Lesson 13
Inductance, Magnetic energy
/force /torque
楊尚達 Shang-Da Yang
Institute of Photonics Technologies
Department of Electrical Engineering
National Tsing Hua University, Taiwan
Outline
„
„
„
„
Inductance
Magnetic energy
Magnetic force
Magnetic torque
Sec. 13-1
Inductance
1.
2.
Self & mutual inductances
Evaluation procedures
Definition-1
v
Closed loop C1 carrying current I1 will create B1
v v
⇒ flux: Φ11 = ∫ B1 ⋅ ds , flux linkage: Λ 11 = N1Φ 11
S1
v
B1
If I1′ = rI1
v
, by B = ∫
C'
v v Only
μ 0 I dl ′ × a R depend on
geometry
2
4π
R
v
v
v v
′ = ∫ B1′ ⋅ ds = rΦ11 ,
⇒ B1′ = rB1 , Φ11
′ = rΛ11
Λ11
S1
⇒ “Self-inductance”
of the loop C1:
Λ 11
L11 =
I1
Definition-2
v
In the presence of another loop C2, B1 will
pass through C2, ⇒ mutual flux linkage: Λ 12 = N 2 Φ 12
v v
where Φ12 = ∫ B1 ⋅ ds ∝ I1
S2
v
v
B1
B1
⇒ “Mutual-inductance”
between the 2 loops:
Λ 12
L12 =
I1
Depend on geometry & material.
Comment
v
v
v
dl ′
μ 0 N1 I1 dl1
,
A
⇒
=
∫C ' R(rv, rv′) v 1 4π ∫C1 R
B1
v
v v
N 2 Φ12 N 2
v N2
=
∇ × A1 ⋅ ds =
A1 ⋅ dl2
L12 =
∫
∫
I1
I1 S 2
I1 C 2
v v μ0 I
A(r ) =
4π
(
)
v
A1
R
v v
μ 0 N1 N 2
dl1 ⋅ dl2
=
4π ∫C1 ∫C2 R
v v
μ NN
dl1 ⋅ dl2
L21 = 0 1 2 ∫ ∫
C2 C1
4π
R
⇒ L12 = L21
Evaluation of inductance (Method 1)
1. Assume current I flowing on the loop.
v
2. Find B by Ampere’s law or Biot-Savart law:
v v
v
v v
μ 0 I dl ′ × a R
∫CH ⋅ dl = I , B = ∫C ' 4π R 2
v v
3. Find Λ (∝ I ) by Λ = N ∫ B ⋅ ds
S
Λ
4. Find L by L = , independent of I
I
Evaluation of inductance-reference figure
v v
v
μ 0 I dl1 × aR
B1 = ∫
2
C1 4π
R
R
v
aR
v
dl1
Evaluation of inductance (Method 2)
1. Assume current I flowing on the loop.
v
v
2. Find H and B by Method 1
3. Find the stored energy
v v
1
Wm = ∫ (H ⋅ B )dv ∝ I 2
2 V′
1 2
4. Find L by Wm = LI
2
Example 13-1: Solenoid inductor (1)
Consider a hollow solenoid with cross-sectional
area S, n turns per unit length. Find the
inductance per unit length L.
1. Assume current I flowing on the loop.
2. By Ampere’s law: B = μ 0 nI
Example 13-1: Solenoid inductor (2)
3. For unit length (l=1), Λ = n ⋅ Φ = n ⋅ (μ 0 nI ) ⋅ S
Λ n(μ 0 nI )S
2
L
=
=
=
n
μ0 S
4. By definition:
I
I
Example 13-2: Two concentric coils (1)
Consider two coils C1, C2 with N1, N2 turns and
lengths l1, l2. They are wound concentrically on
a thin cylindrical core of radius a with
permeability μ. Find the mutual inductance L12.
1. Assume C1, C2 have currents I1, I2
Example 13-2: Two concentric coils (2)
N1
2. By Ampere’s law, uniform field B1 = μ
I1
l1
3. Flux linkage of C2 due to C1: Λ12 = N 2 ⋅ Φ12 = N 2 ⋅ B1 ⋅ S
N1 N 2 2
4. By definition: L12 = μ
πa
l1
Sec. 13-2
Magnetic Energy
1.
