Version 013 – MT2 – chiu – (57425) 1

This print-out should have 17 questions.Multiple-choice questions may continue onthe next column or page – find all choicesbefore answering.

001 10.0 pointsConsider the setup where the potential func-tion is given by: V = 4xz + 2y − 5z2. FindEz at the point: 〈2 m, 2 m, 2.6 m〉.1. 18.02. 11.03. 49.04. 90.05. 91.06. 8.07. 38.08. 69.09. 59.010. 2.0

Correct answer: 18 N/C.

Explanation:

let : x = 2 m ,

y = 2 m , and

z = 2.6 m .

The z-component of the electric field is

Ez = − ∂V

∂z= −(4 x− 10 z)

= 10(2.6 m)− 4(2 m) = 18 N/C .

002 10.0 points

The circuit shown above consists of a bat-tery, a wire, and carries a conventional currentI. At the center of the semicircle, what is ~B?You may assume L ≫ h and that the positivedirection is out of the page.

1. −µ0

2π

2I

R(1 + π)

2.µ0I

4π

(

π

R+

2

h

)

3.µ0I

2π

(

π

R+

2

h

)

4. −µ0

2π

I

R

(

π +2

h

)

5. −µ0I

4π

(

π

R+

2

h

)

correct

6.µ0

4π

I

R(1 + π)

7. −µ0I

2π

(

π

R+

2

h

)

8.µ0

2π

I

R

(

π +2

h

)

Explanation:Examining the figure, we see that the semi-

circular section and the lower straight wirecontribute to |B|. The upper straight wire

portions have d~l ‖ r̂, so do not contribute.Likewise, the short segment of length h ≪ Lmay be neglected. Consequently, we have asuperposition of the magnetic fields of a half-loop and a straight wire of length L,

|~B| = 1

2

µ0

4π

2πI

R+

µ0

4π

2I

h

=µ0I

4π

(

π

R+

2

h

)

.

By the RHR, the direction must be into thepage, so negative.

003 10.0 pointsA total current of 69 mA flows through aninfinitely long cylinderical conductor of radius6 cm which has an infinitely long cylindrical

hole through it of diameter r centered at −r

2along the x-axis as shown.

x

y

What is the magnitude of the magnetic fieldat a distance of 10 cm along the positive x-

Version 013 – MT2 – chiu – (57425) 2

axis? The permeability of free space is 4 π ×10−7T ·m/A . Assume the current density isconstant throughout the conductor.1. 1.07117e-072. 5.67193e-083. 1.53164e-074. 5.41667e-085. 9.98555e-086. 1.40444e-077. 1.48615e-078. 3.40364e-089. 1.18788e-0710. 9.3464e-08

Correct answer: 1.48615× 10−7 T.

Explanation:Basic Concepts: Magnetic Field due to a

Long Cylinder

B =µ0 I

2 π r.

Principle of Superposition.Our goal is to model the given situation,

which is complex and lacks symmetry, byadding together the fields from combinationsof simpler current configurations which to-gether match the given current distribution.The combination of the currents in Fig. 2 willdo so if we choose Icyl and Ihole correctly.

x

yr

Icyl =4

3I

x

yr

2

Hole

Ihole = −1

3I

+

Since the current is uniform, the current

density J =I

Ais constant. Then

J =IcylAcyl

= − IholeAhole

Clearly, Acyl = π r2, and Ahole =π r2

4, so

Ihole = −Icyl4

.

Note: The minus sign means Ihole is flowingin the direction opposite Icyl and I, as it mustif it is going to cancel with Icyl to model thehole.We also require I = Icyl + Ihole. We then

have Icyl =4

3I, and Ihole = −1

3I. With these

currents, the combination of the two cylindersin figure 2 gives the same net current andcurrent distribution as the conductor in ourproblem.The magnetic fields are

Bcyl =

µ0

(

4

3I

)

2 π x

Bhole =

µ0

(

−1

3I

)

2 π (x+ r/2),

so the total magnetic field is

Btotal = Bcyl +Bhole

=µ0 I

6 π

4

x− 1

x+r

2

=µ0 I

6 π

3 x+ 2 r

x(

x+r

2

)

=(4 π × 10−7 Tm/A) (69 mA)

6 π

×

3 (10 cm) + 2 (6 cm)

(10 cm)

(

10 cm +6 cm

2

)

= 1.48615× 10−7 T .

keywords:

004 10.0 pointsWhen a dielectric is present in a charged

capacitor, the dielectric is polarized. Thefigure below shows a simple model: a layerof polarized charge forms at the upper andlower surfaces of the dielectric, leading to areduction of the “effective” plate charge anda reduced field ~E ′.

