N
MAGNETIC FORCES ON FLAT SURFACES
Lorentz Force Law: f m = qv × μoH [ ]
F [N/m]
W
F [N/m] = Nqv1 × μo H12 = I1 × μ o H12
W
Let I1 = I2 = I = Ho W
0 for C1
P = F/W = ẑ μ oHo
2
2
"Magnetic pressure"
I
H=0
J
I1
2
[N/m ]
F1
z
y
x
C2
z
Ho
= zˆμoI1I2 / 2W [N/m]
2
1
C1
H
I for C2
F1 [N/m] = I1 × μo <H>
= zˆ I1μo
I2
Currents exert no net force on themselves
Total H method:
∫∫ A J • da =
v∫C H • ds = w
s << W
s
∫∫ A J • da = I2 ≅ 2H12 W
v∫ C H • ds = w
F = ẑμ oI1I2 / 2W [N/m]
I1
c
magnetic pressure
conductor
<H> = Ho/2
y
Ho
L6-1
ROTARY WIRE MOTOR
Single wire loop spinning in uniform H:
F [Nm-1] = I × μo H
T = v∫ r × F ds = r2WIμoH ẑ [Nm]
c
f
N
T = IAμoH ẑ (A is loop area, N = 1 turn)
ds
T = NIAμoH ẑ for N-turn coil
Axial forces from wires at ends cancel
Torque = f(θ):
T[Nm]
0
With commutator
π
2π
T, ẑ
C
S
H
r
W
θ
I
NIμoH(θ)A
θ
Commutators:
Switch currents to maximize torque
Can have N coils and 2N brushes
θ
spring-loaded
carbon brushes
L6-2
MOTOR BACK VOLTAGE
Force f e on electron inside moving wire:
f e = − e ( Ee + v × μo H ) = 0
Open-circuit wire:
−eEe
⇒ Ee = − v × μo H inside
H
−ev × μo H v
Open-circuit voltage: Φ = Ee W = vμoHW [V]
Force balance
Mechanical power output, N turns:
Pm = ωT = ωNIAμoH [W]
I = (V - Φ)/R
Φ = 2NvμoHW = 2NωrμoHW = NAμoHω
Pm = ωN(V - NAμoHω)AμoH/R = ωK1 – ω2K2
ωmax = V/NAμoH
Φp = V/2
Φ+ H
-
Pm(ω)
= 2ωp
W
v,f,x̂
R
Pp
0
-e
motor
ωp
ωmax
ω
+
-
V
I
+
Φ
-
generator
L6-3
RELUCTANCE MOTOR FIELDS
2-Pole Reluctance Motor:
∇ • B = 0 ⇒ Bμ ≅ Bgap ⇒ Hμ ≅
μo
μ
Hgap << Hgap [
> ~104 ]
μ
μo
∫A J • da = NI = v∫c H • ds = v∫c (Hgap + Hμ ) • ds ≅ 2bHgap
(if Hμ v∫c ds << 2bHgap )
gap
b
Therefore Hgap ≅ NI [A / m]
2b
T
μ
μ
μ
μ
rotor
ds
Area A
θ
rotor
stator
stator D
N turns
I +V-
I
+ V -
D
N turns
L6-4
RELUCTANCE MOTOR TORQUE
Fields:
Hgap = NI
2b
T
μ
Magnetic Flux Linkage Λ:
gap
b
μ
rotor
Λ = N∫∫ B•da = NμoHgap A gap
A
= N2μoIA gap /2b (A gap = RDθ)
Λ = LI
⇒ L = N2μoRDθ/2b
1 2
1 Λ2
w m = LI =
2
2 L
θ
stator
I +V-
D
N turns
Set V = dΛ/dt = 0:
dw m
Λ 2 dL-1
Λ2
2b
2
T == =
I
[
∝
], Λ=NμoHgapRθD
2
2
dθ
2 dθ
2 N μoRDθ
dV
1
= μoHgap2 2bD R = Wgap olume [Nm] Torque
2
dθ
We power coil until overlap is maximum, then coast until it is zero
Magnetic pressure = Energy density [J/m3 = N/m2]
L6-5
¾-POLE RELUCTANCE MOTOR
Winding Excitation Plan:
First excite windings A and B,
A
pulling pole 1 into pole B.
Pole area A = constant, temporarily.
When Δθ = π/3, excite B and C.
When Δθ = 2π/3, excite C and A.
Repeating this cycle results in
C
nearly constant clockwise torque.
A
1
μ
4
θ
B
2
B
μ
C
To go counter-clockwise, excite BC, then AB, then CA.
Torque:
Only one pole is being pulled in here; the other excited winding has
either one rotor pole fully in, or one entering and one leaving that
cancel. Many pole combinations are used (more poles, more torque).
L6-6
ELECTRIC AND MAGNETIC PRESSURE
Electric and magnetic pressures equal the field energy densities, J/m3
Both field types only pull along their length, and only push laterally
The net pressure is the difference between two sides of any boundary
1
2
Magnetic pressure Pm = μo | H | [N/m2] or [J/m2]
2
Area
μ = ∞A
μ=∞
Force fm = APm
Force fm = BPm
Force fm = APm
σ=∞
Area
A
Area B
1
2
Electric pressure Pe = εo | E | [N/m2] or [J/m2]
2
Area
σ=∞ A
σ=∞
Force fe = APe
Force fe = BPe
Area B
L6-7
FORCES ON NEUTRAL MATTER
Kelvin polarization force density:
If ∇ × E = 0 = ∇ • E, then:
Field gradiants ⊥ E ⇒ E is curved
Curved E pulls electric dipoles
+ ++
+ +++
ε > εo
-E
- -
into stronger field regions for ε>εo
If ∇ × H = 0 = ∇ • B, then:
Field gradiants ⊥ H ⇒ H is curved
stronger field regions for μ > μo
-
fx/2
x
fx/2
+
-
z
μ > μo
B
+
d
y•
Kelvin magnetization force density:
Curved H pulls current loops into
+
V
-
I
μo
μ > μo
Induced current loop
B1 > B2
B2
f
× x2
B1
•
fx1
L6-8
MIT OpenCourseWare
http://ocw.mit.edu
6.013 Electromagnetics and Applications
Spring 2009
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