Phys 1112 cp add.

Phys 1112 cp add. - Physics 1112 Spring 2009 University of...

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Unformatted text preview: Physics 1112 Spring 2009 University of Georgia Instructor: HBSch¨ttler u Solutions to Additional Conceptual Practice Problems for PHYS 1112 Final Exam Mon. May 4, 2009, 7:00pm-10:00pm CP 4.01: Answer: (A) 1 Physics 1112 Spring 2009 University of Georgia Instructor: HBSch¨ttler u Detailed Problem Statement: A bar magnet is moved either towards or away from a circular conducting ring with either its north-pole (N ) or its south-pole (S) pointing towards the ring, and the respective pole being located either above the plane of the ring (in ”Top” position ”T”) or below the plane of the ring (in ”Bottom” position ”B”), as shown in the Fig. As also shown in Fig., the plane of the ring defines the x-y-plane, and the magnet’s N -S-axis is oriented parallel to the z-axis while the magnet is being moved. Which of the five motion processes of the bar magnet, (A)-(E) described above, will induce a clockwise (cw) current I in the ring? Solution: Above/below the poles of the bar magnet, B points approximately along the z-axis, out of the N - and into the S-pole, with the field strength B ≡ |B| decreasing with increasing distance from either pole. Hence, the magnetic field B inside the ring points in +z-direction when the N -pole is below the ring pointing up (in Bottom position ”B”) or when the S-pole is above the ring pointing down (in Top position ”T”). Likewise, B inside the ring points in −z-direction when the S-pole is below the ring pointing up (in Bottom position ”B”) or when the N -pole is above the ring pointing down (in Top position ”T”). Also, the field strength B inside the ring increases if either pole is moving towards the ring; and the field strength B inside the ring decreases if either pole is moving away from the ring. By Lenz’s rule, the induced current I in the ring produces a secondary field, BI , which opposes the change ∆B of the time-dependent inducing field B(t). Also, the direction of I is related to the BI -direction by a right-hand (RH) rule: if RH thumb is in direction of BI , the RH 4 fingers point in the flow direction of I. So, to produce a BI pointing in −z-direction requires a clockwise I, as defined in Fig.; to produce a BI pointing in +z-direction requires a counter-clockwise I. Hence, ∆B points in +z-direction, and by Lenz’s rule, BI points in −z-direction, and I flows clockwise if Case (1): B points in +z-direction and its strength B increases in magnitude; or Case (2): B points in −z-direction and its strength B decreases in magnitude. Likewise, ∆B points in −z-direction, and by Lenz’s rule, BI points in +z-direction, and I flows counter-clockwise if Case (3): B points in +z-direction and its strength B decreases in magnitude; or Case (4): B points in −z-direction and its strength B increases in magnitude. Thus, in the answer key above, process (A) corresponds to Case (1), process (B) corresponds to Case (3), process (C) corresponds to Case (4), process (D) corresponds to Case (4), and process (E) corresponds to Case (3). So, only process (A) induces a clockwise current, while (B), (C), (D) and (E) all induce a counter-clockwise current in the ring. 2 Physics 1112 Spring 2009 University of Georgia Instructor: HBSch¨ttler u CP 4.02: Answer: (A) The induced EMF amplitude is Eo = ABω where the loop’s angular speed is ω = 2πf , expressed in terms of the rotation frequency f . Hence, at fixed field strength B, Eo ∝ A × f . So increasing A = 10cm2 → A = 30cm2 = 3 × A and increasing f = 500RPM → f = 1500RPM = 3 × f will