For questions 13 to 16, two statements are given – one — CBSE Class 12 Physics 2026 PYQ

Question

Q. For questions 13 to 16, two statements are given - one labelled Assertion (A) and the other labelled Reason (R). Select the correct answer from the codes (A), (B), (C) and (D) below: (A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A). (B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A). (C) Assertion (A) is true, but Reason (R) is false. (D) Both Assertion (A) and Reason (R) are false. Assertion (A) : All atoms have a net magnetic moment. Reason (R) : A current loop does not always behave as a magnetic dipole.

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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.

Note

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 ( $\vec{M}$ ) 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 $\vec{\mu}_{\text{avg}}$ , then:

\vec{M} = N \vec{\mu}_{\text{avg}}

The SI unit of $\vec{M}$ 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.

\vec{M} = \frac{\text{total magnetic dipole moment}}{\text{volume}}

How Magnetization Connects to the Magnetic Field

When a material gets magnetized, it produces its own magnetic field. The total magnetic field $\vec{B}$ inside the material is the sum of:

The external applied field $\vec{H}$ (caused by free currents, like the current in a solenoid)
The material's response — the magnetization $\vec{M}$

The fundamental relation is:

\vec{B} = \mu_0 (\vec{H} + \vec{M})

where $\mu_0 = 4\pi \times 10^{-7} \, \text{T·m/A}$ is the permeability of free space.

Watch out

Do not confuse $\vec{H}$ (magnetic field intensity, or "magnetizing field") with $\vec{B}$ (magnetic flux density). $\vec{H}$ is what you apply; $\vec{M}$ is what the material does; $\vec{B}$ 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 $\vec{M}$ is proportional to $\vec{H}$ for most materials (at least for small fields):

\vec{M} = \chi_m \vec{H}

where $\chi_m$ is the magnetic susceptibility — a dimensionless number that tells you how easily a material magnetizes.

Material Type	$\chi_m$	Behaviour	Example
Diamagnetic	Small and negative ( $\approx -10^{-5}$ )	Weakly repelled by a magnet; $\vec{M}$ opposes $\vec{H}$	Water, copper, bismuth
Paramagnetic	Small and positive ( $\approx 10^{-5}$ to $10^{-3}$ )	Weakly attracted; $\vec{M}$ aligns with $\vec{H}$	Aluminium, oxygen gas
Ferromagnetic	Large and positive ( $\gg 1$ )	Strongly attracted; $\vec{M}$ can be huge and persists even after $\vec{H}$ is removed	Iron, nickel, cobalt

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 ( $\vec{M}$ )?

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 $\vec{M} = 0$ .
When an external field $\vec{H}$ is applied, dipoles align partially → net $\vec{M} \neq 0$ .

Definition:

\vec{M} = \frac{\text{net magnetic dipole moment}}{\text{volume}}

Units: A/m (same as $\vec{H}$ ).

2. The Fundamental Relation: $\vec{B} = \mu_0 (\vec{H} + \vec{M})$

This is the master equation linking the three magnetic fields:

$\vec{B}$ = magnetic flux density (the total field inside the material)
$\vec{H}$ = applied magnetic field (due to free currents)
$\vec{M}$ = magnetization (response of the material)
$\mu_0$ = permeability of free space ( $4\pi \times 10^{-7}$ H/m)

Why this form?

Step 1: In vacuum, there is no material, so $\vec{M} = 0$ . Then:

\vec{B} = \mu_0 \vec{H}

Step 2: Inside a material, the dipoles themselves produce an additional field. The total $\vec{B}$ is the sum of:

The field due to free currents ( $\mu_0 \vec{H}$ )
The field due to bound currents (from aligned dipoles), which is $\mu_0 \vec{M}$

Hence:

\vec{B} = \mu_0 \vec{H} + \mu_0 \vec{M} = \mu_0 (\vec{H} + \vec{M})

Key insight: $\vec{M}$ is not an independent field — it's the material's response to $\vec{H}$ .

3. Magnetic Susceptibility ( $\chi_m$ ) and Permeability ( $\mu$ )

For linear, isotropic, homogeneous materials (most common in exams), magnetization is proportional to the applied field:

\vec{M} = \chi_m \vec{H}

$\chi_m$ = magnetic susceptibility (dimensionless)
$\chi_m > 0$ for paramagnetic materials
$\chi_m < 0$ for diamagnetic materials
$\chi_m \gg 1$ for ferromagnetic materials (but not linear!)

