Quest Review 4 Electric Force, Magnetic Fields KEY

Version 001 – Review 4 Electric Force, Magnetic fields – tubman – (19112) 1

This print-out should have 42 questions.Multiple-choice questions may continue onthe next column or page – find all choicesbefore answering.

AP EM 1993 MC 55 e1001 10.0 points

Two metal spheres that are initially un-charged are mounted on insulating stands, asshown.

X Y−−−−

A negatively charged rubber rod is broughtclose to but does not make contact with sphereX. SphereY is then brought close toX on theside opposite to the rubber rod. Y is allowedto touchX and then is removed some distanceaway. The rubber rod is then moved far awayfrom X and Y.What are the final charges on the spheres?

Sphere X Sphere Y

1. Positive Negative correct

2. Negative Zero

3. Positive Positive

4. Negative Negative

5. Negative Positive

6. Positive Zero

7. Zero Positive

8. Zero Zero

9. Zero Negative

Explanation:When the negatively charged rod moves

close to the sphere X, the negatively chargedelectrons will be pushed to sphere Y. If X

and Y are separated before the rod movesaway, those charges will remain on X and Y,X is positively charged and Y is negativelycharged.

Charging Metallic Objects002 10.0 points

1) Two uncharged metal balls, X and Y,stand on glass rods and are touching.

Y X

2) A third ball, carrying a positive charge, isbrought near the first two.

Y X+

3) Then the first two balls are separated fromeach other,

Y X+

4) and the third ball is finally removed.

Y X

What are the resulting charges?

1. BallX is negative and ballY is positive.

2. Ball X is positive and ball Y is negative.correct

3. Balls X and Y are still uncharged.

4. Balls X and Y are both negative.

5. Balls X and Y are both positive.


Explanation:When a positive ball is moved near a metal-

lic object (X and Y), the positive charge willattract negative charges, causing X to haveexcess positive charge and Y to have excessnegative charge (X and Y are in contact, sothe total net charge on X and Y should bezero).

− ++

Later, X and Y are separated, retainingtheir charges, so when the third ball is finallyremoved, X will have net positive charge andY will have net negative charge.

Acceleration of a Particle003 10.0 points

A particle of mass 40 g and charge 36 µC isreleased from rest when it is 19 cm from asecond particle of charge −12 µC.Determine the magnitude of the initial ac-

celeration of the 40 g particle.

Correct answer: 2688.78 m/s2.

Explanation:

Let : m = 40 g ,

q = 36 µC = 3.6× 10−5 C ,

d = 19 cm = 0.19 m ,

Q = −12 µC = −1.2× 10−5 C , and

ke = 8.9875× 109 N ·m2/C2 .

The force exerted on the particle is

F = ke|q| |Q|d2

= ma

a = ke|q| |Q|md2

= (8.9875× 109 N ·m2/C2)

×∣∣3.6× 10−5 C

∣∣ ∣∣−1.2× 10−5 C∣∣

(0.04 kg) (0.19 m2)

= 2688.78 m/s2 .

AP EM 1998 MC 39 40004 (part 1 of 2) 10.0 points

Two charged particles of equal magnitude(+Q and −Q) are fixed at opposite corners ofa square that lies in a plane (see figure below).A test charge +q is placed at a third corner.

+Q

+q −Q

What is the direction of the force on thetest charge due to the two other charges?

1.

2.

3.

4. correct

5.

6.

7.

8.

Explanation:The force between charges of the same sign

is repulsive and between charges with oppo-site signs is attractive.


+Q

+q −Q

The resultant force is the sum of the twovectors in the figure.

005 (part 2 of 2) 10.0 pointsIf F is the magnitude of the force on the testcharge due to only one of the other charges,what is the magnitude of the net force act-ing on the test charge due to both of thesecharges?

