NAME:
PHGN462: Advanced E&M
Exam 1
October 10, 2008
1. (20) Maxwell “Quickies”:
a. Write down Maxwell’s equations in E, B form.
Solution:
~ ·E
~ = ρ
∇
0
~
~ ×E
~ = − ∂B
∇
∂t
~ ·B
~ =0
∇
~ ×B
~ = µ0 J~ + µ0 0 ∂ E~ .
∇
∂t
b. Define P and M, and give the definitions of D and H in terms of them.
Solution: P (M) is the electric (magnetic) dipole moment per unit volume. D and H are defined:
~
~.
~ = B −M
H
µ0
~ = 0 E
~ + P~
D
c. Define the bound charge density and bound current density in terms of P and M.
Solution:
~ · P~ = −ρb
~ ×M
~ = J~b .
∇
∇
d. Write down Maxwell’s equations in D, H form.
Solution:
~ ·D
~ = ρf
∇
~
~ ×E
~ = − ∂B
∇
∂t
~ ·B
~ =0
∇
~ ×H
~ = J~f +
∇
~
∂D
∂t .
e. What does it mean for P and M if the material is linear?
Solution: If the material is linear, then the response of the material is proportional to the applied field. Specifically,
~ and M
~ = Xm H.
~ Thus, if the material is linear, then D
~ = E,
~ where = 0 (1 + Xe ), and H
~ = B/µ,
~
P~ = Xe E
where
µ = µ0 (1 + Xm ).
f. Write down Maxwell’s equations in E, B form for a linear material.
Solution: Under the conditions in part e, Maxwell’s equations in part d become:
~ ·E
~ =ρ
∇
~
~ ×E
~ = − ∂B
∇
∂t
~ ·B
~ =0
∇
~ ×B
~ = µJ~ + µ ∂ E~
∇
∂t
1
(1)
2. (15) EM Wave “Quickies”:
a. Derive the wave equation for either E or B in free space from Maxwell’s equations.
~ × (∇
~ × A)
~ = ∇(
~ ∇
~ · A)
~ − ∇2 A)
~
(Useful vector identity: ∇
~ × (vec∇ × E)
~ =
Solution: Take the curl of the left hand side of Maxwell’s third equation and use the vector identity: ∇
2
~ ∇
~ · E)
~ − ∇ E.
~ The divergence term vanishes by Maxwell’s first equation where for free space, ρ = 0. Take the curl of
∇(
the right hand side of Maxwell’s third equation, and then use Maxwell’s fourth equation with J~ = 0 (free space), to find
2~
− µ010 ∂∂tE
2 . Thus,
2~
~ = µ0 0 ∂ E .
(0)
∇2 E
∂t2
b. From your wave equation, identify the speed of the wave.
Solution: By dimensional arguments, the speed of the wave is given by c =
√1 .
µ0 0
~ r, t) = E
~ 0 cos(~k · ~r − ωt) is a solution to the wave equation.
c. Show that E(~
~
2~
~
~ (k · ~r) = ~kf 0 ((~k · ~r) to find: ∇2 E
~ = −k 2 E.
~ The time derivatives give: µ0 0 ∂ 2 E
Solution: Use ∇f
∂t2 = −µ0 0 ω E. Using
part b, the wave equation is satisfied if the parameters, k and ω satisfy c = ω/k.
~ and show that E
~ and B
~ are
d. Using Maxwell’s equations, from the electric field of part c, find the magnetic field, B
perpendicular.
~ ×E
~ = −(~k × E
~ 0 ) sin((~k · ~r − ωt) = − ∂ B~ , thus
Solution: Using Maxwell’s third equation, we have: ∇
∂t
~ =
B
Z
~ ~
~ 0 ) sin((~k · ~r − ωt) = (k × E0 ) cos(~k · ~r − ωt).
dt (~k × E
ω
~ is perpendicular to E.
~
Due to the property of the cross product, B
2
(0)
k
x
ER
R
ET
kT
n1
B (in)
R
EI
B (out)
T
n2
kI
BI (out)
z
y (out)
3. (Fresnel Formula basics) Consider the interface between two linear materials with electromagnetic parameters {1 , µ1 },
and {2 , µ2 } which yield indices of refraction, n1 and n2 , respectively. A plane electromagnetic wave of wavenumber, ~kI , is
incident from the left (region 1). (Measure all angles with respect to the z-axis.)
