ǫ 0 = C 2 /N m 2,

Transcription

1 Version 001 review unit chiu This print-out should have 45 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. The long negatively charged rod points 3. Ib, IIb, IIIb correct 4. Ia, IIb, IIIb 5. Ia, IIb, IIIa 6. Ia, IIa, IIIa 7. Ib, IIa, IIIb Figure above shows a portion of long, negatively charged rod. You need to determine the potential difference V A V B due to the charged rod. Use the convention that up is along the y direction. Consider the following statements: Ia. Direction of the path is along the y direction Ib. Direction of the path is along the -y direction IIa. The sign of V A V B is positive IIb. The sign of V A V B is negative Now bring a test charge q from A to B. Consider the following statements: IIIa. The sign of the potential difference V A V B due to the rod depends on the sign of the test charge q. IIIb. The sign of the potential difference V A V B due to the rod does not depend on the sign of the test charge q. Choose the correct choice: 1. Ib, IIb, IIIa. Ib, IIa, IIIa 8. Ia, IIa, IIIb Ib is correct. Since we want to determine V A V B, A is the final point and B is the initial point, so the path is from B to A. IIb is correct. Notice that with the source charge on the rod being negative, it generates a downward electric field. By inspection the vector E is parallel to the path, so V = V A V B < 0. IIIb is correct. The potential difference is a property of E generated by the source charges. It is independent of the sign or magnitude of the test charge q. Flat Parallel Conductors points Two flat conductors are placed with their inner faces separated by mm. If the surface charge density on inner face A is 97 pc/m and on inner face B is 97 pc/m, calculate the electric potential difference V = V A V B. Use = C /Nm. Correct answer: V. et: = C /N m, σ = 97 pc/m = C/m, and d = mm = 0.00 m. The electric field between two flat conductors is E = σ. Plate A is positively charged and therefore at a higher potential than the negatively charged

2 Version 001 review unit chiu plate B, so we know the answer must be positive. The magnitude of V is given by V = Ed = σd = C/m 0.00 m C /N m = V. Intuitive reasoning: et us place a positive charge in between the two plates and release it. It will be repelled by the positive plate and attracted to the negative plate. The natural tendency for a positive charge is to move from high potential to low potential. In other words, the positively charged plate is at a higher potential. This agrees with the math results where V f V i = E l. TVTubeMI17p points In a television picture tube, electrons are boiled out of a very hot metal filament placed near a negative metal plate. These electrons start out nearly at rest and are accelerated toward a positive metal plate. They pass through a hole in the positive plate on their way toward the picture screen, as shown in the diagram. Hot filament Plate E F = ee Plate v =? The high-voltage supply in the television set maintains a potential difference of 145 V between the two plates, what speed do the electrons reach? Use m e = kg and q e = C and assume that this is not relativistic. Correct answer: m/s. The net energy remains constant throughout the whole process. We can use the following train of logic: E = 0 U K = 0 q V K = 0 e V K = 0 K = e V Then K = K f K i = C145 V = J. = 1 mv f 0 K v f = m.3 10 = 17 J kg = m/s. Interestingly, a more realistic accelerating voltage for a modern picture tube television is in the range of 17kV 5kV. In this range, the electrons reach relativistic speeds and cannot be treated classically. Three parallel plate capacitors points Given three parallel conducting plates which are aligned perpendicular to the x-axis. They are labeled, from left to right as plate 1, and 3 respectively. The total charges on the corresponding plates are Q 1 = 5q, Q = 3q and Q 3 = q. The width of the gap between 1 and is d which is the same as the width across and 3. Determinethe V = V 3 V 1. Thatisgoing from the leftmost plate to the rightmost plate. 1. q/ad. 4q/Ad

3 Version 001 review unit chiu q/Ad 4. 9q/Ad 5. 7q/Ad 6. 6q/Ad 7. 3q/Ad 8. 5q/Ad 9. 7q/Ad correct One may regard the 3-plate system as a composite system which involves two capacitor systems with the 1-capacitor followed by the 3-capacitor. The 1-capacitor has charges Q 1 and Q Q 3, i.e charges of 4q and 4q respectively. The 3-capacitor has charges Q 1 Q and Q 3, i.e charges of q and q respectively. The potential difference is V 3 V 1 = E gap,1 d E gap,3 d V 3 V 1 = 5q/Ad V 3 V 1 = 7q/Ad q/ad Digression: Notice that E is pointing to the left. This implies that the potential hill has an upward slope to the right. Going from plate 1 to plate 3, corresponds to moving to the right, which is climbing the potential hill. This implies V 3 V 1 > 0. The change in potential energy points Consider the setup in Figure above. What is the change in potential energy U = U C U D, in moving an electron from D to C? 1. keqs a 1 b 1. keqs a 1 b 3. -keqs a 1 b keqs a 1 b 5. -keqs a 1 b 1 6. keqs a 1 b correct 7. -keqs a 1 b 8. kqs a 1 b 9. keqs a 1 b 10. kqs a 1 b V C V D = kqs x 3 dx 1 = kqs d x = kqs 1 a 1 b. 1 Multiplying eq1 by the electronic charge -e, we arrive at the potential energy difference from D to C is given by U C U D = evavb = keqs a 1 b.

