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CHAPTER 9
FARADAY’S LAW AND
DISPLACEMENT CURRENT
9.1 FARADAY’S LAW
9.1.1 TIME VARYING FIELD – STATIONARY CIRCUIT
9.1.2 MOVING CIRCUIT – STATIC FIELD
9.1.3 TIME VARYING FIELD - MOVING CIRCUIT
9.2 DISPLACEMENT CURRENT
9.3 LOSSY DIELECTRICS
9.4 BOUNDARY CONDITIONS
1
9.0 FARADAY’S LAW AND DISPLACEMENT
CURRENT
Two topics will be discussed :
(i) Faraday’s Law – about the existence of electromotive force
(emf) in the magnetic field
(ii) Displacement current – that exists due to time varying field
That will cause the modification of Maxwell’s equations (in point form
- static case) studied previously and hence becomes a concept basic
to the understanding of all fields in electrical engineering.
9.1 FARADAY’S LAW
Michael Faraday – proved that if the current can produce magnetic
field, the reverse also will be true.
Proven only after 10 years in 1831.
The magnetic field can produce current in a loop, only if the
magnetic flux linkage the surface of the loop is time varying.
Faraday’s Experiment :
Loop
Flux linkage
-
Galvanometer
+
Battery
Loop
Flux
linkage
• Current produced magnetic field and the
magnetic flux is given by :
m  ∫
B  ds
- +
(1)
s
Galvanometer
Battery
• No movement in galvanometer means that the flux is constant.
• Once the battery is put off – there is a movement in the
galvanometer needle.
• The same thing will happen once the battery is put on - but this
time the movement of the needle is in the opposite direction.
Conclusions : The current was induced in the loop
- when the flux varies
- once the battery is connected
- if the loop is moving or rotating
Induced current will induced electromotive voltage or induced emf
Vemf given by :
Vemf   N


  N  B  ds
t
t s
(V )
(2)
where N = number of turns
Loop
Flux
linkage
Equation (2) is called Faraday’s Law
- +
Galvanometer
Battery
Lenz’s Law summarizes the –ve sign is that : The induced
voltage established opposes the the flux produced by the loop.
In general, Faraday’s law manifests that the Vemf can be established
in these 3 conditions :
• Time varying field – stationary circuit (Transformer emf)
• Moving circuit – static field (Motional emf)
• Time varying field - Moving circuit (both transformer emf
and motional emf exist)
9.1.1 TIME VARYING FIELD – STATIONARY CIRCUIT
(TRANSFORMER EMF)
Vemf
B
 N 
 ds (V )
t
s
(3)
B(t) increases
I
Vemf = the potential difference at
terminal 1 and 2.
1
R
From electric field :
Vemf   E  d 

2
(4)
Diagram shows a circle loop with the surface
area S placed in the magnetic field B(t)
If N = 1 :
Vemf
B induced
∂
B
∫
E  d   -∫  ds
∂
t

s
(5)
Vemf
∂
B
∫
E  d   - ∫  ds
∂
t

s
Using Stoke’s theorem :
∂
B
∇ E  ds  -∫  ds
∫
∂
t
s
s
Hence Maxwell’s equation
becomes :
∂
B
∇ E  ∂
t
(7)
(6)
9.1.2 MOVING CIRCUIT – STATIC FIELD
(MOTIONAL EMF)
Moving bar
Force :
Fm  q u  B 
B constant

Fm
Em 
 u  B 
q
u
Diagram shows a bar moving with a
velocity u in a static field B .
Hence :
u  B  d 
Vemf  ∫
Em  d   ∫
Fleming’s Right hand rule
Thumb – Motion
1st finger – Field
Second finger - Current
9.1.3 TIME VARYING FIELD - MOVING CIRCUIT
Both transformer emf and motional emf exist
Vemf
∂
B
u  B  d 
∫
E  d  ∫
 ds  ∫
∂
t
s
Ex. 9.1 : A coducting bar moving on the rail is shown in the diagram.
Find an induced voltage on the bar if :
(i) A bar position at y = 8 cm and B  4 cos106 t zˆ mWb / m2
2
(ii) A bar moving with a velocity u  20 yˆ m / s and B  4 zˆ mWb / m
(iii) A bar moving with a velocity u  20 yˆ m / s and
ˆ mWb / m 2
B = 4 cos(106t - y)z
P
y
0
u
B
6 cm
x
Q
P
Solution :
(i) Transformer case :
∂
B
Vemf  -∫  ds
∂
t
-3
6
6
4
(
10
)(
10
)
sin
10
t dxdy
∫∫
y 0 x 0
 19.2 sin 106 t (V )
u
B
6 cm
x
0.08 0.06

