Board Previous Year Questions

How to Use This Section

These are CBSE Board-pattern questions on Magnetism and Matter, organised by mark value, with model answers phrased the way examiners reward — definition first, formula stated before use, units carried, and diagrams described where required. The chapter reliably contributes 3-5 marks; the favourites are Gauss's law for magnetism, dipole torque/energy, the B-H-M relations, and the dia/para/ferro comparison.

1-Mark Questions (Definitions & Direct)

Q1. State Gauss's law for magnetism. Answer: The net magnetic flux through any closed surface is zero: $\sum \vec{B}\cdot\Delta\vec{S} = 0$. It reflects the non-existence of magnetic monopoles.

Q2. Define magnetisation of a sample. Give its SI unit. Answer: Magnetisation is the net magnetic moment per unit volume: $M = m_{net}/V$. SI unit: A m$^{-1}$.

Q3. Define magnetic susceptibility. Is it dimensionless? Answer: Susceptibility $\chi$ is the ratio of magnetisation to magnetic intensity, $\chi = M/H$. Yes — it is a pure number (dimensionless).

Q4. Write the relation between relative permeability and magnetic susceptibility. Answer: $\mu_r = 1 + \chi$.

Q5. What happens when a bar magnet is cut into two pieces transverse to its length? Answer: We get two smaller magnets, each with both N and S poles. For a uniform magnet cut into two equal shorter pieces, each piece has roughly half the original magnetic moment; isolated poles are never obtained.

Q6. In which orientation is a magnetic dipole in unstable equilibrium in a uniform field? Answer: With $\vec{m}$ antiparallel to $\vec{B}$ ($\theta = 180^{\circ}$): torque is zero but energy is maximum ($+mB$).

2-Mark Questions (Short Answer)

Q7. Show that the potential energy of a magnetic dipole in a uniform field is $U = -\vec{m}\cdot\vec{B}$. Answer: Work done by an external agent against the restoring torque in turning the dipole slowly through $d\theta$ is $dW = mB\sin\theta\,d\theta$. Hence the potential energy is $U = \int mB\sin\theta\,d\theta = -mB\cos\theta + C$. Choosing $U = 0$ at $\theta = 90^{\circ}$ gives $C = 0$, so $U = -mB\cos\theta = -\vec{m}\cdot\vec{B}$.

Q8. Why can two magnetic field lines never intersect? Why must they form closed loops? Answer: (i) The tangent to a field line gives the direction of $\vec{B}$; intersection would give two directions at one point — impossible. (ii) Since monopoles do not exist, magnetic field lines have no points to begin or end on; they must therefore be continuous closed loops, running from S to N inside a magnet and from N to S outside it.

Q9. Distinguish between diamagnetic and paramagnetic substances on the basis of (i) susceptibility, (ii) behaviour in a non-uniform field. Answer: (i) Diamagnetic: $\chi$ is negative, usually small for ordinary materials ($-1 \le \chi < 0$, with $\chi=-1$ for an ideal superconductor); paramagnetic: $\chi$ is small and positive. (ii) Diamagnetic substances move from stronger to weaker field (weakly repelled); paramagnetic substances move from weaker to stronger field (weakly attracted).

Q10. Define magnetic intensity $\vec{H}$. How is the total field inside a material written in terms of $\vec{H}$ and $\vec{M}$? Answer: $\vec{H} = \vec{B}/\mu_0 - \vec{M}$; it represents the contribution of external or free currents, with unit A/m. The total magnetic field inside the material is written as $\vec{B} = \mu_0(\vec{H} + \vec{M})$.

Q11. What is the Meissner effect? Write the values of $\chi$ and $\mu_r$ for a superconductor. Answer: A superconductor expels magnetic field lines from its interior, so $B = 0$ inside the bulk for the superconducting state — perfect diamagnetism, called the Meissner effect. For an ideal superconductor, $\chi = -1$ and $\mu_r = 0$. This explains magnetic levitation demonstrations with superconductors.

