1.1 A magnetic circuit with a single air gap is shown in Fig. 1.24. The core dimensions
are:
Cross-sectional area Ac  1.8  10  3 m 2
Mean core length l c  0.6m
Gap length g  2.3  10  3 m
N  83 turns
Core:
Mean length lc ,
area Ac ,
permeability 
N turns
Air gap
g
Assume that the core is of infinite permeability ( 
  ) and neglect the effects
of fringing fields at the air gap and leakage flux. (a) Calculate the reluctance of the
core Rc and that of the gap R g . For a current of i  1.5 A , calculate (b) the total
flux
 , (c) the flux linkages 
of the coil, and (d) the coil inductance L.
Sol:
已知 

(a)
Ac  Ag  1.8  10  3 m 2
Let
Rc 
lc
lc

0A
wb
μAc u r u o Ac
Rg 
g
2.3  10  3

u o Ag 4π  10  7  1.8  10  3
 1.017 10 6 A
wb
(b)
φ

NI
83  1.5


 0.1224  10  3 wb
6
Rc  R g Rc  R g 1.017  10
(c)
eN
dφ dλ

dt dt
 λ  Nφ  83  0.1224  10  3  1.016  10  2 wb
(d)
L 
L 
N 2 μ o Ag
λ NΦ N 2 φ N 2 φ N 2
N2






F
g
i
F
φR g
Rg
g
N
μ o Ag
λ 1.016  10  2

 0.6773  10  2 H  6.773mH
i
1.5
1.2 Repeat Problem 1.1 for a finite core permeability of μ  2500 μ o .
Sol
(a):
Ac  Ag  1.8  10  3 m 2
lc
lc
0.6


 1.061  10 5 A
wb
μAc u r u o Ac 2500  4π  10  7  1.8  10  3
(解答中Rc  1.597  10 5 A 是錯誤的)
wb
g
2.3  10  3
Rg 

 1.017 10 6 A
wb
u o Ag 4π  10  7  1.8  10  3
Rc 
(b)
φ

NI
83  1.5


 1.10844  10  4 wb
6
Rc  R g Rc  R g ( 1.017  0.1062 )  10
(解答中φ  1.059 10 4 wb是錯誤的)
(c)
λ  Nφ  83  1.10844  10  4  92  10  4 wb  0.92  10  2 wb
(解答中λ  0.92  10 2 wb是錯誤的)
(d)
λ 0.92  10  2
L 
 0.613  10  2 H  6.13mH
i
1.5
(解答中L  5.858mH是錯誤的)
1.3 Consider the magnetic circuit of Fig. 1.24 with the dimensions of Problem
1.1.Assuming infinite core permeability, calculate (a) the number of turns required
to achieve an inductance of 12 mH and (b) the inductor current which will result in
a core flux density of 1.0 T.
Figure 1.24
Sol :
part (a):
L
因為
  N  L 
又
i
    Rc 
而


N
i
lc
0
 Ac
F  0 Ag
F
F
F
F




lc
g
g
Rtot Rc  Rg
g
0


A
 Ac  0 Ag
0
g
N i  0 Ag
又
F  Ni   
所以
N N N i  0 Ag N 2  0 Ac
L


i
i
g
g
可以得到
N
Lg

 0 Ac
g
12  10 3  2.3  10 3
 110 t u r n s
4  10 7  1.8  10 3
part (b):
因為習題1.1忽略了邊緣效應
Ag  Ac
所以

B g  Bc 
Ac
F  N i  H c lc  H g g
又
B H
又
  
所以
FNi
 F
Bg
0

lc 
Bg
0
g
g
Bg
因此可以得到  i 
Bc
0
N
g

Bc g
1  2.3 10 3

 16.6
 0 N 4 10 7  110
A
1.6 The magnetic circuit of Fig. 1.25 consists of a core and a moveable plunger of
width lp, each of permeability μ. The core has cross-sectional area Ac, and mean
length lc, The overlap area of the two air gaps Ag is a function of the plunger
position x and can be assumed to vary as

x 

Ag  Ac 1 
 X0 
You may neglect any fringing fields at the air gap and use approximations
consistent with magnetic-circuit analysis.
a. Assuming that μ → ∞, derive an expression for the magnetic flux density in
the air gap Bg as a function of the winding current I and as the plunger position
is varied (0 ≤ x ≤ 0.8X0 ). What is the corresponding flux density in the core ?
b. Repeat part (a) for a finite permeability μ.
Figure 1.25
Magnetic circuit for Problem 1.6.
Sol :
part (a):
因為
F  N i  2 gH g  H c l c  l p 
又
   , B H
所以
F  N i  2 gH g
因為
Bc 
所以
 Ag
Bc  
 Ac