4 Version 001 review unit chiu Intuitivereasoning onthesign of U: Natural tendency of the motion is from high potential energy to lower potential energy. Since when the electron is released it should move from C to D, so U C > U D. Alternative explanation: et thecenter ofthedipolebeattheorigin. At a distance x along the ˆx direction V dipole x = V q x s V dipole x = V q x s kq xs/ kq xs/ V dipole x = xs/xs/ x s/ So, we have V C x V D x = kqs kqs x a 1 b Check: E is along the -ˆx direction, V C is expected to be lower than V D. 17P70MI 006 part 1 of points. Q R Q R Q R 5. Q R Q correct R We may write down the potential difference as V A,B = V A,B,plastic V A,B,glass V glass is constant in the region between A and B, and hence V A,B,glass = 0. V A,B = V A,B,plastic V B V A = 1 Q 1 r B 1 Q 1 r A Now, we substitute r A = and r B = R. V B V A = Q R A thin spherical shell of radius R 1 made of plastic carries a uniformly distributed negative charge Q 1. A thin spherical shell of radius R made of glass carries a uniformly distributed positive charge Q. The distance between centers is, as shown in the above figure. Find the potential difference V B V A. ocation A is at the center of the glass sphere, andlocationb isjustoutsidetheglasssphere. 1. Q 1 1 R 007 part of points Find the potential difference V C V B. ocation B is just outside the glass sphere, and location C is a distance d to the right of B. 1. Q 1 d 1 Q R R d. Q 1 R d 1 R Q d Q 1 3. d 1 Q R R d

5 Version 001 review unit chiu Q 1 R d 1 Q 1 1 R 1 d 5. Q 1 R d 1 Q 1 1 R 1 d Q 1 6. R d 1 R Q R R d Q 1 7. d 1 Q R R d Q 1 8. R d 1 R Q correct R R d 9. Q 1 R d 1 R Q R R d The potential difference can be expressed as V B,C = V B,C,glass V B,C,plastic V C V B = Q 1 1 r C,glass r B,glass Q r C,plastic where the different distances are: r B,glass = R r C,glass = R d r B,plastic = R r B,plastic r C,plastic = R d V C V B = Q 1 R d 1 R Q R R d 008 part 3 of points Suppose the glass shell is replaced by a solid metal sphere with radius R carrying charge Q. Which of the following statements concerning the new potential difference V B V A is true? 1. The new potential difference is equal to the old potential difference from part 1. correct. The new potential difference is greater than the old potential difference from part The new potential difference is less than the old potential difference from part 1. Since E = 0 everywhere inside the metal sphere, V AB,metal = 0, the same as for the glass shell. Since the plastic shell is unchanged, the potential difference V B V A must be the same. Shell Game part 1 of 10.0 points Consider a system of a metallic ball with net charge q 1 and radius R 1 enclosed by a spherically symmetric metallic shell with net charge q, inner radius R and outer radius R 3. If q is the charge on the outside surface of the shell and q the charge on its inside surface, then q q = q. R 3 R 1 R q q 1 O B q C q Find the potential at C. OB = b and OC = c.

6 Version 001 review unit chiu V C = k q 1 q R 3 k q 1 R k q 1 R 1. V C = k q 1 c 3. V C = k q 1 q c 4. V C = k q 1 c 5. V C = k q c k q 1 k q 1 R b 6. V C = k q 1 q r 3 k q 1 b k q 1 R 1 7. V C = k q 1 q c 8. V C = k q c 9. V C = k q 1 q k q 1 k q 1 R 3 R b 10. V C = k q 1 q correct c C is outside of the entire charge distribution a distance c from the center, so the enclosed charge Q encl = q 1 q can be treated as a point charge, and V C = k q 1 q c 010 part of 10.0 points Determine the potential at B. 1. V B = k q 1 q R 3 k q 1 b k q 1 R 1. V B = k q b 3. V B = k q 1 q R 3 k q 1 R k q 1 R 1 4. V B = k q c 5. V B = k q 1 b 6. V B = k q 1 q c 7. V B = k q 1 q c 8. V B = k q 1 q b k q 1 k q 1 R b. 9. V B = k q 1 b 10. V B = k q 1 q R 3 k q 1 R k q 1 b correct B isbetweentheshellandthesphere. Consider a Gaussian surface through B concentric to the system. et us start from the inside and use superposition to add contributions as we go outward. We are outside of the sphere, so we can treat its charge q 1 as a point charge, and its potential is V 1 = k q 1 b. The inner surface of the shell carries an induced charge of q = q 1 so its potential is V = k q R = k q 1 R. The outer surface of the shell carries a charge of q = q 1 q, so its potential is V 3 = k q R 3 = k q 1 q R 3. Thus the total potential is V B = V 1 V V 3 = k q 1 b k q 1 R k q 1 q R 3. Four Charges in a Square points Three point charges, each of magnitude q, are placed at 3 corners of a square with sides of length. The charge farthest from the empty corner is negative q and the other two charges are positive q. A O B What is the potential at point A?

