70 The surface magnetisation current is
Kmag = M×n = M0z×ρ = M0ϕ. (6.21)
The net magnetisation current around the disc’s circumference is Imag = sKmag = sM0, and so the dipole moment is
m = (πa2)Imagz = πa2sM. (6.22)
This is just the volume multiplied by the magnetisation field.
6–3 Consider a permanent magnet in the form of a short cylinder of radiusaextending along thez axis fromz=−Ltoz= +Land having uniform magnetisation M=M0z. (a) Find B andH at all points(0,0, z) on the cylinder’s axis, and plot B(0,0, z) and H(0,0, z) vs.
z. (b) Discuss whether the result obtained in part (a) obeys Ampère’s law forHin integral form.
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Essential Electromagnetism: Solutions
71
Magnetism of materials
Essential Electromagnetism - Solutions 6 Magnetism of materials
Solution
x z
θ ’
y
∆B (0,0,z) R
z’
dz’ L
L Kmag
a
M
(a) Because the magnetisation field is uniform, there is no magnetisation volume current.
The magnetisation surface current Kmag(r) = M(r) ×�n is zero on the two ends, and Kmag(r) =M0ϕ� on the cylindrical surface.
The Biot-Savart law gives the magnetic field due to an arbitrary surface current distribu-tion:
B(r) = µ0 4π
∫ [Kmag(r′)dS′]×R�
R2 (6.23)
The diagram shows the contribution∆Bto the magnetic field at(0,0, z)due to the surface magnetisation current in a small patch of the surface making up part of the strip of thickness dz′ atz′. The components of∆B due to the surface currents in different patches around the strip which are not in the z direction will cancel each other out. This leaves only a z-component for the contributiondB, due to the entire strip, so that
dB(0,0, z) = µ0 4π
(Kmagdz′)(2πa)
R2 cosθ′�z, (6.24)
= µ0
4π
(M dz′)(2πa) R2
a
R�z, (6.25)
∴ dB(0,0, z) = µ0M a2dz′
2[(z−z′)2+a2]3/2�z. (6.26)
72
Hence, integrating over the entire cylindrical surface, B(0,0, z) =
∫ L
−L
µ0M a2dz′
2[(z−z′)2+a2]3/2�z, (6.27)
= µ0M 2
[ z+z′
√(z+z′)2+a2 ]L
−L
�z, (6.28)
∴ B(0,0, z) = µ0M 2
[ z+L
√(z+L)2+a2 − z−L
√(z−L)2+a2 ]
�z. (6.29)
Now,H=B/µ0−M, and so
H(0,0, z) =
M0
2
[
√ z+L
(z+L)2+a2 −√ z−L
(z−L)2+a2
]
�z (|z|> L)
M0
2
[
√ z+L
(z+L)2+a2 −√ z−L
(z−L)2+a2
]
�z−M0�z (|z|< L)
(6.30)
Band Hare plotted below.
(b) Ampère’s law in integral form is
∮
Γ
H·dr = If,encl. (6.31)
Essential Electromagnetism: Solutions
73
Magnetism of materials
Essential Electromagnetism - Solutions 6 Magnetism of materials
where If,encl. is the net free current through loop Γ. Since there are no free currents the integral must be zero for any closed path. We only knowHon thez axis, and at |r|=∞ where H must be zero because the source of magnetic field, in this case the magnet, is localised near the origin. But we can construct a closed loop which has as part of it the entirez-axis as follows: (0,0,−∞)to(0,0,+∞)to(0,+∞,+∞)to(0,+∞,−∞)and back to(0,0,−∞). The integrand is zero except along thez-axis, so in this case Ampère’s law in integral form is satisfied provided
∫ ∞
−∞
Hz(0,0, z)dz = 0. (6.32)
Examining the plot ofHz vs.z above, it appears that the integral is indeed zero, as could easily be checked by numerical integration.
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74
6–4 A cylindrical rod of radiusaand lengthh≫ais permanently magnetised along its length which coincides with the z direction, i.e. M = M0�z. (a) Find the surface magnetisation currentKmag(r), and use it together with Ampere’s law to findBandHinside and outside the rod (assume h→ ∞). (b) The rod is now bent into a circle of circumference(h+ 2L) such that there is an air gap of width2L < a. PlotB andHalong the axis of the magnet in the vicinity of the air gap for the case ofa= 0.5and L= 0.2.
Solution
K 2a L
2
mag
M z
M
a Γ
z δz
(a) The surface magnetisation current density isKmag(r) =M(r)×n, hence�
Kmag(a, ϕ, z) = M0�z×ρ� = M0ϕ.� (6.33) This surface magnetisation current is similar to the current in a tightly-wound solenoid, and the magnetic field inside the rod can be calculated in the same way using Ampère’s law∮
B·dr=µ0Iencl. From the symmetry of the problemBinside the rod can only be in the�z direction.
