Faraday’s Law
and Linear DC machine
Faraday’s Law (I)
d
d
C E  dl   dt S B  dS   dt
B
dl
C
n
E
S
Electromotive Force (emf)
------ V
emf   E  dl
C
Total Magnetic Flux

------ Wb
 B  dS
S
Faraday’s Law (II)
d
d
C E  dl   dt S B  dS   dt
B
dl
C
n
E
S
• The emf generated around a closed contour C is
related to the time rate of change of the total
magnetic flux through the open surface S bounded
by that contour.
Faraday’s Law (III)
d
d
C E  dl   dt S B  dS   dt
• Lenz’s Law: The emf induced in the contour is of a polarity
that tends to generate an induced current
whose magnetic flux tends to oppose any
change in the original magnetic flux.
i
E
B
--Vind
+++
B
Vind
d

dt
Faraday’s Law (IV)
Top view
B
B
--Vind
N turns
+++
Vind
d
N
dt
+
Vind
-
Inductor Operation 1 (Like Generator)
R
i
Bext
--Vind
i
+++
R
i
+
Vind
-
 B  dS   B
S
Vind  N
i
Bfrom i
N turns
Note: B is total magnetic flux density. B  Bexternal  Bfrom i

Bext
S
ext
 dS   Bfrom i  dS   ext   from i
Bfrom i
S
d
d  ext
d  from i
d  ext
d from i
d  ext
di
N
N
N

N
L
dt
dt
dt
dt
dt
dt
dt
i is induced current:
Vind
i
R
Dynamical equation:
di
R
N d ext (t )
 i
dt
L
L
dt
Inductor Operation 2 (Like Motor)
R
+
~ V (t )
s
-
i
R
--Vindd
+++
B
B
i
+
+
i - ~ V (t ) V
ind
s
-
i
N turns
Note: i is total current or actual current in operation
Vs (t )  Vind Vs (t ) Vind
i


Magnetic flux density B comes from i.
R
R
R
d d
di
Vind  N
Dynamical equation:
dt

dt
V (t )
di
R
 i s
dt
L
L
L
dt
Induced Voltage from a Moving Bar
B
B
+
+
Vind
-
Vind
l
v
v
-
0
  BA  Blx
Vind
x
d
dx

 Bl
 Blv
dt
dt
Magnetic Force – Biot-Savart’s Law
dF  idl  B
B
dF
idl
B
l
i
F
F=Bli
Linear Machine – Case 1 (Motor, No Load)
i
t=0
VB
B
+
R
Vind
0
1. t  0, close the switch, it 0 
Find
VB
, Find  Blit 0
R
l
x
dv
Find  M
, M : mass of bar, v : velocity of bar. The bar will accelerate.
dt
VB  Vind
2. v , Vind  Blv , i 
, Find  Bli 
R
3. When Vind  VB , i  0, Find  0, no force is on the bar.
The machine reaches steady state with constant velocity v.
V
V
From Vind  Blv, we have v  ind  B .
Bl Bl
Linear Machine – Case 2 (Motor, with Load)
i
t=0
B
+
R
VB
Vind
-
Fload
Find
l
x
0
1. t  0, close the switch, it 0 
VB
, Find  Blit 0 ,
R
Fnet  Find  Fload The bar will accelerate if Fnet  0. Fnet  M
dv
dt
VB  Vind
, Find  Bli 
R
3. When Find  Fload or Fnet  0, motor reaches steady state with constant velocity v.
Find
Vind
From Find  Fload  Bli, we get i 
. Vind  VB -iR  Blv, we have v 
.
Bl
Bl
2. v , Vind  Blv , i 
Linear Machine – Case 3 (Generator)
i
B
+
R
Find
Vind
-
l
Fapply
x
0
1. t  0, apply Fapply, the bar will accelerate. Fapply  M
dv
dt
Vind
, Find  Bli , opposite to Fapply or against th e motion.
R
3. When Fnet  Fapply  Find  0, reaches steady state with constant velocity v.
2. v , Vind  Blv , i 
From Find  Fapply  Bli, i 
Fapply
Bl
. From Vind  iR  Blv, v 
Vind
.
Bl
Linear Machine – Case 4 (Mixed - 1)
i (t=∞)
t=0
VB
R i(t=0) +
Vind
- F
ind
B
Fload
Fapply
l
(t=∞) Find (t=0)
x
0
Assume Fext  Fapply  Fload  0
VB
, Find  Bli in the direction of Fapply
R
dv
Fnet  Fapply  Find  Fload. The bar will accelerate if Fnet  0. Fnet  M
dt
V V
2. v , Vind  Blv , i  B ind , Find  Bli  .
R
WhenVind  VB , i  0, Find  0.
Since Fext  Fapply  Fload  0, the bar will still accelerate.
1. t  0, close the switch and apply Fapply, it 0 
Linear Machine – Case 4 (Mixed - 2)
B
t=0
VB
R i(t=0) +
Vind
- F
ind
Fload
Fapply
l
(t=∞) Find (t=0)
x
0
3. Vind , Vind  VB , directions of i and Find  Bli changed, device becomes generator.
4. When Find  Fext or Fnet  0, bar reaches steady state with constant velocity v
From Find  Fapply  Fload  Bli, we get it 
Vind
Vind
 VB  it  R  Blv, we have v 
.
Bl
Find

