Phys 11120, page 1

Final Exam solutions - Physics 1120 - Fall, 2007 1. A balloon of mass M has been positively charged ( its charge is +Q) and is held up to the ceiling. The ceiling polarizes, so the balloon sticks to it. The friction coefficient between the balloon and the ceiling is µ, and we will assume that the electric field, E, created by the ceiling is uniform and directed upwards. What is the (minimum) magnitude E of this electric field? A) E = µQ/Mg B) E = Mg/µQ C) E = µMg/Q D) E = Q/Mg E) E = Mg/Q The upward force would be F=QE, the downward force is just Mg. Those are the only forces acting in the vertical direction. (Friction is horizontal, and thus totally irrelevant for this question!) QE = Mg means E = Mg/Q. The next two questions refer to this situation: Two positively charged particles, labeled 1 and 2, are placed a distance R apart in empty space and are released from rest. Particle 1 has mass m and charge +Q; particle 2 has mass 2m and charge +Q/2. Each particle feels only the static Coulomb force due to the other particle. (There is no gravity, no friction, nor any other forces in this problem.) 2. How do the magnitudes of the initial accelerations of the two particles compare? [a1 = magnitude of initial acceleration of particle 1, a2 = magnitude of initial acceleration of particle 2] A) a1 = 2 a2 B) a1 = (1/2)a2 C) a1 = 4 a2 D) a1 = a2 E) None of these. The forces on each must be equal (and opposite) by Newton III. But F=ma, so the smaller mass has the larger acceleration, by the ratio of masses. 3. As the particles continue to move apart after their release, the speed of each particle... A) decreases as time goes by B) increases as time goes by C) stays the same. The force is repulsive, and although it gets smaller and smaller, it is still always repulsive. A positive force means a positive acceleration, forever. Acceleration means increasing speed in this case. SPEED just keeps on getting bigger and bigger!

Phys 11120, page 2

The following two questions refer to this situation: Two negative charges are each located a distance r from the origin, as shown. Note that the upper charge is twice as strong as the one to the right. (-2Q compared to -Q) 4. At the origin, the direction of electric field is ...

A) Up and right at exactly a 45 degree angle to the +x axis. B) Down and left at exactly a 45 degree angle to the -x axis. C) Straight up the page D) The field has no direction, because the field is zero E) None of these: the field points in some other direction than the choices given above.

There are two Field vectors at the origin , one pointing straight right, one pointing straight up with twice the magnitude. The sum of those two arrows is indeed "up and right", but it is NOT at a 45 degree angle! (The two forces would have to be equal in magnitude to get that angle) 5. (Assume, as usual, that Voltage is zero at infinity.) At the origin, given the charge configuration above, the VOLTAGE is .... A) +kQ/r B) -kQ/r C) -2kQ/r D) -3kQ/r E) None of these choices is correct Voltage is just the sum of kQ/r, here that would give -kQ/r-2kQ/r. 6. A small plastic bead has a +q charge on its bottom side and a –q charge on its top side (it’s a permanent electric dipole). The bead is placed in an electric field represented by the field line diagram shown. At the instant shown in the diagram, what is the direction of the net force on the bead due to this Electric field? A) the net force is zero B) right → C) left ← D) down ↓ E) up ↑ We've seen this problem on an earlier exam! The force on the + charge is up (in the direction of the field lines) and STRONG because the field lines are more concentrated there. The force on the - side is down but weaker, so the upward force wins. The net force is up. This is an unstable situation (the bead, if "jiggled" by any OTHER forces, would sooner or later flip around, and only THEN would the force be downward. But at the instant shown, it's unambiguously up)

Phys 11120, page 3

7. Point P is located exactly midway between two charged objects, each of which has identical charge +Q. The object on the left is a thin insulating rod, oriented vertically. This rod has its total charge (+Q) spread out uniformly along it. The object on the right is a simple pointlike charge, +Q. Point P is on the centerline, as shown. What is the direction of the electric field at point P, midway between the rod and the point? A) No direction, because the E-field is zero at point P B) to the right → C) to the left ← D) None of the above: the correct direction is not given. Coulomb's law for a point particle gives E = kQ/d^2 away from the +Q, but the rod on the left is not a point! You must think about "superposition", all the pieces of it contribute rightward E fields which are SMALLER due to i) the pieces being farther away than d, and ii) the vertical components of the E fields will cancel. Thus, the rightward E field from the rod is much smaller than the leftward field from the pt, the result is to the left. 8. You have two objects (labeled #1 and #2), both with charges -Q, fixed on the x-axis at x=-d and x=+d respectively. (Define voltage to be zero off at infinity) There are no other charges anywhere. At the origin, halfway between the charges, what can you say about the magnitude of the electric field, and the voltage?

