Exam III Solution: Chapters 18—201. The anode of a battery

A) has a positive charge, while the cathode has a negative charge.B) has a negative charge, while the cathode has a positive charge.C) has a negative charge. So does the cathode. The anode is more negative, however.D) has a negative charge, as does the cathode. The cathode has more negative charge.E) has a positive charge, as does the cathode. The cathode has the greater positive charge.

2. Three batteries are wired as shown and used to light the bulb. As-sume that the batteries all have the same voltage.A) These batteries are in series.B) These batteries are in parallel.C) These batteries are not in either series or parallel.

3. What is the total voltage across the bulb?A) V = V1 = V2 = V3 C) V = 1/V1 + 1/V2 + 1/V3

B) V = V1 + V2 + V3 D) 1/V = 1/V1 + 1/V2 + 1/V3

4. Rewiring the batteries in parallel to the same bulb results inA) greater voltage across the bulb.B) less voltage across the bulb.C) no change in voltage across the bulb.D) no voltage at all across the bulb, because you have just shorted the circuit!

5. When the switch is closed on the circuit shown on the right, what is the direc-tion of the current?A) Electrons move counterclockwise, so that is the direction of the current.B) Electrons move counterclockwise, so the direction of current is

clockwise.C) Electrons will move clockwise, so the direction of the current is clock-

wise.D) Electrons will move clockwise, so the direction of the current will be

counterclockwise.E) There will be no current unless the switch is left open!

6. Compare current and drift velocity.A) Why? They are the same thing. Do you want me to compare zucchinis and

courgettes next? Or maybe an eggplant with an aubergine?B) Drift velocity measures how fast, on average, charges are moving. Current measures the instantaneous

velocity of a charged particle.C) Current is the quantity of charge past a certain point in a given interval of time. Drift velocity is the num-

ber of charges that travel a certain distance in the same interval of time.D) Current is the quantity of charge passing a certain point during a time interval. Drift velocity meas-

ures the average speed of the individual charges.E) Current and drift velocity combine to make a single vector. The drift velocity is the magnitude: how fast

charges are moving. The current is the direction in which the charges move.

7. What is the internal resistance of the continuously variable blow dryer that I have just invented? Here’s the data, you figure it out:A) 0.5Ω C) 2.0Ω E) 10ΩB) 1.0Ω D) 5.0Ω

8. I noticed that when I tested currents higher than 20A, the graph was no longer linear. The slope seemed to be increasing.A) You clearly need to do something about your data taking skills. The

graph, if it starts linear, will stay that way.B) As the blow dryer heats up, the resistance is going to decrease. The slope

should actually be decreasing.C) Resistance increases as temperature increases, so it is not surprising that the slope would increase.

PHYS 1420: College Physics II Fall 2006

9. To increase the resistance of a wire, which change will have the greatest effect?A) Double the length.B) Double the diameter.

C) Double the area.D) Decrease the temperature by 10°C.

10. Electric power can be written as P = IV, and if we use Ohm’s Law to make substitutions, P = I2R or P = V2/R. If these expressions are equivalent, why is Joule heat expressed at P = I2R and not V2/R?A) It isn’t. Either expression is perfectly acceptable, and there is no reason to prefer one or the other.B) Because the V2/R expression puts the resistance in the denominator. This is just wrong.C) The current (moving charge) is what actually creates the heat, so it is conceptually consistent to ex-

press the waste heat in terms of what causes it.D) Neither expression is valid; heat is Q = mc∆T, where charges having mass m move at the speed of light (c)

until they raise the temperature by ∆T.

11. Three resistors are wired in series as shown on the right. Each of the resis-tors is different, with R1 having the least and R3 having the most resistance.A) The most current flows through R1, the least current flows through R3.B) The most current flows through R3, the least current flows through R1.C) The same amount of current flows through each of the three resis-

tors.D) No current flows, because the resistors must be equal in order to draw

current from the battery. Unequal resistors cause a short circuit.

12. The voltage across R1

A) is greater than the voltage across either R2 or R3.B) is less than the voltage across either R2 or R3.C) is identical to the voltage across R2 and R3.D) is exactly zero (same for R2 and R3).

