Magnetization and Magnetic Intensity

Magnetic Materials · Maharashtra HSC

Theory

Magnetization

Definition: the ratio of the magnetic moment to the volume of the material is called magnetization. It is denoted by $\vec{M}$ .
If a magnetic specimen of volume $V$ acquires a net magnetic dipole moment $m_{net}$ due to the magnetising field, then

$\vec{M} = \dfrac{m_{net}}{V}$

It is a vector quantity.
SI unit: $\text{A m}^{-1}$ (ampere per metre).
Dimensions: $[M^0 L^{-1} T^0 I^1]$ .
Why most materials show none: the bulk material is made up of a very large number of atoms, each with an inherent magnetic moment, but these moments are randomly oriented and their vector sum is zero. For some materials, such as $\text{Fe}_3\text{O}_4$ , the vector sum is not zero and the material has a net magnetic moment.

Magnetic intensity

The set-up and the diagram

Draw a long solenoid of $n$ turns per unit length carrying a current $I$ , with a rod of magnetic material lying along its axis. Mark the field of the empty solenoid $B_0$ along the axis and the extra field $B_m$ contributed by the magnetised rod in the same direction.

Derivation of the relation between B, H and M

The magnetic field inside the empty solenoid is

$B_0 = \mu_0 n I$

Let $B_m$ be the magnetic field due to the material kept inside the solenoid. The net field inside the rod is

$B = B_0 + B_m \qquad \ldots (1)$

It is observed that $B_m$ is proportional to the magnetization $M$ of the material:

$B_m = \mu_0 M$

where $\mu_0$ is the permeability of free space. Therefore

$B = \mu_0 n I + \mu_0 M \qquad \ldots (2)$

Introduce the magnetic field intensity (magnetising field) $H = nI$ . Definition: magnetic intensity is the ratio of the strength of the magnetising field to the permeability of free space. The point of it is that, unlike $B$ , it does not depend on the material rod placed inside the solenoid.

$B = \mu_0 H + \mu_0 M$

$\therefore\ B = \mu_0 (H + M) \qquad \ldots (3)$

$\therefore\ H = \dfrac{B}{\mu_0} - M \qquad \ldots (4)$

$H$ and $M$ have the same unit, $\text{A m}^{-1}$ , and the same dimensions. The field induced in the material therefore depends on both $H$ and $M$ .

Magnetic susceptibility

If $H$ is not too strong, the magnetization induced in the material is proportional to the magnetic intensity:

$M = \chi H \qquad \ldots (5)$

Definition: magnetic susceptibility is the ratio of the magnitude of the intensity of magnetization to that of the magnetic intensity.
It is a measure of the response of a material to an external magnetic field — how much magnetization is produced in the substance when it is kept in an external field. It is the magnetic analogue of electric susceptibility.
$\chi$ is the ratio of two quantities with the same unit ( $\text{A m}^{-1}$ ), so it is a unitless, dimensionless constant.
Sign: if the atomic dipole moments oppose the applied field, $\chi$ is negative; if they align along the field, $\chi$ is positive. The number of atomic dipole moments getting aligned along the field is proportional to $\chi$ . It is large for soft iron, $\chi > 1000$ .

Permeability

Substituting (5) into (3),

$B = \mu_0(H + \chi H)$

$\therefore\ B = \mu_0(1 + \chi)H \qquad \ldots (6)$

Writing $\mu_r = 1 + \chi$ and $\mu = \mu_0\mu_r$ ,

$B = \mu_0\mu_r H = \mu H \qquad \ldots (7)$

$\mu_r = 1 + \chi = \dfrac{\mu}{\mu_0}\qquad \mu = \mu_0(1+\chi)$

$\mu$ is the magnetic permeability of the material — the analogue of permittivity in electrostatics — with unit $\text{T m A}^{-1}$ , also written $\text{H m}^{-1}$ or $\text{Wb A}^{-1}\text{m}^{-1}$ . $\mu_r$ is the relative permeability, a pure number.
Permeability tells us how many magnetic lines of force pass through a given substance when it is kept in an external magnetic field, and that number indicates how the material behaves in the field.
For superconductors $\chi = -1$ ; substituting in $\mu = \mu_0(1+\chi)$ gives $\mu = 0$ , so no magnetic line passes through a superconductor.

