Version 042 – midterm 03 – chiu – (56565) 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 points

A unipolar generator consists of a copperdisk of radius R rotating in a uniform, steadymagnetic field B perpendicular to the disk,out of the page. (The magnetic field is pro-duced by large coils carrying constant current,not shown in the diagram.) Sliding contactsare made at the center (on the axle) and at therim of the disk, and the wires are connectedto a voltmeter. If the outer rim travels coun-terclockwise at a speed v, choose a correctanswer for each of following questions.The direction of the instantaneous average

velocity of the electrons at any point alongAB

Ia) upwardIb) downward

The direction of the magnetic force on theelectrons at any point along AB

IIa) towards AIIb) towards B

The direction of the conventional currentflow

IIIa) into the + terminal of the voltmeterIIIb) out of the + terminal of the voltmeter

1. Ib, IIb, IIIb

2. Ia, IIb, IIIb

3. Ia, IIa, IIIb correct

4. Ia, IIa, IIIa

5. Ib, IIa, IIIa

6. Ib, IIa, IIIb

7. Ia, IIb, IIIa

8. Ib, IIb, IIIa

Explanation:Counterclockwise rotation of the disk im-

plies an upward instantaneous velocity at anypoint along AB. Hence the answer is Ia.The magnetic force ~F = q~v×~B is pointing

radially inward. Hence, the answer is IIa.Negative charge is accumulated at the cen-

ter of the disk A, or the conventional currentflows out from the + terminal. Hence, theanswer is IIIb.

002 10.0 pointsConsider the set up shown in the figure wherea solenoid has a steadily increasing magneticflux which generates identical induced emfsfor the two cases illustrated.

AB#1#2(1)

i

Case 1: Two identical light bulbs are inseries; the electrical power consumed by bulb1 and bulb 2 is P1 and P2, respectively.

AB#1#2(2) D

F

C

Case 2: Bulb 2 is shorted by a wire whichis connected between the two points C and F ;the electrical power consumed by bulb 1 andbulb 2 is P ′

1 and P ′

2, respectively.

What is the ratioP ′

1

P1

?

Version 042 – midterm 03 – chiu – (56565) 2

1.P ′

1

P1

= 0

2.P ′

1

P1

= 3

3.P ′

1

P1

=1

4

4.P ′

1

P1

=1

8

5.P ′

1

P1

= 1

6.P ′

1

P1

=1

3

7.P ′

1

P1

= 2

8.P ′

1

P1

= 8

9.P ′

1

P1

=1

2

10.P ′

1

P1

= 4 correct

Explanation:Let E and R be the induced emf and resis-

tance of the light bulbs, respectively.For case 1, since the two bulbs are in series,

the equivalent resistance is simply Req = R+R = 2R and the current through the bulbs is

I =E2R

.

Thus for case 1, the power consumed by bulb1 is

P1 =

( E2R

)2

R =E2

4R.

For case 2, since bulb 2 is shorted, thecurrent through bulb 1 is now

I ′ =ER

.

so the power consumed by bulb 1 is

P ′

1 =

( ER

)2

R =E2

R

and the ratio is

P ′

1

P1

= 4 .

003 10.0 points

In an experiment designed to measure themagnitude of a uniform magnetic field, elec-trons are accelerated from rest through a po-tential difference of V = 135 Volts. The elec-trons travel along a curved path because ofthe magnetic force exerted on them, and theradius of the path is measured to be r = 7 cm.If the magnetic field is perpendicular to thebeam, what is its magnitude?Note: The mass of the electron is

9.1× 10−31 kg and its charge is 1.6× 10−19 C.1. 0.0007262. 0.0004353. 0.0007924. 0.0005545. 0.0009716. 0.000567. 0.000678. 0.001359. 0.00052810. 0.000706

Correct answer: 0.00056 T.

Explanation:By the conservation of energy

∆K +∆U = 0

We note that the initial kinetic energy is zero,

and the final kinetic energy is Kf =1

2mev

2.

The change in the potential energy is given byeV .

