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Unformatted text preview: Department of Physics, University of California, Berkeley
Final Examination Physics 713, Lecture 1, Prof. Smoot
8 AM  11AM, 12 May 2006 Discussion Section: ______________________________ Name of TA: _____________________________________ Problem 1 Problem 5 There [‘3
no 5, Score: ___________________________________________ 2. [25 points] Resistor and Voltage Source Network 1» we on“ («Lifer"3)
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[1 through points P, (2) 12
resistor R6. propriate equations to determine the four currents: (1)
through resistor R3, (3) 13 though point Q and (4) ‘14 through Junction Rules: ‘1‘" It: Izl+ls "B? I; 2 «[3413!
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L25 C13*L)Ru' 13R; * 83 + I‘R‘V £120 L/"" \(3 5011.] ugulhif (vi/(n (b) [5 points] Assume that the voltages E1 = 12 V , sources E2 and E3 are replaced with
capacitors, and that all the resistors have the same value R1 = R2 = R3 = R4 = R5 = R6 = 100 Ohms. Solve for the current [1 and the voltage on each of the two capacitors E2 and E3. If 82‘; 83 0N quadh’f § Hm Cn‘rcuul
bahlee is 1‘4 a stead; 511%, km (c) [5 points] Voltage sources E1 and E2 are removed and replaced by conductors and all
the resistors have the same value R1 = R2 = R3 = R4 = R5 = R6 = 100 Ohms. Draw the . c v simplest equivalent circuit. Determine and label the equivalent resistance(s) and current I 1. ‘/ Br. “, FUZFIFL / "L 4 \l‘ R 5mm? K, 1 Rm‘Rz
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V a" : V0i+dly w n A {I 7; £35 901“} A! fbenﬁfrca E RI;2 RI! T0,] S=§=o (d) [5 points] Instead remove the right half of the circuit so that only the left hand loop
circuit remains. The two remaining voltage sources are still E1 = 12 V and E2 = 6 V and
that all the remaining resistors have the same value R1 = R2 = R3 = 100 Ohms. Again draw the simplest possible equivalent circuit with labels for the components and their values
including the current 11. (e) [5 points] Keeping again only the left half of the circuit, at time t = O replace resistor
R3 instanteously with an inductor With L = IOpH. What is the current 11 as a function of time? 8:2'421
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operating frequency? 9 r V: 3 (10% W5 : 7 H946! ”’3; 7'2.“t€§@s‘r\ra x (D. \ l’LLi vw (b) [5 points] If the power of the microwave is about 1000 watts and the microwaves are
generated in a device called a klystron and transmitted to the microwave cooking area
through a wave guide which is one half wavelength wide and one quarter wavelength high. What is the average Poynting vector < [S[ > or Intensity I, and rms Electric ERMS and
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what is the energy density, and rms Electric and Magnetic ﬁeld values in the microwave
cooking area when it is operating? Assume that the Quality factor of the circuit  microwave
cavity (an LC) and food (R) is such that there is about 10 times the energy stored than is
being drawn out per cycle. 19"
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Chinese food boxes with wire handles? How much power could be coupled into the circuit
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water can be superheated so that it has sufﬁcient internal energy to boil but not yet started
boiling because nothing has yet triggered the formation of steam bubbles. When picked
up or disturbed, it will then errupt in a geyser of hot boiling water. For how long would
one have to heat a container containing 100 grams of water starting at 20 C so that it was
overheated and ready to errupt into boiling? Assume that the water’s internal energy is
distributed uniformly. Rep}? mmfmqrt? wbwmﬁr \ 04%,? Egg,“
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conductor with inside radius r2 = 2 mm and they are concentric. They are separated by an
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‘——————— L————> Figure 5: sketch of coaxial cable with dimensions labeled for problem 5. Treat as two
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current I? =1. 2 ; £1221
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cable, if it carries an. EMS current I ? ‘HAI \‘MF‘AR"lu Op "Hi4L I‘m_ (:5 21;:RJL harm J m “re—3m :5 .2641. m unmi Mode/f Jim" 'H'U— CAINe i5 W
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30 1m we 9‘1”“ W3“) (‘5 m (1,“) + <qc§ : Z<ML\ = LIT»... 24 8. [35 points] Tethered Satellite for electrodynamjc power generated from orbit consists of
a major satellite and a secondary satellite lowered on a conducting cable. This concept was
tested on a Shuttle ﬂight in a near equitorial orbit around the Earth. The smaller satellite was lowered 20 km below the Shuttle. (a) [5 points] If the orbital speed is 7’ km/s and the mean Earth’s magnetic ﬁeld is 0.5 x 10‘4
T, what is the voltage difference generated between the Shuttle and secondary satellite? ® Th aemml learn/Lula i944 gwﬁe é:
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z: S¥OOC3V (b) [5 points] What charge must be on the secondary (lower) satellite for this voltage Ap
proximate the satellite as a conducting sphere of radius r— — 2 m with the Shuttle very (nearly
inﬁnitely) far away? First calculate its capacitance. Then calculate the charge. a: (2,111 ® V: {/T—é‘ é? bdmu Q Wed 6O t: ”trial/6L: VWWW/U’qumx 7‘“ 55: 15¢ ></o”‘c 25 (c) [5 points] Approximate both the Shuttle and secondary satellite as tWO conducting spheres
of radii 3 m and 2 In respectively connected by an electrical cable 20 km long and supporting
the voltage difference you found in (a). Sketch the ﬁeld lines for this conﬁguration and label.
