Solutions to Additional Conceptual Practice Problems for PHYS 1112 Final Exam

Physics 1112Spring 2009University of GeorgiaInstructor: HBSchüttlerSolutions to Additional Conceptual Practice ProblemsforPHYS 1112 Final ExamMon. May 4, 2009, 7:00pm-10:00pmCP 4.01:Answer: (A)1

Physics 1112Spring 2009University of GeorgiaInstructor: HBSchüttlerDetailed Problem Statement:A bar magnet is moved either towards or away from a circular conducting ring with eitherits north-pole (N) or its south-pole (S) pointing towards the ring, and the respective polebeing located either above the plane of the ring (in ”Top” position ”T”) or below the planeof the ring (in ”Bottom” position ”B”), as shown in the Fig. As also shown in Fig., theplane of the ring defines the x-y-plane, and the magnet’s N-S-axis is oriented parallel to thez-axis while the magnet is being moved.Which of the five motion processes of the bar magnet, (A)-(E) described above, will inducea clockwise (cw) current I in the ring?Solution:Above/below the poles of the bar magnet, ⃗ B points approximately along the z-axis, outof the N- and into the S-pole, with the field strength B ≡ | ⃗ B| decreasing with increasingdistance from either pole.Hence, the magnetic field ⃗ B inside the ring points in +z-direction when the N-pole is belowthe ring pointing up (in Bottom position ”B”) or when the S-pole is above the ring pointingdown (in Top position ”T”). Likewise, ⃗ B inside the ring points in −z-direction when theS-pole is below the ring pointing up (in Bottom position ”B”) or when the N-pole is abovethe ring pointing down (in Top position ”T”). Also, the field strength B inside the ringincreases if either pole is moving towards the ring; and the field strength B inside the ringdecreases 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 , whichopposes the change ∆ ⃗ B of the time-dependent inducing field ⃗ B(t). Also, the direction of I isrelated to the ⃗ B I -direction by a right-hand (RH) rule: if RH thumb is in direction of ⃗ B I , theRH 4 fingers point in the flow direction of I. So, to produce a ⃗ B I pointing in −z-directionrequires a clockwise I, as defined in Fig.; to produce a ⃗ B I pointing in +z-direction requiresa counter-clockwise I.Hence, ∆ ⃗ B points in +z-direction, and by Lenz’s rule, ⃗ B I points in −z-direction, and I flowsclockwise ifCase (1): ⃗ B points in +z-direction and its strength B increases in magnitude; orCase (2): ⃗ B points in −z-direction and its strength B decreases in magnitude.Likewise, ∆ ⃗ B points in −z-direction, and by Lenz’s rule, ⃗ B I points in +z-direction, and Iflows counter-clockwise ifCase (3): ⃗ B points in +z-direction and its strength B decreases in magnitude; orCase (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) correspondsto Case (3), process (C) corresponds to Case (4), process (D) corresponds to Case (4), andprocess (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 1112Spring 2009University of GeorgiaInstructor: HBSchüttlerCP 4.02:Answer: (A)The induced EMF amplitude is E o = ABω where the loop’s angular speed is ω = 2πf,expressed in terms of the rotation frequency f. Hence, at fixed field strength B, E o ∝ A × f.So increasing A = 10cm 2 → A ′ = 30cm 2 = 3 × A and increasing f = 500RPM → f ′ =1500RPM = 3 × f will increase E o = 20mV → E o ′ = 3 × 3 × E o = 180mV3

Physics 1112Spring 2009University of GeorgiaInstructor: HBSchüttlerCP 4.03:Answer: (D)By Kirchhoff’s loop rule E B + E = 0 where E = −L∆I/∆t is the self-induced EMF in theinductance L. Hence, ∆I/∆t = E B /L = 50V/10H = 5A/s.4

Physics 1112Spring 2009University of GeorgiaInstructor: HBSchüttlerCP 4.04:mrSt1olE---­j ,Q== Lf,LJI(Lasf?Y lJetJ.m'G ' ) (Las er 1> ct 'nil)Side View+a.kes Q 'lv\l~1 "'1a. se" b II , 0vad IU5 "R ­ a,i5ce \re1'-1 ;vecled"5 &{C\ === 4-1(~ 8.0fi ({l-t >SI _ -~-----4___Pr-e» l.tv-{ cJtElq' '" >l (j r v I yIIAnswer: (E)To levitate a massive object of mass m, the downward weight force mg must be matchedby the upward radiation pressure force ΠA, i.e., mg = ΠA. Here A is the cross-sectionalbeam 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 withα = 2 for a 100% reflecting object. Hence, solving for the time-averaged beam intensity:I = cmg/(αA) which impliesI ∝ m/(αA).So changing the massm S = 40pg → m D = 120pg = 3 × m S ;5

Physics 1112Spring 2009University of GeorgiaInstructor: HBSchüttlerand changing the absorption-reflection factorα S = 1 → α D = 2 = 2 × α S ;and changing the exposure cross-section area from a circle of radius R to a square of 4Rsidelength, i.e.,A S = πR 2 → A D = (4R) 2 = (16/π) × A S ,will change the required intensity for levitationI S → I D = 3 × (2) −1 × (16/π) −1 × I S = (3π/32)I Ssince the levitation intensity I is proportional to m, but inversely proportional to α and A.6

