Bài giảng Engineering electromagnetic - Chapter IX: The Steady Magnetic Field - Nguyễn Công Phương
Given A = –yax + xay – zaz = ρaφ – zaz. Verify Stokes’ theorem for
the circular bounding contour in the xy plane; check the result for:
a) The flat circular surface in the xy plane
b) The hemispherical surface bounded by the contour
c) The cylindrical surface bounded by the contour
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Nguy ễn Công Ph ươ ng
Engineering Electromagnetics
The Steady Magnetic Field
Contents
I. Introduction
II. Vector Analysis
III. Coulomb’s Law & Electric Field Intensity
IV. Electric Flux Density, Gauss’ Law & Divergence
V. Energy & Potential
VI. Current & Conductors
VII. Dielectrics & Capacitance
VIII.Poisson’s & Laplace’s Equations
IX. The Steady Magnetic Field
X. Magnetic Forces & Inductance
XI. Time – Varying Fields & Maxwell’s Equations
XII. The Uniform Plane Wave
XIII.Plane Wave Reflection & Dispersion
XIV.Guided Waves & Radiation
The Steady Magnetic Field - sites.google.com/site/ncpdhbkhn 2
The Steady Magnetic Field (1)
1. Biot – Savart Law
2. Ampere’s Circuital Law
3. Curl
4. Stokes’ Theorem
5. Magnetic Flux & Magnetic Flux Density
6. Magnetic Potential
7. Derivation of the Steady – Magnetic – Field Law
The Steady Magnetic Field - sites.google.com/site/ncpdhbkhn 3
The Steady Magnetic Field (2)
• The source of the steady magnetic field may be:
– Permanent magnet
– Electric field changing linearly with time
– Direct current
• Consider the field produced by a differential dc element
in free space only
The Steady Magnetic Field - sites.google.com/site/ncpdhbkhn 4
Biot – Savart Law (1)
dL R12
Id L× a Id L× R 1
dH =R = P
4πR2 4 π R 3 aR12
I1
H: magnetic field intensity (A/m)
The direction of H is determined by the right-hand rule
I d L× a
dH = 1 1R 12
2 π 2
4 R12
IdLa× Id La ×
dHH=R → = ∫ R
4πR2 4 π R 2
The Steady Magnetic Field - sites.google.com/site/ncpdhbkhn 5
Biot – Savart Law (2)
I= Kb K
b
I= ∫ KdN I
IdLK= dS
IdLa× Ka × dS
H =∫R = ∫ R
4πRR2S 4 π 2
The Steady Magnetic Field - sites.google.com/site/ncpdhbkhn 6
z
dL 1
Biot – Savart Law (3) aR
Id L× a
dH = 1R 12 z’ a
2 π 2 z
4 R12 R12
=
dL1 dz ' a z
ρ − 2
aρ z ' a z
=ρ − → = ρaρ
R12 aρ z ' a z aR12 x y
ρ2+ z ' 2 I
Idz'a× (ρ a − z ') a ∞ Idz'a× (ρ a − z ') a
→ = zρ z → = zρ z
dH2 H2 ∫
4(π ρ 2+ z ') 2 3/2 −∞ 4(π ρ 2+ z ') 23/2
×= ×=
az aρ aa ϕ ; z a z 0
∞ ρdz 'a Iρa ∞
→ = I ϕ = ϕ dz '
H2 ∫ ∫
4π −∞ (ρ 2+ z ' 2 ) 3/2 4π −∞ (ρ 2+ z ' 2 ) 3/2
z '=∞
Iρaϕ z ' I
= = a
π πρ ϕ
4 ρ2 ρ 2+ z ' 2 2
z '=−∞
The Steady Magnetic Field - sites.google.com/site/ncpdhbkhn 7
z
dL 1
Biot – Savart Law (4) aR
z’ az
I R
z a H= a ϕ 12
z 2πρ
aφ 2
x ρaρ y
I
z
0 y
ρ aρ
x I
φ
z
=I α − α
H(sin2 sin 1 ) a ϕ x α y
4πρ 2 α1
ρ
The Steady Magnetic Field - sites.google.com/site/ncpdhbkhn 8
Biot – Savart Law (5)
I
H= a ϕ
2πρ
The Steady Magnetic Field - sites.google.com/site/ncpdhbkhn 9
Biot – Savart Law (6)
Id L× a
H = ∫ R
4π R2
K× a dS
H = ∫ R
S 4π R2
J× a dV
H = ∫ R
V 4π R2
The Steady Magnetic Field - sites.google.com/site/ncpdhbkhn 10
z
dH= d H + d H
Ex. 1 Biot – Savart Law (7) 1 2
dH dH
Given a circular hoop of radius a centered about the 2 1
P(0,0, z )
origin in the xy plane carries a constant current I. R2
Find MFI at P?
