1. Magnetic Properties of Materials
Lecture 12
Magnetic Properties of Materials

2. OBJECTIVES
At the completion of the course, students will be able to...
Understand the physics of magnetism, magnetic fields and magnetization in magnetic objects.
Describe the physical basis of different classes of magnetic materials: paramagnetism, diamagnetism and ferromagnetism.
Calculate energy stored in a magnetic field.

3. Magnetic Field Strength, H
Recall applying Ampere’s law to a coil of wire in free space carrying current I . We found that the magnetic field density B at the centre of the coil is given by the formula
where Bo is the magnetic flux density generated by current I. It is convenient to define a quantity called the magnetic field strength H such that
Hence, we can write

4. Similarly, by applying Ampere’s law to an ideal solenoid with air core we found that the magnetic field density inside the coil is given by the formula
where Bo here is the magnetic flux density generated by current NI. In term of the magnetic field strength H we can write
Hence, we obtain
The importance of the new quantity called H will be apparent when we replace the air core with a magnetic material.

5. Worked Example Solution
A 10 cm long solenoid has 400 turns of wire and carries a current of 2.00 A. Calculate the magnetic field strength inside the solenoid.
Solution
N = 400 turns / 0.1m
= 4000 / m
magnetic field strength inside the solenoid.

6. Atomic Magnetism
Now why do some materials have magnetic fields in the absence of electric currents? Consider the atom. An electron “orbiting” an atom is like a small current loop. Let’s consider first a classical model of the atom, with the electron in a circular orbit:

7. Each electron has a charge of 1
Each electron has a charge of 1.6x10-19 C and this charge may be considered to be concentrated in a small sphere.
In addition to orbiting the nucleus each electron also spins on its own axis while it moves along its orbit.
This movement creates dipole moments – there are two, the orbital and spin dipole moments denoted by mo and ms, respectively.

8. Orbital Dipole Moment, mo
Consider an electron with charge of –e moving with a constant velocity u in a circular orbit of radius r. The time it takes to complete one revolution is
This circular motion of the electron constitutes a tiny current loop with current I given by:
The magnitude of the associated orbital magnetic moment m0 is
where: Le = meur is the angular momentum of the electron and me is its mass.

9. Spin Dipole Moment, ms
In addition to orbiting the nucleus each electron also spins on its own axis while it moves along its orbit.
The magnitude of ms (spin magnetic moment) predicted by quantum theory is
where: ħ is Planck’s constant.

10. ms has a magnitude of 9.27 x 10-24 Ampere x metres2 , and mo is either zero or an integral multiple of the value of ms.

11. Types of Magnetic Materials
In the atoms of many elements the electrons are arranged symmetrically so that the magnetic moments due to the spin and orbital motion cancel out, leaving the atom with zero magnetic moment.
However the atoms of more than 1/3rd of known elements lack this symmetry so that they (the atoms) do possess a magnetic moment.  
However in most of these materials the arrangement of the atoms is such that the magnetic moment of one is cancelled out by that of an oppositely directed near neighbour.
Only in five elements are the atoms arranged with their magnetic moments in parallel so that they supplement rather than cancel one another. These are the ferromagnetic elements: iron, nickel, cobalt, dysprosium, and gadolinium.

12. Ferromagnetism
A ferromagnetic substance contains permanent atomic magnetic dipoles that are spontaneously oriented parallel to one another even in the absence of an external field.
Ferromagnetic materials are characterised in that they have a crystal structure divided into magnetic domains usually of microscopic size, in each of which the magnetic moments of the atoms are aligned. The alignment direction differs however from one domain to another.
The direction of alignment of the magnetic moments is normally along one of the crystal axes but the domains may be oriented randomly in three directions, and a single crystal may contain many domains.

13. In the presence of an external H field favourably oriented domains increase in size through ‘domain wall motion’.  Also the atomic dipole moments tend to “rotate” into alignment with H.
Figure Comparison of (a) unmagnetised and (b) magnetised domains in a ferromagnetic material

14. Paramagnetism
Paramagnetism occurs primarily in substances in which some or all of the individual atoms, ions, or molecules possess a permanent magnetic dipole moment.
Paramagnetic materials exhibit magnetism when the external magnetic field is applied. Paramagnetic materials loose magnetization in the absence of an externally applied magnetic field. These materials are weakly attracted towards magnetic field.