For Amperian rectangular loop Γ (see diagram) with one side of length δz inside the rod at cylindrical radiusρ1 and one outside at cylindrical radius ρ2
[Bz(ρ1, ϕ, z)−Bz(ρ2, ϕ, z)]δz = µ0M0δz. (6.34) That this is independent of ρ2 and applies equally toρ2 → ∞ (where B= 0) tells us that B = 0 outside the rod. Again, since the integral is independent of ρ1 the magnetic field inside the rod is constant,
B(ρ < a, ϕ, z) = µ0M0�z = µ0M. (6.35)
Essential Electromagnetism: Solutions
75
Magnetism of materials
Essential Electromagnetism - Solutions 6 Magnetism of materials
Finally, H =
(B µ0
)
−M = 0 (6.36)
everywhere. Note that this result is for an (unrealistic) infinite magnetised rod, and that near the two ends of the rodB would be different, andHwould be non-zero.
(b) To findB andHon the axis of the magnet near the air gap we can use the information thath ≫ aand assume the magnet in this region is approximately straight, with its axis being along thez-axis, and with the air gap extending from z =−L toz = +L. In that case we can imagine that a short magnet of length2L has been removed from an infinite magnetised rod.
Using the principle of superposition, we can getBin the vicinity of the air gap by subtract-ing the field of the short magnet of length2Lfrom the field of an infinite straight magnetised rod with no air gap. For the short magnet we can use the results from Exercise 6–3 for the magnetic field of the short magnet. Thus,
B(0,0, z) = µ0M0�z − µ0M0 2
[ z+L
√(z+L)2+a2 − z−L
√(z−L)2+a2 ]
�z. (6.37) Now,H = B/µ0−M, and so
H(0,0, z) =
− M20 [
√ z+L
(z+L)2+a2 −√ z−L
(z−L)2+a2
]
�z (|z|> L) M0�z − M20
[
√ z+L
(z+L)2+a2 −√ z−L
(z−L)2+a2
]
�z (|z|< L)
(6.38)
Band Hare plotted below. Note that the area under the plot ofHz(0,0, z) vs.z appears to be zero in agreement with Ampère’s law in integral form for the case of no free currents.
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Essential Electromagnetism: Solutions
77
Magnetism of materials
Essential Electromagnetism - Solutions 6 Magnetism of materials
6–5 Consider a permanently magnetised sphere of radiusawith uniform magnetisationM(r) = M0�z. (a) Find the surface magnetisation current density, and use this to find the magnetic dipole moment of the sphere. Compare this with what you expect given the volume of the sphere and the magnetisation field. (b) FindB and Hat the centre of the sphere.
Solution
a
x
y z
Kmag
M B
H
(a) The surface magnetisation current density is
Kmag(a, ϕ, z) = M×n� = M0�z×�r = M0sinθϕ.� (6.39) The magnetic dipole moment of the surface magnetisation current distribution is
m = 1 2
∮ r×Kmag(r)dS. (6.40)
From symmetry arguments, the dipole moment must bem=mz�z where
mz = �z·1 2
∫ 1
−1
a�r×(M0sinθϕ)2πa� 2d(cosθ) (6.41)
= 2πa3M0 2
∫ 1
−1
sinθ�z·(−�θ)d(cosθ) (6.42)
= πa3M0
∫ 1
−1sin2θ d(cosθ) (6.43)
= πa3M0
∫ 1
−1
(1−cos2θ)d(cosθ) (6.44)
= 4
3πa3M0 (6.45)
as expected.
78
(b) We can use the Biot-Savart law to obtain the magnetic field at the centre of the sphere B(0,0,0) = µ0
4π
∮ Kmag(r′)×R�
R2 dS′. (6.46)
From symmetry, this must beB(0,0,0) =B(0,0,0)�z where
B(0,0,0) = �z· µ0 4π
∫ 1
−1
(M0sinθϕ)� ×(−�r)
a2 2πa2d(cosθ), (6.47)
= µ0M0 2
∫ 1
−1sinθ�z·(−�θ)d(cosθ), (6.48)
= µ0M0
2
∫ 1
−1
sin2θ d(cosθ), (6.49)
= µ0M0 2
∫ 1
−1
(1−cos2θ)d(cosθ), (6.50)
∴ B(0,0,0) = 2
3µ0M0. (6.51)
∴ B(0,0,0) = 2
3µ0M (6.52)
and is in the same direction asM.
NowH=B/µ0−M, so thatH(0,0,0) =−13M and it is in the opposite direction toM.
6–6 Consider the hysteresis loops of the magnetically-soft iron-based amorphous alloy and the magnetically-hard alloy of iron, aluminium, nickel and cobalt shown in Chapter 6.
(a) Estimate the work done to bring 1 cm3 of each material through one cycle of the hysteresis loop. (b) Two transformers operating at 50 Hz have magnetic cores of volume 100 cm3 (one of each type of material) and are (unwisely) operated at a current at which saturation occurs. How much power is lost as heat in each case? [You could print the hysteresis plots and estimate the area by drawing over it a grid and measuring by hand sufficient points on the graph to get within say 20% accuracy for the area.]
Solution
Essential Electromagnetism: Solutions
79
Magnetism of materials
Essential Electromagnetism - Solutions 6 Magnetism of materials
Grid lines have been drawn over the hysteresis plots above. The area of the upper half of the hysteresis loop (for positive BM) is identical to that of the lower half. Read off the (horizontal) “widths” inH at “heights”BM = 0,0.1,0.2, . . . (T). The approximate values are tabulated below.
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