Ac
, Hg 
NI
2g
   Bc Ac  B g Ag


x 
 Bg  Bg 1 

X 0 


part (b):


因為
2 gH g  H c lc  l p  NI ; Bg Ag  Bc Ac
又
Bg   0 H g
FNi
 Ag

Ac

所以
; Bc  H c
 l c  l p  
Bc

Bg
0
 2g

 B g  l c  l p 
Bg


 2g

  Ag

  Ac
 Bg 



0



  l c  l p 
2g 




0 


最後得到
Bg 
Ni
 Ag 

 l c  l p 
2 g  Ac 

0


Ni

x 
1 
 l c  l p 
X 0 
2g 

0




 0
 
  2 g   






N i  0



x 
 l c  l p  
1 

X

0 

1.8
Fig 1.24
An inductor of the form of Fig 1.24 has dimensions:
Cross-sectional area Ac=3.6 cm 2
Mean core length lc=15cm
N= 75 turns
Assuming a core permeability of   2100 0 and neglecting the effects of leakage
flux and fringing fields，calculate the air-gap length required to achieve an inductance of
6.0mH
Sol：
  0 N 2 Ac
g  
L

  o
  
  
 4  10 7  75 2  3.6  10 4   1 

2
  
lc  
  15  10  4.17mm
3
6  10


  2100 
1.9
Fig 1.26
The magnetic circuit of fig1.26 consists of magnetic material in a stack of height h. The
ings have inner radius Ri and outer radius Ro. Assume that the iron is of infinite
ermeability（    ）and neglect the effects of magnetic and fringing. For Ri=3.4cm
Ro=4.0cm
h=2cm
g=0.2cm
calculate
（a） The mean core length lc and the core cross-sectional area Ac
（b） The reluctance of the core Rc and that of the gap Rg
（N=65）
（c） The inductance L
（d） Current I required to operate at an air-gap flux density of Bg=1.35T
（e） The corresponding flux linkages of the coil
Sol:
（a） lc  2 ( Ro  Ri )  g =3.568cm
Ac  ( Ro  Ri )h  1.2cm 2
g
0.2  10 2
7 A


1
.
33

10
 0 Ac 4  10 7  1.2  10 4
wb
（b） R g 
Rc 
lc
0
Ac
N2
（c） L 
 0.317 mH
Rc  Rg
（d）
I


Bg Rc  Rg Ac
N
 33.1 A
（e）   NBg Ac =10.53mWb
1.10 Repeat Problem 1.9 for a core permeability of u=750uo
sol :
(a)
l c  2 ( R0  Ri )  g  2 (4.0  3.4)  0.2  3.57cm
Ac  ( R0  R)h  (4.0  3.4)  2  1.2cm 2
(b)
g
0.2  10  2
Rg 

 1.33  10 7
7
4
u 0 Ac 4  10  1.2  10
A / wb
又因鐵心導磁係數 u  750u 0
所以
RC 
lc
3.57  10  2

 3.16  10 5
7
4
uAc 750  4  10  1.2  10
(c )
N2
L
 0.311mH
Rc  R g
(d )
I
B g Rc  R g Ac
N
 33.8 A
(e)
  NBg Ac  65  1.35  1.2  10.5 mWb
A/ w
1.12 The inductor of fig.1.27 has a core of uniform circular cross-section of area Ac
mean length lc and relative permeability ur and an N-turn winding. Write an
expression for the inductance L.
Core:
mean length lc,
area Ac,
relative permeability ur
N-turm
coil
g
Figure 1.27
Sol:

Rc
+
F
Rg

F
Rc  Rg 
解答設 Ac  Ag 是錯誤的
F  N I
u  u0  u r
u A
N  I  u0  Ag N  I  u0  ur  Ac
F
uA
F 0 g F c 