15. Diamagnetic Materials
These materials are those in which individual atoms do not possess any permanent magnetic dipole moments. [Their orbital and spin magnetic moment add vectorially to become zero]. The atoms of such material however acquire an induced dipole moment when they are placed in an external magnetic field.
Diamagnetic materials repel the magnetic field weakly.

16. THE MAGNETISATION PROCESS
Consider a coil with a non-magnetised cast steel specimen inside it. Laboratory measurements show that as the coil current is increased from zero magnetic field B produced in the core increases at a much faster rate than that produced by an air-cored coil.
Magnetic core
Air core Bm Bo H

17. The extra flux is contributed by the magnetisation of the magnetic material. The amount of extra flux produced is dependent on the type of material used for the core.

18. MAGNETIC VECTOR M
It is convenient to think of the total magnetic flux density B as made up of two components: 
B = B0 + BM
where: 
B0 is the applied flux density,
and
BM is the extra flux density due to the material
BM is a result of the partial alignment of the magnetic moments with the magnetising force.

19. Using the relation B0 = 0H and defining a new quantity M such that Bm = 0M, we can write the total magnetic flux density in the material as
B = μ0H + μ0M =μ0(H + M)
where the first term represents the flux contribution by the current I and the second term represents the contribution of the magnetization of the material due to the alignment of magnetic dipole moments within the atoms by the external field.

20. In general, a material becomes magnetized in response to the external field H, hence M can be explained as
M=mH
where m is a dimensionless quantity called the magnetic susceptibility of the material.
For diamagnetic and paramagnetic materials, m is constant at a given temperature, resulting in a linear relationship between M and H.
For ferromagnetic substances the relationship between M and H not only is nonlinear, but also depends on the previous “history” of the material.

21. We can write B in terms of H and M as follows:
B = μ0(H + M) = μ0(H + mH) = μ0(1 + m)H
Define a new quantity  such that
B = μH. Then
μ = μ0(1 + m)   (H/m)
Often it is convenient to define the magnetic properties of a material in terms of the relative permeability μr
So, we can write
B = μH = µ0µrH

23. Summary of magnetic responses:B M
diamagnetic
(opposes H) H
<<1, negative B M
paramagnetic
(aligns with H) H
<<1, negative
ferromagnetic
(even without H!)
>1, positive BB M

24. Worked Example Solution
From the graph given, find the magnetisation M of the cast steel material.
Solution
At H = 1000A/m,  
B0 = μ0H = 4x10-7x 103 = T.
From the graph for the cast steel 
Bm =1.2 – B0  1.2T.

25. Worked Example Solution
Given a ferrite material which we shall specify to be operating in a linear mode with B = 0.05 T, let us assume µR = 50, and calculate values for m, M, and H.
Solution
Since µr = 1 + m, we have
Also,
Hence
The magnetization is

26. Worked Example Solution
A tightly wound solenoid is 20.0-cm long, has 400 turns, and carries a current of 4.00 A so that its axial field is in the +z direction. Find B and Bapp at the centre when (a) there is no core in the solenoid, and (b) there is a soft iron core that has a magnetization of 1.2 × 106 A/m.
Solution
Magnetic flux density induced in the air-cored solenoid by current I is

27. (b) Magnetic flux density contributed by iron core with a
magnetization M = 1.2 × 106 A/m,
Total magnetic flux density in the solenoid,

28. Worked Example Solution
A toroid has N turns, carries a current I, has a mean radius R, and has a cross-sectional radius r, where r << R (see figure below). The core of the toroid is filled with iron. When the current is 10.0 A, the magnetic field in the region where the iron is has a magnitude of 1.80 T. (a) What is the magnetization? (b) Find the values for the relative permeability, the permeability, and magnetic susceptibility for this iron sample.
Solution r

29. Solution
The magnetization M is related to the total flux and the applied flux via the of the equation
The magnetic flux density inside the a tightly wound air-cored toroid:
Therefore,
Substituting numerical values into the above equation, we obtain

30. Now,
Substituting numerical values into the above equation, we obtain
Relative permeability

31. M = KM - 1 M = 90 – 1 = 89 From the relation KM = 1 + M, we obtain
Substituting numerical values into the above equation, we obtain
M = 90 – 1 = 89

32. Worked Example Solution
The saturation magnetization Mmax of nickel is 4.7 x 105 A/m. Calculate the magnetic moment of a nickel atom.
Solution
Remember the definition of M:
First find the number of atoms per unit volume.