~ B,
~ ~k} vector triads for
a. For the case where the magnetic field is parallel to the interface (±ŷ), draw on the figure the {E,
the incident, reflected and transmitted waves. Solution shown on the figure above.
b. Write down the general solutions for the electric and magnetic fields, rewriting the magnetic fields in terms of the
appropriate electric field amplitude. Solution: The general solutions in each region are:
~I
E
~ 0I eı(~kI ·~r−ωt) ,
= E
~R
E
~ 0R eı(~kR ·~r−ωt) ,
= E
~T
E
~ 0T eı(~kT ·~r−ωt) ,
= E
~ I = n1 (k̂I × E
~ 0I )eı(~kI ·~r−ωt)
B
c
~ R = n1 (k̂R × E
~ 0R )eı(~kR ·~r−ωt)
B
c
~ T = n2 (k̂T × E
~ 0T )eı(~kT ·~r−ωt)
B
c
Referring to the figure, ~kI = kI (sin θI x̂ + cos θI ẑ), ~kR = kR (sin θI x̂ − cos θI ẑ), and ~kT = kT (sin θT x̂ + cos θT ẑ), while
~ 0I = E0I (cos θI x̂ − sin θI ẑ), E
~ 0R = E0R (cos θI x̂ + sin θI ẑ),E
~ 0T = E0T (cos θT x̂ − sin θT ẑ), where θI = θR was used. The
E
fields in region 1 are the sum of the incident and reflected fields.
c. Referring to your drawing, write down the boundary conditions in sufficient detail such that you could solve for the
transmission and reflection coefficients.
Solution: We need two independent conditions to solve for the ratios, ER = E0R /E0I and ET = E0T /E0I . Using Maxwell’s
first equation for linear material, we have 1 E1⊥ = 2 E2⊥ which here gives: 1 (−E0I + E0R ) sin θI = −2 E0T sin θT . We can
rewrite this using the law of refraction to find: 1 − ER = nn21 12 ET .
For the second equation, we need to use Maxwell’s third equation (Maxwell’s second equation is trivially satisfied since
the magnetic field has no perpendicular component in this problem, and Maxwell’s fourth equation gives the same condition
k
k
as Maxwell’s first equation.) Applying Maxwell’s third equation to the boundary gives E1 = E2 , which here gives: (E0I +
cos θ2
E0R ) cos θI = E0T cos θT . Dividing by the incident amplitude gives: 1 + ER = cos θI ET .
These two can be solved to give (solution not asked for):
ET
=
ER
=
2n2 cos θ1 1
n2 cos θ2 1 + n1 cos θ1 2
n2 cos θ2 1 − n1 cos θ1 2
.
n2 cos θ2 1 + n1 cos θ1 2
3
z
−
−
−
x
+
−
+
+
−
y
+
β
−
+
−
+
+
4. (35) A loop of radius, R, is centered at the origin and lies in the xy-plane. The loop consists of two oppositely-charged half
loops of uniform linear charge density, ±λ0 . Let β(t) be the angle with respect to the x-axis to one of the joints where the
half-loops are joined (as shown). Initially, β(t = 0) = 0; so at this time the loop is oriented with the positive side centered
on the positive y-axis, that is,
(
+λ0 , 0 < θ < π
λ(θ, t = 0) =
−λ0 , π < θ < 2π.
where θ identifies an element of the loop. The loop rotates in the +ẑ direction with an angular velocity which increases
~ is:
linearly in time according to, ω(t) = dβ/dt = αt. The general formula for the retarded vector potential, A,
~ r, t) = µ0
A(~
4π
Z ~ 0
J(~r , tR )d3 r0
,
|~
%R |
where %
~R = ~r − ~r 0 (tR ) with the retarded time given by tR = t − |~
%R |/c.
a. Through what angle will the loop have rotated in time, t?
Solution: Integrating the angular acceleration twice gives, β(t) = 21 αt2 (just like constant linear acceleration).
b. What is the retarded time for a point
√ on the z-axis?
Solution: tR = t − |~r − ~r0 |/c = t − R2 + z 2 /c, independent of time.
~ r 0 , tR ) → Rω(tR )λ(θ, tR )dθθ̂, where λ(θ, tR ) is the linear charge density at t = tR similar
c. For this case, we can write d3 r0 J(~
to the t = 0 case shown above. Find λ(θ, tR ).
Solution: Using the result from part b, the linear charge density at arbitrary time, t, is
(
2
2
+λ0 αt2 ≤ θ < π + αt2
λ(θ, t) =
2
2
−λ0 π + αt2 < θ ≤ 2π + αt2
~ t), for a point on the z-axis.
d. Find the vector potential, A(z,
Solution: Using the result from part c and setting ω(tR ) = αtR , the vector potential is
~ t)
A(z,
=
=
µ0
4π
Z
2π
R(αtR )λ(θ, tR )
√
(cos θŷ − sin θx̂)
R2 + z 2
0
αt2
Z
Z 2π+ αt22R R
µ0 RαtR λ0 π+ 2
√
dθ −
dθ (cos θŷ − sin θx̂)
αt2
αt2
R
4π R2 + z 2
π+ 2R
2
dθ
µ0 RαtR λ0
αt2
αt2
= − √
(cos( R )x̂ + sin( R )ŷ)
2
2
2
2
π R +z
where tR = t −
√
R2 + z 2 /c and φ̂ = − sin θx̂ + cos θŷ was used.
4