13. Replacing the three resistors shown with a single equivalent resistor Rs

A) will increase the total amount of current drawn.B) will decrease the total amount of current drawn.C) will increase the voltage.D) will decrease the voltage.E) will have no effect on either the voltage or the total amount of current drawn.

14. If the devices in the circuit shown are actually light bulbs, which bulb glows the brightest?A) Bulb 3: P = I2R, so the largest resistance dissipates the most power.B) Bulb 1: P = V2/R, so the bulb with the least resistance shines the brightest.C) Bulb 2: no reason, but bulb 2 was starting to feel sort of left out.D) This is a trick question. They all have the same brightness.

15. The three bulbs on the right are wired in parallel to the battery as shown. The bulbs are not identical! As be-fore, R1 < R2 < R3. The voltage across bulb 1A) is greater than the voltage across either bulb 2 or

bulb 3.B) is less than the voltage across either of the other

two bulbs.C) is the same as the voltage across either bulb 2

or bulb 3.

16. Which bulb draws the greatest amount of current?

V1

V2

V3

A) Bulb 1. B) Bulb 2. C) Bulb 3. D) I1 = I2 =I3.

17. Which bulb is the brightest?A) Bulb 1. B) Bulb 2. C) Bulb 3. D) All the same.

18. If we continue to add bulbs to the circuit, wiring each in parallel,A) the total amount of current drawn will increase.B) the total amount of current will decrease.C) the voltage increases.D) the voltage decreases. E) there will be no change in either the current or the voltage.

Exam III Part 01

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19. Adding devices in parallel will have what effect on the equivalent resistance of the circuit?A) No change in Rp. B) Increases Rp. C) Decreases Rp.

20. The circuit on the right must be solved using Kirchhoff’s rules. Which set of equations can be used to correctly solve for the currents flowing through each branch?A) I1 + I3 = I2 C) I1 = I2 + I3

V1 – I1R1 – I2R2 = 0 V1 – I1R1 – V2 + I3R3 = 0V2 – I2R2 – I3R3 = 0 V2 – I2R2 – I3R3 = 0

B) I1 + I2 = I3 D) I1 + I2 + I3 = 0V1 – I1R1 – I2R2 = 0 V1 – I1R1 + V2 – I3R3 = 0V2 – I2R2 – I3R3 = 0 V2 – I2R2 – I3R3 = 0

21. Examining the currents marked on the diagram,A) currents 1 and 2 are probably drawn correctly, but I3 is backwards.B) it’s 1 and 3 that are right; reverse I2!C) I1, I2, and I3 all point in the wrong direction: reverse them all!D) all three currents are pointing in the proper direction.E) there is no way to know what direction any of the currents actually

point. The picture is just a guess without basis in the physics.

22. Every magnet hasA) two poles, either a pair of North poles or a pair of South poles.B) two poles, one negative and one positive.C) a pair of poles, one North and one South.D) at least 50% iron in its composition.

23. The magnetic forceA) causes like poles to repel each other.B) causes like poles to attract each other.

24. A friend hands you a rectangular piece of metal, and claims that it has a single magnetic pole, located at the exact center of the object.A) Not possible; if it only had one pole, that pole would have to be located on one end.B) Not possible; magnetic poles always occur in pairs.C) Possible; magnetic monopoles are actually quite common.D) Winner of the 2007 Nobel Prize for Physics: you! Well, your friend, unless you nonchalantly tell him that

the metal is worthless, offer him a dollar for it, then publish your amazing findings in a reputable peer-reviewed journal.

25. A charged particle (+q) enters the uniform magnetic field B (–y direction) shown on the right, traveling with a horizontal velocity v (+x direction). In what direction will a force be exerted on the particle by the field?A) –x C) +z E) none of these!B) +y D) –z

26. Which of the following changes will not result in an increase in the magnitude of the force on the particle?A) increase the charge on the particle. C) increase the magnetic field strength.B) increase the speed of the particle. D) increase the mass of the particle.

