Electricity and Magnetism II

Electricity and Magnetism II
7.1 Electricity and Magnetism II Griffiths Chapter 7 Maxwell’s Equations Clicker Questions

7.2 In the interior of a metal in static equilibrium the charge density ρ is: zero always never zero. sometimes zero, sometime non-zero, depending on the conditions. Class: CONCEPTUAL Correct Answer: A ______________________________ Physics 3320, Sp12 (MD) Lecture 5 [94], 0, 6 Spring 2012 Comments Good to address in anticipation of Ohm’s law – students often get confused about whether the field is ALWAYS zero inside a conductor. Q: Is “static” the same as “not moving”? Is “static” the same as “time-independent”? ========================== Written by: Mike Dubson, Sp12

7.3 Which of the following is a correct statement of charge conservation? Charge Conservation A) B) C) D) E) None of these, or more than one. Class: MATH/PHYSICS Correct Answer: E ___________________________ Phys 3320, Fa11 (SJP) Lecture #6 0, 36, 24, 8, [[32]] ________________________________________________ Fall 2011 Comments We needed to review this, apparently. Slightly tricky, answer is E: Small variant, this time it’s the integral form. And, BOTH B and C are correct!!Got a student to point out B=C is just the divergence theorem. (When I think about current conservation in integral form, I typically think of it as expressed in B, and that was the slightly more popular student answer too) CB Notes: Steve had been dissatisfied with student responses on the continuity equation on Wednesday, so offered a slightly altered version at the start of class. 55% (20% (B) + 35% (C)) chose a correct expression before discussion, but no students had selected the correct response (E) – that more than one was correct. Students near me were discussing in terms of unit analysis, but I had to prompt them to question whether more than one could be correct. Student was unsure, since Q(enclosed) made him “think about a volume”. After discussion [35% (B), 25% (C), 10% (D), 30% (E)] – problem seems to be in seeing the equivalence of the two expressions via the divergence theorem. ==================================== USED IN: Fall 2008 (Dubson), Spring 2008 (Pollock), Fall 2009 (Schibli) LECTURE NUMBER: Dubson (Week 10, Lecture 25), Pollock (30) STUDENT RESPONSES: 2% 7% 17% [[72%]] 2% (FALL 2008) 0% 0% 6% [[ 94%]] 0% (SPRING 2008) 0% 5% 3% [[92%]] 0% (FALL 2009) INSTRUCTOR NOTES: 94% correct. I asked it again on the *last* day of class, they did worse! (50% correct!). This was the first day back after spring break, wanted to see if they remembered what we had covered 10 days ago! -SJP WRITTEN BY: Steven Pollock (CU-Boulder) 3

7.4 For everyday currents in home electronics and wires, which answer is the order of magnitude of the instantaneous speed of the electrons in the wire? more than km/s m/s mm/s μm/s nm/s Class: CONCEPTUAL Correct Answer: A ________________________________ Phys 3320 Fa11 (SJP) Lecture #5 [[4]], 12, 38, 31, 15 (Original form, A = km/s, then m, mm, mum, nm) Physics 3320, Sp12 (MD) Lecture 8 [[10]], 0, 90, 0, 0 ________________________________________________ Fall 2011 Comments Now modified to be “more than km/s” The thermal velocity (given by Sqrt[ 2 kT/m]) is, very roughly, km/sec, using kT = 1/40 eV. But the fermi velocity in metals would correspond to an energy scale of eV’s, and thusv is an order of magnitude bigger at least. Spring 2012 Comments Students were thinking/talking about the drift velocity (it’s on the mm/s scale) since that was what had been emphasized just prior. ==================================== Written by ERK, 3 Dec 2008 HW/Reading reference

7.5 An electric current I flows along a copper wire (low resistivity) into a resistor made of carbon (high resistivity) then back into another copper wire. In which material is the electric field largest? I In the copper wire In the carbon resistor It’s the same in both copper and carbon It depends on the sizes of the copper and carbon Class: CONCEPTUAL Correct Answer: D _______________________________ Physics 3320, Fa11 (SJP) Lecture #6 29, 50, 4, [[17]] Physics 3320, Sp12 (MD) Lecture 5 First Click: , 64, 11, [14] Second Click: , 9, 0, [89] ________________________________________________ Fall 2011 Comments Good question (with a “trick” answer ). For ordinary situations, B will be correct. This is what you get if you focus on the E=J/sigma formula, sigma of carbon is much lower. But technically it must be D, because if the wire gets small enough, E = J/sigma also goes like I/area, and thus could get arbitrarily large, possibly even larger than in the Carbon (despite the much larger sigma) ( But in normal conditions, E is large in resistors and small in conductors in a given circuit) LA Notes: Students started by saying that current density was constant through the system. Paul suggested current was the constant quantity. When we discussed as a class there was a “Ohh, so it’s D” comment. Spring 2012 Comments Used as a warm up to “Current Density and Charge Conservation” tutorial (#01) “First Click” answer was at the beginning of the tutorial. “Second Click” came after doing the first page of the tutorial, and some discussion of the answer to next concept test. (7.6) ==================================== From ERK, 3 Dec 2008 Ohm’s law

7.6 A copper cylinder is machined to have the following shape. The ends are connected to a battery so that a current flows through the copper. A C B Rank order (from greatest to smallest, e.g. A=C>B) Magnitude of E field Conductivity Current Current Density Used in: Fall 2011, Spring 2012 ___________________________________ Fall 2011 Comments Printed this out, did it as an activity. Took a bit over 5 minutes, followed by ~5 minutes of class discussion. _________________________________ Spring 2012 Comments This activity is the first page of the “Current Density” tutorial (#01). Comments (SJP): Good to ask students “which did you answer first”, since some are easier. E.g. conductivity is a property of material (intensive), so A=B=C Current is conserved, so A=B=C also. Then J = I/A, if I is the same, then B>A>C Finally E = J/sigma, so same ordering as J, B>A>C This latter makes sense if you think of field lines “squeezing down” to fit through the bottleneck! Some possible discussion leading Q’s: Since I=J*A, does that mean the electric flux (E*A) in the circuit wire is the same in all three regions (i.e., is electric flux conserved")? Do the field lines look like streamlines of fluid flow, or do they have have to point directly left-right everywhere? Do J and E diverge in the regions where the area shrinks down, and what does that imply about charge buildup in those regions? What are the boundary conditions at the edge of the wire for the E-field? If you think of E field as analogous to “fluid flow”, and using J = n q v, would you argue that in the necked down region it is n increasing (density of charge carries), or v (or both?) Why? Is there an analogy to “pressure” in the electrical situation? _____________________________________ Conservation of total current seems to be the strongest intuition. Some students respond to fluid analogies, but feel that not all students are necessarily enlightened by these. Surprised to find that there are still some students who think batteries are pushing “extra charge” through the wire. ==================================== © University of Colorado, 2011

Must be exactly parallel to the wall
7.7 Inside this resistor setup, what can you conclude about the current density J near the side walls? V0 V=0 I I J?? Must be exactly parallel to the wall Must be exactly perpendicular to the wall Could have a mix of parallel and perp components No obvious way to decide!? Class: CONCEPTUAL Correct Answer: A ____________________________________ Phys 3320, Fa11 (SJP) Lecture #5 [[65], 15, 19,0 _______\_____________________________ Fall 2011 Comments Ended class with this, a little rushed. I talked them through it, one student explained that if it wasn’t [A] then charge density on surface would be changing with time. I agree – either that, or (if drho/dt=0 in steady state) then current would be “leaking” out into empty space, and you wouldn’t have I(in) = I(out) CB Notes: Think it would be useful to explicitly resolve this issue of instantaneous charge buildup versus d(rho)/dt being zero in steady-state. Spring 2012 Comments This question is the first page to the “Ohm’s Law” tutorial (#02). Could therefore be used in conjunction with the class activity, first introducing with this concept test, then having student complete the remaining tasks. ====================================

7.8 Inside this resistor setup, (real world, finite sizes!) what does the E field look like inside ? V0 V=0 I I E?? Must be uniform and horizontal Must have some nonuniformity, due to fringing effects! Class: CONCEPTUAL Correct Answer: A ___________________________________ Phys 3320, Fa11 (SJP) Lecture #7 [[52]], 48 Fall 2011 Comments An excellent split vote! This is a tough one – if there were no conductor, and you were building up E field in a realistic plate capacitor, there WOULD be fringe effects. But with the conductor there, J must be parallel to the edges, and thus E is too, and (as we then show in class), the uniqueness theorem will tell us that the field here is uniform E (it satisfies the boundary conditions on voltage, including dV/dn=0 on edges, so it must be the solution) The fact that the field is uniform is a bit surprising, it tells us that there MUST be some excess charges on the surface of the resistor (there’s nowhere else they can go!) that “shape” the E field this way. Those surface charges are real, and present in real circuits. Also, that has consequences for E outside, which we haven’t been talking about. E can’t be zero out there (E parallel is continuous!) and it can’t ALSO be parallel to the edge everywhere (because E_perp jumps where there are surface charges, and we just argued there MUST be at least some surface charges to shape the field inside!) LA Notes: One student wanted (the answer) to be A (constant, uniform E) because everything was nicer, but couldn’t defend it. After the discussion, another student indicated his dislike of the Uniqueness Theorem. Spring 2012 Comments Similar questions addressed in “Ohm’s Law” tutorial (#01), but with the case of a cone-shaped resistor instead of cylinder. ====================================

