A proton has spin and magnetic moment just like an electron. Why then its effect is neglected in magnetism of

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Q. A proton has spin and magnetic moment just like an electron. Why then its effect is neglected in magnetism of materials?

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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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A spin magnetic moment scales as $\mu \propto 1/m$ ; the proton is $\sim1836\times$ heavier than the electron, so $\mu_p \approx \mu_B/660$ — about $660$ times smaller. Bulk magnetism is governed by electron moments, so the proton's contribution is negligible.

Why a spin moment depends on mass

Both particles have spin $\tfrac12$ , so each has an intrinsic magnetic moment. For a particle of charge $q$ , mass $m$ and gyromagnetic factor $g$ ,

\mu = g\,\frac{q}{2m}\,S, \qquad S = \frac{\hbar}{2}.

The two particles carry the same magnitude of charge, so the moment is controlled by the charge-to-mass ratio $q/m$ — and hence by the mass.

Comparing proton and electron

The natural units are the Bohr and nuclear magnetons:

\mu_B = \frac{e\hbar}{2m_e}, \qquad \mu_N = \frac{e\hbar}{2m_p}, \qquad \frac{\mu_N}{\mu_B} = \frac{m_e}{m_p} \approx \frac{1}{1836}.

The measured moments are $\mu_e \approx \mu_B$ (electron $g\approx2$ ) and $\mu_p \approx 2.79\,\mu_N$ (proton $g\approx5.6$ , from its composite structure). Therefore

\mu_p \approx 2.79\,\mu_N = \frac{2.79}{1836}\,\mu_B \approx \frac{\mu_B}{660}.

So even with the larger $g$ -factor, the proton's moment is about $660$ times smaller than the electron's.

Watch out

Equal spin quantum number does not mean equal magnetic moment. The moment depends on $q/m$ , and the proton's much larger mass crushes it.

Why it is negligible in materials …

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

GUJCET 2026Set x1 markMCQ
$2.5 \times 10^{-4}$
›Reveal solutionSolution

$m=NIA=0.60$ J T $^{-1}$ .

Magnetic moment of a solenoid: …
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GUJCET 2024Set 131 markMCQ
$\frac{Vs}{Am}$
›Reveal solutionSolution

$\dfrac{\text{V·s}}{\text{A·m}}$ equals $\dfrac{\text{T·m}}{\text{A}}$ , which is the SI unit of the permeability of free space $\mu_0$ (H/m).

Concept. $\mu_0$ has units of henry per metre; $1\,\text{H}=1\,\text{V·s/A}$ , so $\mu_0$ is measured in $\dfrac{\text{V·s}}{\text{A·m}}$ .

Steps. …
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GUJCET 2024Set 131 markMCQ
$8 \times 10^{-5}$
›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. …
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GUJCET 2023Set 091 markMCQ
$3.14 \times 10^{-3}$
›Reveal solutionSolution

Field inside a long solenoid is $B=\mu_0 nI$ ; the radius is irrelevant.

Concept: For a solenoid, $B=\mu_0 n I$ where $n$ is turns per unit length.

$n=\frac{N}{L}=\frac{250}{0.5}=500\ \text{m}^{-1}.$ …
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GUJCET 2023Set 091 markMCQ
$Q. A solenoid has a core of a material with relative permeability 400. The windings of solenoid are insulated from the core and carry a current of 2 A. If the number of turns is 1000 per metre, the magnetic field B inside the solenoid is ______ T. (A) 1.5 (B) 1.0 (C) 1.8 (D) 2.0$
›Reveal solutionSolution

A magnetic core multiplies the solenoid field by relative permeability: $B=\mu_0\mu_r nI$ .

Concept: With a core, $B=\mu_0\mu_r n I$ . …
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GUJCET 2022Set 171 markMCQ
$\mu_0 = 4\pi \times 10^{-7}$
›Reveal solutionSolution

$n=N/l=2000$ turns/m, so $B=\mu_0 nI=6.28\times10^{-3}$ T (radius irrelevant).

Concept: $n=\dfrac{500}{0.25}=2000$ turns/m. …
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GUJCET 2022Set 171 markMCQ
$\mu_0 = 4\pi \times 10^{-7}$
›Reveal solutionSolution

With a magnetic core, $B=\mu_0\mu_r nI=16\pi\times10^{-2}$ T. …

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GUJCET 2021Set 151 markMCQ
$6.28 \times 10^{-3}$
›Reveal solutionSolution

$B=\mu_0 nI$ ; radius is irrelevant for a long solenoid.

Concept: …
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GUJCET 2020Set 071 markMCQ
$2.5 \times 10^{3}$
›Reveal solutionSolution

$H = nI$ , independent of the core; $H = 1000\times2 = 2\times10^{3}$ A/m. …

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GSEB Higher Secondary Certificate (HSC) Examination 2018Set ANNUAL1 markMCQ
$Q. A toroid wound with 100 turns/m of wire carries a current of 3A. The core of toroid is made of iron having relative magnetic permeability of mu_r = 5000 under given conditions. The magnetic field inside the iron is ___.$
(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 = (4*pi x 10^-7)(5000)(100)(3) …

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Q.A proton has spin and magnetic moment just like an electron. Why then its effect is neglected in magnetism of materials?

Concept understanding — Magnetic Materials Magnetization

From a Paperclip to a Magnet: The Intuition

The Precise Definition

How Magnetization Connects to the Magnetic Field

The Three Kinds of Magnetic Materials

Why this formula?

Magnetic Materials & Magnetization: Why the Key Formulas Hold

1. What is Magnetization ( $\vec{M}$ )?

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

Why this form?

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

Derivation of relative permeability $\mu_r$ :

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

Why a spin moment depends on mass

Comparing proton and electron

Why it is negligible in materials …

Method: Comparing Intrinsic (Spin) Magnetic Moments of Different Particles

Steps

Q.A proton has spin and magnetic moment just like an electron. Why then its effect is neglected in magnetism of materials?

Why a spin moment depends on mass

Comparing proton and electron

Why it is negligible in materials …

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