Download Class 8 in Electrodynamics

Class 8 in Electrodynamics
Based on course by Yuri Lyubarsky and Edited By Avry Shirakov
Department of Physics, Ben-Gurion University, Beer-Sheva 84105, Israel
This exercise pool is intended for an undergraduate course in “Electrodynamics 1”.
Electromagnetic Waves
Maxwell’s eq’s:
1
ρf
∇·E=
∇·B=0
∇×E=−
∂B
∂t
∇ × B = µJ + µ
∂E
∂t
If ρf , J = 0 then we have electromagnetic wave eq:
∇2 E =
1 ∂2E
c2 ∂t2
It also can be written as E(r, t) = Re E˜0 ei(k·r−ωt) while E˜0 is a complex amplitude, while the magnetic field is:
B(r, t) =
for k =
n
(k × E)
c
2π
. Once again, the field boundary conditions:
λ
1 E1⊥ − 2 E2⊥ = σf
B1⊥ − B2⊥ = 0
k
k
E1 − E2 = 0
1 k
1 k
B1 −
E = kf × n̂
µ1
µ2 2
Snell’s law:
n1 sin(θ1 ) = n2 sin(θ2 )
Intensity:
I=
1
c0 E 2 =< S >
2
Waves: TE (transverse electric)→ Ez = 0;
TM (transverse magnetic)→ Bz = 0;
TEM (transverse electro-magnetic)→ Ez , Bz = 0 (cannot occur in hollow wave-guides).
2
Resonant Cavities
Placing end plane surfaces of a cylindrical wave-guide, radius R with an infinite conductivity. The cavity is filled with
lossless dielectric with µ, . Find the fundamental mode of the waveguide.
Solution
The reflections at the end surfaces, the ẑ dependence of the fields is same as of standing waves, of the form:
A sin(kz) + B cos(kz)
the plane boundary surfaces are at z = 0 and z = L, so the boundary conditions can be satisfied at each surface when
k=s
π
for s = 0, 1, 2, ...
L
For TM fields, vanishing of Et at z = 0 and z = L
sπz Ez = f (x, y) cos
for s = 0, 1, 2, ...)
L
and the transverse fields are
sπz sπ
Et = − 2 sin
(∇f )t
Lγ
L
Ht =
sπz iω
cos
z × (∇f )t
γ2
L
For TE field, vanishing of Hz at z = 0 and z = L requires that
sπz Hz = f (x, y) sin
for s = 1, 2, 3, ...)
L
and the transverse fields are
sπz iµω
Et = − 2 sin
z × (∇f )t
γ
L
Ht =
sπz sπ
cos
(∇f )t
Lγ 2
L
(the boundary conditions are now satisfied), the constant γ 2 is:
∇2t + γ 2 f = 0 while γ 2 = µω 2 − k 2
γ 2 = µω 2 −
sπ 2
L
and as a result the eigenfrequency is:
sπ 2 1
2
ωλs
=
γλ2 +
µ
L
A practical resonant cavity is a circular cylinder, with an inner radius R and length L. For a TM mode the transverse
wave eq. for f = Ez , with the boundary condition Ez = 0 for ρ = R has the solution:
f (ρ, φ) = E0 Jm (γmn ρ)e±imφ where γmn =
xmn is the n’th root of Jm (x) = 0
x0n = 2.405 5.520 8.654
x1n = 3.832 7.016 10.173
x2n = 5.136 8.417 11.620
xmn
R
3
while m = 0, 1, 2, .. and n = 1, 2, 3, ... The TM mode is of the form
r
x2mn
s2 π 2
1
+ 2
ωmns = √
2
µ
R
L
the resonant frequencies form a discrete set of values, for m = 0, n = 1, s = 0 we get the lowest TM mode, ω010 =
2.405
.
√
µR
The explicit fields are
2.405ρ −iωt
Ez = E0 J0
e
R
r
Hphi = −i
E0 J1
µ
2.405ρ
R
e−iωt
the result is independent of L (length of the cylinder), there is no tuning for the
eigenfrequency.
∂f
x0
For a TE mode the basic solution still implies, but the boundary conditions Hz
|R = 0 makes γmn = mn where
∂ρ
R
0
x0mn is the n’th root of Jm
(x) = 0
x00n = 3.832 7.016 10.173
x01n = 1.841 5.331 8.536
x02n = 3.054 6.706 9.970
again, the TE modes are of the form
s
0
2
xmn
s2 π 2
1
+ 2
ωmns = √
2
µ
R
L
for m = 1, n = 1, s = 1 we get the lowest TE mode, ω111
1.841
= √
µ
r
1+
2.912R2
L2
The explicit field expression are
πz 1.841ρ
Hz = H0 J1
e−iωt
cos(φ) sin
R
L
.
For L > 2.03R the resonance frequency ω111 is smaller than that for TM mode (ω010 ). Then the T E111 mode is the
fundamental oscillation of the cavity.
Problem 0355
An infinitely long thin cylindrical shell (inner radius a, thickness d << a) made of a conductor with the conductivity
σ is placed in a magnetic field Re(B0 eiωt ) parallel to the cylinder axis. Find the magnetic field inside the cylinder.
Solution
The magnetic field outside is:
B=
B0 iωt
e + e−iωt
2
inside the conducting layer the field is
µ0 σ
∂B
= ∇2 B
∂t
where we used Jf = σE and ∇ × (∇ × B) = ∇ (∇ · B) − ∇2 B = −∇2 B.
We guess that the solution for B is of the form :
B(r) = B + (r)eiωt + B i (r)e−iωt (ẑ)
4
plugging in the solution
∂2 +
B = iσωµB +
∂r2
∂2 −
B = iσωµB −
∂r2
while the
B + = C1 eik1 r + C2 e−ik1 r and B − = D1 eik2 r + D2 e−ik2 r while
r
k1 =
µσω
(1 + i) and k2 =
2
r
µσω
(1 − i)
2
from the convergence of the magnetic field
B0
= C1 eik1 (d+a) + C2 e−ik1 (d+a)
2
B0
= D1 eik2 (d+a) + D2 e−ik2 (d+a)
2
as we see, inside the field is the solution of Poison’s eq, with constant boundary conditions.
+ iωt
− −iωt
Bin = Bin
e + Bin
e
Bin = C1 eik1 a + C2 e−ik1 a
On the other hand, in the inner bound:
∇ × B = µσE and
H
E · dl = −
dΦmag
dt
For B +
∇×B=−
∂
Az = ik1 C1 eik1 a − C2 e−ik1 a
∂r
I
Edl = Eφ 2πa
dΦmag
d +
= πa2 Bin
= iωπa2 C1 eik1 a + C2 e−ik1 a
dt
dt
summing up,
ik1 C1 eik1 a − C2 e−ik1 a = −µσiωπa2 C1 eik1 a + C2 e−ik1 a
here we find the radio between C1 and C2 , same mechanism is applied for finding B − .