7 Version 001 review unit chiu V = kq. V = kq 3. V = 1 kq 4. V = kq 5. V = 1 kq correct 6. V = 1 kq 7. V = kq 8. V = kq 9. V = kq 10. V = kq The length of the diagonal is, so 5 mm mm 3 mm 1 mm Calculate V 1 V 4. Correct answer: 400 V. Here we simply add the potential differences we ve already found: V 14 = V 1 V 3 V 34 = 160 V80 V160 V = 400 V. The magnetic field of moving electron points V = i V i = kq k q kq = kq 1. P 5 P 6 P 4 θ θ e θ θ d P 1 P 3 P CapacitorMI17p103 v points An isolated large-plate capacitor not connected to anything originally has a potential difference of 800 V with an air gap of 5 mm. Then a plastic slab 3 mm thick, with dielectric constant 6, is inserted into the middle of the air gap as shown in the figure below. An electron is moving horizontally to the right with speed m/s. Each location is d = 5 cm from the electron, and the angle θ = 31. Give the magnetic field at P 3 using theconventionthatoutofthepageispositive T T

8 Version 001 review unit chiu T T T T T correct T et : v = m/s, q = C, d = 5 cm = 0.05 m, and θ = 31. The magnitude of the field at P 3 is B = µ 0 qvsinθ d = T m/a C = T m/ssin m Using the right hand rule and remembering that q is negative, the magnetic field is out of the page. BDirTwoCompMI18x points Consider the following diagram. N 1 The wire rests on top of two compasses. When no current is running, both compasses point North the direction shown by the gray arrows. When the current runs in the circuit, the needle of compass 1 deflects as shown. In what direction will the needle of compass point? 1. West. North 3. South 4. Southeast 5. Southwest 6. East 7. Northeast 8. Northwest correct At location 1, current flows to the left South to make the compass deflect Northeast. Thus, at location, current flows to the right North. B due to the current, at locationbeneath the wire, is upward toward the top of the page West. Thus the net magnetic field due to the Earth and the current carrying wire is Northwest, so the compass needle will point Northwest. The actual angle will depend on the value of the current. GS6HWMI 015 part 1 of 10.0 points An electron is moving through space with a non-relativistic velocity v = v 0,0,0 and passes through the origin at time t = 0. A magnetic field detector is located at s = a,b,0. Use the Biot-Savart law for a moving charge to calculate B s, the field measured at the detector due to the electron passing through the origin. Use the RHR to verify the vector direction of your answer. 1. B = µ 0 a b ẑ

9 Version 001 review unit chiu B = µ 0 3. B = µ 0 4. B = µ 0 5. B = µ 0 6. B = µ 0 7. B = µ 0 8. B = µ 0 a b ẑ a b ŷ a b ŷ 3/ a b ẑ correct 3/ a b 3/ ẑ a b 3/ ŷ a b ŷ We are given everything needed for the calculation except for ˆr, which is given by ˆr = s s = a,b,0 a b Substituting this into the B-S equation, and note that r = a b B = µ 0 e v 0,0,0 a b a,b,0 a b B = µ 0 a b ẑ 3/ 016 part of 10.0 points Nowcalculate B q,wherethedetectorisnow located at q = a,0,b. 1. B = µ 0. B = µ 0 3. B = µ 0 4. B = µ 0 5. B = µ 0 6. B = µ 0 a b ŷ 3/ a b 3/ ẑ a b ŷ a b ẑ a b ẑ 3/ a b ŷ correct 3/ 7. B = µ 0 8. B = µ 0 a b ŷ a b ẑ Note that there will be no change in the magnitudeof B. Theonlychangethatwilloccur is in the direction. To find the new direction, evaluating the following cross-product yields v 0,0,0 a,0,b = v 0 bŷ The B-S law is applied to an electron with charge e, so an extra negative sign comes into the equation. The new direction will be ŷ = ŷ. Hence, the answer is B = µ 0 a b ŷ 3/ Curved Wire Segment 017 part 1 of 10.0 points Consider a current configuration shown below. A long effectively infinite wire segment isconnectedtoaquarterofacirculararcwith radiusa. Theotherendofthearcisconnected to another long horizontal wire segment. The current is flowing from the top coming down verticallyandflowstotherightalongthepositive x-axis. y I a What is the direction of the magnetic field at O due to this current configuration? O O 1. along the positive y-axis. 135 counterclockwise from the x-axis counterclockwise from the x-axis x

10 Version 001 review unit chiu counterclockwise from the x-axis 5. along the negative y-axis For the straight sections, we apply the formulas derived from the figure below, a counterclockwise from the x-axis 7. perpendicular to and into the page θ 1 θ I 8. along the negative x-axis 9. perpendicular toand out ofthepagecorrect 10. along the positive x-axis From the Biot-Savart law we know that d B Id l ˆr. One should first verify that the magnetic field at O contributed by any infinitesimal current element along the current configuration given is perpendicular to the page, which is coming out of the page. Therefore the resulting magnetic field must also be pointing out of the page. 018 part of 10.0 points et I = 1.6 A and a = 0.94 m. Whatisthemagnitudeofthemagneticfield at O due to the current configuration? Correct answer: T. et : I = 1.6 A, and a = 0.94 m. Consider first the one-quarter of a circular arc. Since each current element is perpendicular to the unit vector pointing to O, we can write B arc = µ π/ 0 I 0 = µ 0 I π a = µ 0I 8a. adθ a B = ẑ µ θ 0I dθsinθ a θ 1 = ẑ µ 0I a cosθ 1 cosθ whereẑ istheunitvectorperpendiculartothe plane of the paper that points to the reader. For the downward y-directed current, θ = π and θ 1 = 0. For the horizontal x-directed current θ = π and θ 1 = π. Thus we find the contribution at the point Oforthemagneticfieldfrom thelong vertical wire is same as in the long horizontal wire, and the sum is equal to B = ẑ µ 0I a 11 = ẑ µ 0I πa, /noindent the same as in a long straight wire. Adding the contributions from the straight sections and the arc, B tot = µ 0I 8a µ0 I πa = µ 0I 1 a 4 1 π = T m/a1.6 A 0.94 m 4 1 π = T. Conceptual 17 Q01

