Q.In a permanent magnet at room temperature
Concept understanding — Magnetic Materials Magnetization
From a Paperclip to a Magnet: The Intuition
You already know that a magnet can pick up iron nails. But what is actually happening inside that nail when it gets near the magnet? And why does a plastic comb, rubbed on hair, pick up tiny bits of paper — but never iron filings?
The answer lies in magnetization — the process by which a material becomes magnetic.
Think of a piece of iron as a chaotic crowd of tiny compass needles. Each needle is an atomic magnetic moment (a tiny magnet, arising from the spin of electrons). In unmagnetized iron, these needles point in random directions. Their magnetic effects cancel out, so the iron as a whole shows no net magnetism.
Now bring a strong magnet close. Its magnetic field acts like a command: "Line up!" The tiny compass needles inside the iron start rotating, aligning themselves with the external field. The more they align, the stronger the iron's own magnetic field becomes. This alignment is magnetization.
Magnetization is not the same as inducing a current. It is a purely magnetic reorientation of atomic dipoles inside a material.
The Precise Definition
Magnetization (M) is the net magnetic dipole moment per unit volume of a material. It tells you how strongly a material is magnetized — how many tiny atomic magnets are aligned, and in which direction.
If a material has N atoms per unit volume, each with an average magnetic moment μavg, then:
M=Nμavg
The SI unit of M is amperes per metre (A/m). Why? Because a magnetic dipole moment has units of A·m², and dividing by volume (m³) gives A/m.
M=volumetotal magnetic dipole moment
How Magnetization Connects to the Magnetic Field
When a material gets magnetized, it produces its own magnetic field. The total magnetic field B inside the material is the sum of:
- The external applied field H (caused by free currents, like the current in a solenoid)
- The material's response — the magnetization M
The fundamental relation is:
B=μ0(H+M)
where μ0=4π×10−7T⋅m/A is the permeability of free space.
Do not confuse H (magnetic field intensity, or "magnetizing field") with B (magnetic flux density). H is what you apply; M is what the material does; B is the total field you measure.
The Three Kinds of Magnetic Materials
Not all materials respond the same way to an external field. The magnetization M is proportional to H for most materials (at least for small fields):
M=χmH
where χm is the magnetic susceptibility — a dimensionless number that tells you how easily a material magnetizes.
| Material Type | χm | Behaviour | Example |
|---|---|---|---|
| Diamagnetic | Small and negative (≈−10−5) | Weakly repelled by a magnet; M opposes H | Water, copper, bismuth |
| Paramagnetic | Small and positive (≈10−5 to 10−3) | Weakly attracted; M aligns with H | Aluminium, oxygen gas |
| Ferromagnetic | Large and positive (≫1) | Strongly attracted; M can be huge and persists even after H is removed | Iron, nickel, cobalt |
Ferromagnetic materials have spontaneous magnetization — their atomic moments align even without an external field, forming magnetic domains. Magnetization in these materials is not linear; it saturates and shows hysteresis.
A Simple Example
Take a long iron rod placed inside a solenoid carrying current I. The solenoid produces a uniform field H inside. The iron rod becomes magnetized: its atomic moments align, producing M in the same direction as H.
If χm for iron is about 5000, then M=5000H. The total field inside the rod becomes:
B=μ0(H+5000H)=μ0(5001)H
That is why an iron core can amplify the magnetic field of a solenoid by thousands of times.
The Bottom Line
Magnetization is the measure of how much a material becomes magnetic when placed in an external field. It arises from the alignment of atomic magnetic dipoles. For linear materials, M=χmH. For ferromagnets, the response is nonlinear, strong, and can be permanent — that is how you get a bar magnet from a piece of iron.
Magnetization and the classification of materials as diamagnetic, paramagnetic and ferromagnetic is a core topic of the NCERT Class 12 Physics chapter on magnetism and matter, tested regularly in CBSE boards and JEE Main. Students searching "diamagnetic paramagnetic ferromagnetic materials class 12 physics difference" will find this magnetic-susceptibility-based comparison matches the standard NCERT table.
Why this formula?
Magnetic Materials & Magnetization: Why the Key Formulas Hold
Let's build this from the ground up — starting with what magnetization physically means, then deriving the formulas step by step.
1. What is Magnetization (M)?
Magnetization is the net magnetic dipole moment per unit volume of a material.
- Inside a material, atoms act like tiny magnetic dipoles (due to electron spin and orbital motion).
- Without an external field, these dipoles point randomly → net M=0.
