Download Phys 6321 Midterm

Phys 6321 Midterm - Solution
March 18, 2013
You must do all problems in the time allocated. You may NOT use any notes or
books other than those provided by the instructor do you own work. You may
NOT receive aid from anyone other that the instructor. Show your work clearly and
concisely so that proper credit can be assigned.
1)
Consider the charging capicator circuit shown in Figure 1 below, The voltage, V0 , is
applied at t = 0. The capacitor is a cylindrical parallel plate capacitor of radius, a,
and separation, d. Use the standard parallel plate approximation for the E field in
the limit of small plate separation and large, a.
a) Find the Electric field as a function of time. The voltage on the capacitor is zero when t = 0
b) Using Maxwell’s equations find the Magnetic field around the circumference of the capacitor. (Note there is no conduction current through the capacitor)
c) Find the power flow into the capacitor from the fields
d) Find the total energy which goes into the capacitor as t → ∞
e) Verify that this equals the energy stored in the capacitor at t = ∞ from
the field energy density.
R
Vo
d
a
C
Figure 1: The charging circuit for the parallel plate capacitor
Solution
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From the circuit equation;
Q
Q
dQ
+ C
V0 = IR + C = R
dt
Q = A + Q0 e−t/RC
V0 = −
RQ0 e−t/RC
+ A/C + Q0 e−t/RC /C = A/C
RC
Now Q = 0 when t = 0
Q = V0 C[1 − e−t/rc ]
I =
dQ
0 C e−t/RC = (V /R) e−t/RC
= VRC
0
dt
Apply Ampere’s law. Note there is no conduction current.
~ × B̃ = µ0 J̃ + µ0 ǫ0 ∂ Ẽ = µ0 ǫ0 ∂ Ẽ
∇
∂t
∂t
E = V /d = (Q)/dC = (V0 /d)[1 − et/rc ]
~ × B̃| = µ0 ǫo (V0 /dRC)e−t/rc
|∇
Integrate over a cross section of the capacitor using Stokes theorem to get the value
of the integral.
R
~ · dA
~ =
curlB
R
~ · d~l = µ0 ǫ0 V0 e−t/RC (πa2 )
B
dRC
B(2πa) = µ0 ǫ0 V0 e−t/RC (πa2 )
dRC
µǫVa
B = 0 0 0 e−t/rc
2dRC
~ = (1/µ0 )E
~ × B,
~ points into the volume of the capacitor and
The Poynting vector, S
is in the direction of power flow into the fields within the volume between capacitor
plates. Integrate over the cylindrical shell enclosing the volume, (2πa)d.
2 2
dW = ǫ0 a V0 [e−t/RC − e−2t/RC ]
dt
dRC
The total energy inserted into the volume is obtained by integrating the above equation from 0 to ∞. This results in;
W = (πǫ0 /2) V0a2 /d
The energy density in the field when t = ∞ is;
2
W = ǫ0 /2 V02 /d2
Integrate this over the volume between the capacitor plates.
W = (πǫ0 /2d) V0a2
Which is the result obtained from Maxwell’s equations above.
2)
Using conservation of relativistic energy and momentum, find the energy of a scattered photon of initial energy, E0 , if it is scattered from an electron of mass, m,
initially at rest, through an angle, θ. (Compton Scattering)
Solution
Apply both conservation of Energy and vector Momentum. The relativistic values of
energy and momentum give;
Eo + me = Eγ + Ee
p~0 = p~γ + ~pe
In the above E0 is the incident γ energy with momentum, p~0 . The electron mass
is me with outgoing energy Ee and momentum, p~e . Note that initially the electron
is at rest with energy equal to its rest mass, me c2 . The final electron energy is
Ee = pc
~ 2 + (me c2 )2 . The gamma has zero rest mass. Now solve these equations for
Ee = pγ c.
Ee2 = (E0 + me )2 + Eγ2 − 2Eγ (E0 + me ) = (pe c)2 + m2e c4
(pe c)2 = (p0 c)2 + (pγ c)2 − p0 pγ c2 cos(θ)
In the above, θ is the recoil angle of the electron with respect to the incident photon
direction. Use Ee2 = (pc)2e + (mc)2e . In the remainder of this development use units
where c = 1. Then substitute and collect terms.
pγ = Eγ =
p0 me
p0 (1 − cos(θ)) + me
3)
Find the TE wave modes in a rectangular wave guide of dimensions 0 ≤ x ≤ b and
0 ≤ y ≤ a, and calculate the group velocity for the lowest mode. Proceed as below.
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a) Assume the wave has the form X (x) Y(y) ei[kz z−ωt] . Find the TE fields
which satisfy the boundary conditions.
b) The solution for the field with appropriate boundary condition gives
the dispersion relation.
c) Find the lowest mode for the dispersion relation and then calculate the
group velocity of the wave.
Solution
For the TE mode the only non-zero ẑ component is Bz . It has the form;
Bz = X (x) Y(y) eikz z−ωt
The above form assumes a solution to the wave equation using separation of variables.
In Cartesian coordinates the wave equation takes the form;
2
∂ 2 Y − [k 2 − (ω/c)2 ]X Y]ei[kz z−ωt] = 0
[Y ∂ X
2 + X
z
∂t
∂t2
Let γ 2 = (ω/c)2 − k 2 . The transverse fields are then;
~ ⊥ Bz
~ ⊥ = ik ∇
B
γ2
~ ⊥ Bz
~ ⊥ = iω ẑ × 2∇
E
γ
Then the boundary condition requires that E parallel to the walls of the guide must
vanish. Therefore; Ey = 0 for x = 0, b and Ex = 0 for y = 0, a. This means;
X ∝ cos(nπx/b)
Y ∝ cos(mπy/a)
Finally, substitution in the wave equation above yields the dispersion relation;
[(nπ/b)2 + (mπ/a)2 = γ 2
The group velocity is Vg = dω . The lowest mode occurs for a > b and n = 0 with
dk
m = 1. Take the derivative of the dispersion relation to obtain;
2
Vg = dω = kc 2
dk
ωγ
4)
A long, uncharged, straight wire carries a current, I, in its own rest frame.
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a) Show that in another frame moving with velocity, V0 , with respect to
the rest frame, the wire has a charge per unit length, λ.
b) Find the value of λ.
(Begin by finding the magnetic field in the rest frame and then use the field transformations)
Solution
The magnetic field around the wire in the rest frame is obtained by Ampere’s law.
~ = µ0 I φ̂
B
2πr
Choose a Cartesian geometry and select a point so that φ̂ = ŷ. The current and the
boost are in the x̂ direction. Then use the Cartesian field transformation equations.
There is no E field in the rest frame, but in the moving frame there is an E field given
by;
Ex = 0
Ey = 0
µ0 I
Ez = γV By = γV 2πr
ẑ
~ points radially away from the wire at all points when the
It is then obvious that E
the initial point is generalized about the wire, Figure 2. If there were a charge per
unit length, λ, on a wire, there would be an electric field generated perpendicular to
the wire and having the value of;
Q/L
λ
E = 2πǫ r = 2πǫ
0
0r
Equating this to the field obtained above by the field transformation, we obtain;
λ = µ0 ǫ0 V γI
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z
Wire
B
I
y
x
Figure 2: The geometry for the transformation in cartesian coordinates
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