7.9 Recall the machined copper from last class, with steady current flowing left to right through it I I In the “necking down region” (somewhere in a small-ish region around the head of the arrow), do you think A) B) Class: CONCEPTUAL Correct Answer: A ______________________________ Fall 2011 Comments PowerPoint screw-up – I didn’t have this available during class, but we discussed it without clicking. It’s a good question Answer is A) Div.J= -drho/dt, and in steady state, this will be zero. So Div.E is also 0. There is no buildup of charge in the bulk in steady state. (There will be a small buildup on the walls, e.g. I expect + charges on the sharply tilted wall portion, to help “neck down” the E field) But, that’s not changing with time, and it’s on the surface, not in the bulk if we have uniform copper. Spring 2012 Comments Similar question addressed in “Ohm’s Law” tutorial (#01). ==================================== © University of Colorado, 2011

7.10 Recall the machined copper from last class, with steady current flowing left to right through it I I In steady state, do you expect there will be any surface charge accumulated anywhere on the walls of the conductor? A) Yes B) No Class: CONCEPTUAL Correct Answer: A _____________________________ Fall 2011 Comments PowerPoint screw-up – I didn’t have this available during class, but we discussed it without clicking. It’s a good question Answer is A, got to be! Otherwise, what would make the “shape” of the E field what it actually is? Spring 2012 Comments Surface charges not addressed in “Ohm’s Law” tutorial, but do come up in the “Energy Flow” tutorial (#08) in the context of boundary conditions and energy flowing in a simple circuit. ==================================== © University of Colorado, 2011

Amps/volt V/m m/ohm ohm*m Joules/m
7.11 The resistivity, ρ, is the inverse of the conductivity, σ. The units of ρ are: Amps/volt V/m m/ohm ohm*m Joules/m Class: CONCEPTUAL Correct Answer: D ________________________________ Fall 2011: Not Used Spring 2012: Not used ==================================== From Charles Rogers

7.12 A steady electric current I flows around a circuit containing a battery of voltage V and a resistor R. Which of the following statements about is true? It is zero around the circuit because it’s an electrostatic field It is non-zero around the circuit because it’s not an electrostatic field It is zero around the circuit because there is no electric field in the battery, only in the rest of the circuit It is non-zero around the circuit because there is no electric field in the battery, only in the rest of the circuit! None of the above Class: CONCEPTUAL Correct Answer: A ________________________________ Fall 2011 Comments Note: NOT given to class in Fa11; this is a modified version to improve upon the previous slide. Fa11 had a different answer for “C” – which was really just a rephrasing of “A” (it is zero around the circuit because the electric field is time-independent). (I’m not sure what to do about the distractors – worth thinking about.) Phys 3320, Fa11 (SJP) Lecture #7 [21], 29, 50, 0,0 – with original answer to “C”, as mentioned above. ==================================== Originally from ERK, 3 Dec 2008 Modified by SJP Fall 2011 EMF

7.13 EMF is the line integral of the total force per unit charge around a closed loop. The units of EMF are: Farads Joules. Amps, (that’s why current flows.) Newtons, (that’s why it’s called emf) Volts Class: CONCEPTUAL Correct Answer: E _____________________________ Physics 3320, Fa11 (SJP) Lecture 7 (start of class question) 7, 15, 4, 4, [[70]] Physics 3320, Sp12 (MD) Lecture 7 0, 9, 6, 0, [[86]] ________________________________________________ Fall 2011 Comments Simple review, some confusion about “f” meaning force/charge. LA Notes: The students knew that f had units of force per charge, so it was like E, they concluded the correct answer was volts. Spring 2012 Comments Some students just knew that EMF is like a voltage, didn’t have to think about actual units. ==================================== Written by: Charles Rogers

7.14 Imagine a charge q able to move around a tube which makes a closed loop. If we want to drive the charge around the loop, we cannot do this with E-field from a single stationary charge. + q Can we drive the charge around the loop with some combination of stationary + and – charges? Yes No Class: CONCEPTUAL Correct Answer: B ______________________________ Physics 3320, Sp12 (MD) Lecture 7 8, [[92]] Spring 2012 Comments Used as lead-in to next on EMF (7.15). Students had good explanations, no difficulties. ============================= Written by: Mike Dubson, Sp12 +

1 2 3 4 How many of the following statements are true?
7.15 A circuit with a battery with voltage difference ΔV is attached to a resistor. The force per charge due to the charges is E. The force per charge inside the battery is f = fbat + E How many of the following statements are true? 1 2 3 4 Class: MATH/PHYSICS Correct Answer: D ______________________________ Physics 3320, Sp12 (MD) Lecture 7 0, 24, 16, [[59]], 0 _____________________________ Spring 2012 Comments Lots of good discussion around this one (they really need to break definition of EMF down into separate contributions (inside/outside battery) to easily see which doesn’t work) – confusing to students about what’s going on inside the battery, where the electrostatic field points opposite the direction of current in the wire. “Battery force” (as Griffiths calls it) in terms of chemistry is challenging explanation for them. Important to distinguish between EMF and line integral of E around the loop. Q. What does a voltmeter measure? ============================= Written by Mike Dubson, Sp12 B A

C) ? I would need to do a nontrivial calculation to decide
7.16 Is there a nonzero EMF around the (dashed) closed loop, which is partway inserted between two charged isolated capacitor plates. EMF=0 here EMF≠0 here C) ? I would need to do a nontrivial calculation to decide Class: MATH/PHYSICS Correct Answer: A __________________________________ Phys 3320, Fa11 (SJP) Lecture #7 [[71]], 29 Physics 3320, Sp12 (MD) Lecture 8 18, [[79]], 3 Fall 2011 Comments Some discussion here about what is the nature of the forces in this problem, some students brought up “magnetic fields” (??) another other wondered about whether the capacitor might be discharging (despite my “isolated” in the question) I spent class time talking about DOING that line integral, first “casually” (with the ideal E field of a capacitor) and getting a NONzero answer (negative, if I integrate CW), and then sketching the field lines and realizing the fringe fields give a POSITIVE addition which cancels that negative to get 0. Answer: A) This is an electrostatic field, curl(E)=0, there can be no nonzero line integral of E dot dl here. (Which in fact shows there MUST be fringe fields outside a capacitor) Baily: Students at table nearest me recalled there would be a fringing field, but thought it would be weaker [wouldn’t be enough to cancel out the contribution from between the plates]. After listening to their discussion for a while, I asked if there were any forces at work other than the electrostatic field, and then asked about the closed-loop line integral of an electrostatic field. Student 1 admitted this was a strong argument for the EMF being zero, and commented that (now that he thought about it) it did seem “kind of strange” that the field in a capacitor could drive a current all by itself. Rehn: At first, the students’ inclination was that the EMF was 0 around the loop, but they had no reasoning to back up why. After that, they began to think that E dot dl was non-zero on the left vertical segment, but 0 on the right vertical segment. I think they eventually came to the conclusion that there was an EMF because they thought the E field was uniform in the capacitor and 0 outside. Ragole: One student argued EMF was non-zero because it wasn’t completely in the capacitor. I pushed him to follow that argument and he ended up saying it would mean E would have a curl, which didn’t make sense to him. ====================================

7.17 One end of rectangular metal loop enters a region of constant uniform magnetic field B, with initial constant speed v, as shown. What direction is the magnetic force on the loop? B v L Class: CONCEPTUAL Correct Answer: D _____________________________ Phys 3320, Fa11 (SJP) Lecture #7 11, 37, 5, [[47]] Physics 3320, Sp12 (MD) Lecture 7 0, 17, 11, [[47]], 25 ______________________________________ Fall 2011 Comments Ended class with this, not quite enough time! Repeated at start of Lecture #8 0, 21, 4, [[75]] Can solve this two ways, thinking about Lorentz forces, or thinking about “Flux rule”. (We did it both ways, the “Flux Rule” way requires careful understanding of sign conventions, and consistency between the direction of your EMF line integral, and your corresponding choice of direction for dArea. ) I worked through with them the discussion of the direction of current, and then (see next slide, 7.18) reposed the clicker question (which is about force) LA Notes: Ragole: Students simply did the Lorentz force law (for charges) and assumed that the force was down. I tried to argue that force on charges in the loop wouldn’t move the neutral loop but after the current direction was put up (7.18), they agreed on the correct answer. Rehn: some students were still struggling slightly with the right hand rule and figuring out the direction of the EMF. _______________________________________ Spring 2012 Comments Introductory physics question, but some students still struggle! ==================================== ERK, 5 Dec 2008 Motional emf Up the “screen”  Down the “screen”  To the right  To the left  The net force is zero B

7.18 One end of rectangular metal loop enters a region of constant uniform magnetic field B, with initial constant speed v, as shown. What direction is the magnetic force on the loop? B v L Class: CONCEPTUAL Correct Answer: D _________________________________ Phys 3320, Fa11 (SJP) Lecture #8 Followup, same question but now with current shown 0,8,4, [[88]], 0 Fall 2011 Comments At this point, just need to think about ILB forces. Discuss forces on top and bottom wires (cancelling), and on back end (zero). Discuss PHYSICAL nature here – it’s “fighting the change”. This is an eddy current. Pose the reverse question, what if you try to leave the field? Can show a demo of eddy currents here. (I also discussed the Prius regenerative braking, just informally) ==================================== ERK, 5 Dec 2008 Motional emf Up the “screen”  Down the “screen”  To the right  To the left  The net force is zero B

Yes, current will flow CW Yes, current will flow CCW No
7.19 One end of rectangular metal loop enters a region of constant uniform magnetic field B, out of page, with constant speed v, as shown. As the loop enters the field is there a non-zero emf around the loop? B v L B MODIFIED FROM FALL 2011 Class; CONCEPTUAL Correct Answer: A ________________________________ Physics 3320, Sp12 (MD) Lecture 7 [[100]], 0, 0 Spring 2012 Comments Intro #&M question. =============================== Originally from ERK, 5 Dec 2008 Also used by Steve Pollock, Fall 2011 Modified by Mike Dubson, Spring 2012 Motional emf Yes, current will flow CW Yes, current will flow CCW No

2L2 vB2/R (right) B) 2L2 vB2/R (left) C) 0
7.20 A rectangular metal loop moves through a region of constant uniform magnetic field B, with speed v at t = 0, as shown. What is the magnetic force on the loop at the instant shown? Assume the loop has resistance R. B w v L Class: CONCEPTUAL Correct Answer: C _________________________________ Physics 3320, Fa11 (SJP) Lecture #8 Physics 3320, Sp12 (MD) No Clicker Data Fall 2011 Comments I showed this, let them vote by hand, we did not click. Majority of hands went up for zero (but not all) Answer is C, no change in flux. Can also point out that any Lorentz force on free charges on one end are balanced by same force on free charges on the other end. LA Notes: Again students showed some difficulty when trying to use the right hand rule ==================================== Written by: ERK, 10 Dec 2008 Flux rule for motional emf B 2L2 vB2/R (right) B) 2L2 vB2/R (left) C) 0 D. Something else/not sure...