A) |E| = 0, V = 0 B) |E| is not zero, V = 0 C) |E| = 0, V is not 0 D) |E| is not zero, V is not 0 E) Not enough information to decide. The E fields cancel, but the voltages add up to a negative result. 9. An ohm-meter measures the resistance between points a and b in the arrangement of 4 resistors shown. All four resistors are identical, with R=6.0 Ω. What is the measured resistance between a and b? A) The meter would read infinity, because this is not a complete circuit. B) The meter would read 0 Ω, this is not a complete circuit. C) 24. Ω D) 8.0 Ω E) 6.5 Ω The three R's in parallel add up to R/3, and then you have one more R in series, thus 4R/3 = 8 Ohm

Phys 11120, page 4

The next TWO problems refer to the circuit at right. There are two ideal batteries (voltage V each), and three identical ideal bulbs. Initially the capacitor is uncharged, and the switch is open. 10. Immediately after the switch is closed, the absolute value of the voltage difference across bulb #1 is equal to: A) 2V B) 4V/3 C) V D) 2V/3 E) zero Just after the switch closes, the capacitor acts like a "short circuit", so the total circuit has two batteries in series (V+V) and two bulbs in series. The voltage difference across bulbs 1 and 2 together is thus 2V, but across just ONE of the two bulbs is half that, V. 11. A very long time after the switch is closed, the absolute value of the potential difference across the capacitor is equal to A) 2V B) 4V/3 C) V D) 2V/3 E) zero Now no current can flow through the capacitor (after a long time, it's charged up!) so you have 3 identical bulbs. Total voltage is V+V = 2V, so it must be 2V/3 across each individual bulb. This is the voltage across each of them, in particular, it's the voltage across the 3rd one (which is the same as the voltage across the capacitor)

12. An ideal battery is attached to three identical ideal bulbs, as shown in the circuit to the right. A switch in parallel with bulb 3 is originally open and is then closed. When the switch is closed, what happens to the brightness of bulbs 1 and 2? A) #1 brightens, #2 brightens B) #1 brightens, #2 stays same C) #1 stays same, #2 stays same D) #1 stays same, #2 brightens E) None of these is correct When you close the switch, the resistance of the circuit goes down, so more current comes out of the battery. But wait, bulb 1 is directly across the battery, the current through it can NOT CHANGE (because it always has "V" volts across it!) So that extra current must go through bulb 2. So 1 stays the same, 2 brightens.

Phys 11120, page 5

13. This circuits shown on the right has five identical ideal light bulbs (labeled 1-5). Rank the 5 bulbs from brightest to dimmest (hint: look carefully at the figure!) (A) 1=5>2>3=4 (B) 1>2>3=4>5 (C) 1>2>5>3=4 (D) 1>2=3=4>5 (E) None of these is correct! All the current goes through 1, it's definitely the brightest. 3=4 (they're in parallel, same ΔV across each means same brightness! ). 5 gets MORE current than 3 or 4 individually, because it gets the SUM of their currents. So 1> 5>3=4. We just need to think about #2, where does it fit in? Let's see: ΔV(2)= the SUM of the ΔV(3) and ΔV(5). But a sum is greater than the parts, so ΔV(2)>ΔV(5). That means 2 is brighter than 5! (But it still has to be dimmer than 1, which had ALL the current). So I've got 1>2>5>3=4. 14. Which schematic diagram best represents the realistic circuit shown on the right The two bulbs on the right are in series with each other. But the two of them are in PARALLEL with the single one on the left. That's exactly circuit A!

E) None of these four schematic circuit choices is equivalent to the "realistic" one!

