1 MAGNETOSTATIC FIELD (MAGNETIC FORCE, MAGNETIC MATERIAL AND INDUCTANCE) CHAPTER 8 8.1 FORCE ON A...
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Transcript of 1 MAGNETOSTATIC FIELD (MAGNETIC FORCE, MAGNETIC MATERIAL AND INDUCTANCE) CHAPTER 8 8.1 FORCE ON A...
1
MAGNETOSTATIC FIELD(MAGNETIC FORCE, MAGNETIC MATERIAL
AND INDUCTANCE)
CHAPTER 8
8.1 FORCE ON A MOVING POINT CHARGE
8.2 FORCE ON A FILAMENTARY CURRENT
8.3 FORCE BETWEEN TWO FILAMENTARY CURRENT
8.4 MAGNETIC MATERIAL
8.5 MAGNETIC BOUNDARY CONDITIONS
8.6 SELF INDUCTANCE AND MUTUAL INDUCTANCE
8.7 MAGNETIC ENERGY DENSITY
8.1 FORCE ON A MOVING POINT CHARGE
EQFe Force in electric field:
BUQFm Force in magnetic field:
Total force:
me FFF BUEQF or
Also known as Lorentz force equation.
EQ
BUQ
Charge Condition
Field
Stationary -
Moving
E
EQ
B Field
BUEQ
Combination
E Band
EQ
Force on charge in the influence of fields:
8.2 FORCE ON A FILAMENTARY CURRENT
The force on a differential current element , due to the uniform magnetic field, :
__
dlIB
BlIdFd
____
dlBIBdlIF
0ldBIF
It is shown that the net force for any close current loop in the uniform magnetic field is zero.
Ex. 8.1: A semi-circle conductor carrying current I, is located in plane xy as shown in Fig. 8.1. The conductor is under the influence of uniform magnetic field, . Find:
(a) Force on a straight part of the conductor.
(b) Force on a curve part of the conductor.
0ˆByB
r
I
B
x
ySolution:
(a) The straight part length = 2r. Current flows in the x direction.
(N) 2ˆˆ)2(ˆ 001 IrBzByIrxF
BdlIF __
(N) 2ˆsinˆ 0
0
0
0
2
IrBzdrBIz
BldIF
(b) For curve part, will be in the –ve z direction and the magnitude is proportional to sin
Bdl__
12 FF Hence, it is observed that and it is shown that
the net force on a close loop is zero.
r
I
B
x
y
8.3 FORCE BETWEEN TWO FILAMENTARY CURRENT
P1(x1,y1,z1)P2(x2,y2,z2)
Loop l1
Loop l2
__
11 dlI
2
__
2 dlI
I1
I2
12ˆRa R12
x
y
z
The magnetic field at point P2 due to the
filamentary current I1dl1 :
P1(x1,y1,z1)P2(x2,y2,z2)
Loop l1
Loop l2
__
11 dlI
2
__
2 dlI
I1
I2
12ˆRa
R12
x
y
z
212
112
4
ˆ x 12
R
aldIHd R
(A/m)
BldIFd x (N)
2
12
11222
4
ˆ x x 12
R
aldIldIFdd Ro
222212
11222 x
4
ˆ x x
1
12 BldIR
aldIldIFd
l
Ro
We have :
where is the force due to I2dl2 and due to the magnetic field of loop l12Fd
Integrate:
2 1
12
212
11222
4
ˆ x x
l l
Ro
R
aldIdlIF
2 1
12
2212
1212 x
xˆ
4 l l
Ro ldR
ldaIIF
2
s
22 dsBJF s
dvBJFv
222
For surface current :
For volume current :
Ex. 8.2: Find force per meter between two parallel infinite conductor carrying current, I Ampere in opposite direction and separated at a distance d meter.
