Mid-term - Mid-Term Exam on Ch 22, 23 and 24 1. Proton A...

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Unformatted text preview: Mid-Term Exam on Ch 22, 23 and 24 1. Proton A moves across the US. from south to north while proton B moves from east to west. How are the magnetic forces on protons A and B related? (a) the force on proton A is greater than the force on proton B (b) the force on proton A is less than the force on proton B (c) the forces on the two protons are equal. ((1) The force on proton A is upwards while that on proton B downwards (e) The force on proton A is downwards while that on proton B is upwards This question was designed to test understanding of the equation for the magnetic force on a moving charge (thefirst on the sheet provided), 15 = qfi >< ll. The magnitude of the force is given by F = gen? sin 9, where Bis the angle between the velocity and the magnetic field. First, draw a diagram with an arrow pointing up the page to represent north and an arrow to the left representing the direction from east to west. It looks like: North (magnetic field) Proton A Proton B T he earth 's magnetic field points north. Use the right hand rule for the cross product. The force on proton A is zero because sing = 0. T he force on proton B is downwards. T he only answer that fits is (b). if you had remembered from one of the CALM problems that the earth ’sfield usually has a small vertical component, you would have obtained the same magnitude of the force on proton B (i. 8. q v B), but there would now have been a small westward force on proton A. Answer (b) is still the only one that fits, however. This problem is essentially identical to conceptual problem number 5 on page 742 of Walker’s book. 2. Two perpendicular wires carry currents as shown in the figure below. How are the magnitudes of the magnetic fields at points A and B related? (a) The field at A is greater than the field at B (b) The field at A is less than the field at B (c) The field at A is equal to the field at B (d) Both fields are zero (e) The fields at A and B have opposite signs This problem is number 43 on page 746 of Walker ’5 book. First, using the right hand rule for field direction, you see that the fields due to the two wires reinforce one another at point A (ie they are both out of the page), while the twa fields tend to cancel at point B. Therefore, the magnitude of the field is greater at point A than at point B and solution (a) is correct. Thefiela’ at A is out of the page while that at B is into the page. Thus, answer (e) is also correct. 3. Referring to the figure for question 2, what is the magnitude of the magnetic field at point A: (a) 0.013 mT (b) 0.0 ml" (c) 0.032 mT (d) —0.032 mT (e) 0.05 mT T he formula for the field at distance dfrorn a current carrying wire follows directly from Ampere '3 law: B = #01 / 27rd . Since point A is equidistant from the wires, we have: B = ,tio (4.5 + 6.2)/2a(0. l 6). This works out to give answer (a) 4. The current in a solenoid with 22 turns per centimeter is 0.50 A. The solenoid has a radius of 15 cm. A long, straight wire runs along the axis of the solenoid, carrying a current of 16 A. What is the magnitude of the net magnetic field at a distance of 0.75 cm from the axis of the solenoid? (a) 2 mT (b) 1.45 mT (c) 1.81 mT (d) 0 (e) 2.9 m1“ This is problem 77 on page 748 of Walker ’s book. T o solve it, you need the formula for the magnetic field in a solenoid (given on the formula sheet) and the formula for the field due to a long wire that was used in problem 3. T he field due to the solenoid doesn ’t depend on the distance from the solenoid axis but the field due to the wire does. 1” he field due to the solenoid is parallel to the solenoid axis while that due to the wire is perpendicular to the axis. This means that the two fields have to be added in quadrature (ie square the values of the fields, add them and take the square root) ‘ 3 sateum‘a = MAO“ I T 1: 161(2200)0.5 2 I-BEZVI'O gm}; 2 floI/anct: tetra? Ira/(u; 0.00%} = n’Zéerm‘” ..