Electrodynamics
In the square ABCD there is a uniform
magnetic field pointing into the page.
A resistor R is connected to a
conducting wire abef, so forming a
loop.
The loop is being pulled right at a
constant velocity of vh
So charges in the segment ab are
102
moving right with velocity vh, and so
they experience a force,
~
F~1 = Q(~v × B)
~ point downwards.
~v × B
The mobile -ve charges then move
upwards, giving a downwards current.
The magnetic field exherts a force on
this current.
Z
~
F~2 = (I~ × B)dl
~
= I~ × B
Z
dl
~
= I~ × Bh
103
F = IBh, pointing left
Work done = force * distance
Differentiate . . .
Power = force * velocity
Power applied to the loop = −IBhvh
Compare this power with the electric
power in the circuit.
Electric Power = V I
Conservation of energy gives
⇒ V I = −IBhvh
104
⇒ V = −Bhvh
~
Voltage is the line integral of E.
Z b
a
~ = V = −Bhvh
~ emf • dl
E
= Eemf h = −Bhvh
⇒ Eemf = −Bh
Total magnetic flux through the loop
=Φ=
Z
~
~ • da
B
= Bhs
105
dΦ
ds
= Bh
dt
dt
= −Bhvh
= −E = −Eemf h
So
Z
~ = −Eemf h
~ emf • dl
E
dΦ
=−
dt
Applying the curl theorem,
=
Z
d
~
~
(∇ × Eemf ) • da = −
dt
106
Z
~
~ • da
B
Interchanging limit operations
=−
Z
~
∂B
∂t
!
~
• da
Equating integrands
~
d
B
~ emf = −
∇×E
dt
A changing magnetic field induces an
electric field.
~
∂B
In magnetostatics,
=0
∂t
107
Electrodynamics Before Maxwell
Gauss’ Law
ρ
~
∇•E =
0
(1)
~ =0
∇•B
(2)
(no name)
Faraday’s Law
~
∂
B
~ =−
∇×E
∂t
(3)
Ampere’s Law
~ = µ0J~
∇×B
Lorentz’s Force Law
~ + ~v × B)
~
F~ = Q(E
108
(4)
(5)
Let’s check the self-consistency of these
equations.
Take the div of (3),
~ = −∇ •
∇• ∇×E
~
∂B
∂t
!
The LHS is zero because the div of a
curl is always zero, so
~ =0
∇• ∇×E
The RHS is
~
∂B
∂t
−∇ •
109
!
∂ ~
=−
∇•B
∂t
Using (2)
=0
as expected.
110
Next, let’s take the div of (4)
~ = µ0∇ • J~
∇• ∇×B
The LHS is zero, since the div of a curl
is zero.
The RHS is, using the continuity
equation,
∂ρ
~
µ0∇ • J = −µ0
∂t
and this is not, in general, zero.
This is a contradiction.
In magnetostatics, ρ does not change
with time, so ∂ρ
∂t = 0.
111
Ampere’s Law is correct only in
magnetostatics.
It is a reasonable approximation if
things are not changing too rapidly,
roughly up to radio frequencies.
112
Another Perspective
Ampere’s Law in integral form is
Z
~ = µ0Ienc
~ • dl
B
and by the curl theorem
=
Z ~
~ • da
∇×B
Consider a capacitor which is being
charged up
113
One get’s different answer’s, depending
on the surface one integrates over
It seems that some sort of current
must be flowing between the two plates
of the capacitor, a ”current” which is
not due to moving electrical charges
114
Maxwell’s Repair
The continuity equation is
∂ρ
~
∇•J =−
∂t
Using (1)
∂
~
= − (0∇ • E)
∂t
~
∂E
0
∂t
= −∇ •
!
Hence,
∇•
~
∂
E
J~ + 0
∂t
115
!
=0
The difficulty is that, since the div of a
curl is zero, (4) gives
∇ • J~ = 0
By adding the extra ”stray” term to
(4), self-consistency is recovered.