2.
Energy of assembling current loops
Energy of magnetic fields
Energy of assembling current loops-One loop (1)
Closed loop C1 with self-inductance L1. If the
loop current i1 increases from 0 to I1 slowly
(quasi-static), an emf of:
di1
dΦ 11
= L1
v1 = −
dt
dt
will be induced on C1 to oppose
the change of i1 (Faraday’s law,
Lenz’s law).
Energy of assembling current loops-One loop (2)
The work done to overcome the induced v1
and enforce the change of i1 is:
∞
∞
0
0
W1 = ∫ v1 (t )i1 (t )dt = ∫
I1
di1
1
L1
i1dt = L1 ∫ i1di1 = L1 I12
0
dt
2
which is stored as magnetic
energy:
1
W1 = L1 I 12
2
one loop
Energy of assembling current loops-Two loops (1)
Insert loop C2 with self-inductance L2, mutual
inductance L21. If we maintain i1=I1, while i2
increases from 0 to I2 slowly, an emf of:
dΦ 21
di2
v21 = −
= L21
dt
dt
will be induced on C1 in
an attempt to change i1
away from I1
Energy of assembling current loops-Two loops (2)
The work done to maintain i1 = I1 is:
∞
∞
0
0
W21 = ∫ v21 (t ) I1dt = ∫
I2
di2
L21
I1dt = L21I1 ∫ di2 = L21I1 I 2
0
dt
Energy of assembling current loops-Two loops (3)
dΦ 22
di2
= L2
Meanwhile, an emf of: v2 = −
dt
dt
will be induced on C2 to oppose the change
of i2 (from 0 to I2).
The work done to
overcome v2 and enforce
the change of i2 is:
W22
1
2
= L2 I 2
2
Energy of assembling current loops-Two loops (4)
The total magnetic energy stored in the
system of two current loops is:
1
1
2
W2 = L1 I 1 + L21 I 1 I 2 + L2 I 22
2
2
two loops
Energy of assembling current loops-N loops
The total magnetic energy stored in the
system of N current loops carrying currents I1,
I2, ….., IN , is:
1 N N
Wm = ∑∑ L jk I j I k
2 j =1 k =1
By L12 = Λ12 I1 , the flux (linkage) of loop Ck due
to all the N current loops:
N
Φ k = ∑ L jk I j ,
j =1
1 N
⇒ Wm = ∑ I k Φ k
2 k =1
Energy of continuous current distributions-1
Decompose a system of continuous current
v v
distribution J (r ) in a volume V' into N elementary
current loops Ck , each has current ΔIk and
filamentary cross-sectional area Δak
v v
B, A
v v
Φ k = ∫ B ⋅ ds
Sk
v v
= ∫ A ⋅ dl k
Ck
Energy of continuous current distributions-2
1 N
Wm = ∑ I k Φ k =
2 k =1
v
v
ΔI k ⋅ d l k = J Δa k
(
v v
1 N
ΔI k ∫ A ⋅ dl k
∑
Ck
2 k =1
v v
v
v
d l k = J Δa k d l k = J Δ v k
)
(
)
v v
1 N
⇒ Wm = ∑ ∫ A ⋅ JΔvk ,
2 k =1 Ck
v v
1
Wm = ∫ (A ⋅ J )dv
2 V′
Comments
1 N
We = ∑ QkVk
2 k =1
1 N
Wm = ∑ I k Φ k
2 k =1
1
We = ∫ (ρV )dv
2 V′
v v
1
Wm = ∫ (A ⋅ J )dv
2 V′
Electrostatics
Magnetostatics
Energy of magnetic fields-1
In real applications (especially electromagnetic
waves), sources are usually far away from the
region of interest, only the fields are given
source
v v
J (r )
ROI
R→∞
v v
H,B
Energy of magnetic fields-2
v v
v
v
1
1
(1) Wm = ∫ (A ⋅ J )dv = ∫ A ⋅ (∇ × H )dv
2 V′
2 V′
v
v
J = ∇× H
contain all the
source currents
V′
v v
J (r )
By vector identity:
v v
v v
v
v
∇ ⋅ (A × H ) = H ⋅ (∇ × A) − A ⋅ (∇ × H )
v v v
v
= H ⋅ B − A ⋅ (∇ × H )
v v
v v
1
1
(2) Wm = ∫ (H ⋅ B )dv − ∫ ∇ ⋅ (A × H )dv
2 V′
2 V′
Energy of magnetic fields-3
v v
v
Q ∫ A ⋅ ds = ∫ (∇ ⋅ A)dv,
S
V
⇒∫
V′
V ′ S′
v v
J (r )
v v
v v v
∇ ⋅ (A × H )dv = ∫ (A × H )⋅ ds
S′
v v v
v v
1
1
(3) Wm = ∫ (H ⋅ B )dv − ∫ (A × H )⋅ ds
2 V′
2 S′
I1
I2
Energy of magnetic fields-4
Obs. pt.