____._._-,. Mama...“ _ 3 l; — E“ r‘ a ——- hind. ' some...‘n + “M.” l- ‘1-51 10 1 ll 5. A magnetic field is oriented at an angle of 32° to the normal of a circular loop of diameter 22 cm. If the magnetic flux through the loop is 4.8 x 10'5 T.m2, what is the strength of the magnetic field? (a) 1.26 mT (b) 1.49 mT (c) 0.37 mT (d) 4.80 mT (6) 14.9 mT This is a straightforward application of the definition of magnetic flux given on the formula sheet. k...— —-'1. ¢ : Co:- 9 :: $.T(0-“>1603g20: SVE'ZZU‘KlD Bx 2.22ux to“2 := Li-‘c‘: v: ISS- ‘Q‘; Laggxlszs I «I 6. An EMF is induced in a conducting loop of wire 1.12 m long as its shape is changed from square to circular. Find the average magnitude of the induced EMF if the change in shape occurs in 4.25 s and a local 120 mT magnetic field is perpendicular to the plane of the loop. (a) 600 mV (b) 61.5 mV (0) 0.31 mV ((1) 0.6 mV (6) 6.0 mV This is another application of the formula for magnetic In this case it is the area of the loop that is changing. The change in magnetic flux is obtained by subtracting the area of the circle from that of the square and multiplying by the magnetic field (which is perpendicular to the loop so cos 6 = 1). Substituting the change influx and the time taken to change it into Faraday ’3 law for induced emf gives the answer (d). AWN : C mouth (a Quote, W: “2/2.: A Mg- TF2 r: In” , -A = “12L L 0‘ L 4” AW“ 5W“ 2? Le: flarm '” AQWMM = ? "ngg? W‘L @ : m6 afigasl At}: -‘ ENAA ) = 042x 2.“.th 152“ TML . 25% v (0—3 LALOiJL/Lced e Q: ~—{\l Egg ‘ N51 At; (Fag; Al: ’ *3 '7? 5= '7’ Smflo V? ‘6 Oltclwtlguv 7. The circuit below, with a set of identical bulbs A, B and C is allowed to come to a steady state with the switch closed. Which of the options given best describes the situation immediately after the switch is opened? (a) All the bulbs go out (b) Bulbs A and B stay the same and bulb C gets dimmer (c) Bulbs A and B get brighter and bulb C goes out ((1) Bulbs A and B get dimmer and bulb C goes out (e) Nothing changes This is like the problem that was showrz in class. The light bulbs are identical so, when the switch has been closed for a long time, the current through bulbs A and B (let ’s call it I) is halfthat ofbulb C (ie. 2 [passes through bulb C). When the switch is opened, the inductor resists change in the current through it, so current I Continues to flow through bulbs A and B. The some current now flows through bulb C, so it gets dimmer. The answer is therefore (b) 8. The figure below shows a current-carrying wire and a circuit containing a resistor R. A constant current is first established in the wire and then increased steadily. How does the direction of the current in the circuit change when the current changes from constant to increasing? (a) The current is clockwise in both cases (b) The current is counterclockwise in both cases (0) The current changes from zero to counterclockwise (d) The current changes from zero to clockwise (e) The current is zero in both cases This is problem 22 on page 779 of Walker ’s book. When a steady current flows through the wire, no emf is induced in the closed circuit because there is no change in magnetic flux and a change in magnetic flux is required to generate an induced emf. This observation eliminates answers (a) and (b) immediately. When the current in the wire is steadily increasing, there is a steady increase in the magnetic flux (due to the current in the long wire) that threads through the closed circuit. This observation eliminates answer (e). The magnetic flux through the circuit due to the wire is out of the page (right hand rule). Since the current is increasing, this flux is also increasing. By Lenz is low, the emf induced in the circuit will oppose this change. This means that a current must flow in the circuit to produce a magnetic field into the page in the middle of the circuit. A clockwise current is needed for this (right hand rule), so the answer is (d). 