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In reality, each individual molecule of thedielectric is polarized, becoming a dipole (seethe illustration Fig 17.42 in M&I v.II). Thesum of the dipole contributions gives the elec-tric field due to the polarization of the dielec-tric Epol. The simple model shown above isjustified by assuming that the effects of thepolarized charges in the interior of the dielec-tric largely cancel, leaving only the outermost

layer of dipole charge at the top and bottomsurfaces to contribute.

Given a dielectric constant κ, plate charge Q,and plate area A, determine the magnitude ofthe polarized charge Qpol.

1. κQ

2. Q

(

1 +1

κ

)

3.Q

κ

4. Q(κ− 1)

5. Q

(

κ− 1

κ+ 1

)

6. Q

(

1− 1

κ

)

correct

Explanation:First, note that we can apply the su-

perposition principle inside the dielectric:~E ′ = ~EQ+~EQpol

. We are told a layer of chargeforms on the outer surfaces of the dielectric,

so we can treat the dielectric as though it werecomposed of parallel plates with charge Qpol;we know the electric field for this charge dis-tribution — it is (Qpol/A)/ǫ0. Using this facttogether with E ′ = E/κ, we obtain:

~E ′ = ~EQ + ~EQpol

Q

κAǫ0=

Q

Aǫ0− Qpol

Aǫ0Q

κ= Q−Qpol

Qpol = Q− Q

κ= Q

(

1− 1

κ

)

.

005 10.0 pointsThe figure represents two long, straight, par-allel wires extending in a direction perpen-dicular to the page and carrying currents ofequal magnitude. The current in the left wireruns out of the page and the current in theright runs into the page.

a b c

What is the direction of the magnetic fieldcreated by these wires at location a, b and c?(b is midway between the wires.)

1. down, down, up

2. up, up, down

3. down, zero, up

4. up, down, up

5. down, up, down correct

6. up, zero, down

Explanation:By the right-hand rule the left wire has a

clockwise field and the right wire a counter-clockwise field.

At the leftmost point, the field due to theleft wire points down, while that due to the

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right wire points up. Since

Bwire =µ0

2π

I

r,

The field due to the left wire is larger, so thesuperposition of the fields produces a down-ward field. At the center point, both fieldspoint up. At the rightmost point, the fielddue to the left wire points up, while that dueto the right wire points down. Since the rightwire is closer, it dominates, so the net field isdown.

006 10.0 points

Figure above shows a portion of long, neg-atively charged rod. You need to determinethe potential difference VB − VA due to thecharged rod. Use the convention that up isalong the +y direction.Consider the following statements:

Ia. The sign of VB − VA is positiveIb. The sign of VB − VA is negative

Now bring a test charge q from A to B.Consider the following statements:IIa. The sign of the potential difference

VB − VA due to the rod depends on the signof the test charge q.IIb. The sign of the potential difference

VB − VA due to the rod does not depend onthe sign of the test charge q.

Choose the correct choice:

1. Ia, IIb correct

2. Ia, IIa

3. Ib, IIa

4. Ib, IIb

Explanation:Ia is correct. Notice that with the source

charge on the rod being negative, it generatesa downward electric field. By inspection thevector E is antiparallel to the path, so ∆V =VB − VA > 0.IIb is correct. The potential difference

is a property of E generated by the sourcecharges. It is independent of the sign or mag-nitude of the test charge q.

007 10.0 pointsConsider the following diagram.

N 1

2

The wire rests below two compasses. Whenno current is running, both compasses pointNorth (the direction shown by the gray ar-rows). When the current runs in the circuit,the needle of compass 1 deflects as shown. Inwhat direction will the needle of compass 2point?

1. North

2. Southeast

3. West

4. East

5. South

6. Northwest

7. Northeast correct

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8. Southwest

Explanation:At location 1, current flows to the left

(South) to make the compass deflect North-west. Thus, at location 2, current flows to theright (North). ~B due to the current, at loca-tion 2 above the wire, is downward toward thebottom of the page (East). Thus the net mag-netic field due to the Earth and the currentcarrying wire is Northeast, so the compassneedle will point Northeast. The actual anglewill depend on the value of the current.

