PHYSICS 218
SOLUTION TO HW 3
Created: September 12, 2004 10:42pm
Last updated: October 22, 2004 3:06pm
1. (a) Instead of looking up the units in a table directly, we will argue from what we know. As
0.5CV 2 is the energy stored in a capacitance we can deduce [C] = F = J/V 2 and similarly as
0.5LI 2 is the energy of the magnetic field due to a steady current we have [L] = H = J/A2 .
Putting this together we find
r
r
1
V 2 A2
m
= m
=
(1)
2
L1 C 1
J
s
Where we have used P = U I → W = V A.
Similarly we can find with R = U/I → Ω = V /A
"r #
r
L1
V2
=
= Ω
C1
A2
(2)
Now for 0 we know uE = 0.50 E 2 and thus [0 ] = J/(mV 2 ) using F = qE → [E] = N/C =
V /m, where the latter equality stems from U I = P → N m/s = V As = V C/s. Similarly
we have uM = 1/(2µ0 )B 2 → [µ0 ] = V 2 s2 /(Jm) using from the Lorentz force law F =
~ + ~v × B)
~ → [B] = s/m[E] = V s/m2 . Together
q(E
r
1
mV 2 Jm
m
=
(3)
=
√
2
2
0 µ0
J V s
s
In an analogous manner
r
µ0
0
=
r
mV 2 V 2 s2
=
J Jm
r
V 4 s2
= Ω.
V 2 A2 s2
(4)
You can also derive this latter units from [0 ] = [C]/m = F/m and [µ0 ] = [L]/m = H/m.
2. (a) Differentiating the first equation with respect to x yields
∂ 2 v(x, t) 2
∂ ∂i(x, t)
∂i(x, t)
∂x = −L1
− R1
/
∂t ∂x
∂x
(5)
It is important to note that we were allowed to permute the order of the two derivations in
the second term. This is possible because we assume that all first and second derivatives are
smooth functions of x, t as well as i(x, t) is1 Now inserting the second equation for ∂i/∂x yields
the telegrapher’s equation.
∂ 2 v(x, t)
∂v(x, t)
∂ 2 v(x, t)
=
L
C
+ (R1 C1 + G1 L1 )
+ R1 G1 v(x, t)
1
1
2
2
∂x
∂t
∂t
(6)
(b) To find a wave equation for i(x, t) we use the same method, but differentiate the second
equation first, replacing the ∂v(x, t)/∂x according to the first one. We find
∂ 2 i(x, t)
∂i(x, t)
∂ 2 i(x, t)
=
L
C
+
(R
C
+
G
L
)
+ R1 G1 i(x, t)
1
1
1
1
1
1
∂x2
∂t2
∂t
The equation for i(x, t) has exactly the same structure as the one for v(x, t).
1
(7)
This is known as a theorem of Schwartz and is generally valid. If all up to the n-th derivatives are smooth,
one may permute the order as one wishes. Usually in Physics we assume that our solutions satisfy this smoothness
condition. However this is not trivial. You can win one million dollars if you are able to proove that all solutions of
the Navier-Stokes-equation in Hydrodynamics have this property.
1
PHYSICS 218
SOLUTION TO HW 3
(c) Inserting the assumed solution v(x, t) = V0 ejωt+γx we find after dividing by v(x, t)
γ 2 = −ω 2 L1 C1 + (R1 C1 + G1 L1 )jω + R1 G1 .
(8)
(d) We can write this equation as
γ 2 = ω 2 L1 C1 (j 2 + j(
G1
R1 G1
R1
+
)− 2
)
ωL1 ωC1
ω L1 C 1
(9)
and completing the square we have
γ 2 = ω 2 L1 C1 × [(j +
R1 1
G1 1 2 1 R1 2
G1 2
R1 G1
+
) − ((
) +(
) )+ 2
]
2 ωL1
2 ωC1
4 ωL1
ωC1
2ω L1 C1
(10)
We can now argue that from our assumption R1 ωL1 and G1 ωC1 we can drop the last
two terms. Thus we finally end up with
r
r
R 1 C 1 G 1 L1
+
(11)
α =
2
L1
2
C1
(e) In terms of Z0 this is
α =
R1 1
G1
+
Z0
2 Z0
2
3. (a) For the numerical values given we find
C1
cT
L1
√
−1
−6
−12
2.56 × 10 H/m 4.34 × 10
F/m
0 µ0 = c = 2.998 × 108 m/s
(12)
Z0
p
L1 /C1 = 768.02Ω
L
(b) Calculating R1 = ρ A
= 3.99 × 10−5 Ω/m we have α = (2R1 )/2 × 1/Z0 + G1 /2 × Z0 =
1.04 × 10−7 1/m. For the distance from Niagara Falls to Ithaca the amplitude shrinks by
exp(−1.04 × 10−7 × 2.5 × 105 ) = 0.9743 and thus the power dissipated is about 2.57%. You
have probably noticed the extra factor of 2 in our calculation for α. This is due to the fact
that we actually have the same resistance in both wires of the transmission line. You can
understand this from the DC limit, the current does need to go the distance back and forth,
resulting in a factor of 2.