9. A solenoid is 1.5 m long and has 490 turns per meter. What is the diameter of the solenoid if it stores 0.31 J of energy when it carries a current of 12 A? a) 11 cm b) 5.5 cm 0) 16.5 cm (1) 25.5 cm e) 110 cm This is an application of the formula for the energy stored in a magnetic field given on the formula sheet. The total energy stored in the field is the energy density times the volume of the solenoid. The only unknown is the diameter of the solenoid. _ __ *7 _3 éscflemolefi 1 luv/1J— ; h“ x ‘0 ' {+10- 12: 7-3é‘h‘lm T 1. ' '5 Ago-«3% us: Lg: : _1_ figs‘eMfigd : (?_33qx[0 ‘2. ll 2.. s z flo ’HMMJ? 10. A 3.0 V battery is connected in series with a 35 mH inductor, a 110 Ohm resistor and an open switch. How long after the switch is closed will the current be equal to 0.22 mA? at) 254.5 s b) 0.31 s c) 0.26 ms (1) 2.6 its 6) 0.26 pS Here, one applies the formula given on the formula sheet for the current in an inductor after a switch in the circuit is closed T: i(\-e‘“‘/") = :10— «Ho/ooze) .- R HO 3 ' - - -3-|f-r2flt we O~ZZX £03: 2423Mol— 232,1“022 3 _7_ . ~2 -3.iuz_a.+. we --'2-'%':T><to ; —2.3z’+xto C 3 ‘ “ l X‘o Olqlqs XIO' r: 8 31%. —t : 2,5gxm“; 2.5%.; 11. A step up transformer has 25 turns on one coil and 500 turns on the other. If the transformer is to produce an output voltage of 4800 V with a 12 111A current what input voltage is needed? (a) 240 V (b) 96000 V (c) 0.6 UN ((21) 0.6 V (e) 120 V Since the transformer is a step—up transformer (ie the output or secondary voltage is higher than the primary or input voltage), the 25 turns are on the primary side and the 500 turns on the secondary side. The answer (a) then folloWS immediately from an application of the transformer equation given on the formula sheet. Note that this problem is identical (including the addition of the irrelevant information about the input current) to problem 7 on the CALM homework for Ch 23. .__._ _ _____ :5) VP: 2QQV VS NS‘ 300 QEQQ 12. A “75-watt” light bulb uses an average power of 75 W when connected to a conventional, 120-V household electrical supply. What is the maximum power used by the bulb at any instant in time? (a) 75 W (b) 150 W (0) 106.1 W ((1) 300 W (e) 100 W 2 rm: T he average power used by the light bulb is just I R where R is the resistance of the light bulb. The maximum ower consum tion is 12 R. Since we know that a domestic P P max electric supply provides a sinusoidal voltage, we know that Im = J21 m5 . Therefore the answer to the question is (b) 13. When an RC circuit with a resistance of 15 Ohms is connected to a generator of frequency 60 Hz and rrns voltage 120 V it has a power factor of 0.37. What is the value of the capacitance? (a) 103.8 uF (b) 141 uF (c) 95 uF (d) 70.4 pF (6) 7.04 uF T he formulae for the power factor and the impedance Were given on the formula sheet. 12- in: MW e is“ Worms (405141“: 13 4 ' /(|2c)".1¢)1 " -s '4 e l ,. ‘ :3; C: J.— .,. 70a sun 3 (’6 @2074" no: Hit 14. A manufacturing plant uses 2.22 kW of electrical power provided by a 60 Hz ac generator with an rms voltage of 485 V to run a number of hi gh—inductance electric motors. The plant’s total resistance is 25 kOhms and its power factor is 0.54. What capacitance, connected in series with the power line, should be installed to increase the power factor to unity? (a) 2.1 uF (b) 0.068 uF (c) 103 .3 F (c) 21 uF (d) 6.8 HF We are given the power factor for the manufacturing plant. Using the formula on the equation sheet, we can write the power factor in terms of the resistance (given) and the inductance (unknOWn) of the plant. This allows us to solve for the inductance. When the power factor is unity (ie one), R = Z from the definition of the power factor. Thus, in order to achieve a power factor of unity, the added capacitive reactance has to be equal to the inductive reactance already calculated. This allows us to solve for the added capacitance. Pew“ Costiw V’f/fi :- 0631, e 153L101 /.J_-_ :91 % =' itéficichfl... 