$, so $I_{enc} = nIL$.

Ampère's law: $\oint \vec{B}\cdot d\vec{l} = \mu_0 I_{enc}$.

$$BL = \mu_0 nIL \Rightarrow \boxed{B = \mu_0 nI.}$$

Assumptions: (i) Solenoid is long enough that end effects are negligible. (ii) Turns are tightly and uniformly wound. (iii) The field outside is treated as zero.

---

Q2. Cyclotron — full description

Describe the construction and working of a cyclotron with a labelled diagram. Derive the expression for the maximum kinetic energy of the accelerated particles. State two limitations.

Answer.

Construction.

• Two hollow semicircular conducting dees ($D_1$ and $D_2$) placed in an evacuated chamber.
• A strong uniform magnetic field $\vec{B}$ is applied perpendicular to the plane of the dees.
• An alternating high-frequency voltage is applied across the dees.
• An ion source at the centre injects positive charges.

Principle. Inside each dee, the magnetic field bends the charge into a semicircle. In the gap between the dees, the electric field (reversed at the right moment) accelerates the charge.

Working.

1. A positive ion released at the centre is accelerated across the gap by the electric field.
2. It enters $D_1$ and moves in a semicircle of radius $r_1 = mv_1/(qB)$.
3. It re-enters the gap, where the AC source has reversed polarity — so it is accelerated again.
4. Each crossing increases the speed and the radius (a spiral path).
5. The frequency of the AC voltage matches the cyclotron frequency $f_c = qB/(2\pi m)$ — the resonance condition.

Maximum kinetic energy. When the radius reaches $R$ (the dee radius), $v_{max} = qBR/m$, so

$$K_{max} = \frac{1}{2}mv_{max}^2 = \frac{q^2 B^2 R^2}{2m}.$$

$$\boxed{K_{max} = \frac{q^2 B^2 R^2}{2m}.}$$

Limitations.

• Cannot accelerate electrons — their relativistic mass increases rapidly, breaking the resonance condition.
• Cannot accelerate neutral particles (no charge to feel $\vec{B}$ or $\vec{E}$).

---

Q3. Moving coil galvanometer — full description + conversion

Describe with a diagram the construction and working of a moving coil galvanometer. Derive the expression for current sensitivity. How can it be converted into (a) an ammeter and (b) a voltmeter?

Answer.

Construction.

• A rectangular coil of $N$ turns wound on a non-magnetic frame.
• Placed in a radial magnetic field produced by cylindrical pole pieces and a soft-iron core.
• Suspended on a phosphor-bronze strip (or pivoted on jewelled bearings).
• A spring provides a restoring torque proportional to deflection.
• A mirror or pointer indicates the angular deflection.

Working. When current $I$ flows through the coil, the magnetic force produces a torque $\tau_d = NIAB$ (constant because the field is radial). The spring provides a restoring torque $\tau_r = k\phi$. At equilibrium:

$$NIAB = k\phi \Rightarrow \phi = \frac{NAB}{k}I.$$

The deflection is proportional to the current — linear scale.

Current sensitivity:

$$\boxed{S_I = \frac{\phi}{I} = \frac{NAB}{k}.}$$

Voltage sensitivity: $S_V = \phi/V = NAB/(kG)$, where $G$ is coil resistance.

(a) Conversion to ammeter. A low shunt $S$ is connected in parallel so that most of the current bypasses the galvanometer:

$$S = \frac{I_g G}{I - I_g}.$$

The combination has very low net resistance, suitable for series insertion.

(b) Conversion to voltmeter. A high resistance $R$ is connected in series so that only a tiny current $I_g$ flows for the full-scale voltage $V$:

$$R = \frac{V}{I_g} - G.$$

The combination has very high net resistance, suitable for parallel placement across the element whose voltage is to be measured.

---

Q4. $B$ on the axis of a circular loop

Derive the expression for the magnetic field on the axis of a circular current-carrying loop at a distance $x$ from its centre. Hence find the field at the centre.

Answer.

Setup. A loop of radius $R$ carries current $I$. Consider a point P on the axis at distance $x$ from the centre.

Each current element $Id\vec{l}$ on the loop is at perpendicular distance $r = \sqrt{R^2 + x^2}$ from P, and $d\vec{l}\perp\hat{r}$.

Biot-Savart magnitude:

$$dB = \frac{\mu_0}{4\pi}\frac{I\,dl}{r^2} = \frac{\mu_0}{4\pi}\frac{I\,dl}{R^2 + x^2}.$$

Symmetry argument. $d\vec{B}$ has components along the axis ($dB_x$) and perpendicular to it ($dB_\perp$). When we integrate around the loop, the perpendicular components cancel pairwise; only $dB_x$ survives.

$dB_x = dB\sin\alpha$ where $\sin\alpha = R/\sqrt{R^2 + x^2}$ (geometry from the element to P).

$$dB_x = \frac{\mu_0}{4\pi}\frac{I\,dl}{R^2 + x^2}\cdot\frac{R}{\sqrt{R^2 + x^2}} = \frac{\mu_0 I R\, dl}{4\pi (R^2 + x^2)^{3/2}}.$$

Integrate around the loop: $\int dl = 2\pi R$.

$$\boxed{B_x = \frac{\mu_0 I R^2}{2(R^2 + x^2)^{3/2}}.}$$

Direction along the axis, given by the right-hand curl rule.

Special case — centre of the loop: Set $x = 0$:

$$B_{centre} = \frac{\mu_0 I R^2}{2R^3} = \frac{\mu_0 I}{2R}.$$

---

Q5. Trajectory of a charged particle in $\vec{B}$ — three cases

Discuss the motion of a charged particle in a uniform magnetic field for all three cases: (a) $\vec{v}\parallel\vec{B}$, (b) $\vec{v}\perp\vec{B}$, (c) $\vec{v}$ at an arbitrary angle to $\vec{B}$. Derive relevant expressions.

Answer.

Case (a): $\vec{v}\parallel\vec{B}$.

$\vec{F} = q\vec{v}\times\vec{B} = 0$. The particle continues in a straight line at constant speed. No deflection.

Case (b): $\vec{v}\perp\vec{B}$.

The magnetic force, of magnitude $qvB$, is always perpendicular to $\vec{v}$ — providing the centripetal force for uniform circular motion:

$$qvB = \frac{mv^2}{r} \Rightarrow r = \frac{mv}{qB}.$$

Period:

$$T = \frac{2\pi r}{v} = \frac{2\pi m}{qB},$$

independent of $v$. The particle moves in a circle in the plane perpendicular to $\vec{B}$.

Case (c): $\vec{v}$ at angle $\theta$ to $\vec{B}$.

Decompose: $v_\parallel = v\cos\theta$ (along $\vec{B}$), $v_\perp = v\sin\theta$ (perpendicular to $\vec{B}$).

• $v_\parallel$ is unaffected (Case a applies to this component) — uniform translation along $\vec{B}$.
• $v_\perp$ causes circular motion (Case b) with $r = mv\sin\theta/(qB)$.

The combination is a helix, with:

$$\boxed{r = \frac{mv\sin\theta}{qB},\quad T = \frac{2\pi m}{qB},\quad \text{pitch}\ p = v\cos\theta\cdot T = \frac{2\pi m v\cos\theta}{qB}.}$$

The helix winds around magnetic field lines — the principle behind the Earth's auroras (charged solar-wind particles spiral along field lines and funnel into the polar atmosphere).