- When an external field H is applied, dipoles align partially → net M=0.
Definition:
M=volumenet magnetic dipole moment
Units: A/m (same as H).
2. The Fundamental Relation: B=μ0(H+M)
This is the master equation linking the three magnetic fields:
- B = magnetic flux density (the total field inside the material)
- H = applied magnetic field (due to free currents)
- M = magnetization (response of the material)
- μ0 = permeability of free space (4π×10−7 H/m)
Why this form?
Step 1: In vacuum, there is no material, so M=0. Then:
B=μ0H
Step 2: Inside a material, the dipoles themselves produce an additional field. The total B is the sum of:
- The field due to free currents (μ0H)
- The field due to bound currents (from aligned dipoles), which is μ0M
Hence:
B=μ0H+μ0M=μ0(H+M)
Key insight: M is not an independent field — it's the material's response to H.
3. Magnetic Susceptibility (χm) and Permeability (μ)
For linear, isotropic, homogeneous materials (most common in exams), magnetization is proportional to the applied field:
M=χmH
- χm = magnetic susceptibility (dimensionless)
- χm>0 for paramagnetic materials
- χm<0 for diamagnetic materials
- χm≫1 for ferromagnetic materials (but not linear!)
Derivation of relative permeability μr:
Substitute M=χmH into the master equation:
B=μ0(H+χmH)=μ0(1+χm)H
Define:
μr=1+χm(relative permeability)
μ=μ0μr(absolute permeability)
Thus:
B=μH
Why this matters: It shows that the material simply scales the applied field by a factor μr.
4. Why χm Has Different Signs (Physical Reasoning)
| Material Type | χm | Why? |
|---|---|---|
| Diamagnetic | χm<0 (small, ~10−5) | Applied field induces opposing dipole moments (Lenz's law at atomic level). M opposes H. |
| Paramagnetic | χm>0 (small, ~10−3) | Permanent atomic dipoles align partially with H. Thermal agitation fights alignment. |
| Ferromagnetic | χm≫1 (nonlinear) | Strong quantum-mechanical exchange coupling aligns dipoles spontaneously even without H. |
5. The Curie Law for Paramagnets (Temperature Dependence)
For paramagnetic materials, susceptibility depends on temperature:
χm=TC
where C is the Curie constant.
Why?
- Thermal energy (kBT) randomizes dipole alignment.
- Applied field H tries to align them.
- The competition leads to M∝TH.
From M=χmH, we get χm∝1/T.
Exam tip: Curie law holds for high temperatures and low fields. At very low T, saturation occurs.
6. Summary of Key Formulas (with "why")
| Formula | Why it holds |
|---|---|
| B=μ0(H+M) | Total field = free-current field + bound-current field |
| M=χmH | Linear response approximation (for small fields) |
| μr=1+χm | Direct substitution into B=μ0μrH |
| χm=C/T (Curie law) | Thermal agitation vs. field alignment |
Final takeaway: Magnetization is the material's voice — it tells you how the internal dipoles respond to an external magnetic nudge. The formulas are just a mathematical translation of that physical conversation.
A permanent magnet is a ferromagnet, so each molecule already carries a non-zero magnetic moment (a wrong). Its magnetism comes from domains — regions of aligned moments. In a real permanent magnet at room temperature the domains are only partially aligned (thermal agitation and pinning prevent perfect saturation), giving a strong but sub-saturation net moment. Neither the individual molecular moments nor the domains are perfectly aligned (b and d wrong).
Correct option: (c) domains are partially aligned.
A permanent magnet is a ferromagnetic material whose net magnetisation comes from magnetic domains. At room temperature these domains are only partially aligned, not perfectly. Correct option: (c).
Concept understanding. In a ferromagnet the atoms/molecules carry permanent magnetic moments that couple through the exchange interaction into domains — small regions in which the moments point the same way. In an unmagnetised sample the domains point in random directions and cancel. Magnetising the material makes the domains grow/rotate toward the field, leaving a net moment when the field is removed. Room temperature (≈300 K) is far below the Curie temperature of common magnets (iron Tc≈1043 K), so a large net magnetisation survives — but thermal agitation and domain-wall pinning keep it below saturation.
Testing each option.
- (a) In a ferromagnet each molecule has a non-zero magnetic moment; that is the very origin of the effect. Wrong.
- (b) The molecular moments are not all perfectly aligned — thermal energy tilts and randomises them, and only within a domain do they roughly agree. Wrong.
- (c) The correct picture: the material's magnetisation is produced by domains that are partially aligned, giving a strong but sub-saturation moment. Correct.
