🎓 Class 12PhysicsCBSETheoryCh 5 – Magnetism and Matter⏱ ~14 min
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આ MCQ મોડ્યુલ આના પર આધારિત છે: Magnetisation Magnetic Intensity
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આ મૂલ્યાંકન આના પર આધારિત હશે: Magnetisation Magnetic Intensity
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Magnetisation Magnetic Intensity
5.5 Magnetisation and Magnetic Intensity
Every macroscopic chunk of matter contains an enormous number of atomic magnetic dipoles - electron orbits and electron spins. Normally these dipoles point in random directions and cancel out. When an external magnetic field is applied, the dipoles tend to align (or, in some materials, anti-align), and the substance acquires a net magnetic moment. We define the magnetisation as the dipole moment per unit volume.
\[\vec M = \dfrac{\vec m_{net}}{V}\]
Inside a magnetised material the total field B has two contributions: the externally applied field B0, and the field produced by the aligned atomic dipoles themselves. To handle both cleanly we introduce the magnetic intensity H by
\[\vec B = \mu_0(\vec H + \vec M)\]
Equivalently \(\vec H = \vec B/\mu_0 - \vec M\). H has the same units as M (A/m). Physically: H is set by the free currents (e.g. the wire winding of a solenoid); M is the response of the material; B = μ\(_0\)(H + M) is the total field actually felt at a point.
5.5.1 Magnetic susceptibility
For most ordinary (linear, isotropic) materials, M is proportional to the applied H:
\[\vec M = \chi\,\vec H\]
The dimensionless constant χ is called the magnetic susceptibility. Its sign and magnitude classify the material:
χ < 0 (small, negative) ⇒ diamagnetic
χ > 0 (small, positive) ⇒ paramagnetic
χ ≫ 0 (very large, positive, often nonlinear) ⇒ ferromagnetic
5.5.2 Permeability and relative permeability
Combining B = μ\(_0\)(H + M) with M = χH:
\[\vec B = \mu_0(1 + \chi)\,\vec H = \mu\,\vec H\]
where μ = μ\(_0\)(1 + χ) is the permeability of the material. Define the dimensionless relative permeability:
\[\mu_r = \mu/\mu_0 = 1 + \chi\]
μ\(_r\) measures by how much the material amplifies (or weakens) the magnetic field compared with vacuum.
Quick comparison with electrostatics:
Electric
Magnetic
P (polarisation)
M (magnetisation)
D = ε₀E + P
H = B/μ₀ - M
P = ε₀ χe E
M = χ H
K = 1 + χe (dielectric const.)
μr = 1 + χ (relative permeability)
Fig 5.6 Solenoid with a material core. The free current produces H; the core acquires magnetisation M; total field B = μ₀(H + M).
Worked Example 5.5 - From χ to B in a solenoid core
Example 5.5 L3 Apply
A solenoid has n = 1000 turns/m carrying I = 2.0 A. Find H, then B inside the solenoid (i) in air (χ ≈ 0), (ii) with an aluminium core (χ ≈ 2.3 × 10⁻⁵), (iii) with a soft-iron core (μr ≈ 4000).
H = nI = 1000 × 2 = 2000 A/m (set by the free current alone).
(i) Air: μr ≈ 1; B = μ₀H = (4π × 10⁻⁷)(2000) ≈ 2.51 × 10⁻³ T.
(ii) Aluminium: μr = 1 + 2.3×10⁻⁵; B is barely changed (paramagnetic enhancement of ~0.002%).
(iii) Soft iron: B = μ₀ μr H = 4000 × 2.51 × 10⁻³ ≈ 10 T - a thousand-fold amplification, the basis of every electromagnet.
Worked Example 5.6 - Magnetisation from B
Example 5.6 L3 Apply
Inside a paramagnetic rod, B = 1.0 mT and H = 750 A/m. Find M and χ.
From B = μ₀(H + M) ⇒ M = B/μ₀ - H.
B/μ₀ = 1.0 × 10⁻³/(4π × 10⁻⁷) ≈ 795.8 A/m.
M = 795.8 - 750 ≈ 45.8 A/m. χ = M/H ≈ 0.061.
(That's a fairly strong paramagnet - real values for materials like aluminium are ~10⁻⁵.)
5.5.3 Susceptibility values
Typical susceptibilities at room temperature:
Material
Type
χ (dimensionless)
μr
Bismuth
Diamagnetic
−1.7 × 10⁻⁵
≈ 0.99998
Copper
Diamagnetic
−9.8 × 10⁻⁶
≈ 0.99999
Water
Diamagnetic
−9.0 × 10⁻⁶
≈ 0.99999
Aluminium
Paramagnetic
+2.3 × 10⁻⁵
≈ 1.00002
Liquid oxygen (90 K)
Paramagnetic
+3.5 × 10⁻³
1.00350
Iron (soft)
Ferromagnetic
~ 5500
~ 5500
Mu-metal
Ferromagnetic
~ 100 000
~ 100 000
Interactive: Predict B from H and χ L3 Apply
Adjust H (set by your solenoid) and χ (set by your core material) and see how the resulting B and M change.
M = 2.0 A/m | B = 2.51 mT | μr = 1.001
Tip: try χ = 5000 to see iron-like amplification of B. The same H, but with iron, gives B in tens of teslas.
Activity 5.3 - Iron core dramatically boosts a solenoidL4 Analyse
Wind ~50 turns of insulated copper wire around a hollow plastic tube. Connect to a 1.5 V cell.
With air inside, hold a paper clip just below the tube end - count the maximum number of clips picked up.
Now slip an iron nail into the tube and repeat.
Predict: by what factor does the number of clips lifted change when the iron core is inserted?