Rg  Rc
g
lc
g
lc





1
1

 u0  N  I  Ac   
 g lc 

ur 

因此

L
I

N u0  N 2  Ac

l
I
g c
ur
另解:
另解
F  N I
u  u0  ur

u 0 Ag
N  I  u 0  Ag N  I  u 0  u r  Ac
uA
F
F
F c 

R g  Rc
g
lc
g
lc




Ag Ac 

 u0  N  I 

 g
lc 


ur 

因此
L

I

2
N u 0  N  Ag u 0  N 2  Ac


lc
I
g
ur
1.13 The inductor of fig. 1.27 has the following dimensions:
Ac  1.0cm 2
l c  15cm
g  0.8mm
N  480turns
Neglecting leakage and fringing and assuming  r  1000
the inductance.
Sol:
L  N 
then
i
Rc  R g 

and
lc
 0  r Ac

N i
Rc  Rg
g
 0 Ag
 0 N 2 Ac
 0 N 2 Ac
N2
L


 30.5mH
lc
Ac g
lc
Rc  Rg

g
r
Ag
r
calculate
1.14 The inductor of problem 1.13 is to be operated from a 60-Hz voltage source. (a)
Assuming negligible coil resistance, calculate the rms inductor voltage
corresponding to a peak core flux density of 1.5 T. (b) Under this operating
condition, calculate the rms current and the peak stored energy.
Sol:
part(a)
et    N
d t 
dt
and
 t   B peak A cos t
et   NB peak A cos t
E
2fNABpeak
2
 19.2Vrms
part(b)
I rms 
Vrms
 1.67 Arms
L
W peak  0.5L( 2 I rms ) 2  8.5mJ
1.16 A square voltage wave having a fundamental frequency of 60Hz and equal positive
and negative half cycles of amplitude E is applied to a 1000-turn winding
surrounding a closed iron core of 1.25×10 3 m 2 cross section.
a. Sketch the voltage, the winding flux linkage , and the core flux as a function of time
b. Find the maximum permissible value of E if the maximum flux density is not exceed
1.15T
一電壓方波的頻率為 60Hz，正負振幅大小為 E，使用在線圈為 100 閘截面積為
1.25  10 3 cm 2 的鐵心上
<A>.畫出電壓曲線的週期以及  的波形曲線
E
T = 1/60

t

Φ=BA= 1.25  10 3 cm 2 ×1.15
λ=NΦ=1.4375
<B>找出最大的電壓假設 Bpeak=1.15T
Emax=4fNAcBpeak=4×60×1000×1.25m×1.15=345V
1.17 An inductor is to be designed using a magnetic core of the form of that of
fig1.29.The core is of unif
orm cross-sectional area Ac=5.0cm 2 and of mean length lc=25cm.
a. Calculate the air-gap length g and the number of turn N such that the inductance is
1.4mH and so that inductor can operate at peak currents of 6A without saturating.
Assume that saturation occurs when the peak flux density in the core exceeds
1.7Tand that, below saturation ,the core has permeability μ=3200μ0
For an inductor current of 6A,use Eq3.21 to calculate (i) the magnetic stored energy
in the air gap and (ii) the magnetic stored energy in the core. Show that the total
magnetic stored energy is given by Eq1.47
A.閘數的計算 N 
 NI  0 lc
LI
 99turns 空隙長度的計算 g  0

AcBsat
Bsat

B.從 3.21 來進行計算 Wgap 
Wtot=Wgap+Wcore=0.252J
和 1.47 比較 W=L/2*I2=0.252J
AcgB 2 sat
AclcB 2 sat
 0.207 J Wcore 
 0.045 J
2 0
2
1.19 A proposed energy storage mechanism consists of an N-turn coil wound around
a large nonmagnetic(u=u0) toroidal form as shown in Fig.1.30.As can be seen
from the figure, the toroidal form has circular cross section of radius a and
torridal radius r, measured to the center of the cross section. The geometry of
this device is such that the magnetic field can be considered to be zero
everywhere outside the toroid. Under the assumption that a << r, the H field
inside the toroid can be considered to be directed around the toroid and of
uniform magnitude
H
Ni
2r
For coil with N = 1000 truns, r = 10 m, and a = 0.45 m:
a. Calculate the coil inductance L.
b. The coil is to be charged to a magnetic flux density of 1.75 T. Calculate the
total stored magnetic energy in the torus when this flux density is achieved.
c. If the coil is to be charged at a uniform rate(i.e.,di/dt = constant), calculate
the terminal voltage required to achieve the required flux density in 30 sec.
Assume the coil resistance to be negligible.
Figure 1.30 Toroidal winding for Problem 1.19.
Sol:
(a):
L
L
(b):
u 0 Ag N 2
g
u oa 2 N 2
 12.7 mH
2r
Core volume Vcore  (2r )a 2  40.0 m 3 . 因此
B2
W  Vc o r( e )  4.87 J
2u 0
(c):因為 T = 30 sec,
di ( 2rB ) /( u 0 N )