Kf = eV

Version 042 – midterm 03 – chiu – (56565) 3

1

2mev

2 = eV

v =

√

2eV

me

v =

√

2(1.6× 10−19 C)(135 V)

(9.1× 10−31 kg)

v = 6.89003× 106 m/s

By equating the magnitude of the magneticand centripetal forces we have

e v B =mev

2

r

eB =mev

r

The magnetic field is given by the expression

B =mev

er

B =(9.1× 10−31 kg)(6.89003× 106 m/s)

(1.6× 10−19 C)(7 cm)

B = 0.00056 T

004 10.0 points

A wire loop of radius R carrying a counter-clockwise (when viewed from the right) cur-rent I is moving to the right along the x-axisat a speed v. It passes around a second, sta-tionary wire loop of radius a where a ≪ R.Choose the plot that correctly displays thequalitative behavior of the induced currentI(t) in the stationary loop. On the plots, the+I axis represents clockwise current, whilethe −I axis represents counter-clockwise cur-rent (when viewed from the right on the +x-axis).

1. VI:

2. V:

3. IV:

4. I:

correct

5. VIII:

Version 042 – midterm 03 – chiu – (56565) 4

6. VII:

7. III:

8. II:

Explanation:I is the correct choice. Before the current

loop passes the wire ring, the flux throughthe ring points to the right and is increas-ing; to resist this change in flux, a clockwise(positive) current is induced in the ring. Af-ter the loop passes the ring, the flux stillpoints to the right but is decreasing, induc-ing a counter-clockwise (negative) current inthe ring. To get a sense of the functional formof the change in flux, note that we are entitledto use

Bloop ∝ 1

(x2 +R2)3/2

since a ≪ R. (Since we are only inter-ested in the qualitative behavior, we canleave off the constant factors and concern our-selves only with proportionality.) Since thearea of the ring is not changing, dΦB/dt =Aring dBloop/dt. Again ignoring the constantfactor Aloop, we calculate dBloop/dt,

dBloop

dt∝ d

dt

(

1

(x2 +R2)3/2

)

dBloop

dt∝ −3

2

2x

(x2 +R2)5/2dx

dt

dBloop

dt∝ − 3xv

(x2 +R2)5/2

Examining this result, we see that the pres-ence of x in the numerator means our plotmust pass through 0, while to either side oft = 0 we should expect some polynomial-like behavior. Combined with what we knowabout the sign of the current for times t < 0and t > 0, I is the only correct option.

005 10.0 points

Consider the three configurations shownin the figure. In each case there is a longsolenoid viewed end-on with current flowingclockwise, where the current is a decreas-ing function of time: dI(t)/dt < 0. Mid-way along each solenoid is a piece of neu-tral wire arrranged as shown in the figure.Choose the response that correctly identi-fies the direction of the induced current inthe neutral wires, where CW=clockwise andCCW=counter-clockwise. You may assumethe magnetic field outside the interior of eachsolenoid ≈ 0.

1. I: A:CW, B:no current, C:CW correct

2. VI: A:CW, B:CW, C:CW

3. V: A:CW, B:CW, C:CCW

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4. IV: A:CCW, B:no current, C:CW

5. VII: A:CCW, B:CCW, C:CW

6. VIII: A:CCW, B:CCW, C:CCW

7. III: A:CCW, B:no current, C:CCW

8. II: A:CW, B:no current, C:CCW

9. IX: A:CW, B:no current, C:no current

10. X: A:CCW, B:no current, C:no current

Explanation:I is correct. In each case, the solenoid

produces a magnetic field in the shaded regionthat points into the page but is decreasing.The induced current opposes the change influx, therefore the neutral loops in A and Cacquire a CW current. In B the loop is notclosed, so no current flows.

006 10.0 points

A bar magnet (the shaded cuboid object inthe figure) is placed as shown in the figure,with its axis horizontal and about half of itslength inside the cubical surface of side a.What can you say about the magnetic flux

φB =

∫

~B.d~A through the cubical surface

shown in the figure? (Note: One half of themagnet is shaded lighter only to show that itis inside the cube).A: φB > 0, because there is only one pole

of the magnet inside the surface.B: φB < 0, because there is only one pole

of the magnet inside the surface.C: φB can be either positive or negative,

depending on which pole of the magnet ispresent inside the cube.D: φB = 0 according to Gauss’s law of

magnetism.E: EitherC or D can be true depending on

situation.