1 tray Cosiia;l we Med 65% 1 2. . .a'z. -_..". ‘tt? mt» 1— re C: J- 3mm '2: Q '* (“L {13:} g ‘1'? {Am boll. 01;;er capaciva 2’: absox 103 = \lQa-tgwi—Y :j(2§00051+ (now-icy Le L: L033 x102" H :- ..L- : mid—L—M I ‘ f [S m" C colt. (moral schouxlo‘ 6 8 X‘ F: 2 0.068/1.‘ 15. An RLC circuit has a resonant frequency of 125 Hz. What is the resouant frequency if the resistance, inductance and capacitance in the circuit are doubled? (a) 62.5 Hz (b) 250 Hz (0) 88.4 Hz ((1) 177 Hz (e) 125 Hz T he formula for the resonant frequency of an RLC circuit is given on the equation sheet. T he frequency depends only on L and C, not R. We are given that L and C each increase by a factor of 2. Inspection of the formula for resonant frequency shows that this causes the resonant frequency to drop by a factor of two, so the answer is (a) 16. An RLC circuit has a capacitance of 0.29 uF and a resistance of 10 Ohms. What inductance will produce a resonant frequency of 95 MHZ? (a) 5.78 pH (b) 9.68 pH (c) 36.3 pH (d) 173 H (e) 9.68 mH This is an application of the formula for the resonant frequency given on the equation sheet. Of course, the information about the resistance is superfluous since the resonant frequency doesn 'I depend on resistance. dill-av (0“ 0:29 K L411} L: (1.92 pii 17. The figure below shows a current carrying loop in a magnetic field. If you look at the coil towards the negative y direction to assess rotations of the loop about the y axis and towards the negative x direction to assess rotations about the X axis, which of the following statements best describes the torque on the loop caused by the magnetic field. (a) There is no torque causing the loop to rotate about the y axis (b) The magnetic torque causes the coil to rotate counter-clockwise about y (c) The magnetic torque causes the coil to rotate clockwise about y (d) The magnetic torque causes the coil to rotate counter-clockwise about x (e) The magnetic torque causes the coil to rotate clockwise about it This problem again tests your understanding of the equation for the magnetic force on a moving charge or current. First recognize that there is no magnetic force acting on the horizontal wires in the loop because the current is parallel to the magnetic field and F = [LB sin6with 6 = 0. The force on the right hand side of the loop is into the page and that on the left hand side is out of the page. Since the forCe on each of these sides is uniform along them, there is no net torque about the x axis (ie the force that is applied above the x axis in the diagram cancels the force that is applied below the x axis). These observations eliminate all but possibilities (b) and (c). Looking at the diagram from the top, we see that the directions of the forces are such that the loop will rotate counter-clockwise, so the correct answer is (b). 18. An ac generator with a frequency of 60 Hz and an rms voltage of 120 V is connected to a 35 uF capacitor. The generator produces a sinusoidal waveform. What is the current in the circuit when the voltage across the capacitor is 75 V and decreasing? (a) -2.01 A (b) 2.01 A (c) -1.42 A (d) 1.42 A) (e) —2.88 A This is similar to the phasor problem that was on one of the CALM homeworlcs. Drawing the phasor (see below), we see that when the voltage across the capacitor is positive and decreasing, the current is negative and decreasing. Thus, without doing any calculation at all, We can eliminate answers (1)) and (d). Plugging in the numbers gives answer (a). VG max max Max W" kmm“ Va 2 Vc. SP“th *7/1) ‘t " ‘1' Jana)" MSW“: '20. J; V g wt cu: awe-AA Va 2 $5— \/ ‘ - - °—- u-t—‘flo m; 1?;- -'-‘ ‘20): Vt}. Shh "'l\/3') :3 _ ‘1 mt: |l$.23° ‘KAJL Knee Etc-J” om ugq‘ohdu‘e Suki—ms“ WLJ x4M- out—010°: tee—25.2? ma mhzmflf b84204”; Stan, {4 We. The ‘Si’ “Hoe Cruz“! (OM23 ...
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This note was uploaded on 07/02/2011 for the course PHYS 202 taught by Professor Warren during the Spring '09 term at Indiana.

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Mid-term - Mid-Term Exam on Ch 22, 23 and 24 1. Proton A...

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