.
Bl
Example
B
t=0
R
Fload
VB
Fapply
l
0
B = 0.1 T
l = 10 m
VB=120 V
R = 0.3 W
x
Find steady state current and speed of the bar if
(1) No external force;
(2) Fload  30N; also plot vsteady for Fload from 0 to 50N
(3) Fapply  30N; also plot vsteady for Fapply from 0 to 50N.
(1) – Case 1, (2) – Case 2, (3) – Case 4
Linear Machine – Transient Analysis (1)
B
t=0
VB
R
+
i
Vind
0
Fload
Fapply
l
Find
x
dv
Fnet  Fapply  Find  Fload , Fnet  M
dt
VB  Vind
Vind  Blv, i 
, Find  Bli.
R
dv
V V
M
 Fapply  Find  Fload  Fapply  Bli  Fload  Fapply  Bl B ind  Fload
dt
R
VB  Blv
( Bl )2
BlVB
 Fapply  Fload  Bl

v  Fapply  Fload 
R
R
R
dv
( Bl )2
1 
BlVB 
Dynamical Equation :

v   Fapply  Fload 

dt
MR
M
R 
Linear Machine – Transient Analysis (2)
B
t=0
VB
R
i
+
Vind
-
0
Fload
Fapply
l
Find
x
dv
( Bl )2
1 
BlVB 
Dynamical Equation :

v   Fapply  Fload 

dt
MR
M
R 
dv
For steady state :
 0, we have :
dt
Fapply  Fload
VB
vsteady 
R

( Bl )2
Bl
Linear Machine – Transient Analysis (3)
Velocity (m/s)
150
B = 0.1 T
l = 10 m
VB=120 V
R = 0.3 W
100
50
0
0
0.2
0.4
0.6
0.8
1
1.2
Time (s)
1.4
1
1.2
Time (s)
1.4
1.6
1.8
2
M = 1 kg
Fapply = 40 N
Fload = 20 N
400
Current (A)
300
200
100
0
-100
0
0.2
0.4
0.6
0.8
1.6
1.8
2
EMF Constant and Force Constant
Vind  Blv
Vind
 Bl
EMF Constant: k E 
v
F  Bli
F
 Bl
Force Constant: k F 
i
kE  kF
How to Solve a Dynamic Equation Using SimuLink?
 dx
 dt  z
 dy
 p
 dt
 dz 
2
1 
 2x
2
1  a sin y
  1 
2 
2 2
dt
1

x
(
1

x
)



2
 dp 
1 
2

1

1

a
sin
y a sin( 2 y )



2
 dt  1  x 
Solve




Assume
x(0)  0; y (0)  0; z(0)  2sin  ; p(0)  2cos 
a  0.01
  30o
dynamic.mdl
Another Example
R
B
i
+
+
~ V (t ) V
ind
s
-
Dynamical equation:
i
Vs (t )
di
R
 i
dt
L
L
R  1W, L  2 H , Vs (t )  5cos(2 t )
i (0)  0