116
Maxwell’s Equations
ρ
~
∇•E =
0
~ =0
∇•B
~
∂
B
~ =−
∇×E
∂t
~
∂
E
~ = µ0J~ + µ00
∇×B
∂t
There’s a more symmetric way of
writting them.
~ =ρ
∇•D
117
~ =0
∇•B
~
∂
B
~ =−
∇×E
∂t
~
∂
D
~ = J~ +
∇×H
∂t
These are called Maxwell’s Equations.
Together with Lorentz’s Law, they
contain all of classical electrodynamics.
The extra term, Maxwell’s correction,
~
∂D
∂t is called the displacement current.
118
This correction resolves the problem
with the capacitor
Ampere’s Law with Maxwell’s
correction in integral form is
Z
~ = µ0Ienc +µ00
~ dl
B•
Z
!
~
∂E
~
•da
∂t
~ = 0 so
For one surface, Ienc = I and E
Z
~ = µ0 I
~ • dl
B
For the capacitor, between the plates,
Ienc = 0 and
σ
Q
E= =
0 0 A
119
∂E
1 dQ
I
=
=
∂t
0A dt
0 A
120
So
µ0 0
Z
µ0 0
~
∂E
∂t
Z = µ0 0
!
∂E
∂t
Z I
= µ0
A
121
da
I
da
0 A
Z
= µ0 I
as required
~
• da
da
Beyond Maxwell, there is Quantum
Mechanics and Special Relativity.
Maxwell does not capture quantum
effects.
Modern quantum mechanics includes
Maxwell’s equations.
(when averaged over very large
numbers of particles).
Maxwell’s equations need a correction
when the velocity of particles
approaches the speed of light.
This correction is also needed for
relativity reasons.
122
From Differential to Integral Form
Each of Maxwell’s equations has an
integral form, as well as a differential
form.
Gauss’ Law in differential form is
~ =ρ
∇•D
Integrate both sides over volume
Z V
Z
~ dτ =
∇•D
ρdτ
V
Applying the Divergence Theorem . . .
Z V
~ dτ =
∇•D
123
Z
S
~
~ • da
D
gives
Z
S
~ =
~ • da
D
Z
ρdτ
V
R
Since V ρdτ is the total charge
enclosed in volume V ,
Z
S
~ = Qenc
~ • da
D
i.e. Gauss’ Law in integral form.
Similarly,
Z
S
~ =0
~ • da
B
124
Faraday’s Law in differential form is
~
∂
B
~ =−
∇×E
∂t
Integrate both sides over a surface
Z S
~ =−
~ • da
∇×E
Z
S
~
∂B
∂t
!
Applying the Curl Theorem . . .
Z S
~ =
~ • da
∇×E
I
~
~ • dl
E
gives
I
~ =−
~ • dl
E
Z
125
S
~
∂B
∂t
!
~
• da
~
• da
d
=
dt
Z
S
~
~ • da
B
dΦ
=−
dt
where Φ is the total magnetic flux
cutting through the surface.
I
dΦ
~
~
E • dl = −
dt
i.e. Faraday’s Law in integral form.
126
Similarly, Ampere’s Law is
~
∂
D
~ = J~ +
∇×H
∂t
and the Curl Theorem gives
Z S
~ =
~ • da
∇×H
=
Z
S
~
∂
D
J~ +
∂t
I
!
~
• da
Since
Z
S
~ = Ienc
J~ • da
then
127
~
~ • dl
H
I
Z
d
~
~
~ • da
~
H • dl = Ienc +
D
dt S
128
Maxwell’s Equations
The integral form of Maxwell’s
equations are
Gauss’ LawZ
S
~ = Qenc
~ • da
D
No Name
Z
S
~ =0
~ • da
B
Faraday’s Law
I
dΦ
~
~
E • dl = −
dt
Ampere’s Law
I
Z
d
~
~
~
~ • da
H • dl = Ienc +
D
dt S
129