R
Id L× a 1
dH = 1R 1 0 φ a y
1 π 2
4 R1 I
= ϕ x dL1
dL1 ad a ϕ
= − + ϕ
R1 a aρ z a z → =Iad +
dH1 ( a az z a ρ )
4π (z2+ a 23/2 )
R= z2 + a 2
1 ϕ
−aaρ + z a =Iad −
= z dH2 ( a az z a ρ )
aR1 4π (z2+ a 23/2 )
z2+ a 2
2Ia2 d ϕ π 2Ia2 d ϕ Ia 2
→ = → = = a
dH a z H∫ a z 2 23/2 z
4π (z2+ a 23/2 ) 0 4π (z2+ a 23/2 ) 2(z+ a )
The Steady Magnetic Field - sites.google.com/site/ncpdhbkhn 11
z
Ex. 2 Biot – Savart Law (8)
Find MFI on the z axis? a
2 I
= Ia d
H z, red
2(z2+ a 23/2 )
0
2 a y
= Ia
H z, blue I
2[(z− d )2 + a 23/2 ] x
2
→ =Ia 1 + 1
Hz
2 (za2+ 23/2 ) [( zda − ) 2 + 23/2 ]
∂H ∂2H
z = 0 z =
2 0
∂z = ∂z
z d / 2 zd=/ 2, da =
The Steady Magnetic Field - sites.google.com/site/ncpdhbkhn 12
z
Ex. 3 Biot – Savart Law (9) P (0, 0, z)
Find MFI on the z axis? L/2
2 dz' y
= Ia z'
H a z
2(z2+ a 23/2 )
–L/2
z= z − z', I = Kdz ' x a
2 K
→ = a Kdz '
dH z
2[(z− z ')2 + a 23/2 ]
L /2
a2 Kdz '
→H =
z ∫ −2 + 23/2
z'=− L /2 2[(z z ') a ]
− + +
=K zL/2 + zL /2 =
lim Hz K
22 22 →∞
2 (zL− /2) + a ( zL + /2) + a L
The Steady Magnetic Field - sites.google.com/site/ncpdhbkhn 13
The Steady Magnetic Field
1. Biot – Savart Law
2. Ampere’s Circuital Law
3. Curl
4. Stokes’ Theorem
5. Magnetic Flux & Magnetic Flux Density
6. Magnetic Potential
7. Derivation of the Steady – Magnetic – Field Law
The Steady Magnetic Field - sites.google.com/site/ncpdhbkhn 14
Ampere's Circuital Law (1)
∫ HL.d= I
I
The Steady Magnetic Field - sites.google.com/site/ncpdhbkhn 15
Ampere's Circuital Law (2)
Ex. 1
∫ HL.d= I I
z
ρ
dL
aR dL
=
z’ az HHϕ a ϕ
R12
dL=ρtan( d ϕ ) aϕ ≈ ρϕ d a ϕ
π
ρa 2
x ρ y →H.d L = Hϕ ρ d ϕ
∫ ∫ 0
I 2π
= Hϕ ρ d ϕ
∫0
I
H= a ϕ I
πρ =Hϕ 2πρ = I →Hϕ =
2 2πρ
The Steady Magnetic Field - sites.google.com/site/ncpdhbkhn 16
Ampere's Circuital Law (3)
Ex. 2
I
I
Hϕ =
ρ 2πρ
c
I a
I I
Hϕ = ( a <ρ < b) b
2πρ
ρ 2 ρ= ρ ρ= ρ
ρ < = 2 1 1
a : Iρ I ρ ϕ= − ϕ ϕ= ϕ
a2 →2πρ Hϕ = I 1 1
a2
IHρ= 2πρ ϕ
ρ
→Hϕ = I(ρ < a ) Hϕ
2π a2
The Steady Magnetic Field - sites.google.com/site/ncpdhbkhn 17
Ampere's Circuital Law (4)
Ex. 2
I
Hϕ = ( a <ρ < b)
2πρ
ρ
=ρ <
Hϕ I( a ) c
2πa2 I a
I b
ρ > = + =−=
c : IIenclosed inner conductor I outer conductor II 0
→Hϕ =0 (ρ > c )
The Steady Magnetic Field - sites.google.com/site/ncpdhbkhn 18
Ampere's Circuital Law (5)
Ex. 2
I
Hϕ = ( a <ρ < b)
2πρ
ρ
=ρ <
Hϕ I( a ) c
2πa2 I a
I b
Hϕ =0 (ρ > c )
<ρ <
b c : ρ22− 2 − ρ 2
= + =−b = c
IIenclosed inner conductor I partial outer coductor II I
cb22− cb 22 −
Ienclosed
Hϕ =
2πρ
I c 2− ρ 2
→=Hϕ ( b <<ρ c )
2πρ c2− b 2
The Steady Magnetic Field - sites.google.com/site/ncpdhbkhn 19
Ampere's Circuital Law (6)
Ex. 2
ρ I
HIϕ=(ρ < aH ); ϕ = ( a <ρ < b)
2π a2 2πρ
I c 2− ρ 2
Hϕ=( bcH 0 ( ρ c ) c
2πρ c2− b 2 I a
I b
I
2πa
4a
I
4π a
a 3a
0 2a 3a= b 4a= c
The Steady Magnetic Field - sites.google.com/site/ncpdhbkhn 20
J
Ex. 3 Ampere's Circuital Law (7)
b a
A very long, hollow conductor of radii a & b is located along
z axis, carrying a uniformly distributed current I in z direction.
Find the magnetic field intensity at any point in space?
