physics7B-sp04-final-Smoot-soln

physics7B-sp04-final-Smoot-soln - Solutions 1. [25 points]...

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Unformatted text preview: Solutions 1. [25 points] Short Questions (a) [5 points] Circle T or F for True or False T F (i) The principle of equipartition of energy states that in equilibrium the thermal energy is shared among all active degrees of freedom and is randomly distributed with an average energy of kT/2. T F (ii) If the temperature difference across a conductor triples, the rate at which it transfers heat energy increases by a factor of nine. T F (iii) When a system goes from equilibrium state 1 to state 2, the change in the internal energy is the same for all processes. T F (iv) The internal energy of a given amount of an ideal gas at equilibrium depends only on its absolute temperature. T F (v) For any material that expands when heated, CP is greater than CV . (b) [5 points] T F (vi) The linkage of flux from circuit A to circuit B is the same as the linkage of flux from circuit B to circuit A. T F (vii) All materials are diamagnetic. It is sometimes over ridden by ferromagnetism or paramagnetism. T F (viii) Lenz’s law states that two parallel wires carrying current in the same direction will oppose each other and push each other apart. T F (ix) The wave equation can be derived from Maxwell’s equations. T F (x) Electromagnetic waves are transverse waves. 32 (c) [5 points] A circuit contains a capacitor and resistor in series is connected to an AC source. The circuit schematic is shown in the accompanying figure. A I R B C D C V E Cross-cut line Cross-cut view Figure 4: sketch of circuit for problem 1 (c) The right portion of the figure shows a cross-section through the circuit at the mid-point of the parallel plate capacitor. There are 5 imaginary closed paths (A, B, C, D, E) drawn around portions of the capacitor and the return wire (the wire coming back to the AC source). (i) Rank the absolute value of the B · ds in order (maximum = 1 to minimum = 5). (Break ties by which has the highest mean absolute or rms magnetic field.) Path (i) | B · ds| (ii) | E · ds| A B C D E 5 1(2) 3 4 1 (1) 1 (2) 1 (1) 3 4 5 (ii) If the capacitor is replaced by a solenoid - inductor that produces a uniform magnetic field inside the solid circle (between imaginary circles C and B and perpendicular to them), rank the absolute value of E · ds around the imaginary paths in order (maximum = 1 to minimum =5). 33 (d) [5 points] Circle correct answer (i) If the rms voltage in an AC circuit (made of resistors, capacitors, and inductors) is doubled, the peak current is (A) increased by a factor of 2. from complex Ohm’s law I = V /Z (B) decreased by a factor of √ 2. 2. (C) increased by a factor of √ (D) decreased by a factor of 2. (E) not enough information to determine the change. (ii) If the current in an inductor is doubled, its stored energy will (A) increase by a factor of 2. (B) decrease by a factor of 2. (C) increase by a factor of 4. U = 1 LI 2 2 (D) increase by a factor of 8. (E) not changed. (iii) A positively charge particle is moving northward in a static magnetic field The magnetic force on the particle is toward the northeast. What is the direction of the magnetic field? (A) Upward (B) West (C) South (D) Downward (E) This situation cannot exist. FM = qv × B means F ⊥ v (F) Upward at an angle of 45◦ to East . (iv) If the AC frequency driving an inductor is doubled, the inductive reactance of the inductor will (A) increase by a factor of 2. |XL | = ωL (B) not change. (C) decrease by a factor of 2. (D) increase by a factor of 4. (E) decrease by a factor of 4. (v) An ideal transformer has Np turns on the primary and Ns turns on its secondary. The power dissipated in a load resistance R connected across the secondary is Ps , when the primary voltage is Vp . The current in the primary is then (A) Ps /Vp s (B) Np Pp Ns V Ns s (C) Np Pp V (D) Ns Np (E) Ns Np 2 Ps Vp 2 V2 p R also correct 34 (e) [5 points] Circle correct answer (i) Compasses point north because (A) the north star attracts them. (B) the Earth has an electric charge. (C) there are electric currents in the iron core of the Earth. (D) there are magnetic monopoles near the North Pole. (E) there is a very large bar magnet in the Earth. (ii) The Earth’s magnetic field flips are used for (A) creating new permanent magnets. (B) proving the Earth has a solid iron core.. (C) generate useful power. (D) geologic dating. (E) bird migration. (iii) Which is not a property of EM waves? √ (A) E and B waves have the same velocity vE = VB = c = 1/ 0 µ0 . (B) E and B waves are in phase. (C) EM waves are transverse with E and B perpendicular to the direction of wave motion (D) E and B magnitudes are in the ratio E/B = c. (E) EM waves can self-propagate. They require no medium for propagation. (F) EM waves carry both energy and momentum. (G) The electric field of a wave decreases as one over the square of the distance from the source. (iv) Magnetism comes from (A) magnetic monopoles. (B) moving quanta of light. (C) quantization of charge. (D) moving electric charge. (E) magnesia and lodestones transferred to soft iron. (v) AC is used instead of DC power because (A) it is safer. (B) it carries more power. (C) it makes the use of transformers straight forward. (D) it is higher voltage (E) Westinghouse had deeper-pocket backers than Edison. 35 Solution to Problem 4: Hybrid Car (a) The electric motor will be made up of multiple loops of current rotating in the permanent magnetic field of 1 Tesla. The torque exerted on a current loop in an external magnetic field is rrr = ×B Considering all the turns as separate current loops, the total torque will just be N times this value, or rr r = N ( IA × B) r = NIABsin ˆ where α is the angle between the orientation of the magnetic field and the current loop. The maximum torque occurs when the two are at an angle of 90 degrees to one another, or when sin α = 1. Thus, NAVB (0.3m 2 )(200V )(1 T ) = NAIB = = = 200 N ⋅ m = 147.49 ft ⋅ lb max R (0.3Ω) (b) As the current loop rotates faster (more RPM) the flux through the current loop is changing more rapidly, and this creates a bigger back-emf. This back-emf limits the current supplied to the loop, and thus reduces the torque. So, we should find that torque decreases with increasing RPM. We do the calculation as above, except we make the replacement V → V − emf where emf is the back-emf. What is this value? From Faraday's Law, dΦ B d rr d emf = N = N ( A ⋅ B) = N ( ABcos ) dt dt dt If the loop is rotating with angular frequency ω then we can say = t and d 2 emf = N ( AB cos t ) = NAB sin t = NABf sin t dt 60 where f is the RPM. Thus, the torque is now 2 NA(V − NABf sin t) B NA(V − emf ) B 60 = sin t = sin t R R This will be at a maximum when sin ωt = 1, so 2 NAVB 2 1 − NABf = (200 N ⋅ m) 1 − (0.3m2 )(1T) f = ( 200 − .01 f ) N ⋅ m max = R 60 60 or max = (147.5 − 0.0232 f ) ft ⋅ lb This is a maximum of 147.5 ft-lb at 0 RPM and drops to 0 ft-lb at 6,367 RPM. (c) Using the first hint, and our calculated torque from part (b), we get 2 P= = [(200 − .01 f ) N ⋅ m ] ⋅ f = (20.94 f − .003289 f 2 )W = (.02807 f − .000004411 f 2 )hp 60 This is a parabola. The power is 0 at 0 RPM and 6,367 RPM, and reaches a maximum of 44.6 hp at 3,184 RPM. Using the second hint, we get V − emf R Using our expression for emf found above, we get 2 V− NABf 2 60 P= NABf = (.02807 f − .000004411 f 2 )hp 60 R P = emf ⋅ I = emf (d) The hybrid uses 1 gallon of gasoline in an hour at 60 mph at 60 mpg. All the energy in the gasoline gets turned into heat energy, so the heat energy produced is just the energy stored in a gallon of gasoline, 60kW-hr, or 60min 6 0 s e c E = 60,000 W ⋅ hr = 2.16 × 108 J 1hr 1min The total mass of gasoline is 1kg total mass = 2.16 × 108 J = 4.73kg 4.56 × 10 7 J Since the ratio of carbon atoms to hydrogen atoms is 8 to 18, the percentage of the total mass that is carbon is massC massC 8 ⋅12 96 = = = = .842 massTotal massC + massH 8 ⋅ 12 + 18 ⋅ 1 114 Thus the mass of carbon is massC = .842 massTotal = 3.98kg mass H = .158massTotal = 0.75kg And so the masses of the products are 12 + 16 + 16 44 massC = ⋅ 3.98kg = 14.59kg 12 12 1 + 1 + 16 18 mass H2O = massH = ⋅ 0.75kg = 6.66 kg 2 2 1 mole of ideal gas at STP occupies 22.4L. 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    ¡©§ ¥£ ¡ GE@CA%@976541¢('%$$#" ¨¦¤¢ Grading scheme for Problem 7: The Earth’s Magnetic Field Overview The main idea of this problem is that the Earth’s magnetic field can be approximated by a current ring around the core of the Earth. Getting the exact numerical answers wasn’t too important, but making a reasonable estimate was. One point (per numerical answer) was taken off for lack of units. One point (per mistake) was taken off for numerical mistakes. (a) Core Current Use the Biot-Savart Law (or the formula for magnetic field a distance z above a current loop.) B = Bz = Bz = = r µ0 Idl × ˆ 4π r2 µ0 I sin θ Rc dr sin θ = 2 + R2 2 21 4π Rc (Rc + Re ) 2 e µ0 I sin θ · 2πRc 2 2 4π Rc + Re 2 µ0 IRc 2 + R2 )3/2 2(Rc e Now we can solve for the current. It was fine if you just wrote down the formula for the