y
0
z
Q
y
x
ˆ mWb / m 2
B = 4 cos 106t z
According to Lenz’s law when B increases point P will be at the higher
potential with respect to point Q. (B induced will oppose the increasing B )
(ii) Motional case :
Vemf  ∫
(u  B )  d 
0

( 20 yˆ  4 zˆ )  dx xˆ
∫
x  0.06
 -4.8 mV
Remember : the direction of d 
is opposed the current induced in
the loop.
P
y
0
(iii) Both transformer and motional case :
B  4 cos(106 t - y) zˆ mWb / m2
6 cm
x
Vemf
u
B
z
Q
∂
B
 - ∫  ds  ∫
(u  B )  d 
∂
t
x
0.06 y

-3
6
6
4
(
10
)(
10
)
sin(
10
t - y ' ) dy ' dx
∫∫
x 0 0
20 yˆ  4(10
∫

0

-3
) cos(10 6 t - y ) zˆ  dx xˆ
0.06
y
 240 cos(10 t - y ' ) - 80(10 -3 )( 0.06) cos(10 6 t - y )
6
0
 240 cos(10 6 t - y ) - 240 cos 10 6 t - 4.8(10 -3 ) cos(10 6 t - y )
≈ 240 cos(10 6 t - y ) - 240 cos 10 6 t
y
∂
B
Vemf  -∫  ds  ∫
(u  B )  d 
∂
t
≈ 240 cos(106 t - y ) - 240 cos106 t
From trigonometry :
cos A - cos B  2 sin
Vemf
A B
A- B
sin
2
2
 6 y 
y
 480 sin  10 t -   sin(  )
2 
2

9.2 DISPLACEMENT CURRENT
From continuity of current equation :
and
∂
v
∇ J  ∂
t
A / m3
J  ∇ H
∇∇ H  -
∂
v
∂
∂
D
 - ∇ D  -∇
A / m3
∂
t
∂
t
∂
t
∇∇ H  0  -
∂
v ∂
v

∂
t
∂
t
;
Hence :
∂
∇∇ H  ∇ J  ∇ D
∂
t
and
∇ D   v
∂
∇∇ H  ∇ J  ∇ D
∂
t
∂
D
∇ H  J 
∂
t
where :
R
+
J
= Conduction current density
∂
D
= Displacement current density
∂
t
Hence :
∂
D
∇ H  J c 
 Jc  Jd
∂
t
-
Jc
Jd
Dielectric
C +Q
-Q
Therefore from Faraday’s
law and the concept of
displacement current we
can conclude that both
the magnetic and electric
fields are interrelated.
Maxwell’s Equations
Differential
Form
Integral Form
∂B
∇ E  ∂t
∂D
∇ H  J 
∂t
∂B
E  d   - ∫  ds
∫
∂t
∂D
H

d


I

∫
∫∂t  ds
∇ D   v
D  ds  ∫
 dv
∫
v
s
∇ B  0
Label
Faraday’s Law
Ampere’s Circuital
Law
Gauss’S Law for
Electric Field
v
B  ds  0
∫
Gauss’s Law for
Magnetic Field
s
An integral form of Maxwell’s equation can be found either by using
Divergence Theorem or Stoke Theorem. All electromagnetic (EM) waves
must conform or obey all the four Maxwell’s equations.
Ex.9.2: A parallel plate capacitor having a plate area of 5 cm2 and where the
plates are separated by a distance of 3 mm is connected to a supply voltage,
50 sin 103 t Volt. Calculate the displacement current if the dielectric permittivity
between the plate is   2 0 .
Solution :
D  E  V / d
D
 V
Jd 