d
z
2B
21FI1
I2
I1 = I2 = I
x
y
Solution:
2B at position conductor 2
d
Ix
r
IHB
c
2
ˆ
2
ˆ101
0202
(N/m) 2
ˆ2
ˆ
2
ˆ)ˆ(
2
ˆ
2021
0
1
0
102
102
1
0
22
d
Iy
d
IIy
d
IxdzzI
d
IxdIF
Hence:
Ex. 8.3: A square conductor current loop is located in z = 0 plane with the edge given by coordinate (1,0,0), (1,2,0), (3,0,0) and (3,2,0) carrying a current of 2 mA in anti clockwise direction. A filamentary current carrying conductor of infinite length along the y axis carrying a current of 15 A in the –y direction. Find the force on the square loop.
Solution:
(1,0,0)
(1,2,0)
(3,0,0)
(3,2,0)
x
y
z
2 mA15 A
T ˆ103
104 ∴
A/m ˆ2
15ˆ
26
7-0 z
xHHB
zx
zx
IH
-
Field created in the square loop due to filamentary current :
zxy
aa Rl
ˆˆˆ
ˆˆˆ
Hence:
____
dlBIBdlIF
1
3
0
2
3
1
2
0
63
ˆ1
ˆˆ
ˆ
ˆ3
ˆˆ
ˆ103102
x y
x y
ydyz
xdxx
z
ydyz
xdxx
zF
nN ˆ8
ˆ2ˆ3
1lnˆ
3
2ˆ3ln106
ˆˆlnˆ3
1ˆln106
9
0
2
1
3
2
0
3
1
9
x
xyxy
xyyxxyyxF
(1,0,0)
(1,2,0)
(3,0,0)
(3,2,0)
x
y
z
2 mA15 A
8.4 MAGNETIC MATERIAL The prominent characteristic of magnetic material is magnetic polarization – the alignment of its magnetic dipoles when a magnetic field is applied.
Through the alignment, the magnetic fields of the dipoles will combine with the applied magnetic field.
The resultant magnetic field will be increased.
8.4.1 MAGNETIC POLARIZATION (MAGNETIZATION)
Magnetic dipoles were the results of three sources of magnetic moments that produced magnetic dipole moments : (i) the orbiting electron about the nucleus (ii) the electron spin and (iii) the nucleus spin.
The effect of magnetic dipole moment will produce bound current or magnetization current.
Magnetic dipole moment in microscopic view is given by :
______
dsIdm Am2
___
dmwhere is magnetic dipole moment in discrete and I is the bound current.
In macroscopic view, magnetic dipole moment per unit volume can be written as:
where is a magnetization and n is the volume dipole density when v -> 0.
M
vn
iidm
vM
1
___
0v
1lim A/m
Am-1______
dsnIdmnM
If the dipole moments become totally aligned :
___
idm
ds
___
idm
0
0___
M
dmi
Macroscopic v
Microscopic base
___
idm
0aB0aB
0M
0
0___
M
dmi
Magnetic dipole moments in a magnetic material
tend to align themselves
sdm'___
msm JJ and
8.4.2 BOUND MAGNETIZATION CURRENT DENSITIES
x
y
z
I
aB
M
___
dm
smJ
___
dm
M
M
aB
aB
M
Alignment of within a magnetic material under uniform conditions to form a non zero on the slab surfaces, and a within the material.
sdm'___
aB
smJ0mJ
8.4.3 TO FIND msm JJ and
y
Bound magnetization current :
vInddIm ________
)()()( ldsdnIldsdnIdIm
Am-1______
dsnIdmnM We have:
ldMdIm Hence:
__
ldM = I__
m
s
mm dsJ = I
through the loop l’
on the surface bound by the loop l’
s l s
mm dsMldMdsJ = I
Using Stoke’s Theorem:
Hence:
dIm = Mtan dl'
MJ m (Am-2) is the bound magnetization current density within the magnetic material.