11 Version 001 review unit chiu points 4. µ 0I π The figure represents two long, straight, parallel wires extending in a direction perpendic- R correct h ulartothepage. Thecurrent intherightwire 5. µ 0 I π R 1π runs into the page and the current in the left runs out of the page. 6. µ 0I π π R h a b c What is the direction of the magnetic field created by these wires at location a, b and c? b is midway between the wires. 1. up, zero, down. down, down, up 3. down, zero, up 4. down, up, down correct 5. up, down, up 6. up, up, down By the right-hand rule the right wire has a clockwise field and the left wire a counterclockwise field. B of Asymmetric oop points The circuit shown above consists of a battery and a Nichrome wire and has a conventional current I. At the center of the semicircle, what is the magnitude of the magnetic field? You may assume h. 1. µ 0 I πr π I. µ 0 3. µ 0 R 1π I R π 7. µ 0I 8. µ 0I π π R h π R h Examining the figure, we see that the semicircular section and the lower straight wire contribute to B. The upper straight wire portions have d l ˆr, so do not contribute. ikewise, the short segment of length h may be neglected. Consequently, we have a superposition of the magnetic fields of a halfloop and a straight wire of length, B = 1 µ 0 πi R µ 0 I h = µ 0I π R h AlnicoMagnetMI18x07 01 part 1 of 10.0 points A particular alnicoaluminum, cobalt, nickel, and iron bar magnet magnet A has a mass of 10 g. It produces a magnetic field of magnitude T at a location 0. m from the center of the magnet, on the axis of the magnet. Approximately what is the magnitude of the magnetic field of magnet A a distance of 0.4 m from the center of the magnet, along the same axis? You may assume that the physical dimensions of the magnet are much smaller than the distances involved in the problem. Also, use µ 0 = T m/a. Correct answer: T.

12 Version 001 review unit chiu The expression for the magnitude of the field of a magnetic dipole is Bdipole µ 0 µ = r 3. One way to solve this is to find µ at r = 0. m and then solve for B at r = 0.4 m. Alternately, note that B 1/r 3, so Br 3 is a constant: B 1 r 3 1 = B r T0. m 3 = B0.4 m 3 B = T. Note that doubling the distance causes B to change by a factor of 1/ 3 = 1/8. 0 part of 10.0 points If you removed the original magnet and replaced it with a magnet made of the same material but with a mass of 30 g magnet B, approximately what would be the magnetic fieldat a location0.mfrom thecenter ofthe magnet, on the axis of the magnet? Correct answer: T. Increasing the mass by a factor of 3 causes the dipole moment to increase by a factor of 3. This is like placing 3 identical magnets end to end. As a result, B at r = 0. m increases by a factor of 3, since B µ. Thus, B = T = T. y Whatisthemagnitudeofthemagneticfield at a distance of 11 cm along the positive x- axis? The permeability of free space is 10 7 T m/a. Assume the current density is constant throughout the conductor. Correct answer: T. Basic Concepts: Magnetic Field due to a ong Cylinder B = µ 0I πr. Principle of Superposition. Our goal is to model the given situation, which is complex and lacks symmetry, by adding together the fields from combinations of simpler current configurations which together match the given current distribution. The combination of the currents in Fig. will do so if we choose I cyl and I hole correctly. y I cyl = 4 3 I r x y x Hole r I hole = 1 3 I x Since the current is uniform, the current Off Centered Hole points A total current of 5 ma flows through an infinitely long cylinderical conductor of radius 3 cm which has an infinitely long cylindrical hole through it of diameter r centered at r along the x-axis as shown. density J = I is constant. Then A J = I cyl A cyl = I hole A hole Clearly, A cyl = πr, and A hole = πr 4, so I hole = I cyl 4.

