1FINALS May 10, 2010 Physics 201 Shankar 180 MINS1. At t = 0 when the capacitor is uncharged the switch S in Fig.1 is closed. (i) What areI(t =0+) and I(t = ∞), the currents flowing out of the battery just after the switchis closed and at t = ∞? (ii) Describe qualitatively the currents in the R1and R2Cbranches as time goes by from 0 to ∞. (iii) What is the final charge on the capacitor?10, BOOK IR2VR1SCFigure 1: The switch is closed at t = 0, at which time the capacitor is uncharged.2. (i) Find the force exerted by the small loop on the infinite wire shown in Fig. 1. (ii)What is the torque on the loop about an axis that bisects the loop and is parallel tothe wire ? 15, BOOK IwIaLI12Figure 2: The rectangular loop carries current I2and the infinite wire carries current I1.avlIFigure 3: The rod moves parallel to the wire and perpendicular to its own length.3. In a moving conductor, under equilibrium there should be no net electromagnetic forceon the charges inside. A conducting rod of length l is oriented perpendicular to aninfinite wire carrying current I, with its nearer end a distance a from the wire (Fig.3). It begins to move with a velocity v parallel to the current. What happens to thecharges on this rod when it begins to move ? Which end will be positive? What willbe the equilibrium electric field the charges at the ends set up? Show that the EMFdue to this field is E =µ0Iv2πln(1 +la). 15, BOOK I24. Two infinitely long wires carrying λC/mlie along the x and y axes. Find the field Edue to each one and the total field at the point r = ix + jy. What is V (1, 1) −V (2, 2),the potential difference between the points (1, 1) and (2, 2) when λ =1C/m? 15,BOOK II5. In the circuit below (Fig. 4) the switch has been closed for a very long time and thevoltage across the capacitor is zero. (i) What is the current flowing out of the battery?The switch is opened at t = 0. (ii) Write the equation relating the current I flowingthrough the inductor and the charge Q on the capacitor by going around the loop thatincludes them. What does this tell you about dI/dt at t = 0? (iii) How does thecurrent vary as a function of time in the LC lo op? (iv) What is the maximum currentin the LC loop? 20, BOOK IISVRCLFigure 4:SSS12P1 mdL=Figure 5:6. Light of λ =250nm from a point source S reaches a double slit a distance L =1maway, with the upper slit S1in line with the source and the lower one S2a distanced below, as shown in Fig. 5. Find the smallest value of d that will ensure that thepoint P , located equidistant from the slits, is a minimum. Assume d<<L=1m. 10,BOOK II7. A converging lens of focal length f is placed between an object and a screen lo cateda distance d to the right of the object. Given that 4f<d, find u±, the two possiblelocations for the lens as measured from the object so that a sharp image is formed onthe screen. In each case tell me if the image is real or virtual, inverted or right side upand magnified or demagnified? 15, BOOK III38. (i) Explain why (with a picture or diagram)¯S, the time-averaged intensity of light,defined as the Watts that cross a unit area, is related to the time-averaged energydensity ¯u by¯S =¯uc. 5, BOOK III(ii) A laser with average power 5mW makes a spot 1mm in radius on a screen. Whatis¯S? (Keep just one significant digit for this and the rest of this problem.) If thelight is made of photons of wavelength λ =400nm, how many of them hit unit area inone second? If each one has its momentum reversed on hitting the screen, what is theaverage pressure on the screen? Work with symbols as far as possible. This may savesome time on last part (pressure) if you use previous formula for the part just beforeit. 15, BOOK III9. A particle is in the n = 3 energy state in a box of length L. What is the probabilityfor finding it in the left half of the box? What is the probability for being in the leftone-third of the box? 10, BOOK III10. A particle is in the ground state (lowest energy state) of a box that extends from x =0to x = L. Suddenly the box expands to double the size with the right end now atx =2L. Assume that during this instantaneous expansion the wave function of theparticle remains unchanged. What is the probability that it will be in the first excitedstate of the new box? (Draw some pictures and try to avoid hard work.) 15, BOOKIV11. There are two particles of mass m in a box of length L and their total energy isE =¯h2π2mL2. What individual energy states are the particles in? Can they be fermionsobeying the Pauli Exclusion Principle? 10, BOOK IV12. (i) Explain why ψ(x)=A cosπxLcannot describe a particle of mass m in a ring ofcircumference L. (ii) Argue that ψ(x)=A cos2πxLis however allowed. (iii) Whenmomentum is measured what are the possible outcomes and corresponding relativeand absolute probabilities? (iv) Rescale ψ(x) so that these probabilities add up to 1.(v) Taking this ψ(x) to be the initial state ψ(x, 0), find the state ψ(x, t) at later times.25, BOOK IV4Data SheetElectricity and MagnetismF = q(E + v × B)E =q4πε0r2er(eris unit vector in radial direction)dB =µ0I4πdl × rr3dF = Idl × B�E · dS =QεGauss�B · dS = 0 No monopoles�B · dl = µ0I static case, Ampere�B · dl = µ0I + µ0ε0∂ΦE∂tgeneral caseEM F = −dΦBdtLenz’s LawV (2) − V (1) = −�21E · dr static caseuE=ε0E22electric energy p er unit volumeuB=B22µ0magnetic energy per unit volumeE = cB for an EM wavec =1√µ0ε0Opticsn1sin θ1= n2sin θ21u+1v=1fd sin θ = mλ diffraction grating or double slit5QuantumE =¯hω = hf p =¯hk =hλphotoni¯h∂ψ(x, t)∂t= −¯h22m∂2ψ(x, t)∂x2+ V (x)ψ(x, t)−¯h22md2ψE(x)dx2+ V (x)ψE(x)=EψE(x)ψ(x)=�EψE(x)AEAE=�ψ∗E(x)ψ(x)dx P (E)=|AE|2ψp(x)=eipx/¯h√Lψ(x)=�pApψp(x) Ap=�ψ∗p(x)ψ(x)dx P (p)=|Ap|2p =2π¯hλψn(x)=�2LsinnπxLparticle in a box from x =0tox = Lψ(x, t)=�EψE(x)AE(0)e−iEt/¯hwhere AE(0) are coefficients at t =0CircuitsLdIdt+ RI +1C�I(t)dt = V (t)Ld2Qdt2+ RdQdt+1CQ = V (t)Z = R + i(ωL −1ωC)=|Z|eiφI0=V0ZAverage Power =V0I02cos φEnergy stored =12CV2=Q22CEnergy stored =12LI26Constants and math formulas(1 + x)n= 1 + nx + ... if |x| << 1c =3· 108m/s¯h =10−34J · sh=6.610−34J · se =1.610−19coulombs is the magnitude of charge of electron or proton.14πε0=9· 109Nm2/C2µo4π=10−7N/A2eiθ=cosθ + i sin θz = x + iy = |z|eiθ|z| =�x2+ y2=√zz∗z∗= x − iy = |z|e−iθsin2θ =1 − cos 2θ2cos2θ =1+cos2θ2Volume(sphere)
View Full Document
Unlocking...