The number of clips can jump 100× to 1000×. Reason: H is unchanged (set by NI), but the iron core boosts B by a factor μr. Since the lifting force grows as B², a 100× increase in B means a 10000× increase in force.
Competency-Based Questions L1-L6
A long solenoid of n = 800 turns/m carries 1.5 A. Inside the solenoid, a paramagnetic sample with χ = 5.0 × 10⁻⁴ is placed.
1. The magnetic intensity H inside is: L3 Apply
(a) 600 A/m
(b) 1200 A/m
(c) 2400 A/m
(d) 5000 A/m
(b) H = nI = 800 × 1.5 = 1200 A/m.
2. The magnetisation M of the sample equals: L3 Apply
(a) 0.30 A/m
(b) 0.60 A/m
(c) 1200 A/m
(d) 2.4 × 10⁵ A/m
(b) M = χH = 5.0 × 10⁻⁴ × 1200 = 0.60 A/m.
3. Define magnetic susceptibility and state how it can be used to classify materials. L1 Remember
χ = M/H is the ratio of induced magnetisation to applied intensity. χ < 0 → diamagnetic; small +χ → paramagnetic; large +χ → ferromagnetic.
4. Show that B = μ₀(1 + χ)H and hence μr = 1 + χ. L4 Analyse
From definitions: B = μ₀(H + M). For a linear medium M = χH, so B = μ₀(H + χH) = μ₀(1 + χ)H. Define μ = μ₀(1 + χ) ⇒ μr = μ/μ₀ = 1 + χ.
5. A transformer designer needs a core with μr > 5000 and very low losses. Suggest two material choices and the trade-off involved. L6 Create
Sample answer: (a) Grain-oriented silicon steel - μr ~5000-10000, low hysteresis loss, but limited to power-line frequencies (50/60 Hz). (b) Ferrite (Mn-Zn or Ni-Zn) - μr ~1000-5000, lower saturation B but extremely low losses up to MHz, ideal for high-frequency switching transformers. Trade-off: higher μr usually means higher Bsat but more eddy-current/hysteresis loss; ferrites resolve eddy losses with high resistivity at the cost of lower saturation.
Assertion-Reason Pairs L4 Analyse
Options: (A) Both true, R correct explanation. (B) Both true, R not the explanation. (C) A true, R false. (D) A false, R true.
Assertion: The relative permeability of a vacuum is exactly 1.
Reason: Vacuum has no atoms, so χ = 0 and μr = 1 + χ = 1.
(A). True; the reason correctly explains the assertion.
Assertion: Inserting a soft-iron core inside a solenoid greatly increases B but leaves H unchanged.
Reason: H is determined only by the free current and geometry; B includes the contribution of the magnetised material.
(A). The reason directly explains the assertion.
Assertion: Magnetisation M and intensity H have different SI units.
Reason: M is dipole moment per unit volume.
(D). Both are vectors with the same SI unit (A/m). So the assertion is false; the reason about M is correct.
Frequently Asked Questions - Magnetisation Magnetic Intensity
What is the main concept covered in Magnetisation Magnetic Intensity?
In NCERT Class 12 Physics Chapter 5 (Magnetism and Matter), "Magnetisation Magnetic Intensity" covers the core principles and equations students need for board exam success. The MyAiSchool lesson explains the topic with definitions, derivations, worked examples, and interactive simulations. Key formulas and dimensional analysis are included to build conceptual depth and problem-solving skills aligned with the CBSE 2025-26 syllabus.
How is Magnetisation Magnetic Intensity useful in real-life applications?
Real-life applications of "Magnetisation Magnetic Intensity" from NCERT Class 12 Physics Chapter 5 include electronics, communication systems, medical imaging, solar energy, semiconductor devices, and modern technology. The MyAiSchool lesson links every concept to a tangible example so students see physics as a problem-solving framework for the physical world, not as abstract formulas.
What are the key formulas in Magnetisation Magnetic Intensity?
Key formulas in "Magnetisation Magnetic Intensity" (NCERT Class 12 Physics Chapter 5 Magnetism and Matter) are derived step-by-step in the MyAiSchool lesson. Students should memorize the final formula AND understand its derivation for full board marks. Each formula is listed with its dimensional formula, SI unit, applicability range, and common pitfalls. The Summary section at the end of each part includes a quick-reference formula card.
How does this part connect to other parts of Chapter 5?
NCERT Class 12 Physics Chapter 5 (Magnetism and Matter) is structured so each part builds on the previous one. "Magnetisation Magnetic Intensity" connects directly to neighbouring parts via shared definitions, units, and methodology. The MyAiSchool lesson cross-references related concepts with internal links so students can navigate the whole chapter as one connected story rather than disconnected fragments.
What types of CBSE board questions come from Magnetisation Magnetic Intensity?
CBSE board questions from "Magnetisation Magnetic Intensity" typically include: (1) 1-mark MCQs on definitions and formulas, (2) 2-mark short-answer derivations or applications, (3) 3-mark numerical problems with units, (4) 5-mark long-answer derivations followed by application. The MyAiSchool lesson tags each Competency-Based Question (CBQ) with Bloom level (L1-L6) so students know how to study for each weight.
How can students use the interactive simulation effectively?
The interactive simulation in the "Magnetisation Magnetic Intensity" lesson allows students to adjust input parameters (sliders or selectors) and see physical quantities update in real time. To use it effectively: (1) try extreme values to understand limiting cases, (2) compare with the analytical formula, (3) check unit consistency, (4) test special configurations from worked examples. The simulation reinforces conceptual intuition that pure formula manipulation cannot.
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Physics Class 12 Part I – NCERT (2025-26)
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