1. E

2. A

3. D correct

4. B

5. C

Explanation:D is correct. Gauss’s law of magnetism

holds here. Suppose the North pole of themagnet was inside. The field lines due tothis pole point outwards at every point on thesurface of the cube outside of the magnet, butthey all come back inside the surface throughthe magnet, thus making the net flux zero. Ifthe south pole is inside, things work the otherway around, with field lines pointing inwardseverywhere on the surface outside the magnet,and going out through the magnet. In eithercase, the net flux through the surface is zero.

007 (part 1 of 2) 10.0 pointsAssume: The mobile charge carriers are eitherelectrons or holes. The holes have the samemagnitude of charge as the electrons.Note: In the figure, the point at the upper

edge P1 and at the lower edge P2 have thesame x coordinate.A constant magnitude magnetic field points

out of the paper. There is a steady flow of ahorizontal current flowing from left to right inthe x direction.

Version 042 – midterm 03 – chiu – (56565) 6

L

ba

I

P1

P2V

~B

~By

x

A voltmeter (with an internal resistanceless than infinity) is connected to the system,where the contact points are on the upper andlower surfaces and are in the same verticalplane.Choose the correct answer for the case

where the sign of the charge of current carri-ers are either negative (electrons) or positive(holes).

1. The direction of the (conventional) cur-rent through the voltmeter is downward foreither positive or negative charge carriers.

2. The direction of the (conventional) cur-rent through the voltmeter is downward forpositive charge carriers and upward for nega-tive charge carriers.

3. The (conventional) current through thevoltmeter is zero for either positive or negativecharge carriers.

4. The direction of the (conventional) cur-rent through the voltmeter is upward for pos-itive charge carriers and downward for nega-tive charge carriers. correct

5. The direction of the (conventional) cur-rent through the voltmeter is upward for ei-ther positive or negative charge carriers.

Explanation:

~vd × ~B

~vd

~B

~F = −e~vd × ~B

−

The magnetic force on a negative chargepoints downwards, so an excess of positivecharge develops along the upper edge. Theinduced electric field then points downward,so P1 is at a higher potential relative to P2.The Hall current flows in the direction of theinduced field, so it flows through the volt-meter from top to bottom.

+ + + + + + + +

− − − − − − − −

−~F

008 (part 2 of 2) 10.0 pointsDenote the drift velocity by vd and the mag-nitude of the charge of an electron by q.Using the dimensions (a, b, and L) given in

the figure, what is the magnitude of the Hallvoltage?

1. VHall = 2 vdB b

2. VHall = 2 vdB a

3. VHall = q vdB b

4. VHall = 2 vdBL

5. VHall = vdB b correct

Version 042 – midterm 03 – chiu – (56565) 7

6. VHall = q vdB a

7. VHall = vdBL

8. VHall = q vdBL

9. VHall = vdB a

Explanation:The force due to the induced electric field

must balance the force due to the magneticfield, so

q EHall = q vdB .

The magnitude of the Hall voltage is

VHall = EHall b = vdB b ,

where b is the height of the conducting slab.

009 10.0 points

A square “loop” of wire of side length s =28 cm encompasses a circular region of radiusr = 10 cm that contains a magnetic field|~B| = 0.0003 T oriented at an angle θ = 15

to the perpendicular as shown; outside thisregion, B = 0. Calculate the magnitude ofthe magnetic flux |ΦB| through the squareloop.1. 7.17367e-062. 6.17275e-063. 6.66074e-064. 1.12009e-055. 9.93678e-066. 9.10363e-067. 5.79819e-068. 8.21603e-069. 4.47063e-0610. 1.67649e-05

Correct answer: 9.10363× 10−6 Wb.

