Summary
NCERT Class 12 Physics Chapter 5, Magnetism and Matter, covers the properties of bar magnets, Gauss's law for magnetism, magnetisation, magnetic intensity, and the classification of materials as diamagnetic, paramagnetic, or ferromagnetic.
Chapter 5 of NCERT Class 12 Physics Part I explores magnetism as a fundamental subject. It begins with bar magnets and their field lines, showing that a bar magnet behaves like a magnetic dipole and is analogous to a current-carrying solenoid. The chapter explains Gauss's law for magnetism — the net magnetic flux through any closed surface is zero — reflecting that magnetic monopoles do not exist. It introduces magnetisation M, magnetic intensity H, magnetic susceptibility χ, and relative permeability μr. Materials are classified as diamagnetic (χ negative), paramagnetic (χ small and positive), or ferromagnetic (χ very large), with detailed discussion of domain theory and the Meissner effect in superconductors.
Key points & formulas
- 01A bar magnet acts as a magnetic dipole; its axial field is B = μ₀·2m / (4πr³) and equatorial field is B = –μ₀m / (4πr³) for r >> l.
- 02Gauss's law for magnetism states the net magnetic flux through any closed surface is zero (∮ B·dS = 0), because isolated magnetic monopoles do not exist.
- 03Magnetisation M is the net magnetic moment per unit volume; the total field in a material is B = μ₀(H + M), where H is the magnetic intensity.
- 04Magnetic susceptibility χ relates M and H by M = χH; relative permeability μr = 1 + χ and permeability μ = μ₀μr.
- 05Diamagnetic materials (e.g., bismuth, copper, water) are repelled by magnets (χ < 0); paramagnetic materials (e.g., aluminium, oxygen) are weakly attracted (χ small and positive).
- 06Ferromagnetic materials (e.g., iron, cobalt, nickel) have domain structures and very large χ; hard ferromagnets form permanent magnets while soft ferromagnets lose magnetisation when the field is removed.
Frequently asked questions
01What is the torque experienced by a bar magnet in a uniform magnetic field?
The torque on a bar magnet of magnetic moment m placed in a uniform field B is τ = m × B, with magnitude τ = mB sinθ, where θ is the angle between m and B. The corresponding potential energy is U = –m·B, with minimum energy at θ = 0° (stable equilibrium) and maximum at θ = 180° (unstable equilibrium).
02Why does cutting a bar magnet always produce two smaller magnets and never an isolated north or south pole?
Magnetic monopoles do not exist. A bar magnet is analogous to a solenoid made of circulating current loops. Cutting it — whether along or across its length — simply produces two smaller dipoles, each with its own north and south pole, because the field lines always form continuous closed loops.
03How are diamagnetic, paramagnetic, and ferromagnetic materials distinguished?
They differ in the sign and magnitude of magnetic susceptibility χ. Diamagnetic materials have –1 ≤ χ < 0 and are weakly repelled by magnets (e.g., bismuth, copper). Paramagnetic materials have small positive χ and are weakly attracted (e.g., aluminium, oxygen). Ferromagnetic materials have χ >> 1, possess spontaneously magnetised domains, and are strongly attracted to magnets (e.g., iron, cobalt, nickel).
04Is the NCERT Class 12 Physics Chapter 5 PDF free to download?
Yes, the NCERT Class 12 Physics Part I Chapter 5 PDF is completely free to download on cbseprepmaster.com.
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This is the complete Physics Part I Chapter 5 as published by NCERT — every diagram, solved example, and exercise included, free. Browse all CBSE Class 12 textbooks.
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