27. Which of the following will not result in the reversal of the direction of the force on the particle?A) change the charge from +q to –q.B) change the velocity to the –x direction.C) change the magnetic field to the +y direction.D) reverse both the velocity and the field directions at the same time.E) you cannot reverse the direction of the force. None of these changes would be effective.

28. A charged particle (+q) enters a uniform electric field E (+y direction). It is traveling in the +x direction with a constant speed. In what direction should you orient a uniform magnetic field B so that the particle will con-tinue to travel undeflected in a straight line?

v

A) +x B) –x C) +y D) +z E) –z

Exam III Part 01

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29. What should be the magnitude of the magnetic field be for the particle to remain undeflected?A) B = E B) B = vE C) B = E/v D) B = qE E) B = E/q

30. The mass spectrometer shown consists of a ve-locity selector and a de-tection chamber. The path of a positively charged ion is shown. At which of the indicated positions would you place your detector? B

31. Once you have deter-mined whether to place the detector at A or B, how will you fine-tune the position?A) The more massive

the ion is, the greater the radius of its curved path. It will strike farther from the axis.

B) The greater the charge on the ion, the greater the radius of its curved path.C) The stronger the magnetic field B2, the greater the radius of the curved path.D) The slower the velocity of the ion, the greater the radius of the curved path.E) None of these parameters affect where the ion strikes. The only relevant parameter is d, the separation of

the selector plates shown on the figure. Greater plate separation means greater radius.32. If you are using the same mass spectrometer to analyze negative ions, what adjustment do you need to make?

A) You have to change the direction of B1 to make it out of the page. Keep E pointing down.B) You have to change the direction of E (point it up). Leave B1 alone.C) Leave the velocity selector alone. Move your detector: the particle will curve in the opposite direc-

tion.D) You have to both reverse the direction of B2 and move the detector to the opposite side.E) Make no adjustments at all; the sign of the charge on the ion makes no difference. You might have to

nudge the detector a smidge up or down if the ion is more or less massive, but that’s all.Two long, straight parallel wires are oriented parallel to the y–axis. The current flowing through each wire points in the +y direction. Assume that I1 = 2I2. The plane shown (xz–plane) is perpendicular to the plane of the page (xy–plane).33. At the point on the xz–plane exactly at the midpoint between the wires, the

total magnetic field B = B1 + B2

A) is B = 3µoI/πd, in the +z direction.B) is B = 2µoI/πd, in the –z direction.C) is B = 2µoI/πd, in the +z direction.D) is B = µoI/πd, in the +z direction.E) is B = µoI/πd, in the –z direction.

34. At point P, the direction of the total magnetic field is

A

x x x x xx x x x xx x x x xx x x x xx x x x xx x x x xx x x x xx x x x xx x x x xx x x x x

B

B2

P

A) +x B) –x C) -y D) +z E) –z35. What is the direction of the force on wire 2 because of the magnetic field created by wire 1? Wire 2 is

A) pulled directly towards wire 1 (–x direction). D) pushed directly away from wire 1 (+x direction).B) pushed out of the plane of the paper (+z direction). E) pulled into the plane of the paper (–z direction).C) neither pushed nor pulled nor poked nor tugged nor anything. No force on wire 2 because wire 1!

36. We have rearranged our wires, and now they are perpendicular to each other. What is the direction of the total magnetic field at A?A) The field points out of the page (+z).B) The field points into the page (–z).C) The field has two components in the +x and +z directions.D) The field has two components in the +y and +z directions.E) The field has two components in the +x and +y directions.

Exam III Part 01

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37. Tell me about the force on the horizontal wire because of the magnetic field created by the vertical wire.A) The entire wire is pulled up in the +y direction.B) The entire wire is pushed out of the plane of the page (+z).C) The entire wire is pulled into the plane of the paper (–z).D) The left half is pulled down (-y) while the right half is pulled up (+y). The wire spins!

38. The double–wound solenoid shown carries 5A of current on the outer windings, and 15A of current on the inner. The mag-netic field inside the solenoidA) is zero.B) points straight to the left.C) points straight to the right.D) points straight out of the

page.