13 Version 001 review unit chiu Note: TheminussignmeansI hole isflowing inthe directionopposite I cyl and I, as it must if it is going to cancel with I cyl to model the hole. We also require I = I cyl I hole. We then havei cyl = 4 3 I, andi hole = 1 I. With these 3 currents, the combination of the two cylinders in figure gives the same net current and current distribution as the conductor in our problem. The magnetic fields are 4 µ 0 3 I B cyl = B hole = πx µ I πxr/, so the total magnetic field is B total = B cyl B hole = µ 0I 4 6π x 1 x r = µ 0I 3xr 6π x x r = 107 Tm/A5 ma 6π keywords: 311 cm3 cm 11 cm 11 cm 3 cm = T. BreakingcoilMI points A current-carrying solenoidal coil of length is uniformly wound with N turns. Suppose the coil is now cut into half, and the original current is run through each of the resulting new solenoids with half the number of turns N/ as the original coil and half the length / as the originalcoil. We are now left with: IaOnecoilwithonlyanorthpoleandthe other with only a south pole. Ib Two smaller coils, each with a North end and a South end. Ic Two coils that don t make any magnetic field when a current runs through them. If the magnetic field created inside of one of the new coils far from the ends is B, and thatcreatedbytheoriginalcoilisb allother parameters being the same, then which of the following relations is true? IIa B = B IIb B = B IIc B = B 8 IId B = B 1. Ia, IIa. Ic, IIc 3. Ib, IIb 4. Ic, IIa 5. Ic, IId 6. Ia, IIb 7. Ia, IId 8. Ic, IIb 9. Ib, IIc 10. Ib, IIa correct Magnetic strength of a coil is proportional to N/, where N is the total number of turns andisthelengthofthecoil. When carrying a current, each smaller coil still acts like a magnetic dipole, so must have a North and a South pole. Hence, Ib is correct.

14 Version 001 review unit chiu Since magnetic field strength is proportional to N/ and this ratio does not change when the coil is divided, we still have B = B. Hence, IIa is correct. Cable part 1 of 10.0 points The figure below shows a straight cylindrical coaxial cable of radii a, b, and c in which equal, uniformly distributed, but antiparallel currents i exist in the two conductors. a i out b c i in O F E D C r 1 r r 3 r 4 Which expression gives the magnitude of themagneticfieldintheregionr 1 < catf? 1. Br 1 = µ 0i πr 1. Br 1 = µ 0ir 1 πc correct 3. Br 1 = µ 0ir 1 πb 4. Br 1 = µ 0ir 1 b πr 1 a b 5. Br 1 = µ 0ia r 1 πr 1 a b 6. Br 1 = µ 0ia b πr 1 r 1 b 7. Br 1 = µ 0i πr 1 8. Br 1 = µ 0ir 1 πa 9. Br 1 = Br 1 = µ 0ia r1 b πr 1 a b Ampere s aw states that the line integral B d l around any closed path equals µ 0 I, where I is the total steady current passing through any surface bounded by the closed path. Considering the symmetry of this problem, we choose a circular path, so Ampere saw simplifies to B πr 1 = µ 0 I en, where r 1 is the radius of the circle and I en is the current enclosed. For r 1 < c, B = µ 0I en πr 1 µ 0 i πr 1 πc = πr r µ 0 i 1 c = πr 1 = µ 0ir 1 πc. 06 part of 10.0 points Which expression gives the magnitude of the magnetic field in the region c < r < b at E? 1. Br = µ 0ir b πr a b. Br = µ 0i πr correct 3. Br = µ 0ia r b πr a b 4. Br = µ 0ir πc 5. Br = 0 6. Br = µ 0ir πb 7. Br = µ 0ia b πr r b

15 Version 001 review unit chiu Br = µ 0ia r πr a b 9. Br = µ 0i πr 10. Br = µ 0ir πa For c < r < b, B = µ 0I en πr = µ 0 i πr = µ 0i πr. Ampere s law and a long solenoid points Consider a solenoid with the setup shown in the figure. The total number of turns is N within a total length. Apply Ampere s law for the path along the boundary of the rectangle shown, where the number of turns within the gray area is given by N. Choose the correct pair of statements below. Ia. Bd = µ 0 N I. Ib. Bd = µ 0 N I. Ic. Bd = µ 0 N I. Id. Bd = µ 0 N I. Ie: Bd = µ 0 I. IIb. B outside the solenoid is negligible compared to B inside. 1. Ia, IIb. Ic, IIa 3. Ib, IIa 4. Id, IIa 5. Ic, IIb correct 6. Ie, IIb 7. Ie, IIa 8. Ib, IIb 9. Id, IIb 10. Ia, IIa By inspection the correct Ampere s law expression along the path specified is given by: Bd = µ 0 N I, where the condition that B outside is negligible was used. So the correct answer is the pair: Ic and IIb. This leads to B = µ 0 N I N = µ 0 I, d which is the expected answer. Exam3.P points The figure shows a large number N of closely packed wires, each carring a current I out of the page. The width of this shweet of wires is. Find B in the region between two parallel current sheets with equal currents running in opposite directions. IIa. B outside the solenoid has the same magnitude as B inside and is opposite in direction.

