Transcript Document

Magnetic Field
Basic Concepts:
A current carrying wire produces a magnetic field
in the area around it.
A time changing magnetic field induces a voltage in a
coil (wire) if is passes through the coil (transformer)
A moving wire in the presence of a magnetic field has a
voltage induced in it (generator)
A current-carrying wire in the presence of a magnetic
field has a force induced on it (motor).
I
The Basic Law:
Ampere’s Law
i
H : Magnetic field intensity (A-turns/m)
Ampere’s Law in a very simple form
H.l   Ii
i
i
Ampere’s Law in a coil
H
N .i
lc
N : turns
i : Current (A)
lc: mean length (m)
H : Magnetic field intensity (A-turns/m)
Important: H is the effort that a current is putting to produce a
magnetic field.
The magnetic field will depend on the material of the core.
Now we define:
B   .H
B : Magnetic flux density (T=Wb/m2) , (Tesla= Webers per
square meter)
Where :

and :

: Magnetic permeability (H/m) , (henrys per meter)
  B. A
: Magnetic flux
The permeability of free space is:
0  4 107 H / m
Now we define relative permeability of different
materials:

r 
0
The higher the relative permeability, the more
flux. For steel it is 2000-6000. Materials with
high values of relative permeability
In summary we can write:
  B. A  HA 
NiA
lc
 N .i.
A
lc
Defining new variables and constants:
F  N .i
lc

A
1 A
 
 lc
Then we have:
: magnetomotive force (A-turns), then we have:
: reluctance (A-turns/Wb)
: permeance (Wbs/ A-turn)
  F
F   .
In summary in a magnetic field we have:
F  
lc

A
Compare this formula with Ohm’s law in electric circuits:
v  Ri
R
l
 A
Example 1:
Given : i=1 A, N=100, lc=40 cm, A= 100 cm2

Calculate : F, H, B, ,  and
F  N .i  100
H
F 100

 250
lc 0.4
B  .H 
4  107  5000 250  1.57

lc
0.4

 6366.385
7
A 4  10  5000 0.01
1
   1.57  10 4

  B. A  1.57  0.01  0.0157
r
=5000
Example 2
r  4000
B  0.5T
Assume 5% increase in
effective
Cross-sectional area
for fringing effect
r  4000
Calculate total reluctance and current
N .i  F   .c   .g
N .i  Bc . Ac .c  Bg . Ag .g
lg
lc
N .i  c H c . Ac .
  g H g . Ag .
c Ac
 g Ag
Example 3
r  1500
Assume 4%
increase in
effective
Cross-sectional
area
for fringing
effect
Calculate the flux density in each of the legs
lc

A
The similarities lead us to Magnetic Circuits theory.
This means we can use the rules we had for series and
parallel resistances.
Fringing effect means that when we have
an air gap, the effective cross sectional area
of the air gap is larger than the crosssectional area of the iron-core
What do we do with the
fringing effect?
1- Add the air gap length
to each dimension
2- Assume the effective
area is 5% more than the
cross-sectional are of the
iron-core