CBSE Weightage:

Class 12 — Magnetism and Matter

Class 12 — Magnetism and Matter — chapter strategy, formulas, PYQs, and traps

5 min read

Chapter Overview & Weightage

Magnetism and Matter is one of the smaller chapters in Class 12 Physics — typically 4 to 5 marks on the CBSE board paper. It builds on Magnetic Effects of Current and lays the groundwork for Electromagnetic Induction. The questions are mostly conceptual or formula-direct, making this a high-yield chapter for marks-per-effort.

The chapter has three big themes: bar-magnet behaviour, Earth’s magnetic field, and classification of magnetic materials (dia-, para-, ferromagnetic).

YearCBSE Weightage
20245 marks
20234 marks
20224 marks
20215 marks
20204 marks

Key Concepts You Must Know

  • A bar magnet behaves like a magnetic dipole with dipole moment m=m2L\vec{m} = m \cdot 2L pointing from S to N pole.
  • The magnetic field of a bar magnet at a point on its axis and at its equatorial plane has different formulas — both must be memorised.
  • Earth’s magnetic field has three components: declination (DD), dip angle (θ\theta), and horizontal component (BHB_H).
  • Materials are classified as diamagnetic, paramagnetic, or ferromagnetic based on relative permeability.
  • Curie’s law: for paramagnetic substances, susceptibility decreases with temperature: χ1/T\chi \propto 1/T.
  • Above the Curie temperature, ferromagnets become paramagnetic.

Important Formulas

Baxis=μ04π2mr3B_{\text{axis}} = \frac{\mu_0}{4\pi} \cdot \frac{2m}{r^3}

For a point on the axis at distance rLr \gg L from the centre.

Beq=μ04πmr3B_{\text{eq}} = \frac{\mu_0}{4\pi} \cdot \frac{m}{r^3}

The axial field is twice the equatorial field at the same distance — a classic 1-mark question.

τ=m×B,τ=mBsinθ\vec{\tau} = \vec{m} \times \vec{B}, \quad |\tau| = mB\sin\theta

When the dipole is parallel to the field, torque is zero (stable equilibrium for θ=0\theta = 0).

U=mB=mBcosθU = -\vec{m} \cdot \vec{B} = -mB\cos\theta

Minimum at θ=0\theta = 0 (most stable), maximum at θ=π\theta = \pi.

BH=Bcosθ,BV=Bsinθ,tanθ=BVBHB_H = B\cos\theta, \quad B_V = B\sin\theta, \quad \tan\theta = \frac{B_V}{B_H}

where θ\theta is the angle of dip.

χ=CT\chi = \frac{C}{T}

where CC is the Curie constant. Ferromagnets above Curie point: χ=C/(TTc)\chi = C/(T - T_c).

Solved Previous Year Questions

PYQ 1 (CBSE 2023, 3 marks)

A short bar magnet of magnetic moment 0.5A m20.5\,\text{A m}^2 is placed with its axis at 30°30° to a uniform magnetic field of magnitude 0.16T0.16\,\text{T}. Find the magnitude of torque.

Solution:

τ=mBsinθ=0.5×0.16×sin30°=0.5×0.16×0.5=0.04Nm\tau = mB\sin\theta = 0.5 \times 0.16 \times \sin 30° = 0.5 \times 0.16 \times 0.5 = 0.04\,\text{Nm}

PYQ 2 (CBSE 2022, 2 marks)

Distinguish between diamagnetic, paramagnetic, and ferromagnetic substances with one example each.

Solution:

TypeSusceptibility (χ\chi)Relative Permeability (μr\mu_r)Example
DiamagneticSmall, negativeSlightly less than 1Bismuth, copper
ParamagneticSmall, positiveSlightly greater than 1Aluminium, oxygen
FerromagneticLarge, positiveMuch greater than 1Iron, nickel, cobalt

PYQ 3 (CBSE 2024, 5 marks)

(a) State Curie’s law. (b) The susceptibility of a paramagnetic material at 300K300\,\text{K} is 1.2×1041.2 \times 10^{-4}. Find its susceptibility at 200K200\,\text{K}.

Solution:

(a) Curie’s law states that the magnetic susceptibility of a paramagnetic substance is inversely proportional to its absolute temperature: χ=C/T\chi = C/T.

(b) Since χT=\chi T = constant:

χ2=χ1T1T2=1.2×104×300200=1.8×104\chi_2 = \chi_1 \cdot \frac{T_1}{T_2} = 1.2 \times 10^{-4} \times \frac{300}{200} = 1.8 \times 10^{-4}

Difficulty Distribution

  • Easy (definitions and direct formula): 2 marks — almost guaranteed.
  • Medium (numerical with one substitution): 2-3 marks.
  • Hard (multi-step or matter-classification reasoning): rare in CBSE, common in coaching tests.

This chapter is one of the safest places to score because the formulas are few and the concepts are crisp.

Expert Strategy

CBSE often pairs a numerical with a definition in the same question. Always answer the definition fully — examiners reward complete sentences with technical terms over one-word labels.

Remember the factor of 2 between axial and equatorial fields. JEE Main and CBSE both test this every year. Mnemonic: “Axis is doubled, equator is single.”

For dip-angle problems, draw the right triangle with BHB_H horizontal, BVB_V vertical, and BB as hypotenuse. Geometry settles every confusion.

Common Traps

Trap 1 — Confusing axial and equatorial formulas. Axis: 2μ0m4πr3\frac{2\mu_0 m}{4\pi r^3}. Equatorial: μ0m4πr3\frac{\mu_0 m}{4\pi r^3}. The factor of 2 disappears on the equator.

Trap 2 — Using degrees in radians or vice versa. sin30°=0.5\sin 30° = 0.5, sin(30)=0.988\sin(30) = -0.988 if your calculator is in radians. Always check the mode.

Trap 3 — Mis-identifying the field direction at the equator. On the equatorial plane, the field is opposite to the dipole moment. On the axis, it is parallel.

Trap 4 — Confusing magnetic moment mm with pole strength mm. Both are sometimes called mm. Magnetic moment has units A m². Pole strength has units A m. They are related by m=mpole2L\vec{m} = m_{\text{pole}} \cdot 2L.

Trap 5 — Forgetting that susceptibility is dimensionless. χ\chi has no units. If your answer comes out with units, recheck.