008 10.0 pointsA current-carrying solenoidal coil of length

L and radius R, L ≫ R, is uniformly woundwith N turns. Suppose the coil is now cut inhalf, resulting in two new solenoids with halfthe number of turns (N/2) as the original coil.The coils are connected to separate circuits sothat current flows through them in the samedirection as through the original coil. Whenbrought close to each other so that the cutends face each other:

(Ia) The coils repel each other.(Ib) The coils are attracted to each other.(Ic) The coils do not interact magnetically.

If the magnetic field created inside of oneof the new coils (far from the ends) is B′, andthat created by the original coil is B (all otherparameters being the same), then which ofthe following relations is true?

(IIa) B′ = B

(IIb) B′ =B

2

(IIc) B′ =B

8(IId) B′ = 2B

1. Ic, IIb

2. Ia, IIa

3. Ia, IId

4. Ic, IIa

5. Ic, IId

6. Ib, IIa correct

7. Ib, IIb

8. Ia, IIb

9. Ic, IIc

10. Ib, IIc

Explanation:When carrying a current, each smaller coil

still acts like a magnetic dipole, so must havea North and a South pole. With the cur-rent flowing the same direction as throughthe original coil, the cut ends of the new coilsmust be of opposite polarity — if they werenot, then one coil would have either two northpoles or two south poles, an impossibility. Ibis the correct choice.

Since magnetic field strength is propor-tional to N/L and this ratio does not changewhen the coil is divided, we still have B′ = B.Hence, IIa is correct.

009 10.0 points

Consider the setup in Figure above. Whatis the change in potential energy ∆U = UD −UC , in moving a proton from C to D?

1. 2k e q s

(

1

a2− 1

b2

)

2. k q s

(

1

a2− 1

b2

)

3. -2k e q s

(

1

a2− 1

b2

)

4. -k e q s

(

1

a2− 1

b2

)

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5. -2k e q s

(

1

a− 1

b

)

6. -k e q s

(

1

a− 1

b

)

7. k e q s

(

1

a− 1

b

)

8. 2k e q s

(

1

a− 1

b

)

9. k e q s

(

1

a2− 1

b2

)

correct

10. k q s

(

1

a− 1

b

)

Explanation:Letting the +x direction be to the right,

VD − VC = −∫ D

C

(

2 k q s

x3(−x̂)

)

• d~x

= 2 k q s

∫

d

( −1

2 x2

)

= 2 k q s

(

− 1

2 b2+

1

2 a2

)

. (1)

Multiplying eq(1) by the proton charge e,we arrive at the potential energy differencefrom C to D is given by

UD−UC = e(V (b)−V (a)) = k e q s

(

1

a2− 1

b2

)

.

Intuitive reasoning on the sign of ∆U : Nat-ural tendency of the motion is from highpotential energy to lower potential energy.When the proton is released it should movefrom D to C, so UD > UC .

010 10.0 pointsYou are given three parallel conducting

plates of cross-sectional area A that are per-pendicular to the x-axis. They are labeled,from left to right, as plates 1, 2 and 3, respec-tively. The corresponding plate charges areQ1 = −q, Q2 = 3q and Q3 = −2q. The widthof the gap between both plates 1 and 2 and 2and 3 is d.Determine the ∆V = V3 − V1.

1. −9(q/A)d

2ǫ0

2.(q/A)d

ǫ0

3. −5(q/A)d

ǫ0

4. −3(q/A)d

ǫ0

5.4(q/A)d

ǫ0

6.6(q/A)d

ǫ0

7.5(q/A)d

ǫ0

8. −(q/A)d

ǫ0correct

9.3(q/A)d

ǫ0

Explanation:One may regard the 3-plate system as a

composite system which involves two capaci-tor systems with the 12-capacitor followed bythe 23-capacitor.The 12-capacitor has charges Q1 and Q2 +

Q3, i.e charges of −q and q respectively.The 23-capacitor has charges Q1 +Q2 and

Q3, i.e charges of +2q and −2q respectively.The potential difference is

V3 − V1 = −~E12 • d(x̂)− ~E23 • d(x̂)

=(q/A)d

ǫ0− 2(q/A)d

ǫ0

= −(q/A)d

ǫ0.

