Download Notes 34 3318 Magnetic Circuits

ECE 3318
Applied Electricity and Magnetism
Spring 2016
Prof. David R. Jackson
ECE Dept.
Notes 34
1
Magnetic Circuits
B B
N turns
I
A = cross-sectional area of core
L = length of core
2
Magnetic Circuits (cont.)
Apply Ampere’s law:
 H  dr  I
encl
Assuming that the magnetic
field is constant:
 NI
HL  NI
C
C
N turns
Note:
This assumption
will be accurate if
the core has a high
permeability and a
constant crosssectional area.
I
3
Magnetic Circuits (cont.)
The magnetic flux in the core is:
 L 
A
 0  r A 
  BA   0 r H  A  0 r  HL   HL 

  NI 
L
L


A


 0 r 
1
Recall : HL  NI
C
Hence we have

N turns
I
NI
 L 




A
 0 r 
4
Magnetic Circuits (cont.)
Define the reluctance of the core:
R 
L
0  r A
We then have:
 
Vm
R
where
N turns
Vm  NI
I
5
Magnetic Circuits (cont.)
Here is a circuit model (“magnetic circuit”):
I 
V  Vm  NI
+
-
R =R 
L
0  r A
6
Magnetic Circuits (cont.)
The concept of reluctance allows us to easily solve
more complicated magnetic circuit problems.
A = cross-sectional area of core
L1
N turns
Lg
Air gap
L2
L2   L1  Lg  / 2
I
Square-shaped core
Each segment of the structure is a reluctance.
7
Magnetic Circuits (cont.)
This is the circuit model of the previous structure.
R1
R1
V  NI
R2
Rg
+
-
R2
R1 
R2 
Rg 
L1
0 r A
L2
0  r A
Lg
0 A
R1
8
Magnetic Circuits (cont.)
Another example is shown here.
A = cross-sectional area of core
L1
N turns
L1
Lg
Air gap
L1
L2
I
Each segment of the structure is a reluctance.
9
Magnetic Circuits (cont.)
This is the circuit model of the previous structure.
R1
R1
R1
V  NI
R2
R1
Rg
+
-
R2
R1
R1
10