Faraday’s Law
Faraday’s Law (I)
d
dΦ
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
Φ=
------ Wb
∫∫ B ⋅ dS
S
Faraday’s Law (II)
d
dΦ
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
dΦ
∫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
dΦ
=
dt
Faraday’s Law (IV)
Top view
B
B
--Vind
N turns
+++
Vind
dΦ
=N
dt
+
Vind
-
Linear DC Machine
Induced Voltage from a Moving Bar
B
B
+
+
Vind
-
Vind
l
v
v
-
0
Φ = BA = Blx
Vind
x
dΦ
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
l
VB
, Find = Blit =0
R
x
dv
, M : mass of bar, v : velocity of bar. The bar will accelerate.
Find = M
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.
From Vind = Blv, we have v =
Vind VB
= .
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
dv
=M
dt
VB − Vind
2. v ↑, Vind = Blv ↑, i =
↓, Find = Bli ↓
R
3. When Find = Fload or Fnet = 0, motor reaches steady state with constant velocity v.
From Find = Fload
Find
Vind
.
. Vind = VB -iR = Blv, we have v =
= Bli, we get i =
Bl
Bl
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 the 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
Vind
.
= iR = Blv, v =
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 − Vind
↓, Find = Bli ↓ .
2. v ↑, Vind = Blv ↑, i = B
R
When Vind = VB , i = 0, Find = 0.
1. t = 0, close the switch and apply Fapply , it =0 =
Since Fext = Fapply − Fload > 0, the bar will still accelerate.
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 F=
0, bar reaches steady state with constant velocity v
ext or Fnet
From Find = Fapply − Fload = Bli, we get it =∞ =
V
Vind =
VB + it =∞ R =
Blv, we have v =ind .
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 Ω
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