Faraday : time - varying magnetic field →
electric current with induced voltage (EMF : electromotive force)
N = number of turns (6 in above figure);
ψ = magnetic flux through a single loop of coil
λ = Total magnetic flux


ψ = B · d S
λ=Nψ
There is an E - field between the plates of the battery
(For certain cases λ α I or λ = L I, L = inductance)
Vemf = -
dλ
dt
= -L
di
dt
=-
Ndψ
dt
Direction of current opposes magnetic field
** Aside **
2
lect19new.nb
Power Engineering : Why is 50 Ω Coaxial Line so Special Anyway ?
Long wire, length = Δz,

E≈
ρL
2 π εo ρ
b

ρ=
Q
2 π εo ρ Δz
b
V = - E.ⅆ ρ = - E ⅆ ρ =
a
Hϕ =
I
2πρ
a

ρ
Q
2 π εo Δz
ln[b / a]
3
lect19new.nb
Inductance per unit length =
LL =
1
 
L / Δz =
  B.d S =
I Δz
1
1
 
  μo H.d S =
  μo Hϕ dz dρ =
I Δz
I Δz
b
1
I
1
I
μo
dz dρ =
dz dρ =
ln[b / a]
  μo
  μo
I Δz
2πρ
I Δz
2πρ
2π
a
Capacitance per unit length CL = C / Δz =
Zo = Sqrt[LL / CL ] =
ln[b / a]
2π
Q
V
=
Sqrt[μo / εo ] =
Q 2 π εo Δz
ln[b / a] Δz Q
ln[b / a]
2π
=
2 π εo
ln[b / a]
ηo
Power Handling
Capacity : Dielectric breakdown will occur in the region between
the two conductors if the electric field exceeds a certain
critical value :
E
V
P=
V2
Zo
=
=
1
ρ ln[b / a]
-> E =
V
ρ ln[b / a]
→ Emax =
V
a ln[b / a]
2 π (Emax a )2 ln[b / a]
ηo
To find a to give maximum power
∂P
= 0. solve for a →
∂a
b / a = 1.65 -> Zo = 30 Ω
For minimum attenuation α = R / (2 Zo ) + G Zo / 2,
where R is the series resistance per unit length,
and G is the shunt conductance per unit length;
minimum α occurs when b / a = 3.6 -> Zo = 77 Ω
Zo = 50 Ω is a compromise
4
lect19new.nb
**
Vemf = -
dλ
dt
= -L
di
dt
=-
Ndψ
dt
Direction of current opposes magnetic field
 

E = Eemf + Ee
Charge accumulation causes Ee ,
E field due to charge accumulation on battery terminals

1) Ee does not maintain a steady current

2) Eemf does provides the steady current


 Ee · d l = 0 conservative


 Eemf · d l = Vemf
non - conservative
Vemf


Nd ∫ B · d S
Ndψ


=  Eemf · d l = =dt
dt
5
lect19new.nb
N=1


d∫B·d S 



, B=μH
 Eemf · d l = dt
Stokes Law, we get the Maxwell ' s equation

dB

∇ × Eemf = dt
bfield = {0, 0, .004 Cos[10^6 t]}; l = .08; h = .06;
bfield points in the same direction as the normal to the loop
ψ = Integrate[.004 Cos[10^6 t], {y, 0, .08}, {x, 0, .06}]
0.0000192 Cos[1 000 000 t]
vemf = -D[ψ, t]
19.2 Sin[1 000 000 t]
p1 = Plot[.004 Cos[10^6 t], {t, 0, 2 π 10^-6},
Ticks → {{0, π 10^-6, 2 π 10^-6}, {.1}}, PlotLabel → "B"];
6
lect19new.nb
i = vemf / R
p2 = Plot[19.2 Sin[1 000 000 t], {t, 0, 2 π 10^-6},
Ticks → {{0, π 10^-6, 2 π 10^-6}, {.004}}, PlotLabel → "I"];
GraphicsRow[{p1, p2}]
B
I
π
π
π
1 000 000
500 000
1 000 000
Induced current is in direction such that its magnetic field oppposes
the change in magnetic field. (Lenz' s Law)
Displacement Current :

dD
causes magnetic field;
dt
this is the current through a capacitor,
due to polarization in an insulator
7
lect19new.nb

  dD
∇ ×H = J+
dt

J : conduction current
Vs = 50 Sin[10^3 t];