I1>0, I2=0 B) I1= I2 > 0 C) I1= -I2 > 0
7.21 Consider two situations: 1) Loop moves to right with speed |v| 2) Magnet moves to left with (same) speed |v| What will the ammeter read in each case? (Assume that CCW current => positive ammeter reading) B A Class: CONCEPTUAL Correct Answer: B ____________________________________ Phys 3320, Fa11 (SJP) Lecture #8 0, [[92]], 8, 0, 0 Fall 2011 Comments I needed to clarify what situation 2 is (some students thought the WIRE LOOP) moves to the left, and were thus voting C. Correct answer is B. First, need to work out that situation 1 is positive in this case (by same methods as previous questions, either Lorentz forces or flux rule/Lenz’ law) Then, discuss the “magnet moves left”. One student pointed to flux rule and said it’s the SAME change in flux, so should be same current. Another pointed out that “a co-moving observer” would say they are the same experiment. I had to clarify a little, point out that co-moving observers might disagree on E and B fields, but cannot disagree on which way an Ammeter needle moves! LA Notes: Everyone I saw had no trouble figuring it out. One student and I had a discussion about these two cases being able to refer to the same situation, depending on the reference frame you are in. I1>0, I2=0 B) I1= I2 > 0 C) I1= -I2 > 0 D) I1= I2 = 0 E) Something different/not sure

The magnetic Lorentz force. an electric force.
7.22 Faraday found that EMF is proportional to the negative time rate of change of B. EMF is also the line integral of a force/charge. The force is The magnetic Lorentz force. an electric force. the strong nuclear force. the gravitational force. an entirely new force. Class: CONCEPTUAL Correct Answer: B ________________________________________ Phys 3320, Fa11 (SJP) Lecture #8 42, [[58]], 0, 0, 0 Fall 2011 Comments Despite being an excellent split, the conversation was muted, and I had a hard time getting them to speak their ideas. One student finally gave nice discussion – first, voting FOR magnetic because we’ve just seen examples with it, and without any E field. Another argued for magnetic because “EMF makes charges move”. (So we had to discuss case where there’s an EMF but, say, large resistance and thus no motion) Original student then voted for electric after all, because he realized that our “experiment 3” case had NO motion of anything, no relative motion. (Charlie pointed out that even a changing B field must mean some motion of charges SOMEWHERE). I really had to talk through the “E” story, nobody spontaneously went for that. Not sure if this makes sense to them or not, not much feedback at this point. LA Notes: The discussion of the nature of the force was relatively quiet and they concluded it was E or B but no real discussion of which it was. ==================================== From Charles Rogers Chuck argues that Faraday’s position is “B”, but that “E” is perfectly worth consideration (but we identify it as electric due to special relativity – two fields behave the same, so might as well give them the same name). I might argue that in some situations e.g. with motional emf, the force/charge can in fact be magnetic, not electric. But then, the problem clearly talks about “the situation with time varying B fields... so I think I’m good with the answer as given here: B is best, E is possible (but rejected by physicists as unnecessary)

A time changing B creates an electric field via Faraday’s Law:
7.23 A time changing B creates an electric field via Faraday’s Law: Now I have no idea how to find E. This law suggests a familiar way to find E in all situations. This law suggests a familiar way to find E in sufficiently symmetrical situations. I see a path to finding E, but it bears no relation to anything we have previously seen. Class: CONCEPTUAL Correct Answer: N/A/ Survey ____________________________ Phys 3320, Fa11 (SJP) Lecture #8 4, 17, 67, 8, 4 Fall 2011 Comments No “correct answer”, but this is a great slide to get them from Faraday’s law back to Ampere’s law. In my class, when I asked for what they MEANT by answer C, the first student said “Gauss’ law”. So, we talked about that, someone thought maybe the “divergence theorem”. So we had a lot of Maxwell’s equations on the board, and then I was able to get Ampere’s law out of a few of them… This is a novel point, they know the MATH in a different physical situation. I used this as the intro to my activity, “Faradays Law”, which only had ~10 minutes, so we’ll need to continue it into a second class. LA Notes: Students argued that since they knew DelxE they could find E, but they were a bit hesitant to discuss how. ==================================== Question From Charles Rogers

A) Into the screen B) Out of the screen
7.24 A stationary rectangular metal loop is in a region of uniform magnetic field B, which has magnitude B decreasing with time as B=B0-kt. What is the direction of the field Bind created by the induced current in the loop, in the plane region inside the loop? B w Class: CONCEPTUAL Correct Answer: B ______________________________ Phys 3320, Fa11 (SJP) Lecture 9 (Start of class question) 26, [[67]], 4,4, 0 Physics 3320, Sp12 (MD) Lecture 8 3, [97], 0, 0, 0 Fall 2011 Comments The flux is out, but decreasing, so to “fight the change in flux”, we will try to induce our own field which is outwards. That requires a CCW current flow, and an outwards B(induced) Discussed/asked about fact that technically it might be zero, if this loop had very large resistance,( but it did say metal...) Made sure they see that EMF is still nonzero and independent of resistance, but induced current itself depends on R. LA Notes: Heard no talking during this question. However, after Steve closed it and started asking around for different reasons for their answers, one student said that he was thinking about it like you were pulling the loop out of the B field, and it would give a similar result. Students prefer a “fight the change” method for answering. ==================================== ERK, 10 Dec 2008 Flux rule for induced emf B A) Into the screen B) Out of the screen C) To the left D) To the right E) other/??

A) Into the screen B) Out of the screen
7.25 A stationary rectangular metal loop is in a region of uniform magnetic field B, which has magnitude B decreasing with time as B=B0-kt. What is the direction of the field Bind created by the induced current in the loop, in the plane region inside the loop? B w Just showing the induced current for previous one (7.24) B A) Into the screen B) Out of the screen C) To the left D) To the right E) other/??

Yes, current will flow CW Yes, current will flow CCW No
A rectangular metal loop is moving thru a region of constant uniform magnetic field B, out of page, with constant speed v, as shown. Is there a non-zero emf around the loop? 7.26 B v B MODIFIES 7.18 Class: CONCEPTUAL Correct Answer: C _________________________________ Physics 3320, Sp12 (MD) Lecture 7 17, 8, [[75]] Spring 2012 Comments Intro E&M question, students still not 100% ======================================= Originally from: ERK, 5 Dec 2008 Also used by Steve Pollock, Fa11 Modified by Mike Dubson, Sp12 Motional emf Yes, current will flow CW Yes, current will flow CCW No

7.27 On a piece of paper, please write down your name, Stokes’ theorem, and the Divergence Theorem… Used in Fall 2011 (SJP) Lecture #9 Pop “quiz” (not actually graded) – I had them do it, then draw a line, then do it again with their neighbors, and then hand it in. This was NOT nearly as easy as I thought it should be, gave them several minutes on their own, and they *wanted* several more minutes to talk it through! In the end, we put solutions on board and told them they should have a “picture” in their heads that goes along with them, perhaps e.g. using a familiar Maxwell equation to help them remember it...