t
d t
 I d  J d  ds

S ∂
V
∂
t
5  10  4
3
3
 2 0
(
10
)
50
cos
10
t
-3
3  10
 147.4 cos 10 3 t nA
d
This example is to show the use of Maxwell’s equation and the inter
relation of electric field and magnetic field.
Ex.9.3: Given a magnetic medium with characteristics
has H  2 cos(t - 3 y ) zˆ A / m . Find  and E .
Solution :
∇ H  J 
∂
D
∂
t
;
  0,   20 ,   5 0
 0
∂
E
1
∇ H  
 E   (  H )dt
∂
t

∂
x
∇ H  ∂
0
∂
∂
Hz
∂
Hz
∂
y
∂
z ∂
xˆ yˆ
∂
y
∂
x
0 Hz
 6 sin( t - 3 y ) xˆ
∴E 
-
1
6 sin( t - 3 y )dt xˆ
∫

1
5 0
6 cos(t - 3 y ) xˆ
1
ˆ
6 sin(ωt - 3y)dt x
∫
ε
1
ˆ
= 6 cos(ωt - 3y) x
5ωε0
∂
H
1
∇ E  -
, →H - ∫
(∇ E ) dt
∂
t

∴E =
∂
∂
∂
∂
Ex
∂
Ex
∂
x
∂
y
∂
z
∇ E 

yˆ 
zˆ
∂
z
∂
y
Ex
0 0
- 3(6)

sin( t - 3 y ) zˆ
5 0
H 

1 18
sin( t - 3 y ) dt zˆ
∫
 5 0
18
10 0 0
Hence :
cos(t - 3 y ) zˆ
Compare :
H  2 cos(t - 3 y) zˆ A / m
18
100 0
2
   2.846 108 rad / s
E  - 476.8 cos(2.846 108 t - 3 y) xˆ (V / m)
9.3 LOSSY DIELECTRICS
Main function of dielectric material is to be used as an insulator.
For a perfect dielectric :
 0
Compare (1) and (2) :
  j   j

'
  - j

  '
'
    1 - j '    - j ' '
 

Hence Maxwell’s equation :
∇ H  (  j ' ) E
∇ H  jE
For lossy dielectric :
(1)
 0
∇ H  (  j ) E
'
where :
(2)
 /  ' = loss tangent
Loss tangent is the ratio of the magnitude of the
conduction current density to the magnitude of
the displacement current density

Ic
Id
A lossless capacitor has a loss tangent of zero.
For lossy capacitor, an equivalent circuit can be replaced by its
equivalent resistance in parallel with a perfect capacitor as
shown in the diagram :
+
+
+
+ +
Jd 
Jc
-
+
-
D
t
Ic
- -ve - -
V
and
I c  GV 
R
R
Id
C
Leakage
current
From page 110 & 111
I d  jCV
V  E  dl
R 
I
 J  ds
Hence loss tangent :
Ic
Id
1/ R
s / d





 '
'
'
C  s / d 

''
Q
C

V
 D  ds
 E  dl
Ex.9.4: Find the average power loss per unit volume for a capacitor
having the following properties; dielectric constant 2.5 loss tangent
0.0005 for an applied electric field intensity of 1 kV/m at frequency
500 MHz.
Solution :

Loss tangent  0.0005 
 '
→   (0.0005)( 2 )(500 106 )( 2.5 0 )
 3.476 10-5 S / m
V2
V2
P

2R 2 d
s
 1 E 2sd
2
P
∴
 1 E 2  1 (3.476 10-5 )(103 ) 2
2
2
volume
 17.38 (W / m3 )
9.4 BOUNDARY CONDITIONS
Boundary conditions for time varying field are the same as
boundary conditions in electrostatics and magnetostatics fields.
Tangential components for
E1t  E2t
Normal components :
E and H :
; H1t - H 2t  J s
D1n - D2n   s ; B1n  B2n
If medium 2 is perfect conductor :
Hence :
E H 0
E1t  0 ; H1t  J s
D1n   s ; B1n  0