4. (a) Knowing the transmission speed to be
r
1
c
=√
vT =
= 0.83c
κM µ 0 κE 0
κE
(13)
we have κE = 1.45.
(b)
C1 =
2πκE 0
= 5.36 × 10−11 F/ m = 16.33pF/ft
ln b/a
(14)
This does match the specification within 1% and is thus a reasonable result.
(c)
L1 =
µ0
ln b/a = 3.01 × 10−7 H/m = 0.092µH/ft
2π
2
(15)
PHYSICS 218
SOLUTION TO HW 3
(d) Calculating the impedance of the cable from these values leads to
s
r
L1
ln b/a2
=
× Z0 = 0.199 × 376.99Ω = 74.94Ω
Z =
C1
κE
(16)
This result agrees perfectly with the nominal impedance of 75Ω.
p
5. (a) The skin depth is given by δ = 2ρ/(ωκM µ0 ) which in our case for ν = 60Hz gives δ = 0.011m.
(b) The skin depth tells us how deep the electromagnetic wave penetrates into the conductor,
thus in which layers of the conductor the electric current actually flows. The current is
proportional to the amplitude of the electric field. As the amplitude decreases as e−δx the
current is confined to be near the surface. More than 99% of the current flow within 5δ. Thus
if the skin depth is small compared to the radius of the conductor the effective resistivity is
not given by the usual formula R = ρL/A, as the waves are not able to penetrate the whole
cross-section of the conductor. However in our case it would however not be reasonable to
make the conductor much bigger as it would not only increase the surface area, but as well
the weight of the conductor. This poses an engineering problem, as it makes the transmission
lines more difficult to build.
(c) In this case we find δ = 6 × 10−6 m = 6µm.
(d) As we expect from the formula, the skin depth is much smaller for higher frequencies. Furthermore we understand that the inner steel conductor is only coated with copper and not
entirely consisting of copper: The electric and magnetic fields do not penetrate beyond the
coating. In fact they do neither penetrate the whole coating nor the whole aluminium foil.
Thus skin depth is going to be an important effect.
~=
6. (a) Knowing that the energy flux of a electromagnetic wave is given by its Poynting vector S
~ ×B
~ = 1/(µ0 c)|E|
~ 2 × k̂ we find, introducing a factor of 1/2 from the time average of
1/µ0 × E
linearly polarzed waves,
1
A2 s3
E02 = 2.66 × 10−3 E02 2 = 1.35kW/m2
2µ0 c
m kg
and from this
r
m2 kgW
V
= 1019.9
2
2
3
m As
m
where we have used W = V A. Consequently we have
E0 = 1019.9 ×
B0 = 3.40 × 10−6 T .
(17)
(18)
(19)
(b) We find the total power to be 3.82 × 1026 W .
(c) This is equal to converting 1.33 × 1017 kg per year into energy,
(d) which is about 6.65 × 10−14 the total mass of the sun.
(e) Using hpi = hsi/c we find that the radiation pressure is hpi = 4.5 × 10−6 N/m2 .
(f) Comparing this to the minimum sound level we are able to hear, we may be surprised to find
that this pressure is less than one order of magnitude smaller.
3
PHYSICS 218
SOLUTION TO HW 3
(g) Knowing that the pressure will fall off as 1/r 2 we can use the numerical value we know for
the earth to write
FR = πa2 × 4.5 × 10−6
a2
N REarth 2
17
(
)
=
3.18
×
10
N
m2
r
r2
(20)
(h) We find
4π
1
N m2 a3
FG = GMSun a3 ρ 2 = 5.59 × 1020
× 2ρ
3
r
kg
r
(21)
(i) Setting FG = FR we find
3.18 × 1017 N
3
a2
24 N a
=
3.19
×
10
r2
m r2
a = 9.99 × 10−8 m
(22)
(23)
Knowing that a comet’s tail consists mostly of gas and dust we see that the radiation pressure
can become larger than the gravitational attraction to the sun. Thus we find an explanation
for the comets tail being diverted from its actual path of the comet itself by the radiation
pressure of the sun2
2
This argumentation is disputed by some. It seems that the tail in fact might consist of two parts. One the dust
particles behaving as our argument suggests under the influence of both the radiation pressure and the gravitational
field. The other consisting of gas, which is even more diverted from the orbit due to the protons coming from the sun
(the so called solar wind) exerting a pressure on it.
4