Explanation:The area of the circular region is given by

A = π r2 = π(10 cm)2 = 0.0314159 m2

The magnitude of the flux is given by

|ΦB| = |~B|A cos θ

|ΦB| = (0.0003 T)(0.0314159 m2) cos(15)

|ΦB| = 9.10363× 10−6 Wb

010 10.0 pointsA very long thick wire of radius R carries acurrent I, as in the following figure.

I1 R

Find the magnitude of the magnetic fieldinside the wire, a distance r from the centerof the wire, where r < R.Assume current density is the same at every

point inside the wire.

1.µ0

4π

Ir

R2

2.µ0

4π

IR

r2

3.µ0

4π

Ir2

R

4.µ0

2π

Ir

R2correct

5.µ0

2π

IR

r2

6.µ0

2π

Ir2

R3

7.µ0

2π

Ir2

R

8.µ0

2π

IR2

r3

9.µ0

4π

IR2

r3

10.µ0

4π

Ir2

R3

Explanation:

Version 042 – midterm 03 – chiu – (56565) 8

The current density (current/area) is thesame throughout the wire. Sketch an Ampe-rian loop within the wire with radius r. Then,the current through this loop i divided by areaof this loop is equal to the total current perunit area.

i

π r2=

I

π R2

⇒ i = Ir2

R2

The magnetic field is tangential to the Am-perian loop. Apply Ampere’s law by inte-grating counterclockwise around the loop iflooking at the loop from the right end. Thenthe current flows out of the surface and ispositive.

∮

~B · d~ℓ = µ0 Iinside path

B (2 π r) = µ0 Ir2

R2

⇒ B =µ0

2π

I r

R2.

011 10.0 points

A solid metal cube of edge length d =1.2 cm moving in the positive Y direction at aconstant velocity v = 6 m/s. The cube movesthrough a uniform magnetic field B = 0.01 Tthat points in the +z direction.Which face has the higher potential? Also

find the difference between lower and higherpotential. (Express your answer in units ofmV.)

1. back face, 0.792 mV

2. front face, 0.576 mV

3. top face, 0.864 mV

4. left face, 0.684 mV

5. right face, 0.54 mV

6. left face, 0.504 mV

7. bottom face, 0.648 mV

8. top face, 0.9 mV

9. right face, 0.72 mV correct

10. down face, 0.828 mV

Explanation:Since the box is metallic, it has freely mov-

ing conducting electrons. As the cube movesupward with velocity ~v, electrons inside themetallic cube will also move with it. Sincethey are moving perpendicular to the mag-netic field ~B, they will feel a net magneticforce according to the ~F = q(~v× ~B) equation.The magnetic force will thus push electronsto the left face creating a net potential dif-ference between the right and left face of themetal cube, ultimately setting up an electricfield between right and left faces of the solidcube. This will leave positive charges on theright face, which will be at a higher potential.The potential difference across the right

and left faces of the cube is given by

V = Ed

When equilibrium for the mobile charges isestablished, the electrostatic and magneticforces on an electron are balanced.

FE = FB

eE = evB

E = vB

In terms of B, the potential is given by

V = vBd

Version 042 – midterm 03 – chiu – (56565) 9

V = (6 m/s)(0.01 T)(1.2 cm) = 0.72 mV

012 (part 1 of 2) 10.0 pointsConsider the figure shown below. The switchis initially set at position b. There is no chargenor current around the right loop while atposition b. At t = 0 the switch is set toposition a.

L

R

C

E

S b

a

What is the current at t = 0?

1. I0 =ER

(

1− e−1)

2. I0 =E√2R

3. I0 = 0 correct

4. I0 =ER

5. I0 = E R

6. I0 = ∞

7. I0 =ELR

e−1

8. I0 =ELR

(

1− e−1)

9. I0 =ELR

10. I0 =ER

e−1

Explanation:The differential equation for an LC circuit

is given by

E − I R − LdI

dt= 0 ,

whose solution is

I(t) =ER

(

1− e−t/τL)

.

Hence at t = 0 we have I = 0. (The back emfacross the inductor achieves its greatest valueat t = 0).