A) counter-clockwise B) clockwise C) zero.
7.28 A loop of wire is near a long straight wire which is carrying a large current I, which is decreasing. The loop and the straight wire are in the same plane and are positioned as shown. The current induced in the loop is A) counter-clockwise B) clockwise C) zero. I to the right, but decreasing. loop Class: CONCEPTUAL Correct Answer: B _______________________________ Physics 3320, Sp12 (MD) Lecture 8 3, [97], 0 Spring 2012 Comments Intro E&M problem – students have to think for a minute, but get this. ============================== Written by Mike Dubson, Sp12

Nonzero, but need more information for value
7.29 The current in an infinite solenoid with uniform magnetic field B inside is increasing so that the magnitude B in increasing with time as B=B0+kt. A small circular loop of radius r is placed NON-coaxially inside the solenoid as shown. What is the emf around the small loop? (Assume CW is the direction of dl in the EMF loop integration) B r Class: CONCEPTUAL Correct Answer: A ________________________________ Phys 3320, Fa11 (SJP) Lecture #9 [52], 48, 0, 0, 0 Fall 2011 Comments Signs are hard here – but the EMF in this problem is CW, so the answer is unambiguously A. (The setup is identical to the in-class activity we had just done, so they had a sketch of E in their posession) Notes: Class was split after discussion on whether the induced EMF in the off-center loop was positive or negative. ==================================== Originally from: ERK, 10 Dec 2008 Flux rule for induced emf kπr2 -kπr2 Zero Nonzero, but need more information for value Not enough information to tell if zero or non-zero

The current in an infinite solenoid with uniform magnetic field B inside is increasing so that the magnitude B in increasing with time as B=B0+kt. A small circular loop of radius r is placed outside the solenoid as shown. What is the emf around the small loop? (Assume CW is the positive direction of current flow). 7.30 r B Class: CONCEPTUAL Correct Answer: C ____________________________________ Phys 3320, Fa11 (SJP) Lecture #9 0, 0, [100], 0, 0 Fall 2011 Comments We got 100% on this one, but the discussion is still good – and in particular, the question of “how can the answer be zero if we just computed a nonzero E out there? And, it even *looks* “curly” (in that it curls around the origin) but we’re saying here it has zero curl out there... LA Notes: Ragole: Students said there was no change in flux through the loop (which did not contain the solenoid) and thus there would be no EMF. Rehn: One student thought that EMF was not 0 outside, but was then convinced by another student that it was. This student's argument was that since Curl.E = 0 outside, any closed line integral must also be 0. This seems to be a persistent problem, from what I have seen. ==================================== ERK, 10 Dec 2008 Flux rule for induced emf kπr2 -kπr2 Zero Nonzero, but need more information for value Not enough information to tell if zero or non-zero

Yes No Need more information
7.31 The current in an infinite solenoid of radius R with uniform magnetic field B inside is increasing so that the magnitude B in increasing with time as B=B0+kt. If I calculate V along path 1 and path 2 between points A and B, do I get the same answer? B Path 1 B A Path 2 Class: MATH/PHYSICS Correct Answer: B __________________________________ Phys 3320, Fa11 (SJP) Lecture #10 18, [[82]], 0,0 Fall 2011 Comments Followup to a homework problem (which asks a closely related question) I heard several students say, probably because of that homework, that they “knew the answer but didn’t understand it”. My explanation involves going to the case with NO solenoid, where we know that the integral from A to B is path independent, and thus the loop integral all the way around is zero. So, A to B and B to A (the other path) CANCEL each other, making A to B and A to B (other path) EQUAL each other. That’s when the line integral IS zero. But now the line (loop) integral is NOT zero, and so those two integrals cannot still cancel. Important and difficult point here. Voltage is no longer a well defined scalar function of position. V(B) doesn’t mean ANYTHING any more. But, voltage *as defined in this question*, that is, as a line integral along a SPECIFIC PATH, is certainly still defined, and in real life, you can still use voltmeters to measure it (as long as your wires don’t go “around” a region with changing flux!) So we aren’t tossing voltage as a concept, but we have to be more careful now that we’ve left electrostatics. And, voltage conceived as “potential energy at a point” is now very problematic, there is no well defined “potential energy” ( if you go around a loop like this, you get a sort of “Escher’s infinite staircase” paradox) Notes: Baily: My impression is that no-voting students couldn’t necessarily justify their answer. Think there is some general confusion about the differences between voltage, EMF and electric potential. Paul brings up how it still kind of mystifies him that Curl.E is zero everywhere around the loop, yet the line integral is non-zero – OK, but this seems like the same student difficulty with the field being non-zero because the curl is zero. Rehn: After that, a student up front eventually asked (in front of whole class) if there was an analogy of potential to potential energy that could be used in non-conservative fields. (I don't remember exactly how his question went, but he started by talking about how in previous classes, we have always treated PE and V as very similar, and he was hoping for some sort of analogy to that in the case where Curl.E =/= 0. Steve answered by saying it was a little like one of those MC Escher drawings. Ragole: The students argued that the integral of E*dl around a closed loop was only zero when E had no curl so the voltmeter measurements should differ. ==================================== Originally from ERK, 10 Dec 2008 Flux rule for induced emf R Yes No Need more information

Not enough information
7.32 The current in an infinite solenoid with uniform magnetic field B inside is increasing so that the magnitude B in increasing with time as B=B0+kt. A small circular loop of radius r is placed coaxially inside the solenoid as shown. Without calculating anything, determine the direction of the field Bind created by the induced current in the loop, in the plane region inside the loop? B r Class: CONCEPTUAL Correct Answer: A _______________________________ Fall 2011: Not Used ==================================== ERK, 10 Dec 2008 Flux rule for induced emf Into the screen Out of the screen CW CCW Not enough information

meters m/sec sec/m2 m2 Volts
7.33 Faraday found that EMF is proportional to the negative time rate of change of B. To make an equality, the proportionality factor has units of: meters m/sec sec/m2 m2 Volts Class: MATH/PHYSICS Correct Answer: D _______________________________________ Fall 2011: Not used ===================================== From Charles Rogers Answer is D, m^2 (it’s usually written EMF = dPhi/dt, where Phi is the integral of B over an area) Alternatively, note that E has the same units as v x B from force laws, and thus voltage (which is E*distance) must have the same units as v x B * distance, which is (B/sec) * m^2

7.34 The current in an infinite solenoid with uniform magnetic field B inside is increasing so that the magnitude B in increasing with time as B=B0+kt. A small circular loop of radius r is placed coaxially inside the solenoid as shown. What is the emf around the small loop? (Assume CW is the positive direction of around the loop). B r Class: MATH/PHYSICS Correct Answer: A ______________________________________ Fall 2011: Not Used ==================================== Originally from: ERK, 10 Dec 2008 Flux rule for induced emf kπr2 -kπr2 Zero Nonzero, but need more information for value Not enough information to tell if zero or non-zero

Not enough information
7.35 The current in an infinite solenoid with uniform magnetic field B inside is increasing so that the magnitude B is increasing with time as B=B0+kt. A circular loop of radius r is placed coaxially outside the solenoid as shown. In what direction is the induced E field around the loop? B r Class: CONCEPTUAL Correct Answer: B ________________________________ Fall 2011: Not used ==================================== From ERK, 10 Dec 2008 Flux rule for induced emf CW CCW The induced E is zero Not enough information

7.36 If the arrows represent an E field, is the rate of change in magnetic flux (perpendicular to the page) through the dashed region zero or nonzero? Class: MATH/PHYSICS Correct Answer: A _____________________________________ Physics 3320, Fa11 (SJP) Lecture #10 Pre-class: [[44]], 56 After Discussion: [[52]],48 Physics 3320, Sp12 (MD) Lecture 12 No Clicker Data ==================================== After the pre-class vote, I started talking through Faraday’s law, summarized and clarified the problem, told them it was split, and added ~2 more minutes of discussion. New vote was [[52]]/48, it’s still tough! Curl(E) is zero everywhere, except at the origin, in this problem. Students felt that “E is curly, so it should have a curl”. One student correctly pointed out that you can TAKE the curl with front flyleaf and get zero. Nobody spontaneously came up with any argument besides that one. (Although the previous class, we had CALCULATED this E field for the problem of E outside a large solenoid with ramping up current, and discussed this same point) I show them the next slide [7.37], and ask them to walk through what the LINE integral is, but this required a lot of guidance. This is a subtle and challenging question for them, not sure they all got it. (I asked them about the slide following (7.38) with E = c phihat, and we talked through that as well.) I also pointed out the relation to the coulomb field (which has no divergence anywhere… except the origin) LA Notes Ragole: When asked to take the curl of the E field given, the student argued that it looked like the outside of a solenoid, so it should be 0. He didn’t recall the formula for curl in cylindrical, so he couldn’t easily verify his intuition. Rehn: Initially the students sitting near me went for answer b (can't remember what that said). They were all thinking that E could not be a conservative field because there was a changing flux and it looked "curly.” However, one student I was sitting next to felt that he had to look up a formula in Griffiths before he could be sure. He used polar coord's and decided that Curl.E was 0. I also discussed using the polar coordinate form with a student, and we talked about the fact that for the z component of curl, and we tried to look at it graphically to think about why that was 0. This was harder to do, not sure if it was effective or not. INSTRUCTOR NOTES: Answer is A, zero. The 1/s azimuthal field is the “magic” one that has no curl (away from the origin) WRITTEN BY: Steve Pollock (CU Boulder) dΦ/dt=0 B) dΦ/dt ≠ C) ???

7.37 If the arrows represent an E field, is the rate of change in magnetic flux (perpendicular to the page) through the dashed region zero or nonzero? Fall 2011: Thought this diagram might help them (refers to 7.36), I walked through it (but did not re-click) dΦ/dt=0 B) dΦ/dt ≠ C) ???

dΦ/dt =0 B) dΦ/dt is non-zero C) Need more info...
7.38 If the arrows represent an E field (note that |E| is the same everywhere), is the rate of change in magnetic flux (perpendicular to the page) in the dashed region zero or nonzero? Class: MATH/PHYSICS Correct Answer: B ______________________________ Fall 2011: Did not click, but discussed Contrast with , where spatial dependence is 1/s. ==================================== INSTRUCTOR NOTES: Answer is B, nonzero. If I consider the loop (which appears in the next slide), the line integral around the loop is manifestly NOT zero (E has same magnitude everywhere, but length is bigger on the outer path). Nonzero line integral means flux (enclosed) must be nonzero. -SJP WRITTEN BY: Steven Pollock (CU-Boulder) dΦ/dt = B) dΦ/dt is non-zero C) Need more info...