 2.92  10 3
dt
T
di
v  L  37 V
dt
A / sec
1.20 Figure 1.31 shows an inductor would on a laminated iron core of rectangular
cross section. Assume that the permeability of iron is infinite. Neglect magnetic
leakage and fringing in the two air gaps(total gap length = g). The N-turn
winding is insulated copper wire whose resistivity is ρΩ･m. Assume that the
fraction fw of the winding space is available for copper; the rest of the space is
used for insulation.
a. Calculate the cross-sectional area and volume of copper in the winding
space.
b. Write an expression for the flux density B in the inductor in the terms of the
current density Jcu in the copper winding.
c. Write an expression for the copper current density Jcu in terms of the coil
current I, the number of turns N, and the coil geometry.
d. Derive an expression for the electric power dissipation in the coil in terms of
the current density Jcu.
e. Derive an expression for the magnetic stored energy in the inductor in terms
of the applied current density Jcu.
f. From parts(d) and (e) derive an expression for L/R time constant of the
inductor. Note that this expression is independent of the number of turn in
the coil and does not change as the inductance and coil resistance are
changed by varying the number of turns.
Figure 1.31 Iron-core inductor for Problem 1.20
Sol:
(a):
Acu  f w ab;
(b):
B  u0 (
(c):
Jcu 
J cu Acu
)
g
NI
Acu
Volcu  2ab( w  h  2a)
(d):
Pdiss  Volcu ( J cu2 )
(e):
Wmag  Vol gap (
(f):
1
( ) LI 2
Wmag
2Wmag u 0 whAcu2
L
 2



1
1
R
Pdiss
gVolcu
2
( ) RI
( ) Pdiss
2
2
B2
B2
)  gwh(
)
2u 0
2u 0
1.21 The inductor of Fig. 1.31 has the following dimensions:
a=h=w=1.5cm b=2cm g=0.2cm
The winding factor (i.e., the fraction of the total winding area occupied by
conductor) is f w =0.55. The resustivity of copper is 1.73  10 8   m . When the
coil is operated with a constant dc applied voltage of 35V, the air-gap flux density
is measured to be 1.4T. Find the power dissipated in the coil current, number of
turn, coil resistance, inductance, time constant, and wire size to the nearest
standard size,(Hint : Wire size can be found from the expression
Awire


AWG  36  4.312 ln 
8 
 1.267  10 
Where AWG is the wire size, expressed in terms of the American Wire Gauge, and
Awrie is the conductor cross-sectional area measured in m 2 .)
Sol:
Acu  f wab  0.55  1.5  2  1.65cm2
Volcu  2abw  h  2a   2  1.5  21.5  1.5  2  1.5  36cm3
a.
Pdiss  2Wmag 
 B 2  gVolcu
gVolcu

 

2
gwh
2
0 whAcu2
 20  0 whAcu
 1.42 
1.73  1010  36

 2  0.2  1.5  1.5  

 113.57W 
7 
7
2
 8  10  4  10  1.5  1.5  1.65
b.
I
Pdiss 113.57

 3.24 A
V
35
c.
J cu 
NI

Acu
N
Pdiss
113.57

 1350
Volcu
1.73  1010  36
J cu Acu 1350  1.65

 687.5  687匝
I
3.24
d.
R
Pdiss 113.57

 10.82 
I2
3.242
 
L
 6.18m sec
R
e.
f.
根據查表可知，Wire size=23AWG
1.22 The magnetic circuit of Fig. 1.32 has two windings and two air gaps. The core can
be assumed to be of infinite permeability. The core dimensions are indicated in
the figure.
a. Assuming coil 1 to be carrying a current I1 and the current in coil 2 to be zero,
calculate (i) the magnetic flux density in each of the air gaps, (ii) the flux linkage of
winding 1, and (iii) the flux linkage of winding 2.
b. Repeat part (a), assuming zero current in winding 1 and a current I 2 in winding 2.
c. Repeat part (a), assuming the current in winding 1 to be I1 and the current in
winding 2 to be I 2 .2
d. Find the self-inductances of windings 1 and 2 and the mutual inductance between the
windings.
Sol:
a. Assuming coil 1 current I1 and coil 2 current I2=0A .
(i.)求各氣隙之磁通密度 B
F0 A
A
g
F A  N i
 B1  1 0 1  0 1 1
A1 g1
g1
B