Vs = -  E · d l = E d
d = .003;
E1 = 50 Sin[10^3 t] / .003
16 666.7 Sin[1000 t]
ϵo = 8.854 × 10^-12;
D1 = 2 ϵo E1
2.95133 × 10-7 Sin[1000 t]
8
lect19new.nb
area = 5 × 10^-4;
I1 = area D[D1, t]
1.47567 × 10-7 Cos[1000 t]
C dV / dt = I
C1 = I1 / D[Vs , t]
2.95133 × 10-12
C = ϵ area / d
2 ϵo area / d
2.95133 × 10-12
Maxwell ' s Equations
Differential form :

∇ · D = ρcharge ,

 -∂B
∇ ×E =
∂t

∇ ·B = 0

  ∂D
∇ ×H = J+
∂t
Integral Form :
9
lect19new.nb




 D · d S = Qenclosed ,  B · d S = 0 no magnetic monopole


∂∫ B · d S


Faraday' s Law
E · dl=∂t





 ∂D
H
·
d
l
=
· d S Ampere ' s Law
J
·
d
S
+


∂t


 
D = ϵ E = ϵo E + P


B=μH



Jc = σ E + ρv u

σ E = current through a conductor

ρv u = current through free space
Boundary conditions
E1 t = E2 t , H1 t - H2 t = k
(k = surface current density)
D1 n - D2 n = ρ s surface charge density
B1 n - B2 n = 0


Perfect conductor, σ = ∞, E = 0, H = 0
10
lect19new.nb
Force


 
F = q E+ q u×B
 -∂ρ
∇ ·J =
∂t
Maxwell ' s Equations in the frequency domain

∇ · Ds = ρvs ,

∇ · Bs = 0


∇ × Es = -I ω Bs



∇ × Hs = Js + I ω Ds
Electromagnetic wave propagation
wave equation
Waves in one dimension :
∂2 A
∂t2
- u2
∂2 A
∂z2
=0
u = velocity of a wave
u
f
T
λ
β
=
=
=
=
=
f λ = ω / β,
frequency = 1 / T
period
wavelength
2 π / λ = wave number
∂(As Exp[I ω t])
∂t
-> I ω As Exp[I ω t]
wave equation in complex notation
(I ω)2 As - u2
∂2 As
∂z2
=0
11
lect19new.nb
A = f (z - u t) or A = f (z + u t)
Let f (z - u t) = Sin[z - u t]
∂2 f (z - u t)
∂t2
∂2 f (z - u t)
∂z2
So
= -u2 Sin[z - u t]
= -Sin[z - u t]
∂2 f (z - u t)
∂t2
- u2
∂2 f (z - u t)
∂z2
=0
E & M wave propagation
Let ρcharge = 0

∇ · Es = 0,

∇ · Hs = 0


∇ × Es = -I ω μ Hs




∇ × Hs = σ Es + I ω ϵ Es = (σ + I ω ϵ) Es


∇ × ∇ × Es  = -I ω μ ∇ × Hs
From vector calculus



∇ × ∇ × A  = ∇∇ · A  - ∇2 A
∇2 =
∂2
∂x2
∂2
+
∂y2
∂2
+
∂z2

Since ∇ · E = 0



∇2 Es = I ω μ ∇ × Hs = I ω μ (σ + I ω ϵ) Es 
Let σ = 0 (wave propagation with no attenuation)
12
lect19new.nb


∇2 Es = (Iω)2 μ ϵ Es
We have the wave equation, u = 1 
μϵ