A Moving Conductor
in a Static Magnetic Field
UNIVERSITY OF TECHNOLOGY, SYDNEY
FACULTY OF ENGINEERING
+a
48531 Electromechanical Systems
B
Electromagnetic Induction
Topics to cover:
1. A Moving Conductor in a Static Magnetic Field
2. Faraday’s Law
3. Maxwell’s Equations
A Moving Conductor in a
Static Magnetic Field - Cont.
Cont.
At equilibrium, which is reached very
rapidly, the net force on the free
charges in the moving conductor is
zero, or Fe=−Fm . The voltage between
terminals a and b is then
a
a
b
or
a
Vab = ∫ (u × B) • dl
b
++
+
Fe
l
F
u
-b
Calculate the open circuit voltage between the brushes on a Faraday's
disc as shown schematically in the diagram below.
Fm
a
Fe
F
• dl = ∫ m • dl
q
q
b
b
Vab = −∫ E • dl = −∫
l
F
A Moving Conductor in a
Static Magnetic Field - Example
+a
B
Consider a piece of conductor of length
++
l moving with a velocity u in a static
(non-time varying) magnetic field. A
m
force Fm=qu×B will cause the freely
movable electrons in the conductor to
+
drift toward one end of the conductor
e
and leave the other end positively
charged. This separation of the positive
-and negative charges creates a
Coulombian force of attraction: Fe=qE .
The charge separation process continues until the electric and magnetic
force balance each other and a state of equilibrium is reached.
u
ωr
Shaft
--
-b
This is known as the flux cutting or motional
electromotive force (emf). Symbol e is used
specifically to denote emf.
r2
r1
Brushes
v
B
+
S
N
S
N
A Moving Conductor in a
Static Magnetic Field - Example
Faraday’s Law
Solution:
Choose a small line segment of length dr at position r (r1< r < r2) from
the center of the disc between the brushes. The induced emf in this
elemental length is then
de = Budr = Bωr rdr
where u=rωr. Therefore,
r2
When the magnetic flux linking a circuit changes, an emf is induced in the
circuit. Faraday's Law states that the emf equals the rate of change of
flux. Mathematically,
e=−
where the minus sign indicates the direction of the induced emf is such
that any current produced by it tends to oppose the change in flux, which
is known as the Lenz's Law. More generally, when there N circuits or N
turns, the Faraday's law can be written as
r2
r2
r22 − r12
= ωr B
e = ∫ de = ∫ Bωr rdr = ωr B
2
2
r1
r
1
Calculate the induced emf in a small rectangular coil of N turns
moving with a velocity v in a magnetic field of flux density B=
azBmsinαx.
dx
x
e = −N
dφ
dλ
=−
dt
dt
where λ=Nφ is the total flux linkage.
Faraday’s Law - Example
Faraday’s Law - Example
y
dφ
dt
Solution:
Consider a strip dA=bdx, and therefore dφ=BdA. The linking flux is then
a
determined from
x +
c
∫
φ =
dA=bdx
xc −
B
= −
b
2
B m s in α x ⋅ b d x
a
2
a
xc +
Bmb
[ c o s α x ] x a2
α
c −
2
Bmb 
a
a


cosα  xc +  − cosα  xc −  


2
2
α 
2b
 αa 
s in 
=
 B s in α x c
 2  m
α
= −
xc
O
a/2
a/2
x
(a s c o s (u + v ) − c o s ( u − v ) =
− 2 s in u s in v )
Faraday’s Law - Example
Faraday’s Law - Example
From the above expression, we know the magnitude of emf is
and so for N turns the flux linking the coil is
2 Nb
 αa 
sin 
λ =
 B sin α x c
 2  m
α
dλ
dλ dxc
=−
dt
dxc dt
 αa 
E m = − 2 bN sin 
B v
 2  m
Consider
e=−
Now,
2 bN
dλ
 αa 
sin 
=
 B cos(α x c )α
 2  m
α
dx c
Therefore,
and
v =
dxc
dt
 αa 
= 2 bN sin 
 B cos(α x c )
 2  m
If B= azBmsin(ωt−αx), what is the induced emf?
As
 αa 
e = − 2 bN sin 
 B v cos(α x c )
 2  m
Faraday’s Law - Example
we have
When sin(αa/2) = ±1 or αa = (2n−1)π, (n=1,2,……), it reaches the
maximum, and therefore, the magnitude of emf reaches the maximum,
when a = π / α.
dλ
dt
2 bN
 αa 
sin 
=−
 B cos(ω t − α x c )(ω − α v )
 2  m
α
e=−
ω

 αa 
= − 2bN  − v  sin 
 B cos(ω t − α x c )
α

 2  m
When v = ω/α, the induced emf e = 0.
λ=
2bN
 αa 
sin  Bm sin(ωt − αxc )
 2
α
=
2bN
 αa 
sin  Bm sin(ωt − αvt )
 2
α
Ideal Transformer
A transformer is an alternating current device that transforms
voltages, currents and impedance. For ideal transformers, we
assume that the permeability of the core approaches infinity, and
that all power losses in the windings and the core are ignored.
Thus, by the Ampere’s law, we have
N1i1 − N2i2 =
or
φ
l
φ ⇒0
µA
i1 N2
=
i2 N1
i1
v1
i2
N1
N2
µ→∞
v2
Ideal Transformer - Cont.
Cont.
Maxwell’s Equations
Since the power losses are ignored, the input power must be equal
to the output power:
v1i1 = v2i2
Integral Form
v1 i2 N1
= =
v2 i1 N2
or
dφ
∫ E• dl = − dt
For resistance transform, we can write
2
∇×E = −
C
∂D
2
v N  v N 
R = 1 =  1  2 =  1  RL
i1  N2  i2  N2 
∫ H • dl = I + ∫ ∂t • ds
'
L
i1
For impedance, we have
Differential Form
C
i2
∂B
∂t
∇×H = J +
S
∫ D • ds = Q
∇•D = ρ
∫ B • ds = 0
∇•B = 0
∂D
∂t
Significance
Faraday’s law
Ampere’s law
Gauss’ law
S
2
N 
ZL' =  1  ZL
 N2 
R'L
v1
N1
N2
v2
RL
S
No isolated
magnetic
charge