Magnetic Circuits
Outline
•  Ampere’s Law Revisited
•  Review of Last Time: Magnetic Materials
•  Magnetic Circuits
•  Examples
1
Electric Fields
· dA
=
o E
S
Magnetic Fields
· dA
=0
B
ρdV
V
S
= Qenclosed
GAUSS
GAUSS
FARADAY
d
E · dl = −
dt
C
AMPERE
· dA
B
S
dλ
emf = v =
dt
2
· dl
H
C d
· dA
=
J · dA +
E
dt S
S
Ampere’s Law Revisited
In the case of the magnetic field we can see that ‘our old’ Ampere’s law can
not be the whole story. Here is an example in which current does not gives
rise to the magnetic field:
= 0??
B
B
Side
View
I
I
I
I
B
B
Consider the case of charging up a capacitor C which is connected to very long wires.
The charging current is I. From the symmetry it is easy to see that an application of
Ampere’s law will produce B fields which go in circles around the wire and whose
magnitude is B(r) = μoI/(2πr). But there is no charge flow in the gap across the capacitor
plates and according to Ampere’s law the B field in the plane parallel to the capacitor
plates and going through the capacitor gap should be zero!
This seems unphysical.
3
Ampere’s Law Revisited (cont.)
If instead we drew the Amperian surface as sketched below,
we would have concluded that B in non-zero !
B
B
Side
View
I
I
I
I
B
Maxwell resolved this problem by adding a term to the Ampere’s Law.
In equivalence to Faraday’s Law,
the changing electric field can generate the magnetic field:
· dl =
H
C
d
J · dA +
dt
S
4
· dA
E
S
COMPLETE
AMPERE’S LAW
Faraday’s Law and Motional emf
What is the emf over the resistor ?
vΔt
dΦmag
emf = −
dt
L
v
B
In a short time Δt the bar moves a
distance Δx = v*Δt, and the flux
increases by ΔФmag = B (L v*Δt)
ΔΦmag
= BLv
emf =
Δt
There is an increase in flux through the circuit
as the bar of length L moves to the right
(orthogonal to magnetic field H) at velocity, v.
from Chabay and Sherwood, Ch 22
5
from Chabay and Sherwood, Ch 22
Faraday’s Law for a Coil
The induced emf in a coil of N turns is equal to
N times the rate of change of the magnetic flux on one loop of the coil.
Moving a magnet towards a coil produces a
time-varying magnetic field inside the coil
Rotating a bar of magnet (or the coil)
produces a time-varying magnetic field
inside the coil
Will the current run
CLOCKWISE or ANTICLOCKWISE ?
6
Complex Magnetic Systems
DC Brushless
Reluctance Motor
Stepper Motor
· dl = Ienclosed
H
C
· dA
=0
B
Induction Motor
f = q v × B
S
We need better (more powerful) tools…
Magnetic Circuits: Reduce Maxwell to (scalar) circuit problem
Energy Method:
Look at change in stored energy to calculate force
7
Magnetic Flux
Magnetic Flux Density
Magnetic Field Intensity
Φ [Wb] (Webers)
B [Wb/m2] = T (Teslas)
H [Amp-turn/m]
due to macroscopic
currents
due to macroscopic
& microscopic
= μo μr H
= μo H
+M
= μo H
+ χm H
B
Faraday’s Law
emf = −
emf =
N C · dl
E
dΦ mag
dt
and
8
Φmag =
· n̂dA
B
Example: Magnetic Write Head
Ring Inductive
Write Head
Shield
GMR
Read
Head
Recording Medium
Horizontal
Magnetized Bits
Bit density is limited by how well the field can be localized in write head
9
Review: Ferromagnetic Materials
B
r
B
C
−H
s
B
B,J
0
C
H
H
Initial
Magnetization
Curve
H>0
r
−B
hysteresis
s
−B
0
H
Slope = μ
i
Behavior of an initially unmagnetized
C
Hr : coercive magnetic field strength material.
Domain configuration during several stages
B S : remanence flux density
of magnetization.
B : saturation flux density
10
Thin Film Write Head
Recording Current
Magnetic Head Coil
Magnetic Head Core
Recording Magnetic
Field
How do we apply Ampere’s Law to this geometry (low symmetry) ?