16 1. B = µ 0N I d. B = µ 0dI 3. B = µ 0N I d 4. B = µ 0dI 5. B = µ 0N I 6. B = µ 0dI 7. B = µ 0N I d 8. B = µ 0N I 9. B = µ 0N I correct From the diagram it is clear that there is cancellation of the vertical components of magnetic field contributed by two wires to the left and the right of the observagtion location. Therefore the direction of the magnetic field must be to the left above the current sheet and to the right below : Version 001 review unit chiu Therefore, B d l = B top w = µ 0 I inside path N = µ 0 wi, since there are N/ current-carrying wires per meter, and a width w of the enclosing path. So the top sheet contribution along is given by B top = µ 0N I. Based on the superposition principle, the contributions between the two sheets are pointing in the same direction, they should add. So we have the resultant magnetic field B = B top = µ 0N I. Moving a Charge points It takes 143 J of work to move. C of charge from the negative plate to the positive plate of a parallel plate capacitor. What voltage difference exists between the plates? We first work out the contribution to the magneticfieldduetothetopwiresheet alone. Use Ampere s law, and go counterclockwise around the closed rectangular path. Along the sides of the path, B d l = 0, since B is perpendicular to d l. Along the upper part of the path, B d l = B top w. Along the lower part of the path, B d l = B top w. Correct answer: 65 V. et : W = 143 J and q =. C. The voltage difference is V = W q = 143 J. C = 65 V. Delta V points You move from location i at 5,5, m to location f at 5,4,9 m. All along this path is a uniform electric field whose value is E = 900,100,700 N/C. Calculate V = V f V i.

17 Version 001 review unit chiu Correct answer: 5000 V. Recalling that f V = E d l, i and noting that E is uniform so that it may come outside the integral, V may be calculated simply from E l. V = E l = [E x f x i x E y f y i y E z f z i z ] = 5000 V ong Cylindrical Insulator 031 part 1 of points Consider a long, uniformly charged, cylindrical insulator of radius R and charge density 1.5 µc/m 3. R.4 cm What is the electric field inside the insulator at a distance.4 cm < R from the axis? The volume of a cylinder with radius r and length l is V = πr l. The value of the permittivity of free space is C /N/m. Correct answer: N/C. et : ρ = 1.5 µc/m 3 = C/m 3, r =.4 cm = 0.04 m, and = C /N/m. Consider a cylindrical Gaussian surface of radius r and length l much less than the length of the insulator so that the component of the electric field parallel to the axis is negligible. r R l The flux leaving the ends of the Gaussian cylinder is negligible, and the only contributiontothefluxisfromthesideofthecylinder. Since the field is perpendicular to this surface, the flux is Φ s = πrle and the charge enclosed by the surface is Q enc = πr lρ. Using Gauss law, Φ s = Q enc πrle = πr lρ E = ρr ǫ C/m m = C /N/m = N/C. 03 part of points Determine the absolute value of the potential difference between r 1 and R, where r 1 < R. For r < R the electric field takes the form E = Cr, where C is positive. 1. V = 1 CRr 1r 1. V = CRr 1 r 1 3. V = C 1 R r 1 4. V = C R r1 5. V = CRr 1

18 Version 001 review unit chiu V = 1 C R r 1 correct 7. V = C r 1 8. V = Cr 1 9. V = C R r1 10. V = C 1 r 1 R The potential difference between a point A inside the cylinder a distance r 1 from the axis to a point B a distance R from the axis is B R V = E ds = Edr A r 1 since E is radial, so R V = Crdr = C r R r 1 r 1 R = C r 1 R V = C r 1 = 1 C R r part 3 of points What is the relationship between the potentials V r1 and V R? 1. V r1 = V R Intuitive Reasoning: The natural tendency for a positive charge is to move from A to B, so A has a higher potential. MI17p099a 034 part 1 of 10.0 points An isolated parallel plate capacitor of area A 1 with an air gap of s 1 is charged up to a potential difference of V 1i. A second parallel plate capacitor is initially uncharged, has an area A and a gap of length s filled with plastic whose dielectric constant is κ. Connect a wire from the positive plate of the first capacitor to one of the plates of the second capacitor and connect another one from the negative plate of the first capacitor to the other plate of the second capacitor. The initial voltage of the first capacitor before connection is given by V 1i. Denote the final potential differences of the first and the second capacitors to be V 1f and V f respectively and the final field in the gap the second capacitor in presence of the dielectric be E f. After the connection, let the charge acrossthesecondcapacitorbedenotedbyq. Consider the following statements: Ia. V 1f = V f Ib. V 1f > V f IIa. E f = 1 Q κ IIb. E f = 1 κ A Q A. V r1 > V R correct 3. None of these 4. V r1 < V R Since C > 0 and R > r 1, V B V A = V = 1 C R r 1 < 0. Thus V B < V A and the potential is higher at point A where r = r 1 than at point B, where r = R. IIIa. V f = E f s IIIb. V f = E f s 1. Ib, IIa, IIIb. Ib, IIb, IIIb 3. Ia, IIb, IIIa 4. Ib, IIb, IIIa 5. Ia, IIa, IIIa