33. So

34. Therefore,

35. Worked Example
Calculate the induction and magnetization of a diamagnetic material (with μr = ) under an applied field strength of 2.0 x 105 amperes/m.                        
Solution: First the induction: B = μr μ0 H B = ( ) (4 x 10-7) (2.0 x 105) B = weber/m2 Now the magnetization:
M = (μr - 1) H M = ( )(2.0 x 105) M = -10 amperes/m |M| = 10 amperes/m

36. HYSTERESIS LOOP
A great deal of information can be learned about the magnetic properties of a material by studying its hysteresis loop. A hysteresis loop shows the relationship between the induced magnetic flux density (B) and the magnetizing force (H). It is often referred to as the B-H loop. An example hysteresis loop is shown below.

37. The loop is generated by measuring the magnetic flux of a ferromagnetic material while the magnetizing force is changed.
A ferromagnetic material that has never been previously magnetized or has been thoroughly demagnetized will follow the dashed line as H is increased.
As the line demonstrates, the greater the amount of current applied (H), the stronger the magnetic field in the component (B).
At point "a" almost all of the magnetic domains are aligned and an additional increase in the magnetizing force will produce very little increase in magnetic flux. The material has reached the point of magnetic saturation.

38. When H is reduced to zero, the curve will move from point "a" to point "b." At this point, it can be seen that some magnetic flux remains in the material even though the magnetizing force is zero. This is referred to as the point of retentivity on the graph and indicates the remanence or level of residual magnetism in the material. (Some of the magnetic domains remain aligned but some have lost their alignment.)

39. As the magnetizing force is reversed, the curve moves to point "c", where the flux has been reduced to zero.
This is called the point of coercivity on the curve. (The reversed magnetizing force has flipped enough of the domains so that the net flux within the material is zero.) The force required to remove the residual magnetism from the material is called the coercive force or coercivity of the material.

40. As the magnetizing force is increased in the negative direction, the material will again become magnetically saturated but in the opposite direction (point "d").
Reducing H to zero brings the curve to point "e." It will have a level of residual magnetism equal to that achieved in the other direction.
Increasing H back in the positive direction will return B to zero. Notice that the curve did not return to the origin of the graph because some force is required to remove the residual magnetism.
The curve will take a different path from point "f" back to the saturation point where it with complete the loop.

41. From the hysteresis loop, a number of primary magnetic properties of a material can be determined.
Retentivity - A measure of the residual flux density corresponding to the saturation induction of a magnetic material. In other words, it is a material's ability to retain a certain amount of residual magnetic field when the magnetizing force is removed after achieving saturation. (The value of B at point b on the hysteresis curve.)

42. Residual Magnetism or Residual Flux - the magnetic flux density that remains in a material when the magnetizing force is zero. Note that residual magnetism and retentivity are the same when the material has been magnetized to the saturation point. However, the level of residual magnetism may be lower than the retentivity value when the magnetizing force did not reach the saturation level.

43. Coercive Force - The amount of reverse magnetic field which must be applied to a magnetic material to make the magnetic flux return to zero. (The value of H at point c on the hysteresis curve.)

44. 4. Reluctance -Is the opposition that a ferromagnetic material
shows to the establishment of a magnetic field. Reluctance is
analogous to the resistance in an electrical circuit. 

45. 5. Permeability
A property of a material that describes the ease with which a magnetic flux is established in the component.
Quantitatively, it is the ratio of the flux density (B) created within a material to the magnetizing field (H) and is represented by the following equation
µ = B/H
This equation describes the slope of the curve at any point on the hysteresis loop.

46. µ(relative) = µ(material) / µ(air)
The maximum permeability is the point where the slope of the B/H curve for the unmagnetized material is the greatest. This point is often taken as the point where a straight line from the origin is tangent to the B/H curve.
The relative permeability is arrived at by taking the ratio of the material's permeability to the permeability in free space (air). 
µ(relative) = µ(material) / µ(air)
where: µ(air) = 1.256 x 10-6 H/m

47. Worked Example
The following data are obtained for a nickel-iron alloy during the generation of a steady-state ferromagnetic hysteresis loop:
(a)  Plot the data. (b)  What is the remanent induction? (c)  What is the coercive field? (d)  Determine the saturation induction. (e)  Determine the saturation magnetization.