A) B)

C)

D)

Phys 11120, page 6

15. The next two questions refer to this situation: A cubical box of edge length L=2 meters is placed so that its edges are parallel to the coordinate axes, as shown. The space in and around the box is filled with an electric field given by

!

r E = Ai + Bk where A = 1 N/C (newtons per coulomb), B= 3 N/C. (i and k are usual unit vectors, pointing in x and z directions respectively) In SI units, what is the magnitude of the electric flux through the top shaded surface (the surface at z = +2 m)? A) zero B) 3 C) 6 D) 12 E) None of these Flux is E dot Area. Here, the area of the top is 2m*2m (it's a cube!) We only need the component of E which is parallel to the area vector, i.e. the "z component", or B. Thus the answer is B*2*2 = 3 N/C * 4 m^2 = 12 N m^2/C. 16. The total electric charge enclosed by the cubical box is… A) zero B) positive C) negative D) Not enough info to answer the question! This E field is a CONSTANT. It's completely uniform, whatever field lines go in must go out. So the total flux through the closed box is zero, and by Gauss' law, that means the enclosed charge is also zero. 17. In case (I) a positive charge +Q is surrounded by a spherical Gaussian surface of diameter R. In case (II), two charges, each +Q, are enclosed by a spherical Gaussian surface of diameter 2R. How does the electric flux through the surface in case (II) compare with the electric flux through the surface in case (I)?

ΦΕ(through surface II)/ΦΕ( through surface I) = ... A) 1 B) 2 C) 4 D) 8 E) 16 Gauss' law says electric flux is just Q(enclosed)/epsilon0. The size of the sphere doesn't matter...

(I) (II)

+Q +Q

+Q

2R

R

Phys 11120, page 7

18. A long straight wire, carrying a current I to the right, is located below a stationary rectangular conducting loop. The straight wire and loop are in the same plane, and the entire straight wire is moving downward, away from the loop, with speed v, as shown.

The induced current in the stationary rectangular loop is

A) zero

B) clockwise

C) counter-clockwise Lenz' law says "fight the change". The wire makes flux OUT of the page up where the loop is (by the right hand rule), and this is decreasing. The loop will generate its own flux to try to "keep" that flux, that's counterclockwise! 19. The same current I is flowing through two long (ideal) solenoids, labeled A and B, which both have circular cross section. Solenoid B has twice the diameter of A, half the length of A, and half as many turns (coils) as A. What is the ratio of the total magnetic energy contained in solenoid B to that in solenoid A, that is, what is UB/UA? (Hint: for a solenoid B = µo n I ) A) 1 (total stored magnetic energy is same in each) B) 2 C) 4 D) 8 E) None of these is correct We learned in class that total stored energy is proportional to B^2*volume. Here, we have half as many turns but half the length, so "n" is the SAME for both! So the B field is the same. Thus, the answer is merely the ratio of volumes of the two solenoids. That would be area*length, which is 4 times bigger area for B, but half the length, that makes a total of just 2 times the volume. 20. An electron and a proton, both with the same initial velocity

ov

uur,

enter a region with a uniform magnetic field B into the page, as shown. Each one undergoes semi-circular motion in the field and exits the field some distance d from the entry point. (The diagram shows the path for just one of the two particles.) Consider the following statements, and decide if they are T or F. i) The particle shown in the picture must be the negative one (the electron) ii) The distance "d" for the proton will be greater than "d" for the electron. A) Both i and ii are true B) i is true, but ii is false C) i is false, but ii is true D) Both i and ii are false The right hand rule says that a positive particle moving right into a B-field INTO the page (as shown) would curve the other way. So this must be the negative one. We learned that R = Mv/qB, and since a proton has much more mass, all other quantities being the same, the proton would have the larger radius.

I

I

I A

B

d

Phys 11120, page 8

21. Three ideal polaroid filters, labeled 1, 2, and 3, are placed between a source of unpolarized light and an observer. As shown, the pass axes of the three filters are at angles of 0o, 60o, and 90o relative to the vertical. The middle filter (2) is then removed. What does the observer see?

A) No light when filter 2 is present and some light when filter 2 is removed. B) Some light when filter 2 is present and no light when filter 2 is removed. C) No light when filter 2 is present and still no light when filter 2 is removed. D) Some light when filter 2 is present and more light when filter 2 is removed. E) Some light when 2 is present and the same amount of light when filter 2 is removed. Without #2, we have crossed polarizers, and see nothing. WITH #2, we do see some light. 22. A very large (effectively infinite) plastic slab of thickness T has a uniform positive charge density ρ (charge per volume, C/m3). A student wishes to compute the magnitude E of the electric field at a distance d (d > T/2) from the center of the slab. The student writes down Gauss’s Law

!