Effect of filling a core

The field inside a solenoid or toroid without the material is $B_0 = \mu_0 H$ and with it is $B = \mu H$ . Therefore

$\dfrac{B - B_0}{B_0} = \dfrac{\mu H - \mu_0 H}{\mu_0 H} = \dfrac{\mu - \mu_0}{\mu_0}$

Since $\mu = \mu_0(1+\chi)$ , we have $\dfrac{\mu}{\mu_0} - 1 = \chi$ , so

$\dfrac{B - B_0}{B_0} = \chi \qquad\text{and}\qquad \text{percentage increase} = \chi\times 100$

Aluminium ( $\chi = 2.3\times10^{-5}$ ) filling a toroid gives an increase of only 0.0023 per cent; lithium ( $\chi = 2.1\times10^{-5}$ ) gives 0.0021 per cent. The increase for a paramagnetic core is tiny; for soft iron it is enormous.

Standard results

$\vec{M} = \dfrac{m_{net}}{V}$ , unit $\text{A m}^{-1}$ , dimensions $[M^0L^{-1}T^0I^1]$ .
$H = nI$ , unit $\text{A m}^{-1}$ , independent of the core material.
$B = \mu_0(H + M)$ , $H = \dfrac{B}{\mu_0} - M$ .
$\chi = \dfrac{M}{H}$ , dimensionless.
$B = \mu_0(1+\chi)H = \mu H$ , $\mu_r = 1 + \chi = \dfrac{\mu}{\mu_0}$ .
$\dfrac{B-B_0}{B_0} = \chi$ .
With flux data: $B = \dfrac{\Phi}{A}$ , then $\mu = \dfrac{B}{H}$ , $\mu_r = \dfrac{\mu}{\mu_0}$ , $\chi = \mu_r - 1$ , $M = \chi H$ .

Theory Questions Asked on This Subtopic

Real board questions. Answers are at the end of the page.

1
Define magnetization. State its formula and S.I. unit.
Maharashtra HSC board question — asked in March 2013Answer 1 ↓
2
What is magnetization?
Maharashtra HSC board question — asked in October 2013Answer 2 ↓
3
Define magnetization.
Maharashtra HSC board question — asked in March 2015Answer 3 ↓
4
What is magnetization, magnetic intensity and magnetic susceptibility?
Maharashtra HSC board question — asked in March 2022Answer 4 ↓
5
Define and state unit and dimensions of: i. Magnetization ii. Magnetic susceptibility
Maharashtra HSC board question — asked in February 2023Answer 5 ↓
6
Define magnetization. Write its SI unit and dimensions. OR Define magnetization. State its SI unit and dimensions.
Maharashtra HSC board question — asked in February 2024; March 2018Answer 6 ↓

Numericals

Every numerical set on this subtopic. Full working is at the end of the page.

7
[Oct 13] The magnetic susceptibility of annealed iron at saturation is 4224. Find the permeability of annealed iron at saturation. ( $\mu_0=4\pi\times10^{-7}$ SI unit).
Maharashtra HSC board questionAnswer 7 ↓
8
[Oct 14] The magnetic moment of a magnet of dimensions 5 cm × 2.5 cm × 1.25 cm is 3 Am². Calculate the intensity of magnetization.
Maharashtra HSC board questionAnswer 8 ↓
9
[Oct 15] An iron rod of area of cross-section 0.1 m² is subjected to a magnetising field of 1000 A/m. Calculate the magnetic permeability of the iron rod. [Magnetic susceptibility of iron = 59.9, magnetic permeability of vacuum = $4\pi\times10^{-7}$ S.I. unit]
Maharashtra HSC board questionAnswer 9 ↓
10
[July 17] Find the magnetization of a bar magnet of length 10 cm and cross-sectional area 4 cm², if the magnetic moment is 2 Am².
Maharashtra HSC board questionAnswer 10 ↓
11
[Feb 24] A current carrying toroid winding is internally filled with lithium having susceptibility $\chi=2.1\times10^{-5}$ . What is the percentage increase in the magnetic field in the presence of lithium over that without it?
Maharashtra HSC board questionAnswer 11 ↓

Answers & Solutions

Work the questions above first, then check yourself here. Each answer is set out the way it should be written in the paper.