ρ ≤a:H. d L = I =→= 0 H 0
∫ ρ ≤a
a≤ρ ≤ b: H. dI L =
∫ a≤ρ ≤ b
2 2
= πρ Iρ − a
∫ H.d L 2 Hϕ →H = × I a ϕ
πρ 2 2
ρ 2− 2 2 b− a
= a
I≤ρ ≤ I
a b b2− a 2
I
ρ ≥b : H = a ϕ
2πρ
The Steady Magnetic Field - sites.google.com/site/ncpdhbkhn 21
Ampere's Circuital Law (8)
+ − = 3 z
HLHx1 x 2 ( L ) KL y
→ − = 1 3’ y
Hx1 H x 2 K y
x
− = 1’
Hx3 H x 2 K y
→ = K = Kyay
HHx3 x 1 2
1 2’
H= K( z > 0) L
x2 y
→ z
1
H= − K( z < 0)
x2 y h
→ =1 ×
H K a N
2 K = –Kyay
H=× K a (0 <<z h ) 0
N
H =0 (z h ) K = Kyay
The Steady Magnetic Field - sites.google.com/site/ncpdhbkhn 22
The Steady Magnetic Field
1. Biot – Savart Law
2. Ampere’s Circuital Law
3. Curl
4. Stokes’ Theorem
5. Magnetic Flux & Magnetic Flux Density
6. Magnetic Potential
7. Derivation of the Steady – Magnetic – Field Law
The Steady Magnetic Field - sites.google.com/site/ncpdhbkhn 23
Curl (1) H = H0 = Hx0ax + Hy0ay + Hz0az
HL.d= I 4 3
∫ z ∆x
∆ = ∆
(H. L ) 12−Hy ,12 − y
∂ 1∆ 2
H y 1 y
H− ≈ H + ∆ x
y,1 2 y 0 ∂x 2 x y
∂H
→∆ ≈ +1 y ∆∆
(H. L ) 1− 2Hy 0 x y
2 ∂x
∂
∆ ≈ −∆≈− +1 H x ∆∆
()H. L 23−Hx ,23 − () xH x 0 yx
2 ∂y
∂H
∆ ≈− −1 y ∆∆
(H. L ) 3− 4Hy 0 x y
2 ∂x
∂
∆ ≈ −1 Hx ∆∆
(H. L ) 4− 1Hx 0 y x
2 ∂y
The Steady Magnetic Field - sites.google.com/site/ncpdhbkhn 24
Curl (2) H = H0 = Hx0ax + Hy0ay + Hz0az
HL.d= I 4 3
∫ z ∆x
∂H
∆ ≈ +1 y ∆∆
(H. L ) 1− 2Hy 0 x y 1∆ 2
2 ∂x y
y
∂ x
∆ ≈− +1 H x ∆∆
(H. L ) 2− 3Hx 0 y x
2 ∂y
1 ∂H ∂H ∂
∆ ≈− −y ∆∆ → ≈y −H x ∆∆
(H. L ) 3− 4Hy 0 x y ∫ H.d L x y
2 ∂x ∂x ∂ y
∂
∆ ≈ −1 H x ∆∆
(H. L ) 4− 1Hx 0 y x
2 ∂y
≈ ∆ + ∆
∫ HL.d ( H.L )(12− H.L ) 23 −
+∆ +∆
(H. L )34− ( H. L ) 41 −
The Steady Magnetic Field - sites.google.com/site/ncpdhbkhn 25
Curl (3) H = H0 = Hx0ax + Hy0ay + Hz0az
∂H ∂
≈y −Hx ∆∆ 4 3
∫ H.d L x y z ∆
∂x ∂ y x
∫ H.d L = ∆ I
1∆y 2
∆I ≈ J ∆∆ xy
z x y
∂H ∂
→ ≈y −Hx ∆∆≈∆∆
∫ H.d L xyJxyz
∂x ∂ y
H.d L ∂H ∂ H.d L ∂H ∂
→∫ ≈y −≈Hx →∫ =−=y H x
Jz lim J z
∆∆xy ∂ x ∂ y ∆x, ∆ y → 0 ∆∆xy ∂ x ∂ y
H.d L ∂ ∂H
∫ =H z −y =
lim J x
∆y, ∆ z → 0 ∆∆yz ∂ y ∂ z
H.d L ∂ ∂
∫ =HHx − z =
lim J y
∆z, ∆ x → 0 ∆∆zx ∂ z ∂ x
The Steady Magnetic Field - sites.google.com/site/ncpdhbkhn 26
Curl (4)
H.d L ∂H ∂
∫ =y −H x =
lim J z
∆x, ∆ y → 0 ∆∆xy ∂ x ∂ y
H.d L ∂ ∂H
∫ =Hz −y =
lim J x
∆y, ∆ z → 0 ∆∆yz ∂ y ∂ z
H.d L ∂ ∂
∫ =H x −H z =
lim J y
∆z, ∆ x → 0 ∆∆zx ∂ z ∂ x
∫ H.d L
Define ()rotH = lim
N ∆S → 0 ∆
N SN
- SN : planar area enclosed by the closed line integral
- (rot H)N : the component (of rot H) perpendicular to SN
The Steady Magnetic Field - sites.google.com/site/ncpdhbkhn 27
Curl (5)
∫ H.d L
()rotH = lim
N ∆ → ∆
SN 0 SN
∂∂H∂ ∂ ∂ H ∂
=−Hzy +−Hx H z +−y H x
rot H ax a y a z
∂∂yz ∂∂ zx ∂∂ xy
ax a y a z
∂ ∂ ∂
rot H =
∂x ∂ y ∂ z
Hx H y H z
rot HH= ∇×
The Steady Magnetic Field - sites.google.com/site/ncpdhbkhn 28
