Physics 201Final ExamSolutionsI. i) The condition at equilibrium isXF =Qq4π�01y2− mg =0 =⇒ y0=rQq4π�0mg.ii) To find the spring constant due to the electric force, we write y = y0+(y −y0)=y0+∆y where∆ yis the displacement from equilibrium and note that if∆ y is small compared to y01y2=1(y0+∆y)2=1y201(1 +∆ y/y0)2≈1y20„1 −2y0∆y«=1y20−2y30∆y.The constant term cancels the mg from gravity by part (i), so the force on the charge isF = −2Qq4π�0y01y20∆y = −2Qq4π�0y04π�0mgQq∆y = −2mgy0∆y,which means our spring constant k is 2mg/y0. Thenω =rkm=r2gy0.I I. i) Since the electric field in the interior of a conductor is zero, a Gaussian surface drawn in betweenthe two surfaces must enclose zero charge. So the charge on the inner surface is −q. Meanwhile, thesphere was initially uncharged, so the total charge of the shell must be zero. To balance the charge−q on the inner surface, the charge on the outer surface must be q, because no charge resides in theinterior of the shell.ii) Outside the sphere where r>bthe whole thing looks like a point charge q, so the potential is simplyV =q4π�01r,r>b.The whole conducting shell is at a constant potential, soV =q4π�01b,a<r<b.Finally, inside the inner surface of the sphere the potential from the two surfaces adds to the potentialdue to the point charge to giveV =q4π�0„1r+1b−1a«, 0 <r<a.I II. i) By Amp`ere’s law Ienc= 0 implies B =0.ii) Using a circular loop of radius r in Amp`ere’s law gives2πrB = µ0Iπ(b2− a2)π(r2− a2)=µ0Ir2− a2b2− a2=⇒ B =µ0I2πrr2− a2b2− a2.The direction of the magnetic field is always clockwise.iii) We can use Amp`ere’s law again with Ienc= I, or just note that the magnetic field is continuous atthe boundary where b = r so thatB =µ0I2πr.The direction is again clockwise.1IV. a) As the emf increases so does the current, and with increasing current the upward force on the lo opdue to the magnetic field will increase until it counteracts the fall.b) The change in flux is dΦ=Bwv dt, so the induced current isI =ER=BwvR.It’s clear that this current flows counterclockwise in the loop, which means the force due to themagnetic field points upward (the sides do not contribute to this upward force.) The terminal speedis reached when the total force is zero, i.e.,IwB =B2w2vR= mg =⇒ vterminal=mgRB2w2.c) The loop current flows counterclockwise to counteract the increasing flux, by Lenz’s law.V. We will treat the inductor like a solenoid, i.e., the magnetic field points straight upwards. Then if we drawa rectangular Amperian loop through the length l of the inductor and assume that it has N turns,Bl = µ0NI =⇒ B =µ0NIl.The total flux is thenΦ=µ0NIl· Nπ„d2«2=πµ0d2N24lI,and we can read of the inductance asL =ΦI=πµ0d2N24l=⇒ N =s4lLπµ0d2=s4(0.15 m)(5.8 × 10−3H)π(4π · 10−7Wb/A · m)(0.022 m)2=1350.VI. i) At resonance the total impedance of the circuit is Z = R, so the maximum current is simply I = V/R.The capacitor then feels the voltageVC=Iω0C=Vω0RC≤ Vmax=⇒ R ≥1ω0CVVmax.But the resonance frequency isω0=1√LC,so that the value of the resistance must satisfyR ≥VVmaxrLC=32 V400 Vr1.5H250 × 10−6F=6.2Ω.ii) The total impedance at frequencyω=2ω0=2/√LCisZ = R + i„ωL −1ωC«= R + i32rLC=⇒|Z| =rR2+9L4C,so thatI0=V|Z|=VqR2+9L4C=32 Vq(6.2Ω)2+9(1.5H)4(250×10−6F)=0.28 A.VI I. We have1o+1i=1f,o+ i = d,so that1o+1d − o=1f=⇒ o2− do + fd =0.Solving for o giveso =d ±pd2− 4fd2=d2 1 ±r1 −4fd!=70 cm2 1 ±r1 −4(17 cm)70 cm!=41cm, 29 cm.2VI II. This is the same idea behind diffraction from a double slit. The distance from P to the two speakers issD2+„d2± x«2= Ds1+„x ± d/2D«2≈ D"1+12„x ± d/2D«2#= D +(x ± d/2)22D,and the difference is dx/D, which we want to equal to λ/2 for destructive interference (and hence, silence).But λ = c/f where c is the speed of sound, sodxD=c2f=⇒ x =cD2fd=(300 m/s)(10 m)2(3000 Hz)(1 m)=0.5m.IX. i)Z1/20N2sin2πx dx =N22Z1/20(1 − cos 2πx) dx =N24−N24πsin 2πx˛˛˛˛1/20=N24=1 =⇒ N =2.ii) Note that ψ(x)isnot a state of definite energy, and moreover that it is difficult to obtain the answerby inspection.A2=Z1/20√2sin2πx · 2sinπx dx =√2Z1/202sin2πx sin πx dx =√2Z1/20(cos πx − cos 3πx) dx=√2π»sin πx −13sin 3πx–1/20=4√23π,where we have used the identity2sinx sin y =cos(x − y) − cos(x + y).ThereforeP (E2)=|A2|2=329π2.iii)ψ(x)=√2sin2πx.The function sin22πx is symmetric about x =1/2, and each part in turn is symmetric about x =1/4and x =3/4:0 1/4 1/2 3/4 101ThereforeP„14<x<34«=12,by inspecting the wavefunction.3iv) To absorb light from the n = 2 state, it must make a transition to a higher state. The next state isthe n = 3 state, and the difference in energy isE = �ω = E3− E2=9π2�22m−4π2�22m=5π2�22m=⇒ ω =5π2�2m,which impliesf =ω2π=5π�4m.To emit light, there is only one possible transition, to the ground state n = 1. ThenE = �ω = E2− E1=4π2�22m−π2�22m=3π2�22m=⇒ ω =3π2�2m,orf =3π�4m.X. i) By now we are experts at reading such a wavefunction. The possible momenta arep = ±4π�L,each with equal probability 1/2.ii) Both states correspond to the energyE =p22m=8π2�2mL2,and this energy occurs with certainty, i.e., probability 1.iii) Since this is a state of definite energy, we findψ(x, t)=3cos4πxLe−iEt/�,with E as given
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