· dl =
H
C
S
11
J · dA
Electrical Circuit Analogy
Charge is conserved…
i
+
Flux is ‘conserved’…
· dA
=0
B
S
i
i
V
Ф
EQUIVALENT
CIRCUITS
Electrical
Magnetic
φ
i
V
R
12
Electrical Circuit Analogy
Electromotive force (charge push)=
v=
Magneto-motive force (flux push)=
· dl
E
· dl = Ienclosed
H
C
i
+
i
i
V
Ф
EQUIVALENT
CIRCUITS
V
Electrical
Magnetic
i
φ
R
13
Electrical Circuit Analogy
Material properties and geometry determine flow – push relationship
φ
i
OHM’s LAW
R
V
= μ o μr H
B
DC
J = σE
B
Recovering macroscopic variables:
I=
V
E · dA = σ A
l
l
ρl
V =I
= I = IR
σA
A
=σ
J · dA
14
H
N i = Φ
Reluctance of Magnetic Bar
Magnetic “OHM’s LAW”
N i = Φ
l
A
l
=
μA
φ
15
Flux Density in a Toroidal Core
Core centerline
H
μN i
B=
2πR
R
2R
N turn
coil
μN i = 2πRB = lB
l
lB
=Φ
mmf = N i =
μ
μA
(of an N-turn coil)
i
mmf = Φ
16
Electrical Circuit Analogy
Electrical
Magnetic
Voltage v
Current i
Resistance R
Conductivity 1/ρ
Current Density J
Electric Field E
Magnetomotive Force = N i
Magnetic Flux φ
Reluctance Permeability μ
Magnetic Flux Density B
Magnetic Field Intensity H
17
Toroid with Air Gap
Magnetic
flux
Electric
current
A = cross-section area
Why is the flux confined mainly to the core ?
Can the reluctance ever be infinite (magnetic insulator) ?
Why does the flux not leak out further in the gap ?
18
Fields from a Toroid
· dl =
H
C
Magnetic
flux
J · dA
S
Electric
current
= Ienclosed
A = cross-section area
Ni
H=
2πR
+M
= μo H
B
Ni
B=μ
2πR
μA
Φ = BA = N i
2πR
19
Scaling Magnetic Flux
Magnetic
flux
N i = Φ
&
l
=
μA
2πR
=
μA
Electric
current
A = cross-section area
μA
Φ = BA = N i
2πR
Same answer as Ampere’s Law (slide 9)
20
Magnetic Circuit for ‘Write Head’
Core Thickness = 3cm
2cm
l
=
μA
8cm 2cm
0.5cm
core
i
gap
φ
N=500
+
A = cross-section area
Φ≈
21
-
Parallel Magnetic Circuits
10cm
1cm
Gap a
i
10cm
0.5cm
Gap b
A = cross-section area
22
A Magnetic Circuit with Reluctances in Series and Parallel
“Shell Type” Transformer
Magnetic Circuit
1
+
v1
N1 turns
l2
-
l1
+
v2
-
N2 turns
φa
+
3
N1 i1
N2 i2
Depth A
l=l1+l2
l1
1 =
μA
23
φb
2
+ φc
-
φc
l2
2 =
μA
Faraday Law and Magnetic Circuits
Ф
i1
+
sinusoidal
-
+
v1
-
i2
N1
N2
Primary
Load
Secondary
Laminated Iron Core
Flux linkage
+
v2
-
λ = NΦ
Step 1: Estimate voltage v1 due to time-varying flux…
Step 2: Estimate voltage v2 due to time-varying flux…
24
A = cross-section area
dλ
v=
dt
v2
=
v1
Complex Magnetic Systems
DC Brushless
Reluctance Motor
Stepper Motor
Induction Motor
Powerful tools…
Magnetic Circuits: Reduce Maxwell to (scalar) circuit problem
Makes it easy to calculate B, H, λ
Energy Method:
Look at change in stored energy to calculate force
25
Stored Energy in Inductors
In the absence of mechanical displacement…
WS =
Pelec dt =
ivdt =
dλ
i
=
dt
i (λ) dλ
For a linear inductor:
λ
i (λ) =
L
Stored energy…
WS =
26
λ
0
λ λ2
dλ =
L
2L
Relating Stored Energy to Force
Lets use chain rule …
WS (Φ, r)
∂WS dΦ ∂WS dr
=
+
dt
∂Φ dt
∂r dt
This looks familiar …
dr
dWS
= i · v − fr
dt
dt
di
dr
= iL − fr
dt
dt
Comparing similar terms suggests …
∂WS
fr = −
∂r
27
Energy Balance
i·v
electrical
heat
dWS
dt
dWS (λ, r)
∂WS dλ ∂WS dr
=
+
dt
∂λ dt
∂r dt
For magnetostatic system, dλ=0 no electrical power flow…
dWS
dr
= −fr
dt
dt
28
dr
−fr
dt
mechanical
neglect heat
Linear Machines: Solenoid Actuator
x
l
Coil attached to cone
If we can find the stored energy, we can immediately compute the force…
…lets take all the things we know to put this together…
1 λ2
WS (λ, r) =
2 L
∂WS
|λ
fr = −
∂r
29
KEY TAKEAWAYS
· dl =
H
COMPLETE AMPERE’S LAW
C
d
J · dA +
dt
S
Electrical
Magnetic
Voltage v
Current i
Resistance R
Conductivity 1/ρ
Current Density J
Electric Field E
Magnetomotive Force = N i
Magnetic Flux φ
Reluctance Permeability μ
Magnetic Flux Density B
Magnetic Field Intensity H
30
· dA
E
S
RELUCTANCE
l
=
μA
MIT OpenCourseWare
http://ocw.mit.edu
6.007 Electromagnetic Energy: From Motors to Lasers
Spring 2011
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