increase Eo = 20mV → Eo = 3 × 3 × Eo = 180mV 3 Physics 1112 Spring 2009 University of Georgia Instructor: HBSch¨ttler u CP 4.03: Answer: (D) By Kirchhoff’s loop rule EB + E = 0 where E = −L∆I/∆t is the self-induced EMF in the inductance L. Hence, ∆I/∆t = EB /L = 50V/10H = 5A/s. 4 Physics 1112 Spring 2009 University of Georgia Instructor: HBSch¨ttler u CP 4.04: t mrS 1olE---­ Q== Lf j (Lasf?Y lJetJ.m'G ') , ,L ct S ide View J I (Las er 1> 'nil) +a.kes Q 'lv\l~1 II "'1 vad IU5 "R ­ a, i5 e \re1'-1 ;vecl c e a. se" b ,0 d"5 C\ === &{ 4-1(~ 8.0 fi ({l-t cJtElq' '" >l (j r v I I _ >S -~-----4 Pr-e» l.tv-{ y ___ II Answer: (E) To levitate a massive object of mass m, the downward weight force mg must be matched by the upward radiation pressure force ΠA, i.e., mg = ΠA. Here A is the cross-sectional beam area to which the object is exposed (with A-surface perpendicular to beam direction); and Π = αI/c is the radiation pressure, with α = 1 for a 100% absorbing object and with α = 2 for a 100% reflecting object. Hence, solving for the time-averaged beam intensity: I = cmg/(αA) which implies I ∝ m/(αA). So changing the mass mS = 40pg → mD = 120pg = 3 × mS ; 5 Physics 1112 Spring 2009 University of Georgia Instructor: HBSch¨ttler u and changing the absorption-reflection factor αS = 1 → αD = 2 = 2 × αS ; and changing the exposure cross-section area from a circle of radius R to a square of 4R sidelength, i.e., AS = πR2 → AD = (4R)2 = (16/π) × AS , will change the required intensity for levitation IS → ID = 3 × (2)−1 × (16/π)−1 × IS = (3π/32)IS since the levitation intensity I is proportional to m, but inversely proportional to α and A. 6 Physics 1112 Spring 2009 University of Georgia Instructor: HBSch¨ttler u CP 4.05+4.06: Ollf Tu.O 0..,.1. g Side. 'VJ' I, _---.. ---r----r-:-~-r-:--......;--=---~-~"O""'lj ofl'e'1/-eol ~t {V1.f e~s( te'~ Clv-e ~ I 1------_.........:.-­ 2. - 0 'PI I~ -: : : 0 - -~~ _~1iD~ Q~ FytJ rn rr. £- Vir i f1 +-5 : CoS( OiJ)=(f3')/J.. 2 CoS (bOO) = CP 4.05 (=Q1), Answer: (B) For unpolarized light of time-averaged (TA) intensity I0 passing through polarization filter (PF) 1, the TA intensity after the filter is I1 = I0 cos2 θ1 θ . Here, ... θ means averaging over all possible values of the random angle between an incident waves packet’s electric field E0 and transmission axis T1 , denoted by θ1 ≡ (E0 , T1 ). For uniformly random angle θ1 (i.e., any value of θ1 between 0o and 180o is equally likely to be found among incident EM wave packets), cos2 θ1 θ = 1/2 and hence I1 = (1/2)I0 . The wave packets passing through PF1 (i.e. between PF1 and PF2) are then polarized with electric field vector E1 T1 . But then the angle between E1 (incident upon PF2) and T2 is θ2 ≡ (E1 , T2 ) = 90o , since T2 ⊥ T1 . Hence, by Malus’ Law, the TA intensity transmitted 7 Physics 1112 Spring 2009 through PF2 is I2 = I1 cos2 θ2 = I1 cos2 (90o ) = 0. CP 4.06 (=Q2), Answer: (E) University of Georgia Instructor: HBSch¨ttler u As in CP 4.05, the intensity I1 after PF1 is still I1 = (1/2)I0 with E1 T1 . However, the angle between E1 and T2 is now θ2 ≡ (E1 , T2 ) = 90o − 30o = 60o , since T2 is 30o from the horizontal, but E1 T1 , i.e., E1 is vertical. So, I2 = I1 cos2 θ2 = I1 cos2 (60o ) = I1 × (1/2)2 = (1/2)I0 × (1/2)2 = (1/8)I0 ; and thus I2 /I0 = 1/8. Important to note here: It does not matter here if you add or subtract 180o from the ET angle θ, or change the sign of θ, since cos2 (θ) = cos2 (−θ) = cos2 (θ±180o ) = cos2 (180o ±θ). 