And to find Jsm :
smJ
M
n
loop l’
On the slab surface
From the diagram :
smm Jld
dIM
tan
nMJ sm is the surface bound magnetization current density
(Am-1)
8.4.4 EFFECT OF MAGNETIZATION ON MAGNETIC FIELDS
HBJB
JH
oo
;
charge) (free
Due to magnetization in a material, we have seen the formation of bound magnetization and surface bound magnetization currents density.
due to free charges and bound magnetization currents
Maxwell’s equation:
M
BH
o
JH
Define:
m
o
JJB
mJM
MJB
o
JMB
o
)( MHB o
HM m litysusceptibi magnetic m
)1( mo HB
)1( mr
HB
typermeabiliro
Hence:
Magnetization in isotropic material:
Hence:
Ex. 8.4: A slab of magnetic material is found in the region given by 0 ≤ z ≤ 2 m with r = 2.5. If in the slab, determine: 2mWb/m ˆ5ˆ10 yxxyB
Ja )( mJb )( 0zon )( smJdMc )(
Solution:
236
70
kA/m ˆ775.4ˆ1010510
ˆ5.2104
1 )(
zz
zdy
dB
dBx
dBBHJa xy
r
23 kA/m ˆ163.710ˆ775.45.11 )( zzJJJb rmm
kA/m ˆ387.2ˆ775.4
5.2104
10ˆ5ˆ105.1
)(
7
3
0
yxxy
yxxy
BHMc
rmm
nM ˆsmJ
kA/m ˆ 775 . 4 ˆ 2.387
ˆ )ˆ 387 . 2 ˆ 775 . 4(
yy xx
z yx xy
smJ
Because of z = 0 is under the slab region of 0 ≤ z ≤ 2 , therefore zn ˆˆ
(d)
kA/m ˆ387.2ˆ775.4 )( yxxyMc
Ex. 8.5: A closely wound long solenoid has a concentric magnetic rod inserted as shown in the diagram.In the center region, find: in both air and magnetic rod, (b) the ratio of the in the rod to the in the air, (c) on the surface of the rod and within the rod. Assume the permeability of the rod equals
MBHa and , )(B
smJmJ
B
5o .
r
0
0
ab
z
magnetic rod
Solution:
(a) Using Ampere’s circuital law to the closed path P1 - P2 - P3- P4 . If using
path P1 - P2’ - P3’ -P4 - P1, Hz in the rod will be the same as in the air since
Ampere’s circuital law does not include any Im in its Ien term.
P3’P2’
P4
P3P2
P1
dNI
IdHdzzHzdH enz
P
P
z
3
2
)ˆ()ˆ(
sz JNI
H
Hence:
P3’P2’
P4
P3P2
P1
0
0
ab
zmagnetic rod
airin 0
rod in the )ˆ(ˆ
M
HzMzM zmz
)ˆ(0 zHzB
)ˆ(0 zr HzB
in air
in the rod
(since m of air is zero)
5ˆ
ˆ5 )(
zo
zo
air
rod
Hz
Hz
B
Bb
zczsm MrMznMJc ˆ)ˆ(ˆ )(
0)ˆ( zm MzMJ
flux Js
flux Jsm
z
smJ
M
rn ˆˆ
aa
0
b
sJ smJ
Js=Hz=NI/l
50Hz
Bz=0Hz
Jsm=4Hz
Mz
JsJsm
Bz
Hz
Mz=4Hz
Hz=NI/l
Plots of H, B and M, Js and Jsm
along the cross section of the solenoid and the magnetic rod
8.4.5 MAGNETIC MATERIAL CLASSIFICATION
Magnetic material can be classified into two main groups:
Group A – has a zero dipole moment
0dm diamagnetic material eg. Bismuth
9999834.0,1066.1 5 rm
Group B – has a non zero dipole moment
(a) Paramagnetic material - 0dm 0M;
aBWhen is applied, there will be a slight alignment of the atomic dipole moment to produce 0M
00002.1,102 5 rm μχEg. Aluminum -
dmaB
(b) Ferromagnetic material : has strong magnetic moment in the absence of an applied field.