013 (part 2 of 2) 10.0 pointsAfter leaving the switch at position “a” for along time, move the switch from “a” to “b”.There will be current oscillations.The maximum current will be given by

1. Imax = E√

L

C

2. Imax = E√

C

L

3. Imax =ER

correct

4. Imax =ER

√

L

C

5. Imax =ER

√

C

L

6. Imax = E√LC

7. Imax = E√

1

LC

8. Imax =ER

√LC

Explanation:When the switch S is on at a, the loop

equation is given by

E − LdI

dt− IR = 0

After a long time, the current approaches a

constant value, i.edI

dt= 0. In other words the

loop equation becomes

E − IR = 0

I =ER

Now, the switch S is moved from a to b. Theconservation of the total energy in the LCcircuit implies that at t = 0,

UC + UL = (UL)max =1

2L I2max

Version 042 – midterm 03 – chiu – (56565) 10

since UC = 0 at t = 0. Hence, the maximumcurrent in the LC circuit must be equal tothe current in the circuit at t=0, given by theexpression

Imax =ER

014 (part 1 of 2) 10.0 points

In the given figure, +x points to the right,+y points upwards and +z points out of thepage, towards you. A rectangular currentloop mounted on a pivot aligned along thez-axis is shown in the figure (since your lineof sight is in the plane of the loop, you canonly see it edge-on, like a straight line). Theloop carries a current I = 1.5 A. A uniformmagnetic field of ~B = 6 Tj is present in theregion. If the sides of the loop are given asa = 9 cm (not seen in the figure) and b = 4 cmrespectively, and the angle θ = 19 degrees,what is the magnitude of torque acting on theloop at this instant?1. 0.0041372. 0.002933. 0.010554. 0.0028135. 0.0024336. 0.011847. 0.004218. 0.0040029. 0.00925410. 0.004688

Correct answer: 0.01055 Nm.

Explanation:The area of the loop is given by

A = a×b = (9 cm)(4 cm) = 0.0036 m2

The magnitude of the torque is given by

τ = B I A sin θ

τ = (6 T)(0.0036 m2)(1.5 A) sin(19)

τ = 0.01055 Nm

One can also obtain the magnitude of thetorque by the following argument.According to right hand rule, the magnetic

force acting on the near side (the side vis-ible in the figure) and far side of the loopwill be equal and opposite, and act along thesame line in the negative and positive z di-rection, respectively. Hence, there will not beany torque due to these forces. The magneticforce on the left and right sides (the sidesperpendicular to the plane of the figure) willbe along +x and −x directions respectively,and will have the same magnitude F = BIa.However, their lines of action differ by a dis-tance of d = b sin θ. Hence, the magnitude ofthe torque, which is the moment of the force,can be obtained.

τ = Fd = BIab sin θ = BIA sin θ

015 (part 2 of 2) 10.0 pointsFor the same loop as in the previous problem,suppose the magnetic field is a function ofthe y-coordinate, and is given by ~B = B0yj.What will be the net force acting on the loop?Here, ’A’ stands for the area of the rectan-

gular loop.

1. B0IA cos θk

2. B0IA sin θk

3. B0IA sin θj

4. B0IA cos θi

5. B0IA cos θj

6. B0IA sin θi correct

Explanation:

Version 042 – midterm 03 – chiu – (56565) 11

Since we know that magnetic field varieswith the y-coordinate, we expect that the netforce on the loop may no longer be zero. Forany point on the near side, there will be acorresponding point on the far side with thesame y-coordinate. Hence, the forces on thenear and far sides cancel other exactly. How-ever, this cancellation does not work for theleft and right sides of the loop. All the pointson the left side have the same y-coordinate,say, y1, while all points on the right side havea different y-coordinate, y2. If B1 and B2 arethe corresponding magnetic fields at y = y1and y = y2 respectively, then we can calculatethe net force as follows.

~F = B1Iai−B2Iai

~F = (B1 −B2) Iai

~F = (y1 − y2)B0Iai

~F = (b sin θ)B0Iai

~F = B0Iab sin θi

016 10.0 points

You pull on a simple loop circuit of dimen-sions 2d× L consisting of an ideal wire and aresistor. You pull with just enough force sothat the circuit moves at a constant speed v.Initially, half the circuit lies inside a regionof magnetic field B that is directed out of thepage. How much work must you do to com-pletely remove the circuit from the magneticfield? Choose the correct answer, expressedonly in terms of the given quantities.