; 
 B2 
F2 0 A2 0 N 2i2

A2 g 2
g2
(ii.)求 winding 1 之磁交鏈  
  N  N
PS.為時變中之
1  N11;1 
N1i1
g
g
; R1  1 ; R2  2
R1R2
0 A1
0 A2
R1  R2
R1R2
gg
02 A1 A2
g1 g 2
 21 2 

R1  R2 0 A1 A2 0  A2 g1  A1 g 2  0  A2 g1  A1 g 2 
 1 
A A 
N12i10  A2 g1  A1 g 2 
 0 N12i1  1  2 
g1 g 2
 g1 g 2 
(iii.) 求 winding 2 之磁交鏈  
2  N 22 ;2  B2 A2
A 
 2  0 N1 N 2i1  2 
 g2 
b. Assuming zero current in winding 1 and winding 2 current is I2 .
(i.) 求各氣隙之磁通密度 B
F A
Ni
B1  0; B2  2 0 2  0 2 2
A2 g 2
g2
(ii.)求 winding 1 之磁交鏈  
1  N1`2 ;2  B2 A2 ; B2 
0 N 2i2
g2
;2 
0 N 2i2 A2
g2
A 
 1 0 N1 N 2i2  2 
 g2 
(iii.) 求 winding 2 之磁交鏈  
 A2 

 g2 
2  N 22  0 N 2i2 
c. Assuming the current in winding 1 is I1 and winding 2 is I2 .
(i.) 求各氣隙之磁通密度 B
B1 
B2 

A1

A2


F10 A1 0 N1i1

A1 g1
g1
0 N1i1
g2

0 N 2i2
g2
(ii.)求 winding 1 之磁交鏈  
1  N1 B1 A1  B2 A2   N1 1  2 
A A 
A 
 1  0 N12i1  1  2   0 N1 N 2i2  2 
 g1 g 2 
 g2 
(iii.) 求 winding 2 之磁交鏈  
 A2 
A 
  0 N 22i2  2 
 g2 
 g2 
2  N 2 B2 A2  0 N1N 2i2 
d. Find the self-inductances of winding 1 and 2 and mutual –inductances.
 A1 A2 
A 
i1  0 N1 N 2  2 i2

 g1 g 2 
 g2 
1  N1  L11i1  L12i2  0 N12 
 A2 
A 
i1  0 N 22  2 i2
 g2 
 g2 
2  N 2  L21i1  L21i2  0 N1 N 2 
A A 
A 
A 
 L11  0 N12  1  2 ; L22  0 N 22  2 ; L12  L21  0 N1 N 2  2 
 g1 g 2 
 g2 
 g2 
1.23 The symmetric magnetic circuits of Fig.1.33 has three windings.Windings A and B
each have N turns and are wound on the two bottom legs of the core.The core
dimensions are indicated in the figure.a. Find the self-inductances of each of the
windings.
b. Find the mutual inductances between the three pairs of windings.
c. Find the voltage induced in winding 1 by time-varying current i A (t ) and
iB (t )
in windings A and B. Show that this voltage can be use to measure the imbalance
between two sinusoidal current of the same frequency.
Sol:
先求出各區段的磁組
RA 
lA
l
l
g
R1  1 R2  2 Rg 
Ac
Ac
Ac
Ag
帶入 1.30 式(兩個 RA 並聯)
L11 

N2
R1  Rg
R
 R2  A
2
N 2 Ac
l1  l2 
lA
2

R1  R2  R g 

N2
1
l
g
(l1  l2  A 
)
Ac
2
0
g
0
1
g
(l1  l 2 
)
Ac
0
R A // R1  R2  Rg 
Ac
lA

L AA 
g
l1  l2 
l A (l1  l2 
0
g
0
Ac (l1  l2  l A 
)
g
0
)
N2
 LBB
R A  ( R A // R1  R2  Rg )
N 2 Ac (l A  l1  l2 

l A (l A  l1  l2 

Ac

g
0
g
0
)
)  l A (l1  l2 
l A  l1  l2 
g
0
)
g
N Ac
0
g
lA
l A  2(l1  l2 
)
2
0
LAB  LBA 

N 2 ( R1  R2  Rg )
RA ( RA 2( R1  R2  Rg ))
l1  l2 
g
N Ac
0
g
lA
l A  2(l1  l2 
)
2
0
因為兩繞組的磁通方向不一樣故
L1A  L A1   LB1   L1B 