7.39 Regarding Faraday’s Law and the idea that a time-changing magnetic field creates an electric field: Yeah, I’m feeling pretty good about it. I remember all that Lenz stuff. Pretty new to me. Only have some vague memory of Lenz and Faraday. What’s Lenz’s Law? Completely new. Other. Class: SURVEY Correct Answer: N/A _____________________________ Fall 2011: Not used ==================================== From Charles Rogers

Ah! Now I’m pretty sure I can find E.
7.40 A long solenoid of cross sectional area, A, creates a magnetic field, B0(t) that is spatially uniform inside and zero outside the solenoid. B(t) Yes, yes. I already get it. Ah! Now I’m pretty sure I can find E. Still not certain how to proceed. Class: SURVEY Correct Answer: N/A ____________________________________ Fall 2011: Not used Spring 2012: Not used ==================================== From Charles Rogers

7.41 A long solenoid of cross sectional area, A, creates a magnetic field, B0(t) that is spatially uniform inside and zero outside the solenoid. SO: B(t) Class: MATH/PHYSICS Correct Answer: D _______________________________ Fall 2011: Not used Spring 2012: Not used ==================================== From Charles Rogers

Something that you need to calculate for the particular geometry.
7.42 A current, I1, in Coil 1, creates a total magnetic flux, Φ2, threading Coil 2. If instead, you put the same current around Coil 2, then the resulting flux threading Coil 1 is: Something that you need to calculate for the particular geometry. Is equal to the flux through Coil 2 if the geometry is symmetrical. Is always equal to the flux that I1 caused in Coil 2. Causes no net flux in Coil 1. Class: CONCEPTUAL Correct Answer: C _______________________ Fall 2011: Not used Spring 2012: Not used ==================================== From Charles Rogers

The opposite direction as I1 There is no induced current
7.43 The current I1 in loop 1 is increasing. What is the direction of the induced current in loop 2, which is co-axial with loop 1? 2 I1 Class: CONCEPTUAL Correct Answer: B ________________________________ Phys 3320, Fa11 (SJP) Lecture #10 4, [[96]], 0,0 Fall 2011 Comments This is a review, we covered it last class, no problems visible. Good explanations from class. LA Notes: Students used the “fight the change” logic to say it must induce an opposite current. ==================================== Originally from: ERK, 16 Sept 2009 Mutual induction The same direction as I1 The opposite direction as I1 There is no induced current Need more information

The opposite direction as I1 There is no induced current
7.44 The current I1 in loop 1 is increasing. What is the direction of the induced current in loop 2, which lies in the same plane as loop 1? 2 I1 Class: CONCEPTUAL Correct Answer: A ______________________________ Phys 3320, Fa11 (SJP) Lecture #10 [[79]], 18, 4, 0 Fall 2011 Comments Got a good explanation for A, didn’t hear what the “mistake” was for the 22% who got it wrong. LA Notes: Student drew out the lines of a magnetic dipolar around the loop and argued the nearby loop must have a current in the same direction. ==================================== Originally from: ERK, 16 Sept 2009 Mutual induction The same direction as I1 The opposite direction as I1 There is no induced current Need more information

There is no induced current Need more information
7.45 The current I1 in loop 1 is decreasing. What is the direction of the induced current in loop 2, which lies in a plane perpendicular to loop 1 and contains the center of loop 1? I1 Class: CONCEPTUAL Correct Answer: C ________________________________ Physics 3320 Fa11, SJP Lecture #10 0,0,[[100],0 Physics 3320, Sp12 (MD) Lecture 9 80, 20, [0], 0 Fall 2011 Comments Nice! Clear explanation from student, doesn’t seem problematic. Note that the wording of the problem is slightly confusing, I had to go to the screen and very clearly SHOW what I meant (including the final statement that the center of loop 1 lies in the plane of the screen) Starting to talk about “lab practicalities” and stray fields, here. Rehn: This question brought up an interesting point. One student said immediately that the inductance of one from the other was zero, and his reasoning was that there was no flux through the right loop. I then asked him if he could tell that was true from the formulas on the board. He looked at the formula for M_21 = Integral (some stuff with a dl_1 dot dl_2) and then realized that since dl_1 and dl_2 are orthogonal, that implies M_21 = 0. He seemed happy to make that connection with the Neumann Formula. Ragole: Students said it would be zero, since there is always no flux and I asked if that was always true for perpendicular loops or just this set up. They then discussed if the perpendicular loop were not exactly lined up with the center of the current loop, there would be a current. ==================================== Originally from: ERK, 16 Sept 2009 Mutual induction 2 CW CCW There is no induced current Need more information

Not enough information given.
7.46 Two flat loops of equal area sit in a uniform field B which is increasing in magnitude. In which loop is the induced current the largest? (The two wires are insulated from each other at the crossover point in loop 2.) 2 1 Class: CONCEPTUAL Correct Answer: A _________________________________ Phys 3320, Fa11 (SJP) Lecture #10 Fall 2011 Comments I brought a wire to SHOW what I meant by this, which was nice. No time to click, but we talked it through, seemed like it would’ve been fairly high correct Answer is 1 is larger (Flux through 2 is essentially 0) This, and previous one, led in to discussion of practicalities – in the lab, sometimes you WANT mutual inductance, and sometimes you don’t. This is a trick to avoid it. I also mentioned that looping (i.e. coiling) does the opposite, you could DOUBLE the flux by coiling this into two loops. (Afterwards, I realized that in this case with fixed length, doubling it would divide area by 4, and the doubling gets back half of that, but if you can use twice as much wire ==================================== Originally from: ERK, 10 Dec 2008 Flux rule for motional emf B 1 2 They are both the same. Not enough information given.

A loop of wire 1 is around a very long solenoid 2.
7.47 A loop of wire 1 is around a very long solenoid 2. 2 = the flux thru loop 1 due to the current in the solenoid. = the flux thru solenoid due to the current in the loop 1 1 Class: CONCEPTUAL Correct Answer: A ___________________________ Spring 2012: No Clicker Data (should be in Lecture 9 if used) ========================== Written by Mike Dubson Sp12 Which is easier to compute? M B) M21 C) equally difficult to compute

7.48 A current, I1, in Coil 1, creates a total magnetic flux, F2, threading Coil 2: If instead, you put the same current around Coil 2, then the resulting flux threading Coil 1 is: Something that you need to calculate for the particular geometry. Is equal to the flux through Coil 2 only if the geometry is symmetrical. Is always equal to the flux that I1 caused in Coil 2. Is nearly certain to differ from flux that was in Coil 2. Class: CONCEPTUAL Correct Answer: C ___________________________ Fall 2011: Not used Spring 2012: Not used ========================== From Charles Rogers

7.49 A long solenoid of cross sectional area, A, length, l, and number of turns, N, carrying current, I, creates a magnetic field, B, that is spatially uniform inside and zero outside the solenoid. It is given by: B Class: REVIEW MATH/PHYSICS Correct Answer: C ___________________________ Fall 2011: Not used Spring 2012: Not used ==================================== From Charles Rogers Answer is C

7.50 A long solenoid of cross sectional area, A, length, l, and number of turns, N, carrying current, I, creates a magnetic field, B, that is spatially uniform inside and zero outside the solenoid. The self inductance is: Class: MATH/PHYSICS Correct Answer: D ___________________________ Fall 2011: Not used Spring 2012: Not used ==================================== From Charles Rogers

The box of magnetic field The box of electric field
7.51 Consider a cubic meter box of uniform magnetic field of 1 Tesla and a cubic meter box of uniform electric field of 1 Volt/meter. Which box contains the most energy? The box of magnetic field The box of electric field They are both the same Not enough information given Class: APPLICATION Correct Answer: A ______________________________ Phys 3320, Fa11 (SJP) Lecture #11 [[67]], 13, 20 Fall 2011 Comments Some students were trying to “intuit” this, only a couple tried looking up numbers. Some did realize that 1 T is huge (I asked them where one might find 1T in the world, and where one might find 1V/m. They knew 1T was huge, and someone thought of batteries as being of order 1 V, with size scales between ends of <~1 m) I meta-discussed how in this course, units are still unfamiliar, and they need to be building “number sense” to be able to say if solutions to problems are sense or nonsense. I also pointed out that when my homeworks (like last week!) say “a field is increasing at a rate of 1T/sec” they should comment on this, and say “oh really? Can you do that?” ==================================== Originally from: ERK Fall 2009 ERK Notes: Comparing .5 epsilon0 E^2 to .5 B^2/mu0. Both epsilon0 and mu0 are small parameters in MKS units, so the B energy is much higher. Makes sense, 1 T is a HUGE magnetic field, whereas 1V/m is a VERY SMALL electric field. (I think this is an artifact of our choice of unit system, in a more “natural” unit system “1” unit of each field might give the same energy (?) I’m pretty sure in cgs that energy just looks like (E^2+B^2), so indeed in that system “1” unit of each would be equal.