Curl (6)
∂∂H∂ ∂ ∂ H ∂
=∇×=Hz −y +Hx − H z +y − H x
rot HH ax a y a z
∂∂yz ∂∂ zx ∂∂ xy
∂∂∂HH ∂ ∂∂(ρ HH )
∇×=1Hz −ϕρ +− H z + 1 ϕρ − 1
H aaρ ϕ a z
ρϕ∂∂z ∂∂ z ρ ρρρϕ ∂ ∂
∂(sin)Hθ ∂ ∂ ∂ () rH
∇×=1ϕ −Hθ + 1 1 Hr − ϕ
H ar a θ
rsinθ∂∂ θ ϕ r sin θϕ ∂∂ r
∂ ∂
1 (rHθ ) Hr
+ − aϕ
r∂ r ∂ θ
The Steady Magnetic Field - sites.google.com/site/ncpdhbkhn 29
Ex. Curl (7)
Find the curl of the following vectors:
=2 + 2 +
a) A xy ax yz a y xy a z
z sin ϕ
b)B=ρ cos ϕ a + a
z ρ ρ
θ ϕ
=2 θ ϕ + cos sin
c)C r sin cos ar a θ
r2
The Steady Magnetic Field - sites.google.com/site/ncpdhbkhn 30
Curl (8)
∂∂HH ∂∂ ∂ ∂
=∇×=HHHHz −y + xz − +y − x
Curl: rot HH ax a y a z
∂∂yz ∂∂ zx ∂∂ xy
∂V ∂ V ∂ V
Gradient: ∇=V a + a + a
∂xx ∂ y y ∂ z z
∂D ∂D ∂D
Divergence: ∇=.D x +y + z
∂x ∂ y ∂ z
The Steady Magnetic Field - sites.google.com/site/ncpdhbkhn 31
Curl (9)
∂∂HH∂ ∂ ∂ ∂
=∇×=HHz −y +HHx − z +y − x
rot HH ax a y a z
∂∂yz ∂∂ zx ∂∂ xy
The Steady Magnetic Field - sites.google.com/site/ncpdhbkhn 32
Curl (10)
∂∂H∂ ∂ ∂ H ∂
∇×=Hz −y +Hx − H z +y − H x
rot H= H ax a y a z
∂∂yz ∂∂ zx ∂∂ xy
H.d L ∂H ∂
∫ =y −H x =
lim J z
∆x, ∆ y → 0 ∆∆xy ∂ x ∂ y
H.d L ∂ ∂H
∫ =Hz −y =
lim J x
∆y, ∆ z → 0 ∆∆yz ∂ y ∂ z
H.d L ∂ ∂
∫ =HHx − z =
lim J y
∆z, ∆ x → 0 ∆∆zx ∂ z ∂ x
∇×H = J
(the second of Maxwell’s four equations)
The Steady Magnetic Field - sites.google.com/site/ncpdhbkhn 33
The Steady Magnetic Field
1. Biot – Savart Law
2. Ampere’s Circuital Law
3. Curl
4. Stokes’ Theorem
5. Magnetic Flux & Magnetic Flux Density
6. Magnetic Potential
7. Derivation of the Steady – Magnetic – Field Law
The Steady Magnetic Field - sites.google.com/site/ncpdhbkhn 34
Stokes' Theorem (1)
aN
≈ IN ∆S
J N ∫ H.d L ∆S
∆S →J ≈
N ∆S ∆
= S ∆
IN∫ H.d L ∆ S S
= ∇×
JHN( ) N ∆S
∫ H.d L ∆S
→ ≈∇×(H ) =∇× ( H ). a
∆S N N
S
→ ≈∇× ∆=∇× ∆
∫ HL.d∆S ( Ha ). N S ( HS ).
→HL.d = ( ∇× HS ). d
∫ ∫ S
The Steady Magnetic Field - sites.google.com/site/ncpdhbkhn 35
z
Ex. 1 Stokes' Theorem (2)
1
Given H = 6 rsin φar + 18 rsin θcos φaφ A/m.
Verify Stokes’ theorem. 3
∫HL.d= ∫ ( ∇× HS ). d
z S 2
φ = 0.25 π r = 5
y
dr x
y
rd θ = +θ + θ ϕ
x dLa drr rd aθ rsin d a ϕ
rsin θdφ
The Steady Magnetic Field - sites.google.com/site/ncpdhbkhn 36
z
Ex. 1 Stokes' Theorem (3)
1
Given H = 6 rsin φar + 18 rsin θcos φaφ A/m.
Verify Stokes’ theorem. 3
HL.d= ( ∇× HS ). d
∫ ∫ S 2
= +θ + θ ϕ
dLa drr rd aθ rsin d a ϕ φ = 0.25 π r = 5
→ = ++θ θ ϕ
HLHa.d .( drr rd aθ r sin d a ϕ )
∫ ∫ y
=Hdr + Hrdθ + Hrsin θ d ϕ
∫r ∫θ ∫ ϕ x
H dr= H dr + H dr + H dr
∫r ∫1 r ∫ 2 r ∫ 3 r → =
∫ Hr dr 0
= → =
1,2,3:r 5 dr 1,2,3 0
θ
∫ Hθ rd →θ =
∫ Hθ rd 0
Hθ = 0
The Steady Magnetic Field - sites.google.com/site/ncpdhbkhn 37
z
Ex. 1 Stokes' Theorem (4)
1
Given H = 6 rsin φar + 18 rsin θcos φaφ A/m.
