Download Homework Assignment 10 Solution Set

Homework Assignment 10 Solution Set
PHYCS 4420
6 April, 2004
Problem 1 (Griffiths 7.51 - optional)
The only forces that act on the loop are the magnetic forces on the currents
that lie in the external magnetic field. Thus, we can write
d2 x
= IhB
dt2
m d2 x
=⇒ I =
.
Bh dt2
m
Now, what is the source of the current? We can write the total emf in the loop
as the sum of the emf induced by the external field and the emf that is self
induced. We get
dI
dΦB
−L
dt
dt
dx
dI
= Bh
−L
dt
dt m d2 x
dx
d
= Bh
−L
dt
dt Bh dt2
Lm d2 v
= Bhv −
.
Bh dt2
ε=−
In a perfect conductor, however, this total emf should always be zero (the
current flows with negligible field, and no energy is dissipated). Thus,
mL
d2 v
= 2 2v
2
dt
B h
Bh
=⇒ v(t) = v0 cos( √
t)
mL
√
mL
Bh
=⇒ x(t) = v0
sin( √
t).
Bh
mL
So, given some initial velocity v0 with the rectangular loop partially in a uniform
magnetic field B, the loop oscillates harmonically with angular frequency ω0 =
√Bh .
mL
1
Problem 2 (Griffiths 8.1)
In the cylindrical configuration in Example 7.13 we have the fields between
the cylinders as
λ
V
r̂
r̂ =
2π0 r
r ln ab
µ0 I ˆ
~
theta
B(r)
=
2πr
~
E(r)
=
and the fields outside of this region are zero. Thus, the power carried down the
cylinder is
Z
~ · dA
~
P =
S
A
1
=
µ0
Z
1
µ0
Z
=
b
~ × B)
~ · (2πrdrẑ)
(E
a
b
(
a
V µ0 I
2πrdr(ẑ · ẑ)
r ln ab 2πr
= IV.
Fortunately, Ohm’s law is conserved. We get exactly the same result for the
ribbon configuration in Problem 7.58 since
~ = V x̂
E
d
~ = µ0 K ŷ = µ0 I ŷ.
B
w
~ and B
~ are both zero outside of the space between the ribbons, so
Again, E
Z
~ · dA
~
P =
S
A
1 V
I
=
µ0 wd((x̂ × ŷ) · ẑ)
µ0 d w
= IV.
Problem 3 (Griffiths 8.2)
a We’ve already done this part (see HW 9 Solutions, Problem 6).
~ = σ(t) ẑ = It ẑ
E
0
0 πa2
µ0 Ir
~
B(r)
=
θ̂
2πa2
2
b
1
1 2
2
u(r, t) =
0 E +
B
2
µ0
2 2
µ0 I 2 r2
I t
1
+
=
2 0 π 2 a4
4π 2 a4
1 ~
~
E×B
mu0
1 It µ0 Ir ẑ
×
θ̂
=
µ0 0 2πa2
I 2 tr
=
(−r̂)
2π 2 0 a4
~ =
S
Energy is flowing into the gap; this makes sense because we have to do work to
assemble the charges at the gap (the work comes from whatever is driving the
current). Now, to check Equation 8.14...
∂u(r, t)
~ ·S
~
= −∇
∂t 2 2
1 ∂
I t
µ0 I 2 r2
I 2 tr
1 ∂
+
−r 2 4
=−
2 ∂t 0 π 2 a4
4π 2 a4
r ∂r
2π 0 a
2
2
1
I tr
I t
=
2
4
0 π a
r 0 π 2 a4
I 2t
=
0 π 2 a4
c
The total energy in the gap is
Z
U (t) =
u(r, t)dτ
V
Z a 2 2
1
I t
µ0 I 2 r2
=
+
w2πrdr
0 π 2 a4
4π 2 a4
0 2
µ0 I 2 w a4
I 2 t2 w a2
+
=
4
0 πa 2
4πa4 4
2
2
I wt
µ0 I 2 w
=
+
.
2π0 a2
16π
Note the two distinct terms. One comes from the electric field and depends
on time, while the other comes from the magnetic field and does not change
with time (so long as the current is steady). Again, this matches our physical
intuition that the energy that is flowing into the gap is due to the work it takes
to push charges onto the plates.
3
According to the Poynting vector, the power flowing into the gap is
Z
1 ~
~
P =
S · dA
µ
S 0
I 2 ta
2πaw(−r̂) · (−r̂)
=
2π 2 0 a4
I 2 wt
=
.
0 πa2
This should match the time derivative of the total energy in the gap
∂
U (t)
∂t I 2 wt2
µ0 I 2 w
∂
+
=
∂t 2π0 a2
16π
2
I wt
.
=
0 πa2
P =
Problem 4 (Griffiths 8.6 (a) & (c))
a The momentum density is
ρ
~em =
=
=
=
~
µ0 0 S
~ ×B
~
0 E
0 EB(ẑ × x̂)
0 EB ŷ.
This density is constant everywhere between the plates, so the total momentum
between the plates is
p~em = 0 EBAdŷ.
c Since we don’t know the geometry of the field at its edges we can’t know
the value of the electric field. However, we can say something about the total
~ and
line integral of the field along a rectangular loop that is perpendicular to B
coincides with the plates. Since the separation of the plates is small compared
to their width we can neglect the contribution to the closed loop from the space
between the plates. Thus,
I
~ · d~l = − ∂ΦB
E
∂t
Z
Z
~ top plate · d~l + E
~ bottom plate · d~l = −wd ∂B
− E
∂t
where w is the length of one side of the plates.