magnetic field above a current loop. Recognizing that the problem required Biot-Savart and making a decent but incomplete attempt was awarded up to three points, depending on how good the attempt was. = 2 2 2(Rc + Re )3/2 Bz 2 µ0 Rc = I 5.04 × 109 A Note: You absolutely cannot use Ampere’s Law to find the current in part (a). The problem does not have the requisite symmetry for Ampere’s Law to apply, and in any case, almost no one attempted to draw a loop or define what the enclosed current even meant in this situation. No points were awarded for using Ampere’s Law in part (a), or assuming the current was a long straight wire. (b) Magnetic Moment M = IA 2 = I (πRc ) = 1.92 × 1023 A·m2 1 Partial credit was given for using the wrong radius: we want the core radius because that’s where the current is, and the A in the above formula is the area 2 of the current loop itself. Less partial credit was given for using A = 4πRe . (c) Energy Stored in Field If we want to make the approximation L = BA , then we really ought to use the I central magnetic field instead of the surface magnetic field. We can see from part (a) (replace Re with 0) that this central field is given by: Bc = µ0 I 2Rc and so 2 ΦB Bc A Bc (πRc ) µ0 πRc ≈ = = = 6.87 H I I I 2 The problem explicitly asked for L in terms of geometrical quantities and asked for a numerical value of L. Both were necessary to get full credit. However, points were not deducted for using incorrect values for I calculated in earlier parts of the problem, or for using the surface magnetic field instead of the central field. You can’t use the formula for energy density stored in the magnetic field to get the total energy because you don’t know the magnetic field at any point that’s not on the axis of the current loop, and even if you did you’d have to do an impossible integral. However, a small amount of partial credit was awarded for using this method if it was clear that you realized that the method only gets you a rough estimate. The way to get the energy is using the calculated inductance: L= U= 12 LI = 8.73 × 1019 J 2 (d) Time Constant L for an LR circuit R ρL ρ2πRc 2ρ R= = = 2 A πRc Rc τ= τ= L µ0 πRc Rc π 2 = = µ0 σRc = 4.78 × 1012 s = 1.52 × 105 yr R 2 2ρ 4 Points were not deducted for using the incorrect L calculated previously. Points were deducted for using Re instead of Rc . Partial credit was also awarded for attempting to calculate the rate of energy lost by ohmic heating. 2 (e) Dynamo We have to modify Ohm’s Law because magnetic fields exert a force on moving charged particles. To get full credit, it was necessary to mention the Lorentz force law: F = q (E + v × B) A lot of answers got tangled up in magnetic fields producing electric fields, which produce magnetic fields, which induce electric fields, which.... but the answer had to do with forces. Ohm’s Law (in modified form) still works when E is zero, as in the next part of this question. Note: The reason we can usually write E = ρJ is that the velocity of charged particles is typically very small, so we can ignore that term in Ohm’s Law. However, in highly conducting plasmas, E is small (in a perfectly conducting plasma it would be zero), and so the magnetic force on the particle is important. If E = 0, and we assume v and B are perpendicular, then we obtain: v= ρI ρ ρJ ≈ = = 3.64 × 10−7 m ·s−1 B Bc A L Significant partial credit was given for finding v and leaving a factor of sin θ where θ is the angle between v and B. 3 Solution (a) The heaters are connected in parallel, so they all have the same voltage across them and dissipate the same power. The power dissipated in each resistor is 2200 W = 275 W (1) 8 2 Each resistor has 240 rms volts across it, and the power dissipated is given by P = Vrms /R. Solving this for R, we get P1 = 2 Vrms (240 V)2 R= = = 210 Ω P1 275 W The current is found from P = Irms Vrms . (2) P 275 W = = 1.15 A (3) Vrms 240 V (b) In the conventional Sauna, heat is transferred first by conduction between the rock and the air, and then by convection to your skin, and then by conduction again into your skin. In the infrared sauna, heat is transferred directly from the heater to you by radiation. The heaters will emit roughly a blackbody spectrum, so the effective area, Ae is found from: Irms = P = σAe T 4 ⇒ Ae = P 2200 W = = .775 m2 4 σT (5.67 × 10−8 W/m2 -K) (200 + 273 K)4 (4) (5) (c)This problem has several parts, all of which are straightforward plug and chug. For (i) We can find the frequency from f = c/λ, doing this we find a frequency range of: 20 µm > λ > 5 µm ⇒ 1.5 × 1013 Hz < f < 6 × 1013 