Verify Stokes’ theorem. 3
HL.d= ( ∇× HS ). d
∫ ∫ S 2
= +θ + θ ϕ
∫H.d L ∫ Hdrr ∫ Hrdθ ∫ Hr ϕ sin d φ = 0.25 π r = 5
=
∫ Hr dr 0 y
θ =
∫ Hθ rd 0 x
Hrϕsinθϕ d= Hr ϕ sin θϕ d + Hr ϕ sin θϕ d + Hr ϕ sin θϕ d
∫ ∫1 ∫ 2 ∫ 3
1,3:ϕ= const →d ϕ = 0
1,3
→Hrϕsinθϕ d = Hr ϕ sin θϕ d
∫ ∫ 2
The Steady Magnetic Field - sites.google.com/site/ncpdhbkhn 38
z
Ex. 1 Stokes' Theorem (5)
1
Given H = 6 rsin φar + 18 rsin θcos φaφ A/m.
Verify Stokes’ theorem. 3
HL.d= ( ∇× HS ). d
∫ ∫ S 2
= +θ + θ ϕ
∫H.d L ∫ Hdrr ∫ Hrdθ ∫ Hr ϕ sin d φ = 0.25 π r = 5
=
∫ Hr dr 0 y
θ =
∫ Hθ rd 0 x
Hrϕsinθϕ d= Hr ϕ sin θϕ d
∫ ∫ 2
0.25 π 0.25 π
→H.d L = Hrϕ sin θ d ϕ = Hϕ rsin θ d ϕ = Hϕ 5sin(0.22π ) d ϕ
∫ ∫ 2 ∫0 ∫0 2
0.25 π
= 3.19 Hϕ d ϕ
∫0 2
The Steady Magnetic Field - sites.google.com/site/ncpdhbkhn 39
z
Ex. 1 Stokes' Theorem (6)
1
Given H = 6 rsin φar + 18 rsin θcos φaφ A/m.
Verify Stokes’ theorem. 3
HL.d= ( ∇× HS ). d
∫ ∫ S 2
= +θ + θ ϕ
∫H.d L ∫ Hdrr ∫ Hrdθ ∫ Hr ϕ sin d φ = 0.25 π r = 5
0.25 π
= 3.19 Hϕ d ϕ y
∫0 2
x
Hϕ =18 × 5sin(0.22π )cos ϕ = 57.37cos ϕ
2
0.25 π 0.25 π
→H.d L = 3.19 × 57.37cos ϕ d ϕ = 182.84cos ϕd ϕ
∫ ∫ 0 ∫0
=ϕ0.25 π = π =
182.84sin0 182.84sin(0.25 ) 129.27 A
The Steady Magnetic Field - sites.google.com/site/ncpdhbkhn 40
z
Ex. 1 Stokes' Theorem (7)
1
Given H = 6 rsin φar + 18 rsin θcos φaφ A/m.
Verify Stokes’ theorem. 3
HL.d= ( ∇× HS ). d
∫ ∫ S 2
=
∫ H.d L 129.27A φ = 0.25 π r = 5
(∇× H ). d S y
∫S
∂(H sinθ ) ∂
∇×=1 ϕ − Hθ x
H a r
r sin θ∂ θ ∂ ϕ
∂∂ ∂ ∂
1 1Hr (rH ϕ ) 1 (rH θ ) H r
+ − aθ + − a ϕ
rsin θ∂∂ ϕ rrr ∂∂ θ
1 1 1
=()36r sinθθϕ cos cosa + 6 r cos ϕ − 36 r sin θϕ cos a θ
rsinθr r sin θ
The Steady Magnetic Field - sites.google.com/site/ncpdhbkhn 41
z
Ex. 1 Stokes' Theorem (8)
1
Given H = 6 rsin φar + 18 rsin θcos φaφ A/m.
Verify Stokes’ theorem. 3
HL.d= ( ∇× HS ). d
∫ ∫ S 2
=
∫ H.d L 129.27A φ = 0.25 π r = 5
(∇× H ). d S y
∫S
x
=1()θθϕ + 1 1 ϕ − θϕ
∫ 36r sin cos cosar 6 rrd cos 36 sin cos aθ S
S rsinθ r sin θ
=()θϕ +1 ϕ − θϕ
∫ 36cos cosar 6cos 36sin cos aθ d S
S sin θ
The Steady Magnetic Field - sites.google.com/site/ncpdhbkhn 42
z
Ex. 1 Stokes' Theorem (9)
1
Given H = 6 rsin φar + 18 rsin θcos φaφ A/m.
Verify Stokes’ theorem. 3
∫ (∇× H ). d S
z S 2
φ = 0.25 π r = 5
y
dr x
y
rd θ = 2 θ θ ϕ
x dS rsin dd a r
rsin θdφ
The Steady Magnetic Field - sites.google.com/site/ncpdhbkhn 43
z
Ex. 1 Stokes' Theorem (10)
1
Given H = 6 rsin φar + 18 rsin θcos φaφ A/m.
Verify Stokes’ theorem. 3
HL.d= ( ∇× HS ). d
∫ ∫ S 2
=
∫ H.d L 129.27A φ = 0.25 π r = 5
y
x
∇× =()θϕ +1 ϕ − θϕ
∫ (H ). d S ∫ 36cos cosar 6cos 36sin cos aθ d S
S S sin θ
= 2 θ θ ϕ
dS rsin dd a r
→( ∇×HS ).d = 36cosθ cos ϕ aS d = (36cosθ cos ϕ )(5)2 sin θθϕd d
∫S ∫ S r ∫S
The Steady Magnetic Field - sites.google.com/site/ncpdhbkhn 44
z
Ex. 1 Stokes' Theorem (11)
1
Given H = 6 rsin φar + 18 rsin θcos φaφ A/m.