1.↑ back to the question
i. Definition: The ratio of magnetic moment to the volume of the material is called magnetization. It is denoted by $\vec{M}$ . ii. If magnetic specimen of volume 'V' acquires net magnetic dipole moment ' $m_{net}$ ' due to the magnetising field, then $\vec{M}=\dfrac{m_{net}}{V}$ iii. It is a vector quantity. iv. Unit: Am⁻¹ in SI system. v. Dimensions: $[M^0L^{-1}T^0I^1]$
2.↑ back to the question
[From 11.4, Q1]
i. Definition: The ratio of magnetic moment to the volume of the material is called magnetization. It is denoted by $\vec{M}$ . ii. If magnetic specimen of volume 'V' acquires net magnetic dipole moment ' $m_{net}$ ' due to the magnetising field, then $\vec{M}=\dfrac{m_{net}}{V}$ iii. It is a vector quantity. iv. Unit: Am⁻¹ in SI system. v. Dimensions: $[M^0L^{-1}T^0I^1]$
3.↑ back to the question
[From 11.4, Q1 (Definition only)]
i. Definition: The ratio of magnetic moment to the volume of the material is called magnetization. It is denoted by $\vec{M}$ . ii. If magnetic specimen of volume 'V' acquires net magnetic dipole moment ' $m_{net}$ ' due to the magnetising field, then $\vec{M}=\dfrac{m_{net}}{V}$ iii. It is a vector quantity. iv. Unit: Am⁻¹ in SI system. v. Dimensions: $[M^0L^{-1}T^0I^1]$
4.↑ back to the question
i. [From 11.4, Q1 (Definition only)]
i. Definition: The ratio of magnetic moment to the volume of the material is called magnetization. It is denoted by $\vec{M}$ . ii. If magnetic specimen of volume 'V' acquires net magnetic dipole moment ' $m_{net}$ ' due to the magnetising field, then $\vec{M}=\dfrac{m_{net}}{V}$ iii. It is a vector quantity. iv. Unit: Am⁻¹ in SI system. v. Dimensions: $[M^0L^{-1}T^0I^1]$ ii. Magnetic intensity: The ratio of the strength of magnetising field to the permeability of free space is called as magnetic intensity. iii. Magnetic susceptibility: The ratio of magnitude of intensity of magnetization to that of magnetic intensity is called as magnetic susceptibility.
5.↑ back to the question
i. [From 11.4, Q1]
i. Definition: The ratio of magnetic moment to the volume of the material is called magnetization. It is denoted by $\vec{M}$ . ii. If magnetic specimen of volume 'V' acquires net magnetic dipole moment ' $m_{net}$ ' due to the magnetising field, then $\vec{M}=\dfrac{m_{net}}{V}$ iii. It is a vector quantity. iv. Unit: Am⁻¹ in SI system. v. Dimensions: $[M^0L^{-1}T^0I^1]$ [(i), (iv) and (v)] ii. [From 11.4, Q4 (iii)]
i. [From 11.4, Q1 (Definition only)]
i. Definition: The ratio of magnetic moment to the volume of the material is called magnetization. It is denoted by $\vec{M}$ . ii. If magnetic specimen of volume 'V' acquires net magnetic dipole moment ' $m_{net}$ ' due to the magnetising field, then $\vec{M}=\dfrac{m_{net}}{V}$ iii. It is a vector quantity. iv. Unit: Am⁻¹ in SI system. v. Dimensions: $[M^0L^{-1}T^0I^1]$ ii. Magnetic intensity: The ratio of the strength of magnetising field to the permeability of free space is called as magnetic intensity. iii. Magnetic susceptibility: The ratio of magnitude of intensity of magnetization to that of magnetic intensity is called as magnetic susceptibility. Magnetic susceptibility is unitless and dimensionless quantity.
6.↑ back to the question
[From 11.4, Q1]