E•dA"" =Q(enc) /#0 , and sketches the centered,

cylindrical Gaussian surface shown (dashed). The cylinder has length 2d and end caps each of area A=πr2. What is the correct expression for the charge Qenclosed ?

Qenclosed =... A) A T ρ B) 2A ρ C) 2A d ρ D) 2A T ρ E) A ρ . Charge enclosed is just ρ times volume. (Charge/volume times volume!). But we have to be careful, the correct volume is just the volume where there IS charge, which is only inside the slab. From the picture, that's a volume of area A and width T, thus AT is the volume. So the answer is A T ρ

observer light source (unpolarized) polarizer 1 3 2

side view

r

2d

ρ

T

perspective view

area A

Phys 11120, page 9

23. The equipotential surfaces in a region of space are indicated in the diagram below. A proton is released from rest at point X. What does the proton do? (neglect gravity) A) accelerates to the right. B) accelerates to the left C) remains at rest. D) it drifts up (or down) along the line of constant voltage it starts on E) it accelerates down along the line of constant voltage it starts on.

positive charges move from high to low voltage, this particle would accelerate left. 24. A conducting wire loop with a light bulb (forming a simple closed circuit) is near a long, straight wire carrying a large current which is decreasing at a constant, steady rate (i.e. dI/dt = -constant), as shown. The loop and the straight wire are in the same plane. What can you say about the direction of current in the loop, and the brightness of the bulb (which is a visible measure of the current around the loop)? A) Current is clockwise , brightness is decreasing with time B) Current is clockwise , brightness is steady C) Current is counter-clockwise , and brightness is decreasing with time D) Current is counter-clockwise , and brightness is steady E) There is no current flowing anywhere through the loop, the bulb is dark. This is lenz law for direction, Faraday for time dependence. Lenz' law says "fight the change", if B is in but decreasing, you will make a clockwise current to "keep the flux". Faraday says EMF = d(phi)/dt, and phi is proportional to B which is proportional to I. So if dI/dt is constant, then dB/dt is constant, so EMF is constant. That's a CONSTANT, a steady brightness. 25. A long straight wire, carrying a current I, passes through the center of a circular wire loop. The wire loop is in the plane of the page, and the straight wire is perpendicular to the page. The current is going into the page and is increasing. The current induced in the wire loop is .. A) clockwise B) counter-clockwise C) zero. This current makes a B field which circles around it, but none of that "pokes" through this loop - there is NO flux, and thus no change, no current induced.

-10V 0V 10V 20V 30V 40V

X

I increasing

Wire loop

Phys 11120, page 10

26. A 1200W hairdryer (this refers to the average power of the hairdryer) is designed to be plugged into a standard 120V electrical outlet. What is the maximum number of such hair dryers that can be plugged in and turned on without tripping the breaker in a circuit that has a 15 A breaker switch? Hint: A 15A breaker switch will interrupt the current if the total rms current exceeds 15A. A) 80 B) 10 C) 4 D) 2 E) 1 P(ave) = V(rms)*I(rms). Here, I(rms) = P(ave)/V(rms) = 1200/120 = 10 A. That means only ONE hairdryer can be on, a second one makes 20 A which trips the circuit. The next two problems refer to this situation: a transformer at a power station is designed to step up the voltage from 120 V (RMS, AC) to 1200 V (RMS, AC). The voltage oscillates with a frequency of 60 Hz. 27. If the primary (input) side is a coil with 100 turns, how many turns should the secondary (output) side have? A) 10 turns B) 20 turns C) 1000 turns D) 1200 turns E) None of these is correct! It's step up, so we need MORE turns, in the ratio V(out)/V(in) = 1200/120 = 10:1 So we need 10*100 = 1000 turns. 28. If the station delivers an average power of 120 MW (that's MegaWatts), what is the maximum instantaneous current flowing out of the secondary side? (Answer to two place precision) A) 120 kA B) 71 kA C) 100 kA D) 140 kA E) None of these is correct! Power is V*I. On the secondary side, V(rms) = 1200 V, so we have I(rms, secondary) = P(ave)/V(rms) = 120E6/1200 = 1E5 Amps. But that's the rms current, not the maximum. Maximum is Sqrt[2]*rms, or 1.4E5 = 140 kA.