Energy = U, energy density = u = U/V. UA >UB , uA = uB
7.52 Two long solenoids, A and B, with same current I, same turns per length n. Solenoid A has twice the diameter of solenoid B. Energy = U, energy density = u = U/V. UA >UB , uA = uB UA =UB , uA < uB UA >UB , uA < uB UA >UB , uA > uB None of these A B Class: MATH/PHYSICS Correct Answer: A _______________________________ Physics 3320, Sp12 (MD) Lecture 10 [91], 9, 0, 0, 0 __________ Written by Mike Dubson, Sp12

The laws governing electrodynamics which we have derived so far are:
7.53 The laws governing electrodynamics which we have derived so far are: These laws are valid Class: CONCEPTUAL Correct Answer: A _________________________ Fall 2011: Not used Spring 2012: Not used ==================================== ERK Fall 2009 Always if the charges are not accelerating if the fields are static if the currents are steady if the fields are not in matter, but in vacuum

By calling it a ‘Law’, we expect that:
7.54 Ampere’s Law relates the line integral of B around some closed path, to a current flowing through a surface bounded by the chosen closed path. By calling it a ‘Law’, we expect that: It is neither correct nor useful. It is sometimes correct and sometimes easy to use. It is correct and sometimes easy to use. It is correct and always easy to use. None of the above. Class: CONCEPTUAL Correct Answer: C ___________________________ Fall 2011: Not used Spring 2012: Not used ====================================

A complicated partial differential of F The Laplacian:
7.55 Take the divergence of the curl of any (well-behaved) vector function F, what do you get? Always 0 A complicated partial differential of F The Laplacian: Wait, this vector operation is ill-defined! E. ??? Class: MATH Correct Answer: A _______________________________ Physics 3320, Fa11 (SJP) Lecture 14 [[76]], 0, 19, 0, 5 Physics 3320, Sp12 (MD) Lecture 12 [93], 7, 0, 0, 0 ________________________________ Fall 2011 Comments Silent vote was about 55% correct. (One student was confused about order – they realized you can’t take curl of div, and didn’t realize this was backwards) Discussed various ways to “see” this, including brute force (order of partial derivatives doesn’t matter) or “intuitive”, thinking of “del” as a regular vector. (There are other elegant proofs, which I mentioned but I didn’t go into that) Spring 2012 Comments Can be used as introduction to “Maxwell-Ampere, Part1” tutorial (#06A) ====================================

Take the divergence of both sides of Faraday’s law:
7.56 Take the divergence of both sides of Faraday’s law: What do you get? 0=0 (is this interesting!!?) A complicated partial differential equation (perhaps a wave equation of some sort ?!) for B Gauss’ law! ??? Class: MATH Correct Answer: A _______________________________ Physics 3320, Fa11 (SJP) Lecture 14 [[64]], 14, 23, 0,0 Physics 3320, Sp12 (MD) Lecture 12 [80], 13, 0, 7, 0 Fall 2011 Comments LA Notes: The student near me quickly commuted the derivatives on the RHS and got 0=0. We debated for a bit if 0=0 is meaningful. ==================================== Modified from ERK Fall 2009 Answer Is A) Work it out – move div past the d/dt and use “Gauss’ law for magnetic fields” And yes, I think it’s interesting. The left side is zero by MATH, the right side is zero by PHYSICS. So it’s telling us something about the consistency of 2 of Maxwell’s equations. Students came up with this, it was a nice discussion.

Take the divergence of both sides of Ampere’s law:
7.57 Take the divergence of both sides of Ampere’s law: According to this, the divergence of J = A complicated partial differential of B Always 0 ?? Class: MATH Correct Answer: C _______________________________ Physics 3320, Fa11 (SJP) Lecture 14 10, 0, [[86]], 5, 0 Physics 3320, Sp12 (MD) Lecture 12 20, 3, [77], 0 Fall 2011 Comments Answer is C. I have div(J) = -drho/dt on the board, at the start of class, and there was some confusion about whether I was asking what IS the divergence of J, or what do you get *according to Ampere’s law*. (The latter is what I meant) Again, a nice discussion – we’re seeing a breakdown of “conservation of charge” from Ampere, something is wrong! (And, it’s wrong when d(rho)/dt is nonzero, i.e. non-steady state) LA Notes: Students first said that DivJ=d(rho)/dt, but I asked them what the LHS was equal to, so they conclude that DivJ= d(rho)/dt=0 which sounded wrong to them ==================================== Based on ERK Fall 2009

7.58 Our four equations, Gauss’s Law for E and B, Faraday’s Law, and Ampere’s Law are entirely consistent with many experiments. Are they consistent with charge conservation? Yes. And, I’m prepared to say how I know No. And, I’m prepared to say how I know. ???? Class: MATH/PHYSICS Correct Answer: A _______________________________ Fall 2011: Didn’t click, this is what we had just decided. Spring 2012: Not used ==================================== Originally from Charles Rogers

7.59 Ampere’s Law relates the line integral of B around some closed path, to a current flowing through a surface bounded by the chosen closed path. The path can be: Any closed path Only circular paths Only sufficiently symmetrical paths Paths that are parallel to the B-field direction. None of the above. Class: MATH/PHYSICS Correct Answer: A _______________________________ Physics 3320, Fa11 (SJP) Lecture 14 [[95]], 0, 5, 0, 0 Slide was animated. Probably spent more time on this and the next one than I needed to, could “tell” rather than clicking. Just reminding people about the meaning of the integral form of Ampere’s law, and where it comes from here. I did take some time to talk/remind people that “symmetry” is needed to extract B from this equation. It’s always true, but not always useful to find B! LA Notes: Students immediately answered the correct answer (A), but really couldn’t back up why when pressed. ==================================== From Chuck Rogers.

7.60 Ampere’s Law relates the line integral of B around some closed path, to a current flowing through a surface bounded by the chosen closed path. The surface can be: Any closed bounded surface Any open bounded surface Only surfaces perpendicular to J. Only surfaces tangential to the B-field direction. None of the above. Class: MATH/PHYSICS Correct Answer: B _______________________________ Physics 3320, Fa11 (SJP) Lecture 14 24, [[76]], 0,0,0 Some confusion about what “open” vs “closed” means. Had to discuss/review that. (You could also argue that it’s E, it’s any open bounded surface WITH THE SAME BOUNDARY as you use for your loop on the left side, but nobody went for/thought about that. I did bring it up) ==================================== From Charles Rogers

Stoke’s Theorem: line v. surface integral
7.61 Stoke’s Theorem: line v. surface integral Rank order (over blue surfaces) where J is uniform, going left to right: A) iii > iv > ii > i B) iii > i > ii > iv C) i > ii > iii > iv D) Something else!! E) Not enough info given!! Class: MATH/PHYSICS Correct Answer: D _______________________________ Physics 3320, Fa11 (SJP) Lecture 14 5, 5, 15, [[70]] Fall 2011 Comments Answer is D, i=ii=iii=iv. When I presented it, I TOLD them what the “distractors” (below) were, i.e. I mentioned these particular issues. Not much difficulty here, good student explanations/reasoning. Might not have needed quite so much time on this one either, but seems like a relevant review/conceptual underpinning to the next sequence. LA Notes: Students next to me wanted to say that the value of ∫∫J.dA should be equal for all surfaces, but they didn't offer much of an explanation for why. I think that the conversation after the clicker question might have helped to clarify why the flux is the same for each loop, since they got to hear others' explanations of why, when they couldn't exactly generate an explanation on during the question. ============================================ INSTRUCTOR NOTES: I thought it would be harder, D got 90% of the votes, and nobody had anything to argue except "all the same". In the discussion, though, we talked about "how do you know", and discussed what "current through" means (and e.g. how some extra current does enter case iii, but then re-exits). I also pointed out integral B dot dl around the entrance loop would be the same... Ans D: a = b = c = d I picked A distractor to be “largest area to smallest area” (visually?) I picked B distractor to be “most perpendicular to least perpendicular”?? I picked C distractor to be “most area towards the left” ?? -SJP WRITTEN BY: Ward Handley, adapted by Steve Pollock (CU-Boulder) 61

7.62 We are interested in B on the dashed “Amperian loop”, and plan to use to figure it out. What is Ithru here? I B) I/ C) D) Something else E) Not enough information has been given! Class: MATH/PHYSICS Correct Answer: A _______________________________ Physics 3320, Fa11 (SJP) Lecture 14 [[86]], 0,0,0,14 This question is addressed in the “Maxwell-Ampere” tutorial (#06) Fall 2011 Comments There IS perhaps some ambiguity (as we will see when I “clarify” in the next questions) I set this up by reminding them of what we would do if the plates were not there, convincing myself and them that B = mu0 I/ 2 pi r for a simple long wire. ==================================== I I

7.63 We are interested in B on the dashed “Amperian loop”, and plan to use to figure it out. What is Ithru? The surface over which we will integrate J.dA is shown in light blue. I B) I/ C) D) Something else E) ?? Class: MATH/PHYSICS Correct Answer: A _______________________________ Physics 3320, Fa11 (SJP) Lecture 14 [[95]], 0,0,0 1 Fall 2011 Comments So here, I argued there’s no ambiguity – it really is I (and thus, we DO know the B field even in this different problem than before, with a capacitor in the middle of the wire) ==================================== I I

7.64 We are interested in B on the dashed “Amperian loop”, and plan to use to figure it out. What is Ithru? The surface over which we will integrate J.dA is shown in light blue. I B) I/2 C) 0 D) Something else E) ?? Class: MATH/PHYSICS Correct Answer: C _______________________________ Physics 3320 Fa11 (SJP) Lecture 14 Fall 2011 Comments I blew the animation here, showed them the answer. Oh well – we talked it through – convinced ourselves the answer must be ZERO here! So we have a serious problem: Stokes’ theorem insists that the answer must be the same as the previous clicker question, but we’re getting different results. Something is wrong, and it’s not Stokes’ theorem that’s broken (that’s math), so it must be Ampere! ==================================== I I

7.65 A constant current flows into a capacitor. Therefore, the capacitor voltage drop: is zero. is a non-zero constant. C) has a constant change with time. has a constant integral with time. None of the above. Class: CONCEPTUAL Correct Answer: D _________________________ Fall 2011: Not used Spring 2012: Not used ====================================