Verify Stokes’ theorem. 3
HL.d= ( ∇× HS ). d
∫ ∫ S 2
=
∫ H.d L 129.27A φ = 0.25 π r = 5
2
(∇×H ).d S = (36cosθ cos ϕ )(5) sin θθϕ d d y
∫S ∫ S
0.25π 0.22 π x
= (36cosθ cos ϕ )(5)2 sin θθϕd d
∫0 ∫ 0
0.22 π
π 0.25 π
0.25 1 2
= 900 sinθ cos ϕd ϕ = 182.84cos ϕd ϕ
∫0 ∫0
2 0
0.25 π 0.25 π
= 182.84cos ϕd ϕ =182.84sinϕ = 129.27 A
∫0 0
The Steady Magnetic Field - sites.google.com/site/ncpdhbkhn 45
Ex. 2 Stokes' Theorem (12)
Extract ∫ H . d L = I from ∇×H = J
∇×H = J
→∇×(H.S ) d = J. d S
→( ∇×H.S ) d = J.S d = I
∫S ∫ S
HL.d= ( ∇× HS ). d
∫ ∫ S
→∫ HL.d = I
The Steady Magnetic Field - sites.google.com/site/ncpdhbkhn 46
z
Ex. 3 Stokes' Theorem (13)
Given A = –yax + xay – zaz = ρaφ – zaz. Verify Stokes’ theorem for
the circular bounding contour in the xy plane; check the result for:
a) The flat circular surface in the xy plane
b) The hemispherical surface bounded by the contour
c) The cylindrical surface bounded by the contour
0
R y
AL.d= ( ∇ × AS ). d x
∫ ∫ S
π
2 2 2 2
dL= Rd ϕ a ϕ →A.d L = R d ϕ →A.d L = Rdϕ = 2 π R
∫ ∫ 0
∂∂A ∂∂ ∂ A ∂
∇×=−Azy +− AA xz +− y A x =
A ax a y aa zz2
∂∂yz ∂∂ zx ∂∂ xy
(∇×=AS ).d (2 a ). dS a = 2 dS = 2 π R 2
∫flat ∫ flatz z ∫ flat
The Steady Magnetic Field - sites.google.com/site/ncpdhbkhn 47
z
Ex. 3 Stokes' Theorem (14)
Given A = –yax + xay – zaz = ρaφ – zaz. Verify Stokes’ theorem for
the circular bounding contour in the xy plane; check the result for:
a) The flat circular surface in the xy plane
b) The hemispherical surface bounded by the contour
c) The cylindrical surface bounded by the contour
0
R y
2π R2 =AL . d = ( ∇× AS ). d x
∫ ∫ S
(∇ ×AS ).d = (2 a ).( Rdd2 sinθ θ ϕ ) a
∫hemi ∫ hemi z r
= θ
az .a r cos
π / 2
π/ 2 2 π cos2 θ
→∇×(A ).d S = R2 sin 2 θ dd θ ϕ = − 2π R2 = 2π R2
∫hemi ∫θ=0 ∫ ϕ = 0
2 θ =0
(∇ ×A ).d S = ?
∫cy
The Steady Magnetic Field - sites.google.com/site/ncpdhbkhn 48
The Steady Magnetic Field
1. Biot – Savart Law
2. Ampere’s Circuital Law
3. Curl
4. Stokes’ Theorem
5. Magnetic Flux & Magnetic Flux Density
6. Magnetic Potential
7. Derivation of the Steady – Magnetic – Field Law
The Steady Magnetic Field - sites.google.com/site/ncpdhbkhn 49
Magnetic Flux & Magnetic Flux Density (1)
• The magnetic flux density B is defined in free space:
B = µ0H
• Unit: Wb/m2 or T or G (1T = 10000G)
π –7
• Permeability µ0 = 4 10 H/m
• Definition of magnetic flux: Φ = BS.d
∫S
• Electric flux: Ψ =DS.d = Q
∫S
The Steady Magnetic Field - sites.google.com/site/ncpdhbkhn 50
Magnetic Flux & Magnetic Flux Density (2)
• Gauss’ law for the magnetic field:
B.d S = 0
∫S
• The last of Maxwell’s four equations:
∇.B = 0
D.d S = Q = ρ dv
• Maxwell’s four equations: ∫S ∫ V v
∇ = ρ E.d L = 0
.D v ∫
∇× =
E 0 H.Ld= I = J.S d
∫ ∫ S
∇×H = J
B.d S = 0
∇.B = 0 ∫S
The Steady Magnetic Field - sites.google.com/site/ncpdhbkhn 51
Ex. 1 Magnetic Flux & Magnetic Flux Density (3)
2 2
Given the magnetic field B = 3 xy az Wb/m . Determine the magnetic flux crossing
the portion of the xy plane lying between x = 0, x = 1, y = 0, & y = 1.
Φ = B.d S
∫S
1 1 2
= 2 → Φ = ∫ ∫ 3xy dxdy = 0.5 Wb
B3xy a z x=0 y = 0
=
dS dxdy a z
The Steady Magnetic Field - sites.google.com/site/ncpdhbkhn 52
Ex. 2 Magnetic Flux & Magnetic Flux Density (4)
Find the flux between the conductors of the coaxial line.