8 The area vector of the loop A must be normal to the plane of the circuit loop, i.e., parallel to the z-axis. So, let’s choose it to point in −z-direction, i.e., into the paper, i.e. in the +B-direction. The initial flux through the circuit, before changing the field/circuit configuration, is then Φm = BA cos Θ = +BA > 0 since Θ ≡ (A, B) = 0o . The field or circuit configuration change processes (A), (B), (C), (D), (E) and (F) will then result in a time-dependent change First note, for orientation, that for the x- and y-axis shown, the z-axis is out of the paper. 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W & 3 ¦ R P G E 0 1 A 7   0 1 0 ( ¡ # !    ¨ ¨ ¨ ¦ ¢ DYX"2)"uV"5s%qfp8S)i5C`$'fedb`C$`TRDDYX"2UV"TSQI£HFD3C"B9@8$65342)&'%$"©§¤¥£¡ CP 4.07: Physics 1112 Spring 2009 y x 9 University of Georgia Instructor: HBSch¨ttler u Physics 1112 Spring 2009 University of Georgia Instructor: HBSch¨ttler u of Φm , with final values of flux Φm = B A cos Θ , magnetic field strength B and angle Θ ≡ (A , B ), after changing the field/circuit configuration, as follows: (A) Φm = −BA < +BA = Φm with Θ = 180o and B = B; (B): Φm = 0 < BA = Φm with Θ = 90o and B = B; (C): Φm = B A < BA = Φm with Θ = 0o and B < B; (D): Φm = BA = Φm with Θ = 0o = Θ and B = B; (E): Φm = B A > BA = Φm with Θ = 0o and B > B; (F): Φm = 0 < BA = Φm with Θ = 90o and B = B. Therefore, the time-dependent change of flux ∆Φm ≡ Φm − Φm and the resulting induced EMF E = −∆Φm /∆t (with ∆t > 0 here) have the following signs: ∆Φm < 0 and hence E > 0 for processes (A), (B), (C) and (F). By the sign convention for Faraday’s law, a positive EMF E > 0 drives the current I in the loop in right-hand (RH) direction around A (i.e., point RH thumb in A-direction; then 4 RH fingers will point in I-direction). Since A was chosen to point into the paper, ”RH direction around A” means: clockwise. So, current I flows clockwise around the loop and therefore deposits positive charge on the upper capacitor plate (and an equal amount of negative charge on the lower capacitor plate) in processes (A), (B), (C) and (F). ∆Φm > 0 and hence E < 0 for process (E). By the sign convention for Faraday’s law, a negative EMF E < 0 drives the current I in the loop against RH direction around A Hence, current I flows counter-clockwise around the loop and therefore deposits negative charge on the upper capacitor plate (and an equal amount of positive charge on the lower capacitor plate) in process (E). ∆Φm = 0 and hence E = 0 for process (D). Hence, there is no induced EMF and no induced current I in the loop and therefore zero charge being deposited on the either capacitor plate. For an alternative way of solving this, using Lenz’s rule, see posted solution of Homework Problem P08.01 for details. 10 For detailed explanation of this solution, see see posted solution of Homework Problem P08.02. Y˜¬­™4« F) ”« ­2D5@•²™­˜±dˆ™­2¯™™Y˜­™˜4« ª –˜ ” ³´® ’ ³ “˜–˜ ° © ¢ ° “˜™–® ¬– ” y u &a k ! & 0 #  X ! k # € §u`#"`$x¥m‚VxD$!"`53…¥¥y  y &a k#! §uu`"`&$|#$2p%"~A"}#"`68r h # g # # h ! i y y &a k#! A# # h # g #! i §uu`"`&$h"l$|#"2p|h"`¥y { yzX2ˆY#53uD5pxT6Y`S3FY!5¥Dp``w`&$ t 0 W XX g 0n & 1 k 13 !X A A k X YoF"Y162…`$85"˜HnrC5"H4D%h%Yhv§!$2C$D"l`"t`k"`"BD$"`S%XV$'QC$SDX"Y!…`$rTVpRHpgdo6H`20l"`Tf`&$B`i 1 & 3 m 0 X & ! X s (W X n1 t X 1 u # 01 0 X ! X & & a # ! &  X ! ! k 3 & s #  #1 3 1 t X & t  j X q # n  m # & X ! k a j X ! # ! W U & 3 ¦ t ! X r c X ¦ h XW g # ! ¡ c c  a 1 ! A ¦ ! 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X ! k 3 X & #  #1 3 X 1 t X t  Œ‹ Š ˆ†… Fd‚‰x$‡2„ˆƒ „ f d —ƒ • — – • ” ” † ’  ‰ ‡ † y „ ƒ y‚  € y w hv'ge‘b•™˜ƒ–u)§“‘%ˆxVw…HfVxv CP 4.08: Physics 1112 Spring 2009 E R − + 11 Ÿ‘s"H‚•¤¢dSwŸœ›“šz™Y˜@– •”D’ ©¨ § ¦ ¥’ ¥ £ ¡ ž ™ — “ Œ‹ Š ˆ†… Fd‚‰x$‡2„ˆƒ University of Georgia Instructor: HBSch¨ttler u ...
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