19 Version 001 review unit chiu Ib, IIa, IIIa 7. Ia, IIb, IIIb 8. Ia, IIa, IIIb correct For the first capacitor alone, from Gauss law it can be shown that E = 1 Q A 1 = V 1i s 1 Solving for Q leads to: ǫ0 A 1 Q = V 1i After the connection, say that the top plate of capacitor 1 is connected to the top plate of capacitor, and bottom plate of 1 to that of the bottom plate of. By definition: V 1f = V f and the correct answer is Ia. Denote the plate charges to be Q 1 and Q. Analogous to solving for Q, we can similarly solve for the second capacitor with plate charge Q with the presence of the dielectric K, E f = 1 Q κ A Hence, the choice IIa is correct. The correct definition for the potential difference is given by V f = E f s. Hence, IIIb is the correct answer. 035 part of 10.0 points For this part of the problem, we will simplify the problem assuming that the two capacitors have identical geometry, i.e. A 1 = A = A and s 1 = s = s. There is still the dielectric slab with dielectric constant κ in capacitor. Before the connection let the initial voltage of the first capacitor by V 1i. Denote the final potential differences of the first and the second capacitor to be V 1f and V f. After the connection, let the charge across the second capacitor be denoted by Q. Consider the following statements: Ia. V 1f = V f Ib. V 1f > V f s 1 IIa. E f = 1 Q κ A IIb. E f = 1 κ Q A IIIa. V 1f = κ 1κ V 1i IIIb. V f = 1 1κ V 1i 1. Ib, IIb, IIIa. Ia, IIa, IIIa 3. Ia, IIb, IIIa 4. Ib, IIa, IIIb 5. Ib, IIb, IIIb 6. Ib, IIa, IIIa 7. Ia, IIa, IIIb correct 8. Ia, IIb, IIIb The explanation for the choices of Ia and IIa is the same as that given in part 1. Since the two potential differences are the same, we will refer to this common potential difference as V f. In terms of this common potential difference V f the total charge Q can be written as Q = Q 1 Q = ǫ0 A s V f κ ǫ0 A s V f Expressing Q in terms of V 1i we obtain ǫ0 A ǫ0 A ǫ0 A V 1i = V f κ V f s s s On dividing through with the factor of A s we have V 1i = V f κv f = 1κV f

20 V f = Version 001 review unit chiu V 1i 1κ By definition, V 1f = V f. So, IIIb is the correct answer. Potential Diagrams points Consider a conducting sphere with radius R and charge Q, surrounded by a conducting spherical shell with inner radius R, outer radius 3R and net charge Q. Q 4. kq 6R V r R R3R The charge on the inner sphere is Q, concentrated on its surface. The induced charge on the inner surface of the spherical shell is Q, so the charge on the outer surface of the spherical shell is Q net Q inner = QQ = Q. Q What potential vs radial distance diagram describes this situation? Q Q Q on surface Q on surface Q on surface 1.. kq 6R V correct kq 6R V R R3R R R3R r r The potential within a conductor is constant and the electric field within a conductor is zero. The potential for 3R < R < outside the conductors is [ ] QQ Q V r = k = k. r r For R r 3R inside the conducting shell, Q V 3R = V r = V R = k. 3R For R < r < R between the conductors, 3. kq 6R V R R3R r [ Q V r = k r Q R Q ] 3R = kq r 1. 6R

21 Version 001 review unit chiu For 0 < r R inside the conducting sphere, [ 1 V R = V r = V 0 = kq R 1 ] = 7 kq 6R 6R. Mag. field of two moving charges points 9. Ia, IId 10. Ib, IId The E-field at P points in the î direction and is nonzero. Using the Biot-Savart law, we know that the B-field is given by B = q v ˆr. Thus, the B-field is in the ˆk direction for both the charges. B of Moving Charge points Two equal and opposite charges are equidistant from point P and are moving towards eachotherwiththesamespeedv. Thecharges and point P are all in the xy plane. The E-field at point P Ia Is not the zero vector Ib Is the zero vector The B-field at point P IIa Is in the ˆk direction IIb Is in the ˆk direction IIc Is in the ĵ direction IId Is in the ĵ direction IIe Is the zero vector 1. Ia, IIe. Ia, IIb correct 3. Ib, IIb 4. Ib, IIc 5. Ia, IIa 6. Ib, IIe 7. Ia, IIc 8. Ib, IIa The electron in the figure is located at the origin and traveling with velocity v = , ,0 m/s. What is B z at point A = ,0,0m? Correct answer: 3.36 T. This is a straightforward application of the Biot-Savart aw. For B z, we have B z = µ 0 q r v x ˆr y v y ˆr x. Inserting the values of the problem, this becomes B z = µ 0 e v y r x = µ 0 ev y rx = 3.36 T WireAndCompassMI18p part 1 of 10.0 points When you bring a current-carrying wire down onto the top of a compass, aligned with the original direction of the needle and 7 mm above the needle, the needle deflects by 13 degrees, as in the figure below.

22 Version 001 review unit chiu B Earth Wire 13 Assuming the compass needle was originally pointing toward the north, what direction is the conventional current traveling in the wire, and what is the direction of the force on the compass needle due to the magnetic field caused by the current in the wire? 1. North, West Correct answer: Ultimately, we want to use the expression B wire = µ 0 I r to find the current in the wire. We know r, but not B wire yet. To find B wire, we can use what we know about the Earth s magnetic field and the deflection of the compass needle. Consider the following simple diagram: B Earth B wire Needle. North, East 3. South, East correct 4. South, West We just need to think about how the compass needle responds to the presence of the current-carrying wire to answer this question. Since the compass needle deflects to the east, that is the direction of the magnetic force on the needle due to the current in the wire. To decide which way current is traveling in the wire, we use the right hand rule. Your fingers should point toward the east, so in order for your fingers to curl around the wireand point toward the east, your thumb must point toward the south. So the current is traveling toward the south. 040 part of 10.0 points Calculate the amount of current flowing in the wire. The measurement was made at a location where the horizontal component of the Earth s magnetic field is Use B Earth = 10 5 T. µ 0 = Tm/A. From this drawing, we can write down tanθ = B wire B Earth B wire = B Earth tanθ 10 5 Ttan13 = T. Now we simply rearrange the expression above to find the current: I = B wirer µ0 = T7 mm Tm/A = A. FieldOfongWireMI18p points A long current-carrying wire, oriented North- South, lies on a table it is connected to batteries which are not shown. A compass lies on top of the wire, with the compass needle about 3 mm above the wire. With the current running, the compass deflects 9 to the