Eg: metals such as nickel, cobalt and iron.
8.5 MAGNETIC BOUNDARY CONDITIONSMHB and ,To find the relationship between
s
HB /21n
l
2/h
sJ
2/h
Boundary2/h2/h
ab
cd
11 , m
22 , mRegion 2:
Region 1:
HB and To find normal component of at the boundary
Consider a small cylinder as 0h 0__
dsBand use
0
21
21
__
nn
nn
BB
sBsBdsB
2211 nn HH
s
HB /21n
l
2/h
sJ
2/h
Boundary2/h2/h
ab
cd
11 , m
22 , mRegion 2:
Region 1:HB and To find tangential component of
Consider a closed abcd as 0h
and use enc
l
IdlH __
at the boundary
stt
enctt
JHH
IlHlH
21
21
sJ tH1 tH 2where is perpendicular to the directions of and
sn JHHa 2121ˆ
21ˆna=
In vector form :
21ˆna is a normal unit vector from region 2 to region 1
x xy
z
yxz ˆˆˆ
stt J
BB
2
2
1
1
mJM
stt JHH 21
We have:
mv v m IdVJdVM
mlIdlM
__
and
smtt JMM 21
HM m
sm
t
m
t JMM
2
2
1
1
2
22
1
11
m
n
m
n MM
Hence:
We have:
Hence:
2211 nn HH
We have:We have:
Hence:
Hence:
2
2
1
121 or
tt
tt
BBHH
If Js = 0 :
If the fields were defined by an angle normal to the interface
222111 coscos BBBB nn
22 , mRegion 2:
11 , mRegion 1:1
2
x
y
Boundary at y = 0 plane
11 HorB
22 HorB
1B
1tB
1nB
22
2211
1
1 sinsin
BHH
Btt
2
1
2
1
2
1
tan
tan
r
r
(1)
(2)
Divide (2) to (1):
Ex. 8.6: Region 1 defined by z > 0 has 1 = 4 H/m and 2 = 7 H/m in region 2 defined by z < 0. A/m on the surface at z = 0. Given
, find
xJ sˆ80
mT ˆˆ3ˆ21 zyxB 2B
Solution:
Z = 0
#1
#2
1B
1tB
1nB
2B 2tB 2nB
21n z
z12n
xJ sˆ80
z
yx
tH 2
tH11BNormal component
ˆ
ˆ
ˆˆˆˆ3ˆ2
ˆ)ˆ(
12
121211
zBB
z
zzzyx
nnBB
nn
n
A/m ˆ750ˆ500
104
10ˆ3ˆ2
ˆ3ˆ2
6
3
1
11
111
yx
yxBH
yxBBB
tt
nt
A/m ˆ670ˆ500
ˆ80ˆ750ˆ500
ˆ80ˆˆ750ˆ500
ˆ1212
yx
yyx
xzyx
JnHH stt
mT ˆˆ69.4ˆ5.3
ˆ69.4ˆ5.3ˆ670ˆ500107
222
6222
zyxBBB
yxyxHB
nt
tt
Ex. 8.7: Region 1, where r1 = 4 is the side of the plane y + z < 1 . In region 2 , y + z > 1 has r2 = 6 . If find yxB ˆˆ21 22 and HB
Solution:
2/)ˆˆ(ˆ zyn The unit normal:
zyxHB
HzyxH
zyxBBB
BzynB
zyyxB
trt
tt
nt
nn
n
ˆ75.0ˆ75.0ˆ3
ˆ125.0ˆ125.0ˆ5.01
ˆ5.0ˆ5.0ˆ2
ˆ5.0ˆ5.0ˆ2
12
1
2
ˆˆˆˆ2
2202
20
1
111
2121
1
121211 ˆ)ˆ( nnBB n na
y
z
r1 = 4
r2 = 6#1
#2
O
12n
(A/m) ˆ04.0ˆ21.0ˆ5.01
(T) ˆ25.0ˆ25.1ˆ3
02
2
zyxH
zyxB
nt BBB 222
y + z = 1
8.6 SELF INDUCTANCE AND MUTUAL INDUCTANCE
+
_
VL
Coil
I
Magnetic flux
From circuit theory the induced potential across a wire wound coil such as solenoid or a toroid :
Simple electric circuit that shows the effect of energy stored in a magnetic field of an inductor :
dt
dILVL
where L is the inductance of the coil, I is the time varying current flowing through the coil – inductor.