I
Hϕ = ( a <ρ < b)
2πρ
µ
0I d
→B =µ H = a ϕ
0 2πρ
c
Φ = ∫ B.d S I a
S I b
dS= dρ dz a ϕ
d b µ µ
→ Φ = 0I ρ 0Id b
∫ ∫ aϕ.d dz a ϕ = ln
0 a 2πρ 2π a
The Steady Magnetic Field - sites.google.com/site/ncpdhbkhn 53
Ex. 3 Magnetic Flux & Magnetic Flux Density (5)
Find the total flux through the rectangular circuit. I
dρ
a
L
Φ = B.d S ρ
∫S
µ I
B=µ H = 0 a
0 πρ ϕ
2 b
dS= Ld ρ a ϕ
ρ =b µ I µ IL b
→ Φ = 0 Ld ρ = 0 ln
∫ρ =a 2πρ 2π a
The Steady Magnetic Field - sites.google.com/site/ncpdhbkhn 54
The Steady Magnetic Field
1. Biot – Savart Law
2. Ampere’s Circuital Law
3. Curl
4. Stokes’ Theorem
5. Magnetic Flux & Magnetic Flux Densit y
6. Magnetic Potential
7. Derivation of the Steady – Magnetic – Field Law
The Steady Magnetic Field - sites.google.com/site/ncpdhbkhn 55
Magnetic Potential (1)
• The scalar magnetic potential Vm is defined by:
= −∇
H Vm
∇× = → ∇ × = = ∇ × −∇
• H J H J (Vm )
• Curl of gradient of a scalar field is zero, therefore:
=−∇ =
HVm ( J 0)
∇ = →∇ =µ ∇ =
.B 0 .B0 .H 0
→µ ∇−∇ = →∇2 = =
0 .(Vm ) 0 Vm 0 (J 0)
a
V= − H. d L
m, ab ∫b
The Steady Magnetic Field - sites.google.com/site/ncpdhbkhn 56
y
Magnetic Potential (2)
P(ρ, π/4, 0)
a <ρ <b :J = 0
I φ x
= I <ρ < c
H a ϕ (a b) I 1 ∂V a b
2πρ → =−∇V =− m
πρm ϕ ρ∂ ϕ
= −∇ = 2
HVm ( J 0)
∂V I I
→m = − →V = − ϕ
∂ϕ2 π m 2π
I 1
AssumeV = 0 →=V2 n − π ( n =±± 0, 1, 2, ...)
m ϕ =0 mP 2π 4
1
=I n − ( n =±± 0, 1, 2, ...)
8
The Steady Magnetic Field - sites.google.com/site/ncpdhbkhn 57
Magnetic Potential (3)
• Definition of a vector magnetic potential A:
BA= ∇×
• Unit: Wb/m
1 1 1
• HBA= = ∇× →∇×HJA = = ∇×∇×
µ µ µ
0 0 0
• A may be determined by: dL R
P
µ aR
= 0Id L I
A ∫ µ Id L
4π R dA = 0
4π R
The Steady Magnetic Field - sites.google.com/site/ncpdhbkhn 58
Magnetic Potential (4) R=ρ2 + z 2
µ z
= 0Id L P(ρ, φ, z)
dA Id L = Idz az
4π R µ Idz
→ = 0 y
dL= dz a dA a z
z 4π ρ 2+ z 2
R=ρ 2 + z 2 x
φ ρ z
µ
→=0Idz az ==
dAAAz dϕ0 d ρ 0
π ρ 2+ 2 1
4 z →dHA = ∇× d
1 µ
HA= ∇× 0
µ 1 ∂dA
0 = − z a
µ∂ ρ ϕ
0
Idz ρ
→dH = a ϕ
4π (ρ 2+ z 23/2 )
The Steady Magnetic Field - sites.google.com/site/ncpdhbkhn 59
Magnetic Potential (5)
µ Id L
dA = 0
4π R
• In the case of current flow throughout a volume with a
density J, then:
Id L = Jdv
µ Jdv
→A = 0
∫V 4π R
The Steady Magnetic Field - sites.google.com/site/ncpdhbkhn 60
Ex. 1 Magnetic Potential (6)
A very long, straight conductor lies along the z axis, carrying a uniform current I in
the z direction. Find the magnetic potential difference between two points in space?