23 West. At this location, the horizontal component of the Earth s magnetic field is about 10 5 T. Version 001 review unit chiu Whatisthemagnitudeofthemagneticfield at location A, on the table top, a distance.7cm to the East ofthewire, due onlyto the current in the wire? Correct answer: T. let : B earth = 10 5 T, r compass = 3 mm = m, r A =.7 cm = 0.07 m, and θ = 9. The magnetic field due to a wire is B = µ 0 I r Since tanθ = B compass B earth, B compass = B earth tanθ. By writing the equation for the magnetic field of a wire for the compass and for point A and dividing the two equations, we obtain B A = B compass r compass r A = B earth tanθ r compass r A = 10 5 Ttan m 0.07 m = T. number of atoms points A bar magnet is aligned east - west, with its center = 0.3 m from the center of a compass as shown in the above figure. The compass is observed to deflect 50 away from north as shown, and the horizontal component of the Earth s magnetic field is known to be 10 5 tesla. Approximately how many atoms are in the bar magnet, assuming that one atom has a magnetic dipole moment of Am. Correct answer: The magnitude of the B-field is given by B = B E tanθ 10 5 tan50 B = B T The magnetic dipole moment is given by µ = B r 3 = B3 µ0 µ0 µ = m = 3.905Am Theno. ofatomsinthebarmagnetisgiven by N = µ µ atom = Am Am = Conceptual 4 Q points A normal piece of iron produces no external magnetic field. Suppose a piece of iron

24 Version 001 review unit chiu consisted of one very large domain instead of many small ferromagnetic domains. Would this piece of iron produce an external magnetic field? 1. No. It does not create magnetic field at all.. Yes. It creates magnetic field. correct 3. No. It does not create magnetic field because the net magnetic field is zero. In normal iron, each domain acts like a small bar magnet. However, the random orientation of the domains causes cancellation and there would be no cancellation and it would create magnetic field. WireSheetDensityMI points In the figure below, a large number of closely-spaced wires parallel to the z-axis form a sheet of current in the xz plane, with each wire carrying a current I in the z direction out of the plane of the figure. It is known that the magnitude of magnetic field at point P shown in figure is B. What is the algebraic expression for the number density of the sheet number of wires per unit distance along the x-axis? You can assume the dimensions of the sheet to be infinite in both x and z directions implying that there are a large number of these closely-packed wires and they are all very long. In the figure, x points to the right, y points upwards and z points out of the plane of the figure, towards you. B. µ 0 I B 3. µ 0 I 1 4. d B 5. 4µ 0 I 6. 1 d 7. d 8. 4B µ 0 I Using Ampere s law, it is easy to obtain the magnetic field due to the sheet of wire at a point above or below the sheet, in terms of the number density n of the wires. On inverting this algebraic expression, we can get the number density in terms of the magnetic field. We proceed in the following manner. First, we shall choose a rectangular closed path as shown in the figure. Using right hand rule and symmetry, it is straightforward to show that the magnetic field due to the sheet points to left at P and right at Q and has the same magnitude B at any point on the top and bottom sides of the rectangle. Now applying Ampere s law, we get B in the following manner. B. dl = µ 0 I enclosed 1. B µ 0 I correct Bx0Bx0 = µ 0 Inx B = µ 0nI

25 Version 001 review unit chiu From this, the number density can be obtained as n = B µ 0 I. Cylindrical Shell of Current points A long cylindrical shell has a uniform current density. The total current flowing through the shell is 11 ma. The permeability of free space is T m/a. 7 cm 3 cm 17 km gives the easiest solution. Consider a circle of radius r 1 centered around the center of the shell. To use Ampere s law we need the amount of current that cuts through this circle of radius r 1. To get this, we first need to compute the current density, for the current flowing through the shell. J = I A I = πrb πr a 11 ma = π[0.07 m 0.03 m ] = A/m. The current is 11 ma. Findthemagnitudeofthemagneticfieldat a point r 1 = 4.1 cm from the cylindrical axis. Correct answer: nt. et : = 17 km, r a = 3 cm = 0.03 m, r b = 7 cm = 0.07 m, r 1 = 4.1 cm = m, I = 11 ma, and µ b = T m/a. r b The current enclosed within the circle is I enc = π[r 1 r a] J = π[0.041 m 0.03 m ] A/m = A. Ampere s aw, B d s = µ 0 I enc Bπr 1 = µ 0 I enc B = µ 0I enc πr 1 = T m/a π0.041 m A = nt. r a The current I = 11 ma. Since the cylindrical shell is infinitely long, and has cylindrical symmetry, Ampere s aw