39. If you double the number of windings on the outer solenoid without changing the current through either,A) the net magnetic field doubles, but the direction does not change.B) the net magnetic field is cut in half, and the direction is opposite.C) there is an increase in field strength (not a doubling, though). The field direction has also reversed.D) the field strength decreases (not by half, though). The direction of the field remains the same.E) there is no change in the net field. If the currents are unchanged, so is the total field strength.NOTE: I accidentally cut the correct answer! Everyone gets credit for this one! However, when this shows up again on the final exam, you can be certain that I won’t make the same mistake twice, so you should probably know why answer D is correct and the rest of them aren’t.

40. The rectangular loop of wire shown initially lies in the xy plane of the page as shown. The uniform magnetic field shown remains constant in the +x direction, and a current is induced when the loop is con-nected to a voltage source (not shown). What is the the force on each segment of the loop?A) Top: F = IwB (+x) C) Top: F = IwB (+z)

Right: F = IhB (-y) Right: F = IhB (-z)Bottom: F = IwB (–x) Bottom: F = IwB (–z)Left: F = IhB (+y) Left: F = IhB (+z)

B) Top: F = 0 D) Top: F = IwB (+x)Right: F = IhB (+z) Right: F = 0Bottom: F = 0 Bottom: F = IwB (–x)Left: F = IhB (–z) Left: F = 0

41. As a result of this current, what happens to the loop?A) There is a net force on the loop F = ILB in the +x direction, where

L is the length of the loop perimeter: L = 2(h + w). The loop starts to slide to the right, off the page.B) Everything in answer A is correct, except the direction of the net force. The force will oppose the current,

so the loop will start to slide to the left, right into these words.C) There is no net force on the loop, but there is a net torque with magnitude τ = IAB, where A is the

area of the loop: A = h·w. The loop will spin with respect to the y-axis, as the right vertical seg-ment comes out of the plane of the paper and the left segment moves into the plane of the paper.

D) Once again, the previous answer is almost right. The magnitude of the torque will be τ = IAB, but A is half the area of the loop: A = ½(h·w). The direction of the spin in the previous answer is correct.

E) Answer C gets the magnitude right, but the direction wrong. The loop spins so that the top segment comes out of the plane of the paper and the bottom segment is pushed into the plane of the page.

5A 15A

Exam III Part 01

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42. The loop shown is rotated in the uniform magnetic field as shown above. At which of the above positions is the magnetic flux through the loop minimum? C

43. We want to use the magnet shown to induce a current in the closed loop of wire. Which direction should we move the magnet in order to create a current flowing in the direction shown on the figure?A) Directly to the left (-x direction).B) Directly to the right (+x direction).C) Either way, it does not matter.

44. If we keep the magnet stationary, and spin the loop counterclockwise (loop is initially in the yz-plane, spin with respect to z-axis),A) the magnetic flux through the loop increases until the loop is

horizontal (in the xz-plane).B) the magnetic flux decreases until the loop is horizontal.C) the magnetic flux remains constant because the magnetic field remains constant.D) the flux changes, but there is no way to tell if it is increasing or decreasing unless we have the area of the

loop and the number of coils that make up the loop.

45. The metal rod shown falls (-y direction) through the uniform magnetic field (+z direction) under the influence of gravity. Does that bulb light? It isn’t plugged into the wall socket or anything.A) The bulb will light up. B) No it won’t.

46. Describe the flux change.A) ∆Φ/∆t = 0. The bulb does not light, so there is no voltage across the wire.B) ∆Φ/∆t = A(∆B/∆t). The loop area remains constant while the magnetic field

changes with time.C) ∆Φ/∆t = [∆(BA)]/∆t. Both the field and the area are changing simultaneously.D) ∆Φ/∆t = BLv, where L is the length of the rod and v is its speed. Flux remains

constant because the speed is constant. This is why the bulb won’t light.E) ∆Φ/∆t = BLv, but the speed is increasing as the rod falls. Not only does the bulb light, but it actually

gets brighter as the rod continues to fall. I hope you thought of some way to stop it before it smashes into the bulb and shatters it into a million pieces.

47. The coil on the left is wound around an iron core and attached to an AC power supply. When the AC power supply is switched on,A) The bulb lights.B) The bulb will not light, it isn’t connected to the power supply!