i. Definition: The ratio of magnetic moment to the volume of the material is called magnetization. It is denoted by $\vec{M}$ . ii. If magnetic specimen of volume 'V' acquires net magnetic dipole moment ' $m_{net}$ ' due to the magnetising field, then $\vec{M}=\dfrac{m_{net}}{V}$ iii. It is a vector quantity. iv. Unit: Am⁻¹ in SI system. v. Dimensions: $[M^0L^{-1}T^0I^1]$ [(i), (iv) and (v)]
7.↑ back to the question
Given: χ = 4224 To find: Permeability (μ) Formula: $\mu=\mu_0(1+\chi)$ Calculation: From formula, $\mu=4\pi\times10^{-7}(1+4224)$ $=169\times3.142\times10^{-5}$ $=\{\text{antilog}[\log(169)+\log(3.142)]\}\times10^{-5}$ $=\{\text{antilog}[2.2279+0.4972]\}\times10^{-5}$ $=\{\text{antilog}(2.7251)\}\times10^{-5}$ $=531\times10^{-5}$ $=5.31\times10^{-3}\text{ Hm}^{-1}$ Ans: The permeability of annealed iron at saturation is $5.31\times10^{-3}$ Hm⁻¹.
8.↑ back to the question
Given: $l=5$ cm $=5\times10^{-2}$ m, b = 2.5 cm = $2.5\times10^{-2}$ m, h = 1.25 cm = $1.25\times10^{-2}$ m, $M_{net}=3$ Am² To find: Intensity of magnetization ( $M_Z$ ) Formula: $M_Z=\dfrac{M_{net}}{V}$ Calculation: using formula, $M_Z=\dfrac{M_{net}}{l\times b\times h}\qquad(\because V=l\times b\times h)$ $=\dfrac{3}{5\times10^{-2}\times2.5\times10^{-2}\times1.25\times10^{-2}}$ $=\dfrac{3\times10^6}{12.5\times1.25}=\dfrac{30\times10^5}{15.625}$ $=\{\text{antilog}[\log(30)-\log(15.625)]\}\times10^5$ $=\{\text{antilog}[1.4771-1.1938]\}\times10^5$ $=\{\text{antilog}[0.2833]\}\times10^5$ $=1.92\times10^5$ A/m Ans: The intensity of magnetization is $1.92\times10^5$ A/m.
9.↑ back to the question
Given: H = 1000 A/m, χ = 59.9, $\mu_0=4\pi\times10^{-7}$ S.I. unit To find: Permeability (μ) Formula: $\mu=\mu_0(1+\chi)$ Calculation: From formula, $\mu=4\pi\times10^{-7}(1+59.9)$ $=4\times3.142\times10^{-7}\times60.9$ $=\{\text{antilog}[\log(4)+\log(3.142)+\log(60.9)]\}\times10^{-7}$ $=\{\text{antilog}[0.6021+0.4972+1.7846]\}\times10^{-7}$ $=\{\text{antilog}[2.8839]\}\times10^{-7}$ $=765.4\times10^{-7}$ $\therefore \mu=7.654\times10^{-5}\text{ Wb/A-m}$ Ans: The magnetic permeability of the iron rod is $7.654\times10^{-5}$ Wb/A-m.
10.↑ back to the question
Given: $M_{net}=2$ Am², $l=10$ cm $=10\times10^{-2}$ m, area of cross section = 4 cm² = $4\times10^{-4}$ m² To find: magnetisation ( $M_Z$ ) Formula: $M_Z=\dfrac{M_{net}}{V}$ Calculation: Since, V = l × cross sectional area $=(10\times10^{-2}\times4\times10^{-4})$ $=4\times10^{-5}\text{ m}^3$ From formula, $M_Z=\dfrac{2}{4\times10^{-5}}$ $\therefore M_Z=5\times10^4$ A/m Ans: The intensity of magnetisation is $5\times10^4$ A/m.
11.↑ back to the question
The magnetic field inside the solenoid without Lithium $B_0$ is given as, $\mu_0H$ . The magnetic field inside the solenoid with Lithium B is given as, $\mu H$ . Relative change in magnetic field, $\dfrac{B-B_0}{B_0}=\dfrac{\mu H-\mu_0H}{\mu_0H}=\dfrac{\mu-\mu_0}{\mu_0}\qquad....(i)$ $\because \mu=\mu_0(1+\chi)$ $\therefore \dfrac{\mu}{\mu_0}-1=\chi$ $\therefore \dfrac{\mu-\mu_0}{\mu_0}=\chi\qquad....(ii)$ From (i) and (ii), $\dfrac{B-B_0}{B_0}=\chi$ Percentage increase in the magnetic field after inserting Lithium is $\dfrac{B-B_0}{B_0}\times100=2.1\times10^{-5}\times100=0.0021\ \%$ Ans: Percentage change in magnetic field is 0.0021 %.