7.66 A parallel plate capacitor has a constant current delivered to it, leading to a relationship between current and voltage of: Class: MATH/PHYSICS Correct Answer: D ________________________ Fall 2011: Not used Spring 2012: Not used ====================================

7.67 Our four equations, Gauss’s Law for E and B, Faraday’s Law, and Ampere’s Law are entirely consistent with many experiments. Are they RIGHT? Yes. And, I’m prepared to say how I know No. And, I’m prepared to say how I know. What?! Are we philosophers? Class: SURVEY Correct Answer: _________________________ Fall 2011: Not used Spring 2012: Not used ==================================== From Charles Rogers

7.68 Our four equations, Gauss’s Law for E and B, Faraday’s Law, and Ampere’s Law are entirely consistent with many experiements. Are they INTERNALLY CONSISTENT? Yes. And, I’m prepared to say how I know No. And, I’m prepared to say how I know. Ummm…, could we play with them for a bit? Class: SURVEY Correct Answer: _________________________________ Compare with 7.67 Fall 2011: Not used Spring 2012: Not used ==================================== From Charles Rogers

7.69 The complete differential form of Ampere’s law is now argued to be: The integral form of this equation is: Class: MATH Correct Answer: D ________________________________ Physics 3320, Fa11 (SJP) Lecture #15 0,8,0,[[88]],4 _______________________________ Fall 2011 Comments (Note: after ~1 minute it was only about 50% correct, (with C still getting votes too.) I told them to talk to each other, and that only half the class had it right – the room got louder and they definitely converged nicely. Excellent conversation around this one.) D is the correct answer: You integrate both sides over da, and use Stokes’ on the LHS. This emphasizes Stokes’ theorem (once again) and makes you think through where the integral form comes from. Good question, glad we asked it. Notes: Student commented that he’s starting to see a “recurring theme” in asking about Stokes’ Theorem and integral forms of differential equations. Rehn: One student asked me what the units of µ_0 are. I am not sure what he was trying to accomplish by getting the units of µ_0, but he seemed confident about what he was thinking. In addition, one student said that he knew it had to be a line integral on the right side, but I never heard anyone talk about the right side of the equation. I didn't see anyone actually write anything out, and I suspected that they would probably make some mistakes doing it in their heads. Ragole: At first, a few students seemed a bit clueless. One student said he applied Stokes’ Law but in his explanation, skipped the step of integrating both sides and then applying Stokes. Others ended up agreeing with him from “half intuition, half guts.” ==================================== Originally from: ERK Fall 2009 E) Something else/???

7.70 Consider a large parallel plate capacitor as shown, charging so that Q = Q0+βt on the positively charged plate. Assuming the edges of the capacitor and the wire connections to the plates can be ignored, what is the direction of the magnetic field B halfway between the plates, at a radius r? s A. a z B r I I Class: MATH/PHYSICS Correct Answer: A ________________________________ Physics 3320, Fa11 (SJP) Lecture #15 [[81]] 0, 4, 11, 4 _______________________________ Fall 2011 Comments I thought it would be quick, but they really wanted to keep talking about this one. Lots of healthy discussion! Although they converged on the right answer, here again they were not getting it immediately and there was good argument. (E.g, “B” was getting some votes at first, before the peer conversations picked up, and ended getting nothing) In the full-class follow up, we talked about the sign (is it plus or minus?) With the conventions as shown in this picture, it is NOT ambiguous, the answer is “positive phi-hat”, exactly the same as outside the capacitor. (We discussed the “Right handed cylindrical system”, to decide which way +phi must be) LA Notes: Students near me were reasoning in terms of how the charge was building up, which way the changing electric field points, and then applying the right-hand rule, but the easy way of arriving at the answer is that it points in the same way as if the current-carrying wire were continuous. Students near me saw that it pointed out of the board, but couldn’t decide if that was positive or negative phi-hat, they couldn’t say right away whether s-hat crossed into z-hat was positive or negative. ==================================== Originally from: ERK Fall 2009 C. D. Q -Q E. ??? d

7.71 Consider a large parallel plate capacitor as shown, charging so that Q = Q0+βt on the positively charged plate. Assuming the edges of the capacitor and the wire connections to the plates can be ignored, what kind of amperian loop can be used between the plates to find the magnetic field B halfway between the plates, at a radius r? D. A different loop! E) Not enough symmetry for a useful loop a A B s r I I Class: MATH/PHYSICS Correct Answer: B ________________________________ Physics 3320, Fa11 (SJP) Lecture #15 _______________________________ Fall 2011 Comments I animated this – did not bother clicking, but had people should out their ideas. One student said it shouldn’t MATTER, you could use a square loop if you want, so we had a nice discussion of the “truth” of Ampere’s law vs the “usefulness” in integral form, to extract B. So after this question, we had discussed the logic of Ampere’s law, knew the direction and symmetry of B, knew what amperian loop we would choose, and then I set them free to DO it. Gave them ~5 minutes (See next slide – 7.72) ==================================== Originally from: ERK Fall 2009 z C Q -Q d

7.72 Consider a large parallel plate capacitor as shown, charging so that Q = Q0+βt on the positively charged plate. Assuming the edges of the capacitor and the wire connections to the plates can be ignored, what is the magnitude of the magnetic field B halfway between the plates, at a radius r? A. B. C. D. E. None of the above s a z r I I Class: MATH/PHYSICS Correct Answer: E ________________________________ Physics 3320, Fa11 (SJP) Lecture #15 36, 28, 8, 12, [[16]] _______________________________ Fall 2011 Comments Slide is animated, so they have to work it out before seeing answer. Note that in class before this, I had already shown them that E = Q/(A epsilon0) (uniform, neglecting fringe fields) This question was done as an “activity” (I animated the slide so the answers didn’t show up for about 5 minutes, and even then they took another ~3 minutes). I then the answers up so people could click to indicate that/when they were done. Interestingly, given the ~8 minutes for this activity, the could not come up with it correctly. I went through it at the board (see below). Point #1 which caught many was that the amperian loop has radius r, not a, so the left side of Ampere’s law is B * 2pi r, not B*2pi a. Next, we discussed that “mu0 I = 0”. This seemed ok Next, we started working on the double integral of E dot dArea. A LOT of students wanted it to be E * A = E* pi a^2. There were other interesting ideas shouted out, including ratios of r^2 to a^2. Nobody seemed “fooled” by using d, and yet some clicked on answer B (which has right units, and right dependence on r, but wrong denominator in detail) Some students said “But D would be correct if you swapped a for r, right” (which is correct). Another said “B would be correct if you swapped A for d^2, right?” (which is also correct). I pointed out that the “r” dependence is critical, if you go to r=0 and get a nonzero (azimuthal??) field, what would THAT mean? Paul Beale pointed out that it’s also interesting that B is NOT the same inside the capacitor as it would be outside at the same radius, so you must reconsider your simplistic “Biot-Savart” argument, SOMETHING novel is going on here, you can’t totally ignore the capacitor plate elements. LA Notes: Students near me got confused on which area they were working with, the area of the capacitor plates (which appears in the formula for E), or the area from the flux integral (Amperian loop). Might not have had this confusion if the formula for E had been given in terms of the surface charge density, so I think this is a good distractor – most popular choice (A – 36%) corresponded to making this error. 28% chose B, which at least had the correct radial dependence. ==================================== Originally from: ERK Fall 2009 Answer is E, none of these! Use E = Q/(epsilon0 * A) to get B * 2pi r = mu0 dQ/dt pi r^2 / A Thus, B = mu0 beta r/(2*area of plates) = mu0 beta r/(2*pi*a^2) Q -Q d

The laws governing electrodynamics which we have derived so far are:
7.73 The laws governing electrodynamics which we have derived so far are: These laws are valid Class: MATH/PHYSICS Correct Answer: A _____________________________ Phys 3320, Fa11 (SJP) Lecture 16 [[84]], 5, 0,11, 0 _______________________________ Fall 2011 Comments This was our pre-class question. I wanted to make the following points: This semester is about “time dependence”, or *dynamics*. Motion and acceleration are now “fair game”, as they MUST be for this to be useful once we leave statics. These equations are also true in matter (that’s the topic of today’s lecture), we will rewrite/reorganize them to make them easier to solve, but the physics is not changed. I also took this opportunity to review “conservation of charge” (asked them how that’s “buried” in these equations, only one student after ~30 seconds of wait time came up with the trick of taking div(Ampere’s law). So we reviewed that. I emphasized that although even if we added that term to conserve charge, you COULD take the philosophical stance that THESE are the empirical laws consistent with experiments, and thus that conservation of law is a *consequence* of this set of equations. I also pointed out (at Charlie’s prompting) that we will discover that special relativity (and the invariance of c) comes out of these equations too. So, they rock  Answer is A (except, perhaps, in the quantum limit, since it’s a classical field theory) ==================================== Adapted from ERK Fall 2009 Always if the charges are not accelerating if the fields are static if the currents are steady if the fields are not in matter, but in vacuum

zero B) a3 P0 C) P0 D) P0 /a3 E) 2 P0 a2
7.74 The cube below (side a) has uniform polarization P0 (which points in the z direction.) What is the total dipole moment of this cube? zero B) a3 P0 C) P0 D) P0 /a3 E) 2 P0 a2 z Class: MATH/PHYSICS Correct Answer: B ________________________________ Physics 3320, Fa11 (SJP) Lecture 16 Silent vote was roughly 5, [50], 0, 40, 5, i.e. still rather split. Then I told them to converse, and it suddenly flipped to 0, [[90]], 0, 10, 0 _______________________________ Fall 2011 Comments Correct answer is B. On the board I had already defined P = p/volume. See comments below, it’s simple but seems worthwhile. LA Notes: For the dipole moment of the cube, the students initially though a*P, then I asked what the units of P were and they realized the answer was P*a^3 ==================================== Used in 3310: 0% [[92%]] 2% 6% 0% (FALL 2008) 6% [[78%]] 6% 11% 0% (SPRING 2008) INSTRUCTOR NOTES: Started class with this. Asked it silently (78%) then collaborating (90%, and the mistakes were accompanied by moans of “I meant to multiply” when I said the answer) Super simple, but I think it was helpful as a pre-class question just to emphasize units and the IDEA of “P * dt = dipole moment” that we will use in class today. -SJP Dubson: Good question for establishing terminology. WRITTEN BY: Steven Pollock (CU-Boulder) x