29. If you halve the period, T of a traveling electromagnetic wave in vacuum, what happens to the wavelength of that wave? A) wavelength is the same, it is independent of period. B) wavelength increases by a factor of 2. C) wavelength increases by a factor of 4 D) wavelength decreases by a factor of 2 E) wavelength decreases by a factor of 4. We know lambda * frequency = c = constant, and frequency = 1/T. So lambda/T = constant. If T is halved, then lambda is halved too, to keep their ratio the same.

Phys 11120, page 11

30. A coil of wire with N=100 turns and area A = 0.10 m2 is oriented so that its plane is perpendicular to a uniform magnetic field B which is increasing at a rate of 0.010 T/s. The coil is connected to a resistor R = 10Ω. What is the power dissipated in the resistor at the moment when B = 0.1 T? A) 0.1 W B) 10–2 W C)10–3 W D)10–4 W E) None of these. Faraday says EMF = -N d(B*A)/dt. Here, N=100, A = .1, and dB/dt = .01. So EMF = 100*.1*.01 = 0.1 V. But power is V^2/R = (.1V)^2/(10 Ohm) = 1E-3 W. 31. A "coaxial cable" consists of a long inner solid wire with radius R, and an outer (hollow) cylindrical wire with inner radius 2R and outer radius 3R . The inner wire carries total current I into the page and the outer (hollow cylindrical) wire carries the same magnitude current I out of the page. Both wires have uniform current density. What is the magnitude of the magnetic field B at the surface of the outer wire, that is, at r = 3R? A) µ0I/2πR B) µ0I/ 6πR C) µ0I/πR D) zero E) None of these. Ampere's law says the integral of B around a circle is equal to the current "poking through" that circle. Here, there is I(in) and I(out), they go in opposite directions, but have equal magnitude, so the total current "poking through" the page here is ZERO. There is thus ZERO integral of B around that outer wire, and by symmetry, B=0 everywhere. 32. An electron (charge -e) is released from rest in a region where there is a uniform E-field and a uniform B-field, both are pointing along the +x-direction; that is, E=Exi , B = Bx i. What is the path of the electron after its release? A) Left and curving upwards B) Left and curving downwards C) straight left ←, along the –x direction, forever . D) straight right →, along the +x direction, forever. E) Some other motion. The E field will accelerate the electron (with negative charge) to the LEFT. This is "antiparallel" to the B field, which thus has no effect. It just goes left, forever!

Phys 11120, page 12

33. An end-on view of a solenoid is shown to the right. There is a decreasing clockwise current around the solenoid, causing a decreasing uniform B-field in its interior. What is the direction of the electric field at the point x inside the solenoid and directly below the solenoid’s central axis as shown in the diagram? (A) left ← (B) right → (C) up ↑ (D) down ↓ (E) zero This is Faraday's law. E field lines run in circular rings around the center of the solenoid, with a direction given by Lenz' law (fight the change. I is decreasing, so the E field must point clockwise, which is LEFT at the point shown) 34. A small elliptical shaped wire loop is carrying a current I which is increasing in time. Above it and below it are two small circular conducting loops, labeled 1 and 2, as shown. All three loops are FLAT and lie in the plane of the page. The currents induced in loops 1 and loop 2 are A) Both clockwise B) Both counterclockwise C) Both zero D) Clockwise in loop 1, but counterclockwise in loop 2 E) Counterclockwise in loop 1, but clockwise in loop 2. The increasing current makes a field which is OUT of the page inside the ellipse, but "returns" (into the page) at both loop locations. So you have a field into the page, and increasing, and both loops fight that. So both are CCW. 35. Which one of the follow statements is correct (only one is correct!)? A) The magnetic force on a charged particle may change its kinetic energy B) A test charge released from rest will always initially move along a B-field line. C) Where the B-field lines are most dense, the magnitude of the magnetic field is largest D) The net magnetic flux through a closed surface may be non-zero E) B-field lines may cross The others are all false!