Multiplying both sides by the total charge ±Q and dividing by the length
w (taking each individual line integral along the plate of length w) we get the
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total force on the plates. From the right-hand rule, and the fact that the top
plate is negatively charged and the bottom plate is positively charged we see
that the force on both plates is in the y direction. So the total force on the
configuration must be
∂B
ŷ.
F~ = −Qd
∂t
To find the change in momentum of the plates we just integrate over all time
Z
Z
∂B
~
ŷ
F dt =
−Qd
∂t
T
T
∆~
pplates = QdB ŷ
= 0 EBAdŷ.
Total momentum, mechanical plus electromagnetic, is conserved.
Problem 5 (Griffiths 9.3)
Equating the real and imaginary parts of the summed complex amplitudes
gives the following two equations:
A3 sin δ3 = A1 sin δ1 + A2 sin δ2
A3 cos δ3 = A1 cos δ1 + A2 cos δ2 .
Taking the ratio of both sides gives us δ3 as
A1 sin δ1 + A2 sin δ2
δ3 = arctan
.
A1 cos δ1 + A2 cos δ2
To solve for A3 we square both equations and add them. This gives
q
A3 = A21 + A22 + 2A1 A2 (cos δ1 cos δ2 + sin δ1 sin δ2 )
q
= A21 + A22 + 2A1 A2 cos(δ1 − δ2 ).
Problem 6 (Griffiths 9.9 & 9.10)
9.9 Following Equations 9.51 and 9.52 we can just write down the fields.
The only thing we have to do is decide on ~k and n̂ in whatever coordinate system
we use. Let’s stick to cartesian coordinates.
a We start with
~k = −kx̂
n̂ = ẑ
5
where the magnitude of k in terms of ω is k = ωc , with c representing the speed
of light. Therefore,
ω
~
E(x,
y, z, t) = E0 cos(− x − ωt)ẑ
c
E
ω
0
~
B(x,
y, z, t) =
cos(− x − ωt)ŷ.
c
c
b In this case we can directly write
~k = √k {1, 1, 1}
3
but to find n̂ we need to be clever. We are looking for a unit vector that is
perpendicular to both ~k and ŷ. These conditions give us two equations
{nx , ny , nz } · {1, 1, 1}
=⇒ nx + ny + nz
{nx , ny , nz } · {0, 1, 0}
=⇒ ny
=
=
=
=
0
0
0
0
which tells us that ny = 0 and nx = −nz . Finally, the fact that n̂ should be
normalized gives just two possibilities
1
n̂ = √ {±1, 0, ∓1}.
2
Let’s choose the first (positive in the x direction), which gives the direction of
~ as
the polarization of B
1
k̂ × n̂ = √ {−1, 2, −1}.
6
Therefore,
ω
ω
ω
1
~
E(x,
y, z, t) = E0 cos( √ x + √ y + √ z − ωt) √ (x̂ − ẑ)
3c
3c
3c
2
E
ω
ω
ω
1
0
~
B(x,
y, z, t) =
cos( √ x + √ y + √ z − ωt) √ (−x̂ + 2ŷ − ẑ).
c
3c
3c
3c
6
9.10
For a perfect absorber we have
P =
I
1300
N
=
≈ 4.3 × 10−6 2
8
c
2.9979 × 10
m
and for a perfect reflector it’s just twice this value
P ≈ 8.6 × 10−6
N
.
m2
N
Atmospheric pressure is about 105 m
2 , so the radiation pressure from the sun is
−11
−11
about 4 × 10
to 9 × 10
atmospheres.
6
Problem 7 (Griffiths 9.11) Doing it the long way the time average is
1
hf gi =
T
Z
T
A cos(~k · ~r − ωt + δa )B cos(~k · ~r − ωt + δb )dt
0
Z 2π
ω
ω
AB cos(θ − ωt + δa ) cos(θ − ωt + δb )dt
2π 0
AB
=
cos(δa − δb ).
2
=
The complex conjugate method gives
1
1
~
~
Re[f˜g̃ ∗ ] = Re[Ãei(k·~r−ωt) B̃ ∗ e−i(k·~r−ωt) ]
2
2
1
= Re[ÃB̃ ∗ ]
2
1
= AB(cos δa cos δb + sin δa sin δb )
2
AB
cos(δa − δb ).
=
2
It looks like the same amount of effort here, but the second method doesn’t
require integrating in Mathematica or a long series of trig identitites!
Problem 8 (Griffiths 9.14 - optional)
We really only need one boundary condition. The parallel component of
the electric field is always continuous across a boundary. Thus, the (electric)
polarization of the transmitted and reflected waves must be the same as the
polarization of the incident wave. Finally, noting the relationship between ~k
~
and n̂ gives us the polarization of B.
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