 NN1Ac
l A  2(l1  l2 
g
0
 NN1
R A  2( R1  R2  Rg )
)
依 1.41 公式,考慮繞組 A,B 對繞組一的互感
e
d
d
d
d
Li  ( LA1iA  LB1iB )  ( LA1iA  LA1iB )  LA1 (iA  iB )
dt
dt
dt
dt
1.28 The coils of the magnetic circuit shown in Fig. 1.36 are connected in series so that
the mmf’s of paths A and B both tend to set up flux in the center leg C in the same
direction. The coils are wound with equal turns, N1 = N 2 =100. The dimensions are:
Cross-section area of A and B legs=7 cm
Cross-section area of C legs=14 cm
2
2
Length of A path = 17 cm
Length of B path = 17 cm
Length of C path = 5.5 cm
Air gap = 0.4 cm
The material is M-5 grade, 0.0012-in steel, with a stacking factor of 0.94.Neglect
fringing and leakage.
a. How many amperes are required to produce a flux density of 1.2T in the air gap?
b. Under the condition of path (a), how many joules of energy are stored in the
magnetic field in the air gap?
c. Calculate the inductance.
Sol:
由圖 1.10 所示,M-5 要產生 1.2T 需要磁場強度為 H m  14( A )
M
所以所需的磁通密度為 H g 
B
0

1.2
 9.54  105 ( A )
7
M
4  10
依照 F  Ni  Hl 的公式求出電流
I
Hl 14  (17  5.5  0.4)  10 2  9.54  105  0.4  10 2

 38.2
N
100
在乘上堆疊因數 0.94
38.2  0.94  35.9( A)
依公式 1.47 式整理得
W 

2
2L

( N )

N2
2
R
2
g
2
B 2 Ag
0 Ag
2

gAg B 2
2 0
 3.21( J )
其磁交鏈為
  N  NAg B  100  7 104 1.2  0.084(wb)
故電感值為
L   I  0.084  35.9  2.34(mH )
1.33 Using the magnetization characteristics for neodymium-iron-boron given infig1
find the point of maximum-energy product and corresponding flux density and
magnetic field intensity . Using these values .repeat EX1.10 with the alnico 5
magnet replaced by a neodymium-iron-boron magnet. By what factor does this
reduce the magnet volume required to achieve the desired air-gap flux density?
Sol:
從圖 1 可以得知
B = 0.63 T and H = -470 kA/m.
so maximum energy is 2.9.  105 J / m 3
Am =Ag(Bg/Bm)
= 2*[0.8/0.63] cm 2 = 2.54 cm 2
lm = -g(Hg/Hm)=  g ( Bg / 0 Hm)
=  0.2(0.8 / 0 (4.7 *105 )) = 0.27 cm
Thus the volume is 2.54×0.25 = 0.688
which is a reduction by a factor of 5.09/0.688 = 7.4.
samarium-cobalt
Maximum energy
B=0.63T
B=0.47T
H=-470 kA/m
H=-360 kA/m
Fig.1 Magnetization curves for common is permanent-magnet materials
1.34 Figure2 shows the magnetic circuit for a permanent-magnet loudspeaker the voice
coil (not shown)is in the form of a circular cylindrical coil which fits in the air
gap .a samarium-cobalt magnet is used to create the air-gap dc magnetic field
which interacts with the voice coil currents to produce the motion of the voice coil .
the designer has determined that the air gap must have radius R=1.8cm.length
g=0.1cm , and height g=0.1cm , and height h=0.9cm . assuming that the yoke and
pole piece are of infinite magnetic permeability (    ),find the magnet hm and
the magnet radius R m that wall result in an air-gap magnetic flux density of 1.2 T
and require the smallest magnet volume.(Hint : Refer to EX1.10 and to Fig.1.19 to
find the point of maximum energy product for samarium cobalt.)
Sol:
From Fig. 1, the maximum energy product for samarium-cobalt occurs at
B = 0.47 T and H = -360 kA/m
the maximum energy product is 1.69 *105 ( J / m3 )
we want Bg = 1.2 T, Bm = 0.47 T and
Hm = −360 kA/m.
hm = −g (Hg / Hm) = −g (Bg /μ0*Hm)
= 2.65 mm
Am = Ag (Bg / Bm) = 2πRh(Bg/Bm) = 26.0
(R=1.8cm and h=0.9cm)
So smallest magnet volume Rm  Am /   2.87cm
Fig.2 Magnetic circuit for the loudspeaker of problem