S′
S′
R→∞
v
1
H ∝ 2
R
v 1
A∝
R
Energy of magnetic fields-5
v v v 1 v
v
1
2
I 2 = ∫ (A × H )⋅ ds ≈ A( R ) H ( R ) ⋅ 4πR
2 S′
2
1 1
1
2
∝ ⋅ 2 ⋅R ∝ → 0
R R
R
v
⇒ Wm = I1 = ∫ wm (r )dv
V′
1 v v
3
H ⋅ B ( J m ) …energy density
2
Energy of magnetic fields-6
V′
⎛ 1 v v⎞
dWm = ⎜ H ⋅ B ⎟dv
⎝2
⎠
⎛ 1 v v⎞
Wm = ∫ ⎜ H ⋅ B ⎟dv
V′ 2
⎠
⎝
R→∞
Example 13-3: Coaxial cable inductor (1)
Find the stored magnetostatic energy and
inductance per unit length of:
Cylindrical symmetry,
Ampere’s law, ⇒
⎧ v μ0 I
v ⎪⎪aφ 2πa 2 r , if r < a
B=⎨
⎪av μ 0 I , if a < r < b
⎪⎩ φ 2πr
v
v
H = B μ0
Example 13-3: Coaxial cable inductor (2)
⎧ μ0 I 2 2
r ,r < a
⎪
2 4
v
v
1
⎪ 8π a
wm = H ⋅ B = ⎨
2
2
I
μ
⎪ 0 ,a < r < b
⎪⎩ 8π 2 r 2
Energy density:
Differential volume
(L=1): dv = 2πr ⋅ dr
Example 13-3: Coaxial cable inductor (3)
Total stored energy:
μ0 I 2
Wm1 =
4πa 4
Wm 2
μ0 I
=
4π
2
∫
a
0
2
μ
I
r 3 dr = 0 , r < a
16π
μ0 I ⎛ b ⎞
1
∫a r dr = 4π ln⎜⎝ a ⎟⎠, a < r < b
b
2
1 2
2(Wm1 + Wm 2 ) μ 0 μ 0 ⎛ b ⎞
Wm = LI , L =
ln⎜ ⎟
=
+
2
2
8π 2π ⎝ a ⎠
I
internal external
Sec. 13-3
Magnetic Force
1.
2.
Force on current loops
Example: force between parallel wires
Force on current-carrying loops-1
Consider an elemental current-carrying wire
of cross-sectional area S, represented by a
v
differential displacement vector dl
Free charges within the
wire of charge density ρ
v
v
move with velocity u // dl ,
experiencing a force of:
(
v v v
v
dFm = ρS dl (u × B )
)
Force on current-carrying loops-2
vv v v
dl u = u dl
v
v
J = ρu
v v v
v
dFm = ρS dl (u × B )
v v
v v v
= ρS u dl × B = JS dl × B
(
(
v v
v
⇒ dFm = I dl × B
)
For a current loop C :
v v
v
Fm = I ∫ dl × B
C
)
Force on current-carrying loops-3
v
If B is created by another closed loop C2
carrying a current I2, the force exerted on the
loop C1 carrying a current I1 is:
v
F21 = I 1 ∫
C1
v v
dl1 × B21
v v
v
μ0 I dl ′ × aR
B=∫ '
C 4π
R2
v
μ0 I 2
B21 =
4π
∫
v v
dl 2 × a R21
C2
R212
Force on current-carrying loops-4
v
μ 0 I1 I 2
F21 =
4π
∫ ∫
C1
(
C2
)
v
v v
v
dl1 × dl 2 × a R 21
= − F12
2
R21
Counterpart in electrostatics:
Coulomb’s force between
two charges
v
v
F12 = aR12
1
q1q2
2
4πε 0 R12
Example 13-4: Force between two long wires
Find the force per unit length between two
infinitely long, parallel wires separated by d,
carrying currents I1, I2 in the same direction.