3-Mark Questions (Application & Numericals)

Q12. A solenoid of 1000 turns per metre carries a current of 2.0 A. Its core has relative permeability 400. Find $H$, $B$ and $M$ inside the core. Answer:

1. $H = nI = 1000 \times 2.0 = 2 \times 10^3$ A/m (independent of the core).
2. $B = \mu_r\mu_0 H = 400 \times 4\pi\times10^{-7} \times 2\times10^3 \approx 1.0$ T.
3. $M = (\mu_r - 1)H = 399 \times 2\times10^3 \approx 8 \times 10^5$ A/m.

Q13. A bar magnet of magnetic moment 0.32 J/T is placed in a uniform field of 0.15 T. Find the torque when the magnet is at $30^{\circ}$ to the field, and the work needed to turn it from the stable position to this orientation. Answer:

1. $\tau = mB\sin\theta = 0.32 \times 0.15 \times 0.5 = 0.024$ N m.
2. $W = \Delta U = mB(1 - \cos 30^{\circ}) = 0.048(1 - 0.866) \approx 6.4 \times 10^{-3}$ J.

Q14. Write the axial and equatorial fields of a short bar magnet and compare their magnitudes and directions. Answer: Axial: $\vec{B}_A = \frac{\mu_0}{4\pi}\frac{2\vec{m}}{r^3}$, parallel to $\vec{m}$. Equatorial: $\vec{B}_E = -\frac{\mu_0}{4\pi}\frac{\vec{m}}{r^3}$, antiparallel to $\vec{m}$. At the same distance, $|B_A| = 2|B_E|$; both fall as $1/r^3$.

Q15. Explain the formation of domains in a ferromagnet and what happens to them in an external field. Answer: Atomic dipole moments interact cooperatively and spontaneously align over regions called domains (of the order of about 1 mm in size, containing a very large number of atoms). With no external field, the domain magnetisations are randomly oriented, so the bulk sample may show no net magnetisation. In an external field $B_0$, domains aligned with the field grow and other domains tend to reorient along it, producing strong magnetisation.

5-Mark Questions (Long Answer)

Q16. (a) Derive the expression for the torque on a magnetic dipole in a uniform magnetic field, and obtain its potential energy. (b) Discuss the equilibrium orientations. (c) A magnet of moment 0.6 A m$^2$ in a 0.2 T field is rotated from alignment through $180^{\circ}$; find the work done. Answer:

1. (a) Each pole experiences equal and opposite forces in a uniform field — net force zero, but a couple acts. The moment of the couple is $\tau = mB\sin\theta$, i.e. $\vec{\tau} = \vec{m}\times\vec{B}$, directed so as to rotate the dipole towards alignment.
2. Energy: $U = \int mB\sin\theta\,d\theta = -mB\cos\theta = -\vec{m}\cdot\vec{B}$, choosing zero potential energy at $\theta = 90^{\circ}$.
3. (b) $\theta = 0$: $U = -mB$, stable equilibrium; $\theta = 180^{\circ}$: $U = +mB$, unstable equilibrium. Both have zero torque.
4. (c) $W = \Delta U = mB(\cos 0^{\circ} - \cos 180^{\circ}) = 2mB = 2 \times 0.6 \times 0.2 = 0.24$ J.

Q17. (a) Define magnetisation, magnetic intensity and susceptibility, giving SI units. (b) Derive $B = \mu_0\mu_r H$ from $B = \mu_0(H+M)$. (c) Classify materials on the basis of $\chi$ with one example each. Answer:

1. (a) $M = m_{net}/V$ (A/m); $H = B/\mu_0 - M$ (A/m); $\chi = M/H$ (dimensionless).
2. (b) For linear materials, $M = \chi H$, so $B = \mu_0(H + \chi H) = \mu_0(1+\chi)H = \mu_0\mu_r H = \mu H$, where $\mu_r = 1+\chi$ and $\mu = \mu_0\mu_r$.
3. (c) Diamagnetic: $\chi < 0$, e.g. bismuth; paramagnetic: small positive $\chi$, e.g. aluminium; ferromagnetic: very large positive $\chi$, e.g. iron.