a
V= − H. d L
m, ab ∫b
=ρ + ρ ϕ + a I I
dLa dρ d a ϕ dz a z →V = − d ϕ =(ϕ − ϕ )
m, ab ∫b 2π 2π b a
I
H= a ϕ
2πρ
The Steady Magnetic Field - sites.google.com/site/ncpdhbkhn 61
z
L
Ex. 2 Magnetic Potential (7)
Id L R
Find the vector magnetic potential in the plane bisecting a
straight piece of thin wire of finite length 2 L in free space. 0 ρ
ρ
I P( ,0,0)
µ Id L
dA = 0 −L
4π R
=
IdL Idz ' a z
R=ρ 2 + ( z ') 2
L µ µ 2+ρ 2 +
→ρ = 0Idz ' = 0I L L
A( ,0,0) ∫ a z ln az
2 2 4π 2 2
z'=− L 4π ρ + (z ') L+ρ − L
The Steady Magnetic Field - sites.google.com/site/ncpdhbkhn 62
The Steady Magnetic Field
1. Biot – Savart Law
2. Ampere’s Circuital Law
3. Curl
4. Stokes’ Theorem
5. Magnetic Flux & Magnetic Flux Density
6. Magnetic Potential
7. Derivation of the Steady – Magnetic – Field Law
The Steady Magnetic Field - sites.google.com/site/ncpdhbkhn 63
(1)
• Use formulae/definitions:
×
=Id L a R =µ =∇×
HBHBA∫ 0
4π R2
• to show that
µ Jdv
A = 0
∫V 4π R
µ Jdv Id L× a
AH=∫0 → = ∫ R
V 4π R 4π R2
The Steady Magnetic Field - sites.google.com/site/ncpdhbkhn 64
(2)
µ Jdv Id L× a
AH=∫0 → = ∫ R
V 4π R 4π R2
µ J dv
Current element at ( x , y , z ), A at (x , y , z ) →A = 0 1 1
1 1 1 2 2 2 2 ∫V π
4 R12
BH= µ BA∇×
0 →H = =
= ∇× µ µ
BA 0 0
∇ ×A ∇ µ J dv 1 J dv 1 J
→=H 2 2 =× 2 0 1 1 = ∇ × 1 1 = ∇ × 1 dv
2 µ µ∫V π π ∫V 2 π ∫V 2 1
0 04 R 12 4 R12 4 R12
∇×()()SV =∇ S × V + S ( ∇× V )
1 1 1
→=H ∇ ×+J() ∇× J dv
2π ∫V 2 1 211
4 R12 R 12
The Steady Magnetic Field - sites.google.com/site/ncpdhbkhn 65
(3)
µ Jdv Id L× a
AH=∫0 → = ∫ R
V 4π R 4π R2
Current element at ( x1, y1, z1), A at (x2, y2, z2)
1 1 1
→=H ∇ ×+J() ∇× J dv
2π ∫V 2 1 211
4 R12 R 12
∇ × =
2J 1 0
1 1
→=H ∇ × J dv
2π ∫V 2 11
4 R12
1 R a
R=( xx − )(2 +− yy )( 2 +− zz ) 2 →∇ =−12 =− R 12
12 21 21 21 2 3 2
R12 R12 R 12
1 a× J 1 J× a
→H = R12 1 dv = 1R 12 dv
2π ∫V 2 1 π ∫V 2 1
4 R12 4 R12
The Steady Magnetic Field - sites.google.com/site/ncpdhbkhn 66
(4)
µ Jdv Id L× a
AH=∫0 → = ∫ R
V 4π R 4π R2
Current element at ( x1, y1, z1), A at (x2, y2, z2)
1 J× a
→H = 1R 12 dv
2π ∫V 2 1
4 R12
=
J11dv I 1 d L 1
I d L× a
→H = 1 1R 12
2 ∫ π 2
4 R12
The Steady Magnetic Field - sites.google.com/site/ncpdhbkhn 67
(5)
∇×H = J
∇×H = J
B 1
BH= µ →∇×HA =∇× = ∇×∇×
0 µ µ
0 0
BA= ∇×
∇×∇×A = ∇( ∇ .A ) − ∇2 A
∇22 = ∇ + ∇ 2 + ∇ 2
AAAAxx a yy a zz a
∇( ∇.A ) − ∇2 A
→∇×H =
µ
0
The Steady Magnetic Field - sites.google.com/site/ncpdhbkhn 68
(6)
∇×H = J
∇( ∇.A ) − ∇2 A
→∇×H =
µ
0
µ J dv
A = 0 1 1
2 ∫V π
4 R12
∇.()SA = A. () ∇ S + S ( ∇ .A )
µ 1 1
→ ∇.A =0 J . ∇ +( ∇ .J ) dv
22π ∫V 12 211
4 RR12 12
The Steady Magnetic Field - sites.google.com/site/ncpdhbkhn 69
(7)
∇×H = J
µ 1 1
→ ∇.A =0 J . ∇ +( ∇ .J ) dv
22π ∫V 12 211
4 RR12 12
1
(∇.J )dv = 0
∫V 2 1 1
R12
∇1 =R12 = −∇ 1
13 2
RR12R12 12
µ 1
→ ∇.A =0 − J . ∇ dv
22π ∫V 11 1
4 R12
The Steady Magnetic Field - sites.google.com/site/ncpdhbkhn 70
(8)
∇×H = J
µ 1
→ ∇.A =0 − J . ∇ dv
22π ∫V 11 1
4 R12
∇.()SA = A. () ∇ S + S ( ∇ .A )
µ 1 J
→ ∇.A =0() ∇ .J − ∇ . ( 1 ) dv
22π ∫V 111 1
4 RR12 12
The Steady Magnetic Field - sites.google.com/site/ncpdhbkhn 71
(9)
∇×H = J
µ 1 J
→ ∇.A =0() ∇ .J − ∇ . ( 1 ) dv
22π ∫V 111 1
4 RR12 12
∂ρ
∇.J =−v = 0
1 1 ∂t
J.d S= ∇ .J dv
∫S ∫ V
µ J
→∇.A =−0 1 d S = 0
2 2π ∫S 1
4 1 R12
The Steady Magnetic Field - sites.google.com/site/ncpdhbkhn 72
(10)
∇×H = J
∇( ∇.A ) − ∇2 A
→∇×H =
µ
0
µ J dv ∇.A = 0
A = 0 x
x ∫V π ∇2 = − µ
4 R Ax0 J x
ρ dv
V = v →∇ 2 A =− µ J →∇2A =− µ J
∫V πε y0 y 0
4 0 R
ρ ∇2 A = − µ J
∇2V = − v z0 z
ε
0
→∇×H = J
The Steady Magnetic Field - sites.google.com/site/ncpdhbkhn 73
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