Derivation of relative permeability $\mu_r$ :

Substitute $\vec{M} = \chi_m \vec{H}$ into the master equation:

\vec{B} = \mu_0 (\vec{H} + \chi_m \vec{H}) = \mu_0 (1 + \chi_m) \vec{H}

Define:

\mu_r = 1 + \chi_m \quad \text{(relative permeability)}

\mu = \mu_0 \mu_r \quad \text{(absolute permeability)}

Thus:

\vec{B} = \mu \vec{H}

Why this matters: It shows that the material simply scales the applied field by a factor $\mu_r$ .

4. Why $\chi_m$ Has Different Signs (Physical Reasoning)

Material Type	$\chi_m$	Why?
Diamagnetic	$\chi_m < 0$ (small, ~ $10^{-5}$ )	Applied field induces opposing dipole moments (Lenz's law at atomic level). $\vec{M}$ opposes $\vec{H}$ .
Paramagnetic	$\chi_m > 0$ (small, ~ $10^{-3}$ )	Permanent atomic dipoles align partially with $\vec{H}$ . Thermal agitation fights alignment.

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The key idea is that not all atoms have a net magnetic moment — only those with unpaired electrons do. The reason given is also false because a current loop always behaves as a magnetic dipole. Therefore both statements are false.

Understanding the Concept

The magnetic moment of an atom arises primarily from two sources: the orbital motion of electrons (like tiny current loops) and the intrinsic spin of electrons. For an atom to have a net magnetic moment, these contributions must not cancel out completely.

In most atoms, electrons fill orbitals in pairs. Within each pair, the two electrons have opposite spins, so their spin magnetic moments cancel. Similarly, if all orbitals are completely filled, the orbital angular momentum also sums to zero. Only atoms with unpaired electrons — like iron, cobalt, nickel — possess a permanent net magnetic moment. Atoms like helium or neon, with all electrons paired, have zero net magnetic moment.

Now examine the two statements:

Assertion (A): "All atoms have a net magnetic moment." This is false — as explained, atoms with completely filled shells (noble gases, for example) have no net magnetic moment.
Reason (R): "A current loop does not always behave as a magnetic dipole." This is also false. Any current loop, regardless of shape or size, produces a magnetic field that at large distances is exactly that of a magnetic dipole. The magnetic dipole moment of a current loop is $\vec{m} = I \vec{A}$ , where $I$ is the current and $\vec{A}$ is the area vector. This is a fundamental result in electromagnetism.

Watch out

A common mistake is to think that because some atoms are non-magnetic, a current loop might also sometimes fail to be a dipole. But the two ideas are unrelated — a current loop is always a magnetic dipole, while an atom is only magnetic if it has unpaired electrons. …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2026Set 55/1/11 markMCQ
$Q. For questions 13 to 16, two statements are given - one labelled Assertion (A) and the other labelled Reason (R). Select the correct answer from the codes (A), (B), (C) and (D) below: (A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A). (B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A). (C) Assertion (A) is true, but Reason (R) is false. (D) Both Assertion (A) and Reason (R) are false. Assertion (A) : All atoms have a net magnetic moment. Reason (R) : A current loop does not always behave as a magnetic dipole.$
›Reveal solutionSolution
The key idea is that not all atoms have a net magnetic moment — only those with unpaired electrons do. The reason given is also false because a current loop always behaves as a magnetic dipole. Therefore both statements are false.

Understanding the Concept

The magnetic moment of an atom arises primarily from two sources: the orbital motion of electrons (like tiny current loops) and the intrinsic spin of electrons. For an atom to have a net magnetic moment, these contributions must not cancel out completely.

In most atoms, electrons fill orbitals in pairs. Within each pair, the two electrons have opposite spins, so their spin magnetic moments cancel. Similarly, if all orbitals are completely filled, the orbital angular momentum also sums to zero. Only atoms with unpaired electrons — like iron, cobalt, nickel — possess a permanent net magnetic moment. Atoms like helium or neon, with all electrons paired, have zero net magnetic moment.

Now examine the two statements:

Assertion (A): "All atoms have a net magnetic moment." This is false — as explained, atoms with completely filled shells (noble gases, for example) have no net magnetic moment.