- (d) Domains being all perfectly aligned would mean full saturation, which does not hold at ordinary temperature for a real permanent magnet. Wrong.
Correct option: (c) domains are partially aligned. The molecular moments are non-zero (ruling out a) but neither the moments (b) nor the domains (d) are perfectly aligned at room temperature.
Method: Reasoning About Ferromagnetic Domain Alignment
Use this elimination approach for conceptual questions about the state of magnetisation inside a permanent magnet or ferromagnetic sample.
Steps
Step 1: Recall the two-level structure of a ferromagnet
Individual atoms/molecules each carry a nonzero magnetic moment — this is what makes the material ferromagnetic in the first place, and it is never zero. These moments group into domains: regions where neighbouring moments are aligned by the exchange interaction.
Step 2: Distinguish "molecular alignment" from "domain alignment"
A claim that individual molecular moments are all "perfectly aligned" is a much stronger — and generally false — statement than a claim about domains being aligned. Thermal agitation always tilts individual moments somewhat, even within an aligned domain, so treat any "perfectly aligned molecules" option with suspicion.
Step 3: Judge the degree of domain alignment against temperature
At ordinary (room) temperature, below the Curie temperature, domains are real but only partially aligned. Full/perfect alignment (saturation) would need either a very strong external field or a temperature near absolute zero; at room temperature, thermal effects and domain-wall pinning keep the material below saturation.
Step 4 (Applying to this problem): Eliminate the zero-moment and perfect-alignment extremes
Reject any option claiming molecular moments are zero (contradicts ferromagnetism itself) or that alignment is "perfect" (contradicts realistic room-temperature behaviour). The physically correct middle ground — partial domain alignment — is what a real permanent magnet at room temperature shows.
- GUJCET 2026Set x1 markMCQQ.A closely wound solenoid of 800 turns and area of cross section 2.5×10−4 m2 can carry a current of 3.0 A. The magnetic moment associated with it is ______. (A) 60 JT−1 (B) 0.60 JT−1 (C) 6 JT−1 (D) 0.06 JT−1
›Reveal solutionSolution
m=NIA=0.60 J T−1.
Magnetic moment of a solenoid:
m=NIA=800×3.0×(2.5×10−4)=800×7.5×10−4=0.60 J T−1.
✓Final answerOption (B) 0.60 JT−1
ANSWER: (B)
- GUJCET 2024Set 131 markMCQQ.AmVs is the unit of which physical quantity? (A) χm (B) μ0 (C) χc (D) ε0
›Reveal solutionSolution
A⋅mV⋅s equals AT⋅m, which is the SI unit of the permeability of free space μ0 (H/m).
Concept. μ0 has units of henry per metre; 1H=1V⋅s/A, so μ0 is measured in A⋅mV⋅s.
Steps.
[μ0]=mH=mV⋅s/A=A⋅mV⋅s.
The magnetic susceptibility χm is dimensionless and ε0 has different units, so μ0 is correct.
✓Final answer(B) μ0
ANSWER: (B)
- GUJCET 2024Set 131 markMCQQ.A solenoid has a core of a material with relative permeability 400. The windings of the solenoid are insulated from the core and carry a current of 2 A. If the number of turns is 1000 per meter then the value of magnetic intensity will be ________. (A) 8×10−5 Am−1 (B) 2×103 Am−1 (C) 2×10−3 Am−1 (D) 8×105 Am−1
›Reveal solutionSolution
H=nI — it depends only on the current and turns per metre, not on the core.
Concept. Magnetic intensity (magnetising field) in a solenoid is H=nI, independent of the core material's permeability.
H=nI=(1000m−1)(2A)=2000=2×103Am−1.
✓Final answerOption (B) 2×103Am−1
ANSWER: (B)
- GUJCET 2023Set 091 markMCQQ.A solenoid of length 0.5 m has a radius of 1 cm and is made up of 250 turns. It carries a current of 5 A. What is the magnitude of the magnetic field inside the solenoid? (A) 3.14×10−3 T (B) 6.28×10−3 T (C) 62.8×10−3 T (D) Zero
›Reveal solutionSolution
Field inside a long solenoid is B=μ0nI; the radius is irrelevant.
Concept: For a solenoid, B=μ0nI where n is turns per unit length.
n=LN=0.5250=500 m−1.
B=(4π×10−7)(500)(5)=π×10−3≈3.14×10−3 T.