2
2
1CVWE
In a capacitor, the energy is stored in the electric field :
In an inductor, the energy is stored in the electric field, as suggested in the diagram :
+
_
VL
Coil
I
Magnetic flux
switch
Idtdt
dILIdtVW
tt
t
tt
t
Lm 00
00
)( 2
1 2
0
0
JouleLIdILItt
t
HenryI
L
Define the inductance of an inductor :
where (lambda) is the total flux linkage of the inductor
Nm
)( HenryHI
L
I
NL m
Weber turns
Hence :
H
Two circuits coupled by a common magnetic flux that leads to mutual inductance.
I1 Circuit 1N1 turns
Circuit 2N2 turns
I2
1
1212 I
M
Mutual inductance :
12 is the linkage of circuit 2 produced by I1 in circuit 1
For linear magnetic medium M12 = M21
Ex. 8.8: Obtain the self inductance of the long solenoid shown in the diagram.
Solution:
mψAssume all the flux links all N turns and that does not vary over the cross section area of the solenoid.
B flux
N turns
22
22
2
have We
al
IN
Nal
NINaH
HB
NaBNm
l
aN
IL
22
I
SNb
NII
Nac
B
I
N
IL m
2
4
2
Ex. 8.9: Obtain the self inductance of the toroid shown in the diagram.
a
cb
m
0
avel Bads ˆ __
Mean path
Cross sectional area S
Feromagnetic core
N turns
I
Solution:
area sectional cross toroidal-
radiusmean - where2
2
S
bb
SNL
Ex. 8.10: Obtain the expression for self inductance per meter of the coaxial cable when the current flow is restricted to the surface of the inner conductor and the inner surface of the outer conductor as shown in the diagram.
Solution:
a
b
mψ
mψThe will exist only between a and b and will link all the current I
a
b
I
dzdr
r
I
I
dzdrH
IIL
b
a
c
c
b
a
cm
ln2
2
1
0
1
0
Ex. 8.11: Find the expression for the mutual inductance between circuit 1 and circuit 2 as shown in the diagram.
Let us assume the mean path :
2b >> (c-a)
Solution:
Two circuits coupled by a common magnetic flux that leads to mutual inductance.
I1 Circuit 1N1 turns
Circuit 2N2 turns
I2
a
b
c
b
SNN
I
SNbIN
I
Nac
B
I
N
IM m
22
4
21
1
211
1
2
2
12
1
2)12(
1
1212
8.7 MAGNETIC ENERGY DENSITY
Henry I
L
We have :
Joule 2
1
2
1
2
1 22 III
LIWm
BSNINIW mm 2
1
2
1
Consider a toroidal ring : The energy in the magnetic field :
Multiplying the numerator and denominator by 2b :
bSb
NIBWm
222
1
a
cb
m
IS
N turns
toroid theof volume theis )2( and 2
where VbSHb
NI
BHVWm 2
1
Hence :
32 2
1
2
1 JmHBHV
Ww m
m
HBwm 2
1In vector form :
dvHBI
WI
Lv
m 2
12222
Hence the inductance :
Ex. 8.12: Derive the expression for stored magnetic energy density in a coaxial cable with the length l and the radius of the inner conductor a and the inner radius of the outer conductor is b. The permeability of the dielectric is .
(J) ln4
21
8
1
82
1
2
2
22
2
22
22
a
blI
drrlr
IW
dvr
IdvHW
r
IH
b
a
m
vvm
Solution:
a
b
mψ