011 (part 1 of 2) 5.0 pointsConsider a system of a metallic ball with

net charge q1 and radius R1 enclosed by aspherically symmetric metallic shell with netcharge q2, inner radius R2 and outer radiusR3 . If q

′′

2 is the charge on the outside surfaceof the shell and q′2 the charge on its insidesurface, then q′′2 + q′2 = q2 .

Version 013 – MT2 – chiu – (57425) 7

O

q2

q1

B

C

R1

R2

R3

q′2

q′′2

Find the potential at C. OB = b andOC = c .

1. VC= 2 k

q1c

2. VC= k

q2c

3. VC= k

q1 + q2c

4. VC= k

q1 − q2√2 c

5. VC= k

q1 + q2R3

+ kq1c− k

q1R3

6. VC=

√2 k

q2c

7. VC= k

q1c

8. VC= k

q1 + q2c

− kq1R2

+ kq1b

9. VC= k

q1 + q2R3

− kq1R2

+ kq1R1

10. VC= k

q1 + q2R3

correct

Explanation:C is a point inside the outer shell. The

potential inside the shell is constant, so mustbe equal to the potential at the outer surfaceR3. As a consequence,

VC = kq1 + q2R3

,

since the potential at R3 is equal to that of apoint charge of charge q1 + q2 located at theorigin.

012 (part 2 of 2) 5.0 pointsDetermine the potential at B.

1. VB= k

q1 + q2R3

2. VB= k

q1 + q2R3

− kq1R2

+ kq1b

3. VB= k

q1 + q2R3

− kq1b+ k

q1R1

4. VB= 2 k

q1b

5. VB= k

q1b

6. VB= k

q1 + q2R3

− kq1R2

+ kq1R1

correct

7. VB= k

q2b

8. VB= k

q1 + q2b

9. VB=

√2 k

q2c

10. VB= k

q1 − q2√2 c

Explanation:B is located inside the central sphere; again,

we know the potential inside the sphere isconstant and equal to the potential at itssurface. There are several ways to solve thisproblem. We will consider two.

We can apply the superposition princi-ple to the potentials due to each surfacecharge distribution. The surface charges are:q′′ = q1 + q2 at R3, q′ = −q1 at R2, andq = q1 at R1. Remember that the poten-tial difference between any two points insidea spherically symmetric shell of charge is zero,so we must only consider the potential of eachcharge distribution up to but not inside of isradius. Consequently,

VB = Vq′′(b) + Vq′(b) + Vq(b)

= kq1 + q2R3

− kq1R2

+ kq1R1

.

Alternatively, we can recognize that thepotential at B is just the sum of the potentialdifferences from ∞ → R3 and from R2 →R1, since the potential difference inside theconductors is zero. ∆V from ∞ → R3 is justthe potential at R3 due to the total charge,

Version 013 – MT2 – chiu – (57425) 8

while ∆V from R2 → R1 is just due to thecharge q on the surface at R1. Therefore,

VB = Vq′′(R3)− Vq′′(∞) + Vq(R1)− Vq(R2)

= kq1 + q2R3

− 0 + kq1R1

− kq1R2

,

which is equal to the other method, as ex-pected.

013 10.0 points

b

P6

P5

P4

P1

P2

P3

e+

d

θ

θ

θ

θ

An proton is moving horizontally to theleft with speed 5× 106 m/s. Each locationis d = 5 cm from the electron, and the angleθ = 33 ◦. Give the magnetic field at P6 usingthe convention that out of the page is positive.

1. 3.49 × 10−17T

2. −1.74 × 10−17T correct

3. 2.61 × 10−17T

4. −1.16 × 10−17T

5. −2.61 × 10−17T

6. 1.16 × 10−17T

7. −3.49 × 10−17T

8. 1.74 × 10−17T

Explanation:

Let : v = 5× 106 m/s ,

q = 1.6× 10−19 C ,

d = 5 cm = 0.05 m , and

θ = 33 ◦ .

The magnitude of the field at P3 is

B =

∣

∣

∣

∣

µ0

4 π

q v sin(θ)

d2

∣

∣

∣

∣

= (1× 10−7 T ·m/A)(1.6× 10−19 C)

×(5× 106 m/s)sin(33 ◦)

(0.05 m)2

= 1.74× 10−17 T .