Explain what is different than one would have naively thought for the electrostatic case. 1, at declmicl’hc case, 1%
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/ \ use. (d) [5 points] Power was taken from the system by letting a current flow and power experi
ments on the Shuttle and tethered spacecraft. If the current is I and the voltage is V what is
the electrical power provided? How much force IS exerted on the cable? Evaluate for a cur—
rent of 1 ampere Does it speed up or slow down the Shuttle—cable—satellite system? (Hint:
Does it change the altitude? Which way is the force?) What makes the circuit or closed
path? The basically stationary Earth’s upper atmosphere, the iOIlOSpl’leI‘e; is conductive and
completes the circuit. F~=IV = 320603. 1:; IQ l3 7» j'ZOK/03/t .SK/O’q: 1 M,
ad, at is ferruﬁilztwlcr '59 Ha mm. 8% (guild [QUI‘HQ "force “ Q C(MS {er(12., GUI—I It \S‘AaU/c/ dag/gait ~H’L Q/Ttl‘tudq. 7Q 'romlahre acts as Q (ﬁSFQJ‘SJCSA Kidd 1723’“ 71c cQo/mM. 26 Qxd “Tle Vlﬂikl "Karen 6%»! So we Ural ““6 (e) [5 points] Could you run the experiment backwards? That is could the Space Station
use solar cells or fuel cells to produce electrical power and run current down a tether cable
to a secondary satellite and produce a force to match or overcome the residual atmospheric
drag at its altitude and keep its orbit stable? How much electrical power would be required
to keep the Space Station of mass 183,283 kg in its orbit, if its altitude is 353 km and loses
about 2 km per year. If the tether cable is 5 km long, what is the required current and
voltage? The OrE‘rls ll: £va— J r; l1» lowg milq MM, “4. “while {om Charm o4 6».
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27 (f) [5 points] At the Space Shuttle altitude of 400 km the typical electron density in the
ionosphere is around 105c'nf3 or 10117713. The typical temperature (thermal energy) is around
1200 K. Near the second satellite (2—m radius moving 7000 m/s) what is the necessary
drift velocity of the ionized electrons away from the secondary satellite to produce a 1
Ampere current? How does that compare to their thermal velocity? How much higher
would the electron density need to be around the secondary satellite to make the drift
velocity reasonable and comparable to their thermal velocity? Chg dare‘lu ix. ‘0le At l!
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VA‘L N :. nAv= {0”M‘3 m a 7X tMoo Ml: : (RM/0’5 ”(L Cymbal a at}: firm/0'5 Mar/0”": .001 A“?
Th). 1 an»? Curried mam/lac c509. to Invert cl) (for? E9;— ﬂd, wed had
dwells; {000 {as Mater. (g) [5 points] The deployment was almost complete when the unexpected happened: the
tether suddenly broke and its end whipped away into space in great wavy wiggles. The
failure was not caused by excessive tension, but rather that an electric current had melted
the tether. Unwinding of the reel uncovered pinholes in the insulation and escaping air
converted to plasma. The instruments aboard the tether satelite showed that this plasma
diverted through the pinhole about 1 ampere at 3500 volts to the metal of the shuttle and
from there to the ionospheric return circuit. That current was enough to melt the cable.
Why is the current higher at the pinhole point where residual air plasma is escaping rather
than at the surface of the Shuttle? Is the electrical power enough to melt the copper braid
which is the conducting (not strength) portion of the cable? The heat capacity of copper is
about 0.09 Cal/ gm/ K and its melting point is 1084.62 C. How large a piece of copper would
be melted, if the heat were concentrated? [Tu Why (“in t busts at It. Orclu— +0 datum was at
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