Reason (R): "A current loop does not always behave as a magnetic dipole." This is also false. Any current loop, regardless of shape or size, produces a magnetic field that at large distances is exactly that of a magnetic dipole. The magnetic dipole moment of a current loop is $\vec{m} = I \vec{A}$ , where $I$ is the current and $\vec{A}$ is the area vector. This is a fundamental result in electromagnetism.

Watch out

A common mistake is to think that because some atoms are non-magnetic, a current loop might also sometimes fail to be a dipole. But the two ideas are unrelated — a current loop is always a magnetic dipole, while an atom is only magnetic if it has unpaired electrons. …
Start your 14-day free trial to see the full solution →
CBSE 2026Set A1 markMCQ
$Q. Relative permeability is equal to (A) μr = μ0\cdotμ (B) μr = μ/μ0 (C) μr = μ0/μ (D) μr = \sqrt(μ\cdotμ0)$
›Reveal solutionSolution

Relative permeability μr = μ/μ₀ (a dimensionless ratio).

Relative permeability compares the magnetic permeability of a material with that of free space:

$\mu_r=\frac{\mu}{\mu_0}$

…

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CBSE 2026Set ANNUAL1 mark
$Q. Write the relation between magnetisation (M), magnetic intensity (H) and magnetic field (B) of a substance.$
›Reveal solutionSolution

A material's total magnetic field B combines the externally applied field (through H) and the material's own induced magnetisation M.

When a magnetic material is placed in an external magnetising field, it develops a magnetisation M (net magnetic moment per unit volume) in response to the magnetic intensity H present. The resultant magnetic field B inside the material is the sum of the contribution due to H and the contribution due to the material's own mag …

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CBSE 2024Set ANNUAL1 mark
$Q. The resultant magnetic moment produced per unit volume of a substance is called __________.$
›Reveal solutionSolution

Magnetisation M is defined exactly as the net magnetic moment per unit volume of a material.

When a magnetic material is placed in an external field, the atomic dipole moments tend to align, producing a net magnetic moment in the sample. The magnetisation is defined as:

$M = \dfrac{\text{net magnetic moment}}{\text{volume}} = \dfrac{m_{net}}{V}$

…

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CBSE 2022Set I1 markMCQ
$Q. Intensity of a magnetising field (H) is equal to (A) B_0/μ_0 (B) μ_0/B_0 (C) B_0 μ_0 (D) \sqrt(B_0 μ_0)$
›Reveal solutionSolution

The magnetic field and magnetising field are related by B = μ_0 H (in vacuum), so H = B_0/μ_0.

The magnetic intensity (magnetising field) H describes the field produced by free currents alone, independent of the medium. In free space (vacuum) the total field B_0 is related to H by $B_0 = \mu_0 H$ .

Rearranging gives $H = B_0/\mu_0$ , whose SI unit is ampere/metre (A/m).

…

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CBSE 2022Set I1 markMCQ
$Q. Relative permeability is equal to (A) μ/μ_0 = μ_r (B) μ_0/μ = μ_r (C) μ_r = μ\cdotμ_0 (D) \sqrt(μ_0 μ) = μ_r$
›Reveal solutionSolution

Relative permeability μ_r = μ/μ_0.

The relative permeability of a material is the ratio of its absolute permeability μ to the permeability of free space μ₀:

$\mu_r = \dfrac{\mu}{\mu_0}.$

…

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CBSE 2021Set A1 markMCQ
$Q. Which of the following relations is correct for permeability? (A) μ = H/B (B) μ = B/H (C) μ = B.H (D) μ = (B + H)$
›Reveal solutionSolution

Permeability μ = B/H.

Inside a magnetic material the magnetic flux density B is related to the magnetising field intensity H by B = μH, where μ is the (absolute) permeability of the material. Rearranging:

$\mu = \frac{B}{H}$

…

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CBSE 2018Set ANNUAL1 mark
$Q. The resultant magnetic moment of diamagnetic and paramagnetic substance are zero and finite respectively. Why?$
›Reveal solutionSolution
Diamagnetic atoms have fully paired electrons (moments cancel → zero); paramagnetic atoms have unpaired electrons (moments don't cancel → finite).

Each electron in an atom has a magnetic moment due to its orbital motion and its spin. The net atomic moment is the vector sum of all these electron moments.

Diamagnetic substances: every electron is paired with another of opposite spin/orbital moment, so the individual moments cancel completely. The resultant magnetic moment of the atom is therefore zero. (Such materials are only weakly repelled by a field, an effect induced by the applied field itself.) …
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Understanding the Concept

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