Hz (6) For (ii), the average electric field in the sauna will be zero, because the sauna will emit a linear combination of plane waves, each of which will have zero electric field on average, so the net average electric field will be zero. The rms field, however will not be zero. The rms field is found from knowing that the time average pointing vector is equal to the power per unit area emitted from the surface, so < S >= ⇒ Erms = 2 0 Erms c = σT 4 σT 4 = 1034 V/m 0c 1 (7) (8) For (iii), the radiation pressure is P= <S> σT 4 = = 9.46 × 10−6 N/m2 c c (9) For (iv), the total force on a person, would be the integral of the radiation pressure over their body: F= P dA = 1 <S> dA = (.80)(2200 W) = 5.87 × 10−6 N c c (10) (d) The entropy created in the heater is: ∆Sheater = Q (2200 W)(30 × 60c) = = 8372 J/K T 200 + 273 K (11) The entropy change due to the heat transfer from the heater to the people: ∆Stransfer = − Q Theater + Q Tpeople = −8372 J/K + 12260 J/K (12) So adding the results of (??) and (??), we get ∆Stotal = 12260 J/K (13) (e) The temperatures must be converted from Fahrenheight to kelvin. TH = 550◦ F → TH = 561 K (14) TL = 50◦ F → TL = 283 K (15) TL 283 =1− = .495 eCarnot = 1 − TH 561 (16) The Carnot efficiency is then If the efficiency is Carnot, then the engine is reversible, so the entropy change of the universe is zero. (f ) If the efficiency of the engine is η , the work and heat out of the engine are related to the heat in through: W = ηQH (17) 2 QL = (1 − η )QH (18) The rate of change of the entropy of the high temperature reservoir is dSH 1 dQH =− dt TH dt (19) The rate of change of the entropy of the low temperature reservoir is dSL 1 dQL = dt TL dt (20) The rate of change of the entropy of the universe is dST dSH dSL 1 dQH 1 dQL = + =− + dt dt dt TH dt TL dt (21) Inserting (??) and (??) into (??), using the fact that η = .8eCarnot , and doing some algebra, one eventually arrives at dST 1 dW 1 = = 2200 W = 1.94 J/K dt 4TL dt 4 × 283 K (22) Grading Scheme This problem was pretty straightforward, and was broken into many smaller parts.There was not much partial credit to give here - you either knew what formula to use or you didn’t. In general a point was deducted from answers that had the wrong units or no units. Part (a): • +1 for thinking that 2200 W was the power of a single heater and that they were supposed to find the total resistance and current. • +2.5 for the current and +2.5 for the resistance. I was flexible as to whether they put the rms or peak current, as the problem did not state. • -1 for a silly mistake in calculating the current or the resistance. • +2 for calculating the total resistance and total current, but not the current and resistance of one resistor. 3 Part (b): • +1 for each correct identification of a heat flow process. • +1 total for the whole identification part if they explained how conduction, convection, and radiation work, but not how they are working in this sauna. • Some subjective ness was used to determine how many points for the explanations • +0 for just saying ”conduction, convection, and radiation” with no explanation of how they work in the heater. • +2 for finding the effective area. Part (c): • Each part of this problem was worth one point with no partial credit, except for part (ii), which was worth 2 points - one for the average electric field and one for the rms electric field. Some partial credit was given if the person sketched out exactly how to do the problem with equations but didn’t plug in. • +.5 for 7.36 × 10−6 N on part (iv), which forgets that only 80% of the energy is going into the person. Part (d): • +3 for having the wrong answer because you thought each resistor dissipated 2200 W, and you already lost points on (a) for this. • +1.5 for getting the transfer part right, but forgetting to add the entropy ”created” by the heater. Part (e): • +0 points for evaluating the Carnot formula using temperatures not on the absolute scale. This indicates a lack of understanding of how this formula works. • + .5 points for having the correct formula, but evaluating it incorrectly with some attempt to use absolute temperature perhaps from incorrect temperature conversion. • +1.5 for correctly calculating the efficiency • +1.5 for realizing that since the efficiency is Carnot, the change in entropy is zero. 4 Part (f ): • +4 for doing this correctly but thinking that each resistor put out 2200 W and that the total power was 8 × 2200 W. • +.5 for an approximation like ∆S ≈ eQH /T that gets close. 5 ...
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