3. The Equations of Electromagnetism
(at this point …)
q
Gauss’ Law for Electrostatics ∫ E • dA = ε0
Gauss’ Law for Magnetism ∫ B • dA = 0
dΦB
Faraday’s Law of Induction ∫ E • dl = − dt
Ampere’s Law ∫ B • dl = µ0 I
4. The Equations of Electromagnetism
Gauss’s Laws ..monopole..
q
1
∫ E • dA = ε0
2
∫ B • dA = 0 ?
...there’s no
magnetic monopole....!!
5. The Equations of Electromagnetism
Faraday’s Law .. if you change a
.. if you change a
dΦ B magnetic field you
magnetic field you
3 ∫ E • dl = − dt induce an electric
induce an electric
field.........
field.........
Ampere’s Law
4 ∫ B • dl = µ0 I .......is the reverse
.......is the reverse
true..?
true..?
6. ...lets take a look at charge flowing into a capacitor...
...when we derived Ampere’s Law B E
we assumed constant current...
∫ B • dl = µ0 I
7. ...lets take a look at charge flowing into a capacitor...
...when we derived Ampere’s Law B E
we assumed constant current...
∫ B • dl = µ0 I
E
.. if the loop encloses one B
plate of the capacitor..there
is a problem … I = 0
Side view: (Surface
is now like a bag:)
8. Maxwell solved this problem
by realizing that....
Inside the capacitor there must B E
be an induced magnetic field...
How?.
9. Maxwell solved this problem
by realizing that....
Inside the capacitor there must B E
be an induced magnetic field...
How?. Inside the capacitor there is a changing E ⇒
B A changing
x
x x x x
E
electric field
x x x x x induces a
x x
magnetic field
10. Maxwell solved this problem
by realizing that....
Inside the capacitor there must B E
be an induced magnetic field...
How?. Inside the capacitor there is a changing E ⇒
B A changing dΦ E
∫ B • dl = µ 0ε 0 dt = µ 0 Id
x
x x x x
E
electric field
x x x x x induces a where Id is called the
x x
magnetic field displacement current
11. Maxwell solved this problem
by realizing that....
Inside the capacitor there must B E
be an induced magnetic field...
How?. Inside the capacitor there is a changing E ⇒
B A changing dΦ E
∫ B • dl = µ 0ε 0 dt = µ 0 Id
x
x x x x
E
electric field
x x x x x induces a where Id is called the
x x
magnetic field displacement current
dΦ E
∫ B • dl = µ 0 I + µ 0ε 0 dt
Therefore, Maxwell’s revision
of Ampere’s Law becomes....
12. Derivation of Displacement Current
dq d( EA )
For a capacitor, q = ε0 EA and I = dt = ε0 dt .
d(Φ E )
Now, the electric flux is given by EA, so: I = ε 0 dt ,
where this current , not being associated with charges, is
called the “Displacement current”, Id.
dΦ E
Hence: I d = µ0 ε0
dt
and: ∫ B • ds = µ0( I + Id )
⇒ ∫ B • ds = µ0 I + µ0ε 0 dΦ E
dt
13. Derivation of Displacement Current
dq d( EA )
For a capacitor, q = ε0 EA and I = dt = ε0 dt .
d(Φ E )
Now, the electric flux is given by EA, so: I = ε 0 dt ,
where this current, not being associated with charges, is
called the “Displacement Current”, Id.
dΦ E
Hence: I d = µ0 ε0
dt
and: ∫ B • dl = µ0( I + Id )
⇒ ∫ B • dl = µ 0 I + µ 0ε 0 dΦ E
dt
14. Maxwell’s Equations of
Electromagnetism
q
Gauss’ Law for Electrostatics
∫ E • dA = ε0
Gauss’ Law for Magnetism
∫ B • dA =0
dΦ B
Faraday’s Law of Induction ∫ E • dl = − dt
dΦ E
Ampere’s Law
∫ B • dl = µ0 I + µ0ε0 dt
15. Maxwell’s Equations of Electromagnetism
in Vacuum (no charges, no masses)
Consider these equations in a vacuum.....
......no mass, no charges. no currents.....
q
∫ E • dA = ε 0 ∫ E • dA = 0
∫ B • dA = 0 ∫ B • dA = 0
dΦB dΦ B
∫ E • dl = − dt ∫ E • dl = − dt
B • dl = µ0 I + µ0ε 0 dΦ E
dΦ E
∫ dt ∫ B • dl = µ0ε 0 dt
16. Maxwell’s Equations of Electromagnetism
in Vacuum (no charges, no masses)
∫ E • dA = 0
∫ B • dA = 0
dΦ B
∫ E • dl = − dt
dΦ E
∫ B • dl = µ0ε 0 dt
17. Electromagnetic Waves
Faraday’s law: dB/dt electric field
Maxwell’s modification of Ampere’s law
dE/dt magnetic field
dΦ E dΦB
∫ B • dl = µ0ε0 dt ∫ E • dl = − dt
These two equations can be solved simultaneously.
ˆ
E(x, t) = EP sin (kx-ωt) j
The result is:
ˆ
B(x, t) = BP sin (kx-ωt) z
18. Electromagnetic Waves
dΦ E dΦ B
∫ B • dl = µ0ε 0 dt ∫ E • dl = − dt
B
dE E
dB
dt
dt
19. Electromagnetic Waves
dΦ E dΦ B
∫ B • dl = µ0ε 0 dt ∫ E • dl = − dt
v
B
dE E
dB
dt
dt
Special case..PLANE WAVES... E = E y ( x ,t )
j
B = Bz ( x ,t )k
∂ 2ψ 1 ∂ 2ψ
satisfy the wave equation = 2 2
∂x 2 ν ∂t
Maxwell’s Solution ψ = A sin( ωt + φ )
21. F(x) λ
Static wave
F(x) = FP sin (kx + φ)
k = 2π / λ
k = wavenumber
x λ = wavelength
F(x) λ
Moving wave
F(x, t) = FP sin (kx - ωt )
v
ω = 2π / f
ω = angular frequency
x
f = frequency
v=ω /k
22. F
v Moving wave
x F(x, t) = FP sin (kx - ωt )
What happens at x = 0 as a function of time?
F(0, t) = FP sin (-ωt)
For x = 0 and t = 0 ⇒ F(0, 0) = FP sin (0)
For x = 0 and t = t ⇒ F (0, t) = FP sin (0 – ωt) = FP sin (– ωt)
This is equivalent to: kx = - ωt ⇒ x = - (ω/k) t
F(x=0) at time t is the same as F[x=-(ω/k)t] at time 0
The wave moves to the right with speed ω/k
23. Plane Electromagnetic Waves
Ey ˆ
E(x, t) = EP sin (kx-ωt) j
ˆ
B(x, t) = BP sin (kx-ωt) z
Bz
Notes: Waves are in Phase, c
but fields oriented at 900.
k=2π/λ. x
Speed of wave is c=ω/k (= fλ)
c = 1 / ε0µ0 = 3 ×108 m / s
At all times E=cB.