v
v μ 0 I1
,
B12 = − a x
2πd
v
v
1 v
F12 = I 2 ∫ dl2 × B12
0
v
⎛ v μ 0 I1 ⎞
= I 2 ∫ (a z dz )× ⎜ − a x
⎟
0
2πd ⎠
⎝
v μ 0 I1 I 2
= −a y
…attraction force
2πd
1
Sec. 13-4
Magnetic Torque
1.
Example: magnetic force & torque
exerted on a current loop
Example 13-5: Force & torque on current-carrying loops (1)
Consider a circular loop on the xy-plane with
radius b, current I in clockwise sense, and is
v v
v
placed a “uniform” magnetic filed: B = B⊥ + B||
v
v
− a z B⊥ a y B||
Example 13-5: Force & torque on current-carrying loops (2)
The force exerted on a differential current
v
v
v
element dl = −aφ bdφ on the loop due to B⊥ :
(
)
v
v
= I (− a bdφ )× (− a B )
v v
v
dFm = I dl × B , ⇒
b
φ
v
dF⊥
φ
v
= ar IbB⊥ dφ
z
⊥
Example 13-5: Force & torque on current-carrying loops (3)
The force exerted on a differential current
v
v
v
element dl = −aφ bdφ on the loop due to B|| :
(
)
v v
v
dFm = I dl × B , ⇒
v
v
v
dF|| = I (− aφ bdφ )× (a y B|| )
v
v
v
= IbB|| dφ (a x sin φ − a y cos φ )× (a y )
v
= a z IbB|| sin φdφ
Example 13-5: Force & torque on current-carrying loops (4)
v
The total force exerted on the loop due to B⊥ :
v
v
dF⊥ = ar IbB⊥ dφ , ⇒
φ
v
v
2π
F⊥ = ∫ dF⊥
0
2π v
⎡
= IbB⊥ ∫ ar (φ )dφ ⎤ = 0
⎥⎦
⎢⎣ 0
Example 13-5: Force & torque on current-carrying loops (5)
v
The total force exerted on the loop due to B|| :
v v
dF|| = a z IbB|| sin φdφ , ⇒
v
v
2π
F|| = ∫ dF||
0
2π
v
⎛
= a z IbB|| ⎜ ∫ sin φ ⋅ dφ ⎞⎟ = 0
⎠
⎝ 0
Example 13-5: Force & torque on current-carrying loops (6)
v
The total torque exerted on the loop due to B⊥ :
v
v
dF⊥ = ar IbB⊥ dφ , ⇒
φ
v
v
2π
v
T⊥ = ∫ dF⊥ × (− ar b )
0
2π v
v ⎞
⎛
= − Ib B⊥ ⎜ ∫ ar × ar ⎟ = 0
⎠
⎝ 0
2
Example 13-5: Force & torque on current-carrying loops (7)
v
The total torque exerted on the loop due to B|| :
v v
dF|| = a z IbB|| sin φdφ , ⇒
v
v
2π
v
T|| = ∫ dF|| × (− ar b )
0
= Ib B|| ∫
2
= Ib B|| ∫
2
2π
0
2π
0
v
v
[az × (− ar )]sin φdφ
v
(− aφ )sin φdφ
v
v
v
− aφ = a x sin φ − a y cos φ
Example 13-5: Force & torque on current-carrying loops (7)
1 − cos 2φ
2
sin 2φ
2
v
v 2π 2
v 2π
2
⎡
T|| = Ib B|| ⋅ a x ∫ sin φdφ − a y ∫ sin φ ⋅ cos φ ⋅ dφ ⎤
⎥⎦
⎢⎣ 0
0
v
v
2
= a x Iπb B|| = a x mB||
(
)
v v
v v v
T = T⊥ + T|| = T|| = a x (−m z ) B y
In general, ⇒
v v v
T = m× B