Using the right hand rule (and rememberingthat q is positive), the magnetic field is intothe page.

014 (part 1 of 2) 5.0 pointsThree point charges +q are placed at cornersof a square with sides of length L.

+

+

+

A B

LO

What is the potential at point O?

1. V = 2k q

L

2. V = 3√2k q

Lcorrect

3. V = 3k q

L

4. V =

(

2 +1√2

)

k q

L

5. V = 3k q√2L

6. V =

(

2 +1√2

)

k q2

L

7. V =

(

2 +1√2

)

k q

L2

8. V = 2k q

L2

Version 013 – MT2 – chiu – (57425) 9

9. V =k q

L

10. V = 2 ,√2k q

LExplanation:

The distance from any vertex to O isL√2,

so

V =∑

i

Vi = 3k q

(L/√2)

= 3√2k q

L.

015 (part 2 of 2) 5.0 pointsNow place a −q charge at A and move the +qcharge at B to infinity.

−

+ +

B

LO

How much work is required to bring the +qcharge from infinity to point O?

1. W = 3√2k q2

L

2. W =k q2√2L2

3. W = − k q√2 L2

4. W =k q2√2 L

5. W = −3√2k q2

L

6. W = −k q2

L2

7. W =√2k q2

Lcorrect

8. W = 3k q2√2 L

9. W = −k q2

4L

10. W = −√2k q2

L

Explanation:At point 0,

V =∑

i

Vi = 2√2kq

L−

√2kq

L

=√2k q

L,

so the work to move the charge from infinityto B is

WB=

√2k q2

L.

016 10.0 pointsThe figure shows a portion of two ribbonsof width L, each containing a large numberN of closely packed wires. Each wire in theupper ribbon carries a current I into the page;each wire in the lower ribbon carries a current2I out of the page. Use Ampere’s law todetermine ~B at point P . Assume the +xdirection is to the right.

1. B =µ0N I

2L(−x̂) correct

2. B =3µ0N I

2Lx̂

3. B =3µ0N I

Lx̂

4. B =3µ0N I

2L(−x̂)

5. B =µ0N I

L(−x̂)

6. B =µ0N I

2Lx̂

7. B =µ0N I

Lx̂

8. B =3µ0N I

L(−x̂)

Explanation:

Version 013 – MT2 – chiu – (57425) 10

From the diagram, we can see that any pairof wires equidistant from but on either side ofP will generate magnetic fields whose verticalcomponents cancel. Therefore the direction ofthe magnetic field must be to the right abovethe top current ribbon and to the left below(vice-versa for the lower ribbon).

We first work out the contribution to themagnetic field due to the top ribbon alone.Draw a rectangular Amperean loop (of widthℓ and height h) as shown in the figure. Inte-grating clockwise about this 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 = Bupperℓ .

Along the lower part of the path,

∫

~B · d~l = Blowerℓ = Bupperℓ ,

since Bupper = Blower by symmetry.

Therefore,

∮

~B · d~l = 2Btopℓ = µ0 Iinsidepath

= µ0

(

N

L

)

ℓ I ,

where N/L is the current density inwires/meter. So the top ribbon contributes

Btop =µ0N I

2Lx̂ .

at P .

A similar analysis for the lower ribbon re-veals that its contribution at P is given by

Bbottom =µ0N I

L(−x̂) .

Therefore the vector sum at P is equal to:

~B(P ) =µ0N I

2L(−x̂) .

017 10.0 pointsAn isolated large-plate capacitor (not con-nected to anything) originally has a potentialdifference of 800 V with an air gap of 5 mm.Then a plastic slab 3 mm thick, with dielec-tric constant 6, is inserted into the middle ofthe air gap as shown in the figure below.

++++++++

−−−−−−−−

5 mm

1 mm 3 mm 1 mm

1 2 3 4

Calculate V1 − V4.1. 720.02. 700.03. 468.04. 660.05. 572.06. 500.07. 624.08. 687.59. 416.010. 400.0

Correct answer: 400 V.

Explanation:Here we simply add the potential differ-

ences we’ve already found:

∆V14 = ∆V12 +∆V23 +∆V34

= (160 V) + (80 V) + (160 V)

= 400 V .