ρb =0, σb≠0 ρb ≠0, σb≠0 ρb =0, σb=0 ρb ≠0, σb=0
7.75 In the following case, is the bound surface and volume charge zero or nonzero? ρb =0, σb≠0 ρb ≠0, σb≠0 ρb =0, σb=0 ρb ≠0, σb=0 Class: CONCEPTUAL Correct Answer: B ________________________________ Physics 3320, Fa11 (SJP) Lecture 16 10, [[86]], 0,5,0 _______________________________ Fall 2011 Comments Note: On the board I already had written and discussed sigma = P dot nhat, and rho = -div(P) Answer is B. (There is clearly a + sigma on the top, a (small) – sigma on the bottom, and positive div(P) throughout. Looks like positive div(P), and thus negative rho, which is consistent with there being excess + on the outside. ) This went quickly, no big problems. Some students were curious about e.g. the sign of rho (so I pointed out that in this CT, the top surface is very +, but the bottom is only a little -, so the volume should have “net –” to conserve total charge, consistent with the minus sign in –Div(P) One student wanted to know if you could EVER have a situation with rho NOT zero, but sigma = 0. (Like “d”) I puzzled a bit, then pointed out that by the divergence theorem, if div(P) is nonzero, then the volume integral of it is nonzero, thus the SURFACE integral of P dot dA is nonzero, and thus you would HAVE to have a nonzero sigma (to conserve charge, basically). (But then I hemmed, if you had SOME regions with + div(P) and others with –div(P), then the total integral would vanish, and I cannot say for sure what happens on the surface. ) Anyway, seemed like a good question, could even turn it around on students in the future, make THEM come up with an answer! Notes Baily: Think it might be worth emphasizing how the “free” charge is not necessarily mobile (can be painted on, for example) – bound charge is specifically due to the response of the material to an applied field, and the free charge is anything else, which is controlled by the experimenter. Ragole: Dipoles with a divergence: the students weren’t sure about rho_b, one argued it should be zero because the material was neutral. Another argued that P clearly had a divergence. Both agreed there had to be some surface charge. ==================================== In 3310, the response rates were 10% [[80%]] 5% 5% 0% INSTRUCTOR NOTES: : 80% correct. Not sure why it wasn’t 100%, didn’t get good arguments from the class. Some were confused about “where the negative bound volume charge density actually comes from”. Seems like a pretty good example to talk through. (SJP - Note that I redrew the physical dipoles - the original picture is “hidden” behind , they did not look neutral. If the “---” is supposed to represent more spread out electrons, then this looks like a SMALLER dipole moment to me... but ok, the picture on the right works) Note that bound surface charge is less than zero on the bottom and greater than zero on the top -SJP Dubson: I hear a lot of people looking at just the physical dipole picture and not the idealized dipole picture. Doing physical counting instead of looking at formulas. But most people got it right in the end. WRITTEN BY: Steven Pollock (CU-Boulder) Physical dipoles idealized dipoles 75

Everywhere: throughout the volume and on all surfaces
7.76 A solid cylinder has uniform magnetization M throughout the volume in the z direction as shown. Where do bound currents show up? Everywhere: throughout the volume and on all surfaces Volume only, not surface Top/bottom surface only Side (rounded) surface only All surfaces, but not volume Class: CONCEPTUAL Correct Answer: D ______________________________ Phys 3320, Fa11 (SJP) Lecture #17 On the board I had written Maxwell’s equations in matter, along with equations including K_bound = Mxnhat and J_bound = Curl(M) 18, 18, 14, [[41]], 9 - silent (pre class) 5, 9, 0, [[86]], 0 - after 2 minutes of peer discussion _______________________________ Fall 2011 Comments Seemed like a useful and necessary review. Notes Baily: During discussion, students nearby seemed uncertain on the volume bound current, and I called their attention to the curl of M equation, which led them quickly to it being zero. Same argument when Steve asks for input from class, a math answer but not a physical explanation. Ragole: Students near me were initially focused on the volume current and argued it was zero. When they were reminded of the surface bound currents, they all quickly voted for the correct answer, current around the cylinder wall. Rehn: Two students were discussing the question and trying to relate the formula J_B = del x M on the board to the problem. They weren't sure exactly what to do to figure it out. After this, there was time for discussion, and one of the students decided on D, saying that the current residing on the surface was ``simplest," and therefore that had to be the correct answer. I asked why there couldn't be a current throughout the volume. They responded that since there is uniform M, this must bear some similarity to a solenoid, where the current flows in phi-hat direction on the ``surface" to produce uniform B. ==================================== CORRECT ANSWER: D. M cross nhat is only nonzero on the curved outer surface (where it runs AROUND the cylinder, like a solenoidal current). Curl(M)=0 inside, so there is no bulk volume current density in there. I drew the “end on” picture, with a bunch of atoms all lined up, to make the argument that there’s no current in the middle, but there IS an unbalanced surface current running around the outside. Talked about/reminded that surface currents are real, that electrons aren’t running “all the way around” but nevertheless the current is physical and causes B fields inside… In 3310 it was ~88% correct WRITTEN BY: Steven Pollock (CU-Boulder) M

Choose boundary conditions
7.77 Choose all of the following statements that are implied by Choose boundary conditions (I) (II) (III) A) (I) only B) (II) only C) (I) and (II) only D) (I) and (III) only E) Some other combo! Class: MATH/PHYSICS Correct Answer: C ______________________________ Phys 3320, Fa11 (SJP) Lecture #17 8,0, [[58]], 29, 4 _______________________________ Many of the issues that came up while students were discussing this in FA11 are addressed in the “Boundary Conditions” tutorial (#07) Fall 2011 Comments I gave this AFTER I had gone through a parallel blackboard example that started from Gauss’ law, used Divergence theorem, to find the formula for Eperp (that Eperp “jumps” if and only if you have a nonzero surface charge.) Discussed issues of the “shape” of the surface we’re using, and the limit where some side is vanishing fast. Everyone seems to realize that (I) follows (good!) A little confusion about whether this equation tells us about Eparallel or Eperp, (i.e. they’re not coming up with Stokes’ theorem and the appropriate line integral to draw) but also a lot of students were confused (though it’s not showing up in the clicker distribution) about WHY the dB/dt aspect is not mucking up this boundary condition. [Note: This issue is addressed explicitly in the “Boundary Conditions” tutorial (#07)] Spent some class time talking about how ( for my divergence case involving an integral of rho on the right side), the volume integral gives ZERO for charge enclosed in my “tiny box”, unless rho is itself infinite. And that infinite rho CAN effectively happen if you have a sigma (surface charge), it’s a delta function, like squeezing finite charge into an infinitely flat “box”. In THIS problem, however, our dB/dt is presumably NOT going to blow up in any way, there’s no infinitely thin B field in any physics problem, so now the “flux” of the B field through our little Faraday loop is VANISHING, giving us the zero on the right hand side. So, whether or not you have a dB/dt on the right side, we STILL get that E(parallel) is continuous. Notes Baily: Steve closed the discussion down because people seemed stuck, suggested that people guess if they don’t know, so not clear what proportion of that result is guessing, versus what they were actually thinking. Ragole: The student nearest me drew the correct amperian loop, made the arguments that da->0 so the right hand side is zero, but the parallel components of E are still around; he concluded that Eparallel was continuous. I asked about the contributions from the side and he used the argument previously used that the sides vanish fasters than the top and bottom ==================================== INSTRUCTOR NOTES: Adapted from a related question by Chandralekha Singh. Improving students' understanding of magnetism, C. Singh, Proceedings of the Annual Conference of the American Society for Engineering Education (ASEE), AC , 1-16, (2008): 77

Choose boundary conditions
7.78 Choose all of the following statements that are implied by (for any closed surface you like) Choose boundary conditions (I) (II) (III) A) (I) only B) (III) only C) (I) and (II) only D) (I) and (III) only E) Something else! Class: MATH/PHYSICS Correct Answer: D ________________________________ This is also explicitly addressed in the “Boundary Conditions” tutorial (#07) Fall 2011: Didn’t use, this was Singh’s question ==================================== USED IN: Spring 2008 (Pollock) LECTURE NUMBER: 36 STUDENT RESPONSES: 11% 6% 39% [[28%]] 17% INSTRUCTOR NOTES: Start of class with this. I did it silently first, and they were all over the map. (Correct answer D had 28%). Then I let them talk to their neighbors, D moved to 50% (and C and E were equally represented). I'm vote for I and also III, which gives D. -SJP WRITTEN BY: Chandralekha Singh. Improving students' understanding of magnetism, C. Singh, Proceedings of the Annual Conference of the American Society for Engineering Education (ASEE), AC , 1-16, (2008): 78

Electricity and Magnetism II

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