48. Answer:
(a)

49. (b)  The remanent induction is the value of induction, B, when
the field strength, H, is Therefore the remanent induction is 0.92 weber/m2
(c)  The coercive field is the value of field strength, H, when
the induction, B, is Therefore the coercive field strength is -18 amperes/m
(d)  The saturation induction is the maximum value of the induction Therefore the saturation induction is 0.95 weber/m2

50. (e)  The saturation magnetization is determined from the
saturation induction:
B = μo(H + M) Bs = μo(H + Ms)
Solving for Ms:
Ms =(Bs/μo) -  H Ms =[(0.95)/(4 x 10-7)] - (50) MS = 7.56 x 105 amperes/m

51. Worked Example
Assuming the hysteresis loop shown below is traversed at a frequency of 60 Hz, calculate the rate of energy loss (i.e. power loss) for this magnet.

52. Answer:
The rate of energy loss per second is equal to the energy lost per cycle times the number of cycles per second:
(power loss) = (energy lost per cycle) x (cycles per second)
We are given the cycles per second:
frequency = 60 Hz = 60 cycles/second
We need to find the energy lost per cycle. This is just the area inside the hysteresis loop. We can estimate this area by a rectangle that is approximately [(2 x 18) amperes/m] wide by approximately [(2 x 0.93) webers/m2)] high: area = [(2 x 18) amperes/m] x [(2 x 0.93) webers/m2)] area = (36)(1.86) [(ampere.weber)/m3] area = (36)(1.86) [(ampere).(Joules/ampere)/m3] area = (66.96) Joules/m3

53. Substituting this into the formula above for power loss:
power loss = (66.96 J/m3) x (60 cycles/s) power loss = 4.02 kW/m3

54. Shape Of The Hysteresis Loop
The shape of the hysteresis loop tells a great deal about the material being magnetized.
Relative to other materials, a material with a wider hysteresis loop has:
Lower Permeability
Higher Retentivity
Higher Coercivity
Higher Reluctance
Higher Residual Magnetism
Relative to other materials, a material with the narrower hysteresis loop has:
Higher Permeability
Lower Retentivity
Lower Coercivity
Lower Reluctance
Lower Residual Magnetism

55. Worked Example Solution4.
An iron bar magnet having a coercivity of 4000
A/m is to be demagnetized. If the bar is
Worked Example
An iron bar magnet having a coercivity of 4000 A/m is to be demagnetized. If the bar is inserted within a cylindrical wire coil (a solenoid) 25 cm long and having 150 turns, what electric current is required to generate the necessary magnetic field?
Solution
inserted within a cylindrical wire coil (a solenoid
) 25 cm long and having 150 turns, what electric
To demagnetize a magnet having a coercivity of 4000 A/m, an H
field of 4000 A/m must be applied in a direction opposite to that of the magnetization. From the notes, the H field in a solenoid is:
H = NI/L, where L is the length.
Therefore,
current is required to generate the necessary magne
tic field?

56. THE ENERGY IN A MAGNETIC FIELD
To derive an expression for the field energy we'll look at the behaviour of the field within a simple toroidal inductor. We equate the field energy to the electrical energy needed to establish the coil current.
When the coil current increases so does the magnetic field strength, H. That, in turn, leads to an increase in magnetic flux. The increase in flux induces a voltage in the coil. It's the power needed to push the current into the coil against this voltage which we now calculate.

57. We choose a toroid because over its cross-sectional area, A, the flux density should be approximately uniform (particularly if the core radius is large compared with it's cross section). We let the flux path length around the core be equal to l and the cross-sectional area be equal to A. We assume that the core is initially unmagnetized and that the electrical energy (W) supplied to the coil will all be converted to magnetic field energy in the core (we ignore eddy currents).
Faraday's law gives the voltage as
Substituting -

58. We can rewrite the previous equation as
Now,
Ni = MMF and H = MMF/l
Ni = Hl. so
Substituting:

59. Also, from the definition of flux density Φ= AB, so dΦ = AdB
Also, from the definition of flux density Φ= AB, so dΦ = AdB. Substituting
This gives the total energy in the core. If we wish to find the energy density then we divide by the volume of the core material:
Thus,

60. If the magnetization curve is linear then there is a further simplification. Substituting
we obtain
By doing the integration, we obtain

61. Worked Example
An air-filled solenoid has 390 turns of wire, a mean radius of 15.0 cm, and a cross-sectional area of 4.20 cm2.
If the current is 5.60 A, calculate the magnetic field of the solenoid.
Calculate the energy density in the magnetic field.

62. Solution Magnetic field of the solenoid,
Energy density in the magnetic field.

63. END