✓Final answer(A) 3.14×10−3 T
ANSWER: (A)
- GUJCET 2023Set 091 markMCQQ.A solenoid has a core of a material with relative permeability 400. The windings of solenoid are insulated from the core and carry a current of 2 A. If the number of turns is 1000 per metre, the magnetic field B inside the solenoid is ______ T. (A) 1.5 (B) 1.0 (C) 1.8 (D) 2.0
›Reveal solutionSolution
A magnetic core multiplies the solenoid field by relative permeability: B=μ0μrnI.
Concept: With a core, B=μ0μrnI.
B=(4π×10−7)(400)(1000)(2)=0.32π≈1.0 T.
✓Final answer(B) 1.0 T
ANSWER: (B)
- GUJCET 2022Set 171 markMCQQ.A solenoid of length 0.25 m has a radius of 1 cm and is made up of 500 turns. It carries a current of 2.5 A. What is the magnitude of the magnetic field inside the solenoid? (μ0=4π×10−7 SI) (A) 6.28×10−3 T (B) 6.28×10−2 T (C) 6.28×10−4 T (D) 6.28×10−1 T
›Reveal solutionSolution
n=N/l=2000 turns/m, so B=μ0nI=6.28×10−3 T (radius irrelevant).
Concept: n=0.25500=2000 turns/m.
B=μ0nI=(4π×10−7)(2000)(2.5)=2π×10−3≈6.28×10−3 T
✓Final answer(A) 6.28×10−3 T
ANSWER: (A)
- GUJCET 2022Set 171 markMCQQ.A solenoid has a core of a material with relative permeability 400. The windings of the solenoid are insulated from the core and carry a current of 1 A. If the number of turns is 1000 per metre, find magnetic field (B) ________ T. (μ0=4π×10−7 SI) (A) 1.6π×10+2 (B) 16π×102 (C) 16π×10−2 (D) 0.16π×10−2
›Reveal solutionSolution
With a magnetic core, B=μ0μrnI=16π×10−2 T.
Concept: B=μ0μrnI=(4π×10−7)(400)(1000)(1)=4π×10−7×4×105=16π×10−2 T.
✓Final answer(C) 16π×10−2
ANSWER: (C)
- GUJCET 2021Set 151 markMCQQ.A solenoid of length 0.5 m has a radius of 1 cm and is made up of 1000 turns. It carries a current of 10A. What is the magnitude of the magnetic field inside the solenoid? (A) 6.28×10−3 T (B) 2.51×10−2 T (C) 1.71×10−2 T (D) 7.23×10−3 T
›Reveal solutionSolution
B=μ0nI; radius is irrelevant for a long solenoid.
Concept:
n=0.51000=2000 m−1,B=(4π×10−7)(2000)(10)=8π×10−3≈2.51×10−2 T.
✓Final answer(B) 2.51×10−2 T
ANSWER: (B)
- GUJCET 2020Set 071 markMCQQ.The relative permeability in a core of a solenoid is 400. The windings of a solenoid are insulated from the core and carry a current of 2A. If the number of turns is 1000 per meter. Then magnetic Intensity inside the core of solenoid is ______ A/m. (A) 2.5×103 (B) 2×103 (C) 2.5×10−3 (D) 2×10−3
›Reveal solutionSolution
H=nI, independent of the core; H=1000×2=2×103 A/m.
Concept — magnetising field vs. flux density. The magnetic intensity (magnetising field) inside a solenoid depends only on the free current and turn density: H=nI, where n = turns per metre. The permeability affects B=μH, not H itself.
H=nI=(1000m−1)(2A)=2×103A/m
✓Final answer(B) 2×103 A/m
ANSWER: (B)
- GSEB Higher Secondary Certificate (HSC) Examination 2018Set ANNUAL1 markMCQQ.A toroid wound with 100 turns/m of wire carries a current of 3A. The core of toroid is made of iron having relative magnetic permeability of mu_r = 5000 under given conditions. The magnetic field inside the iron is ___.(a) 0.15 T(b) 0.47 T(c) 1.5 x 10^-2 T(d) 1.88 T
›Reveal solutionSolution
The field in the toroid core is B = mu_0 mu_r n I; substituting gives about 1.88 T.
Field inside a toroid with a magnetic core: B = mu_0 mu_r n I, where n = 100 turns/m, I = 3 A, mu_r = 5000.
B = (4pi x 10^-7)(5000)(100)(3) = (4pi x 10^-7)(1.5 x 10^6)
= 4*pi x 0.15
= 1.885 T approximately 1.88 T.
✓Final answer(d) 1.88 T.
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