1. D:L2B2d2

R

2. G:LBdv

R

3. H:L2Bvd

R

4. E:L2B2d

R

5. B:L2B2v

R

6. J: L2B2d

7. A:L2B2vd

Rcorrect

8. I: L2B2vd

9. C: ILBd

10. F: ILB

Explanation:A is the correct choice. First note that as

the circuit moves to the right, the magneticflux through through the circuit decreases.Consequently, a current I must flow counter-clockwise to oppose this change in flux; on thevertical segment of wire inside the magneticfield region, this current is acted upon by themagnetic field, producing a force ILB to theleft.We are told that the circuit moves at a con-

stant speed v; to move at a constant speed,net acceleration must be 0, so the appliedforce must be equal and opposite to the mag-netic force on the current: Fext = ILB. Sinceonly half the circuit is in the field region, thetotal work done to remove the circuit from thefield must be

W = Fd = ILBd

Then, using I =ER

and applying Faraday’s

law,

W = ILBd =ELBd

R

W =

∣

∣

∣

∣

dΦB

dt

∣

∣

∣

∣

LBd

R

Version 042 – midterm 03 – chiu – (56565) 12

W =d

dt(LxB)

LBd

R

W =L2B2d

R

dx

dt

W =L2B2vd

R

where the x introduced in line 3 refers tothe length of the circuit that remains in thefield region.

017 10.0 pointsA bar of negligible resistance and mass of

53 kg in the figure is pulled horizontally acrossfrictionless parallel rails, also of negligible re-sistance, by a massless string that passes overan ideal pulley and is attached to a suspendedmass of 820 g. The uniform magnetic fieldhas a magnitude of 640 mT, and the distancebetween the rails is 96 cm. The rails are con-nected at one end by a load resistor of 96 mΩ.

640 mT 640 mT 640 mT

96 cm

820 g

a

96mΩ

53 kg

What is the magnitude of the terminal ve-locity (i.e., the eventual steady-state speedv∞) reached by the bar? The acceleration ofgravity is 9.8 m/s2 .1. 3.386262. 0.7203243. 11.67424. 20.62145. 73.49636. 5.209517. 2.41388. 2.043669. 1.9774310. 1.48119

Correct answer: 2.04366 m/s.

Explanation:

Let : m = 53 kg ,

M = 820 g = 0.82 kg ,

ℓ = 96 cm = 0.96 m ,

B = 640 mT , and

R = 96 mΩ = 0.096 Ω .

B B B B B

ℓT

T

MFga

R

am

Fm

~Fg = M~g

~Fm = I~ℓ× ~B

~Fnet = (M +m)~a = ~Fg − ~Fm

E = I R = −dΦB

dt

ΦB = ~B · ~AE = B ℓ v .

It follows from Lenz’s law that the mag-netic force opposes the motion of the bar.When the wire acquires steady-state speed,the gravitational force Fg is counter-balancedby the magnetic force Fm.

Fg = M g = Fm = ℓ I B (1)

I =M g

ℓB. (2)

To find the induced current, we use Ohm’s law

and substitute in the induced emf, E = −dΦ

dt

I =|E|R

=1

R

dΦ

dt. (3)

Note: We have ignored the minus sign fromthe induced emf E because we will eventu-ally evaluate the magnitude of the terminalvelocity. The flux is Φ = BA , so

|E| = dΦ

dt= B

dA

dt= B ℓ v , and (4)

Version 042 – midterm 03 – chiu – (56565) 13

I =B ℓ v

R. (5)

Using Eqs. 2 and 5 and noting that v is theterminal velocity v∞

M g

ℓB=

B ℓ v∞R

. (6)

Solving for the magnitude of the terminalvelocity v∞

v∞ =M gR

ℓ2B2(7)

=(0.82 kg) (9.8 m/s2) (0.096 Ω)

(0.96 m)2(640 mT)2

= 2.04366 m/s .

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