Phys 11120, page 13

+Q –Q

+Q

–Q

1 2

1 2

Before:

After:

36. Two charges, labeled 1 and 2, with charges +Q and –Q, are a fixed distance apart. A metal cube, with no net charge on it, is placed between the charges, as shown in the diagram. What happens to the magnitude of the net force on charge 1 when the cube is placed between the charges. A) net force on 1 is unchanged B) net force on 1 decreases to zero C) net force on 1 decreases (but remains non-zero) D) net force on 1 increases E) we cannot decide if we do not know the numerical magnitude of the charge Q. You will polarize that metal cube, making some - charge on the left side, and some + on the right. So there is an ADDITIONAL force on +Q: a RIGHTWARDS force attracting it to the nearby negative side, and a much smaller leftwards force repelling it from the back side. And then there's still the same old rightwards force from -Q that never goes away! So you have the old right force, plus a new BIG right force (and a small left force), the result must be an even bigger right forced than you started with. The next three questions refer to this situation. A circuit consists of a battery, two resistors, an inductor and a switch as shown. Initially the switch is open and has been open for a long time 37. The switch is closed. Immediately after the switch is closed, what is the magnitude of the current that flows through the battery?

A) 0 A B) 0.5 A C) 1. A D) 2. A E) None of these is correct! Right after the switch is closed, the current through the inductor cannot change right away, so it stays zero. Thus, the circuit is just "battery - R2 - back to battery", and thus the current through the battery is 10 V / 10 Ohm = 1 Amp. 38. After a very long time, what is the magnitude of the current through the battery? A) 0 A B) 0.5 A C) 1. A D) 2. A E) None of these is correct! After a long time, the inductor looks like an ideal wire. So the circuit is just two parallel resistors, and R(total) = 5 Ohms, so 2 Amps flows out of the battery. (1 Amp = 10V/10 Ohms through EACH of the two parallel resistors) 39. After being left closed for a very long time, the switch is opened again. Immediately after the switch is re-opened, describe the current flowing the rightmost resistor, labeled R2. A) 1 A, flowing up ↑ B) 1 A, flowing down ↓ C) 2 A, flowing up ↑ D) 2 A, flowing down ↓ E) 0, no current flows through R2 at that instant. The current was 1 Amp flowing DOWN through the inductor in the previous problem - so that's what it still must be immediately after opening the switch. (You cannot instantly change I through an inductor). Where can this current go? It has no choice, it must go up through R2...

Phys 11120, page 14

40. A long solenoid with many turns per length has a uniform magnetic field B within its interior. Consider the imaginary rectangular path of length "c" and width "a" with the bottom edge entirely within the solenoid, as shown. What is the integral of

!

r B • d

r L " around this

rectangular path in the counterclockwise direction? A) 0 B) 2 B c C) 2 B (c+a) D) B a c E) B c This fancy integral just means "add up B dot dL around the loop". On the bottom leg, of length c, you are INSIDE the solenoid, B is parallel to dL, and you get Bc. On the right and left legs, B is perpendicular to dL, you get nothing due to the dot product. Outside, B=0, again nothing... 41. A coil of wire carrying current I can rotate freely about an axis in a uniform magnetic field. The coil is in the plane of the page, and the magnetic field is directed from left to right as shown. If released from rest in the position shown, which way does it rotate? (A) the right side will rotate out of the page (B) the left side will rotate out of the page. (C) the loop will not rotate at all, but it will feel a net force to the right D) the loop will not rotate at all, and it will feel zero net force E) the loop will not rotate at all, and it will feel a net force up the page. The right hand rule applied to each leg in turn gives me that the force on the LEFT current is out of the page, but on the right segment is into the page. That makes a torque, it rotates.

Phys 11120, page 15

42. An electromagnetic plane wave, given by

!

v E (x, y,z) = Eo

ˆ z sin(kx +"t) , propagates to the left.

The figure above represents this wave at a particular instant in time. Four points in space are labeled I,J,K, and L. All four points lie in the x-y plane and have the same x-coordinate. (Note the orientation of the E- and B-fields: the E- field oscillates in the x-z plane; the B-field oscillates in the x-y plane ) For the instant shown, rank the magnitudes of the electric field at the four points, from largest to smallest.

A) EI > EJ > EK > EL

B) EJ > EI > EK > EL

C) EJ > EI = EK > EL

D) EL > EK > EJ > EI

E) EI = EJ = EK = EL

This is a plane wave, the formula has no dependence on y.

E is the SAME everywhere as you move up or down along the y axis, that's why it's a

"plane" wave (it's the same everywhere in any y-z plane)

Have a safe and excellent winter break. Both of us (Prof's Pollock and Gurarie) had a great time teaching

you - we hope to see some of you in upcoming physics classes. Take a look around after this exam and

think about all the places you see Maxwell's equations in action! (Maybe a harder thing to think about

would be - where DON'T you?!)