Physics 1112Spring 2009University of GeorgiaInstructor: HBSchüttlerCP 4.05+4.06:OllfTu.O 0..,.1. gI, _---..Side. 'VJ'ofl'e'1/-eol---r----r-:-~-r-:--......;--=---~-~"O""'lj ~t {V1.f e~s( te'~I - 02.1------_.........:.-­I~ -:::::: 0--Clv-e ~'PIrnrr.-~~ _~1iD~Q~ FytJ £- Viri f1 +-5 :CoS( OiJ)=(f3')/J..CoS (bOO) = 2CP 4.05 (=Q1), Answer: (B)For unpolarized light of time-averaged (TA) intensity I 0 passing through polarization filter(PF) 1, the TA intensity after the filter is I 1 = I 0 〈cos 2 θ 1 〉 θ . Here, 〈...〉 θ means averaging overall possible values of the random angle between an incident waves packet’s electric field E ⃗ 0and transmission axis T 1 , denoted by θ 1 ≡ ̸ ( E ⃗ 0 , T 1 ). For uniformly random angle θ 1 (i.e.,any value of θ 1 between 0 o and 180 o is equally likely to be found among incident EM wavepackets), 〈cos 2 θ 1 〉 θ = 1/2 and hence I 1 = (1/2)I 0 .The wave packets passing through PF1 (i.e. between PF1 and PF2) are then polarized withelectric field vector E ⃗ 1 ‖ T 1 . But then the angle between E ⃗ 1 (incident upon PF2) and T 2 isθ 2 ≡ ̸ ( E ⃗ 1 , T 2 ) = 90 o , since T 2 ⊥ T 1 . Hence, by Malus’ Law, the TA intensity transmitted7

Physics 1112Spring 2009University of GeorgiaInstructor: HBSchüttlerthrough PF2 is I 2 = I 1 cos 2 θ 2 = I 1 cos 2 (90 o ) = 0.CP 4.06 (=Q2), Answer: (E)As in CP 4.05, the intensity I 1 after PF1 is still I 1 = (1/2)I 0 with E ⃗ 1 ‖ T 1 . However, theangle between E ⃗ 1 and T 2 is now θ 2 ≡ ̸ ( E ⃗ 1 , T 2 ) = 90 o − 30 o = 60 o , since T 2 is 30 o from thehorizontal, but E ⃗ 1 ‖ T 1 , i.e., E ⃗ 1 is vertical. So, I 2 = I 1 cos 2 θ 2 = I 1 cos 2 (60 o ) = I 1 × (1/2) 2 =(1/2)I 0 × (1/2) 2 = (1/8)I 0 ; and thus I 2 /I 0 = 1/8.Important to note here: It does not matter here if you add or subtract 180 o from the ⃗ E-T angle θ, or change the sign of θ, since cos 2 (θ) = cos 2 (−θ) = cos 2 (θ±180 o ) = cos 2 (180 o ±θ).8

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Physics 1112Spring 2009University of GeorgiaInstructor: HBSchüttlerof Φ m , with final values of flux Φ ′ m = B ′ A cos Θ ′ , magnetic field strength B ′ and angleΘ ′ ≡ ̸ ( A ⃗′ , B ⃗ ′ ), after changing the field/circuit configuration, as follows:(A) Φ ′ m = −BA < +BA = Φ m with Θ ′ = 180 o and B ′ = B;(B): Φ ′ m = 0 < BA = Φ m with Θ ′ = 90 o and B ′ = B;(C): Φ ′ m = B ′ A < BA = Φ m with Θ ′ = 0 o and B ′ < B;(D): Φ ′ m = BA = Φ m with Θ ′ = 0 o = Θ and B ′ = B;(E): Φ ′ m = B ′ A > BA = Φ m with Θ ′ = 0 o and B ′ > B;(F): Φ ′ m = 0 < BA = Φ m with Θ ′ = 90 o and B ′ = B.Therefore, the time-dependent change of flux ∆Φ m ≡ Φ ′ m − Φ m and the resulting inducedEMF E = −∆Φ m /∆t (with ∆t > 0 here) have the following signs:∆Φ m < 0 and hence E > 0 for processes (A), (B), (C) and (F). By the sign convention forFaraday’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 inI-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 positivecharge on the upper capacitor plate (and an equal amount of negative charge on the lowercapacitor plate) in processes (A), (B), (C) and (F).∆Φ m > 0 and hence E < 0 for process (E). By the sign convention for Faraday’s law, anegative EMF E < 0 drives the current I in the loop against RH direction around ⃗ AHence, current I flows counter-clockwise around the loop and therefore deposits negativecharge on the upper capacitor plate (and an equal amount of positive charge on the lowercapacitor plate) in process (E).∆Φ m = 0 and hence E = 0 for process (D). Hence, there is no induced EMF and no inducedcurrent 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 HomeworkProblem P08.01 for details.10