48. The figure depictsA) a step-up transformer. The iron core and the coils amplify the

primary source voltage, increasing it so that there’s enough secondary voltage to light the bulb.

B) a step-down transformer. The secondary voltage will be smaller than the primary voltage. This is probably the schematic for that low-voltage lighting system that your dad just installed along the front walkway.

49. The current through the primary coilA) is greater than the current through the secondary coil, because there are more coils.B) is less than the current through the secondary coil because the voltage is greater.C) is equal to the current through the secondary coil because the primary power matches the secondary

power.

BA C D

Exam III Part 01

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Problem 01

(5 points) Reduce the five resistors shown to a single equivalent resistance.

Replace R1 and R2 with RA:

€

RA = R1 + R2RA =10Ω+ 5Ω =15ΩReplace R4 and R5 with RB:

€

RB = R4 + R5RB = 5Ω+ 20Ω = 25ΩReplace RA, R3, and RB with RC:

€

1Rc

=1RA

+1R3

+1RB

1Rc

=115Ω

+110Ω

+125Ω

Rc = 4.8Ω(5 points) How much current does the circuit draw? How much power is dissipated?

€

V = IRC10V = I 4.8Ω( )I = 2.1A

€

P = V2

RC

P =10V( )2

4.8Ω( )= 21V

(10 points) Use Ohm’s Law to determine the current through and voltage across each resistor.

Same voltage across RA, R3, and RB: VA = V3 = VB = 10V

€

V = I3R310V = I3 10Ω( )I3 =1A

Same current through R1 and R2: I1 = I2 = IA

€

V = IARA

10V = IA 15Ω( )IA = 0.67A

€

V1 = I1R1V1 = 0.67A( ) 10Ω( )V1 = 6.7V

€

V2 = I2R2V2 = 0.67A( ) 5Ω( )V2 = 3.3V

Same current through R4 and R5: I4 = I5 = IB

€

V = ICRC10V = IC 25Ω( )IC = 0.40A

€

V4 = I4R4V4 = 0.40A( ) 5Ω( )V4 = 2.0V

€

V5 = I5R5V5 = 0.40A( ) 20Ω( )V4 = 8.0V

Exam III Part 01

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Problem 02

Use Kirchhoff’s Laws to determine the current across each of the resistors in the circuit.

Junction Rule

€

I1 − I2 − I3 = 0I1 = I2 + I3Loop 1: Upper Interior

€

V1 − I1R1 − I2R2 +V2 = 02I1 + 4I2 =12Loop 2: Lower Interior (note: you have to be careful with sign because you have to go against the current for some part of the loop!!! I looped ccw)

€

V3 −V2 + I2R2 − I3R3 = 04I2 − 8I3 = 0Loop 3: Complete Outer Perimeter

€

V1 − I1R1 − I3R3 +V3 = 02I1 + 8I3 =12Solve the system of equations:

€

I1 = I2 + I32I1 + 4I2 =122I1 + 8I3 =12

€

I1 = 7228 A = 2.6A

I2 = 4828 A =1.7A

I3 = 2428 A = 0.9A

Exam III Part 01

Page 8

I1

I2

I3

Problem 03

The mass spectrome-ter shown is to be calibrated for singly ionized lithium (Li+). You will, of course, remember that lithium has an atomic 6.94 amu. You will, of course, also remem-ber that an amu =1.67x10-27 kg.

If the electric field in the velocity selector is fixed at E=1.5x103V/m, what magnetic field B1 is required to select ions traveling at 1x105 m/s?

€

FE = qE↓FB = qvB1↑FE + FB = 0qE − qvB1 = 0E = vB11.5 ×103 Vm( ) = 1×105 ms( )B1B1 = 0.015T

If B2 has the same magnitude (note opposite direction) as B1, what is the radius of the deflected ion beam?

€

FB = mac

qvB2 = m v 2

r

r =mvqB2

r =6.94amu( ) 1.67 ×10−27 kg

amu( ) 1×105 ms( )1.6 ×10−19C( ) 0.015T( )

r = 0.48m

Exam III Part 01

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