1. Fnet =F√3

2. Fnet = 0

3. Fnet = F

4. Fnet =2F√3

5. Fnet =2F

3

6. Fnet =3F

2

7. Fnet = 3F

8. Fnet =F√2

9. Fnet =√2 F correct

10. Fnet = 2F

Explanation:The individual forces form a right angle, so

the magnitude of the net force is

Fnet =√

F 2 + F 2 =√2F .

Hewitt CP9 22 R02006 10.0 points

Why does the gravitational force between

the Earth and moon predominate over electricforces?

1. Because the distance between the Earthand the moon is very large.

2. Because the masses of the Earth andmoon are very large.

3. Because there is no electric charge on themoon.

4. Because both the Earth and the moon areelectrically neutral. correct

Explanation:Since the electrical force between the two

objects varies directly as the product of theircharges and the Earth and the moon are elec-trically neutral, the electrical force betweenthem is zero.

AP B 1993 MC 68007 10.0 points

The diagram shows an isolated, positivecharge Q, where point B is twice as far awayfrom Q as point A.

+Q A B

0 10 cm 20 cm

What is the ratio of the electric fieldstrength at point A to the electric fieldstrength at point B?

1.E

A

EB

=8

1

2.E

A

EB

=2

1

3.E

A

EB

=1

1

4.E

A

EB

=1

2

5.E

A

EB

=4

1correct

Explanation:

Let : rB= 2 r

A.


The electric field strength E ∝ 1

r2, so

EA

EB

=

1

r2A

1

r2B

=r2B

r2A

=(2 r)2

r2= 4 .

AP EM 1993 MC 36008 10.0 points

From the electric field vector at a point, onecan determine which of the following?

I. the direction of the electrostatic force ona test charge of known sign at that point;

II. the electrostatic charge at that point;III. the magnitude of the electrostatic force

exerted per unit charge on a test chargeat that point.

1. I only

2. III only

3. I and III only correct

4. None of these

5. I and II only

6. II and III only

7. All of these

8. II only

Explanation:The definition of the electrostatic force is

~E =~F

q, so ~F = q ~E and ~F acts in the same

or opposite direction to ~E depending on thesign of the charge. If we only consider themagnitude F = q E for a unit charge, the

electric field’s magnitude is E =F

q.

Four Charges in a Square Short009 10.0 points

Consider a square with side a. Four charges−q, +q, +q, and −q are placed at the cornersA, B, C, and D, respectively

−

−

+

+

D

A

C

B

aO

What is the magnitude of the electric fieldat the center O?

1. EO =1

2√2

k q

a2

2. EO = 4√2k q

a2correct

3. EO =k q

a2

4. EO = 2√2k q

a2

5. EO = 3k q

a2

6. EO =1√2

k q

a2

7. EO =√2k q

a2

8. EO =1

3√2

k q

a2

9. EO = 3√2k q

a2

10. EO =1

4√2

k q

a2

Explanation:The distance between each corner and the

center isa√2, so the magnitude of each electric

field at D is

E = kq

(a√2

)2= 2 k

q

a2

The two negative charges yield forces point-ing away from them from O and the two posi-tive charges yield forces pointing toward themfrom O with the collinear charges adding al-gebraically:


‖~EA + ~EC‖ = ‖~EB + ~ED‖ = 2E = 4 kq

a2.

E

EA+E

C

EB+ E

D

The Cartesian components of the two vec-tors with the origin at O are

~EA + ~EB = 4 kq

a2

(− 1√

2ı+

1√2

)and

~EB + ~ED = 4 kq

a2

(− 1√

2ı− 1√

2

), so

~E = 4 kq

a2

(− 1√

2− 1√

2

)ı

+

(1√2− 1√

2

)

= −4√2 k

q

a2ı ,

with magnitude −4√2 k

q

a2.

Two Charge Field010 (part 1 of 3) 10.0 points

Two point charges at fixed locations pro-duce an electric field as shown.

A B

Y

X

How would a negative charge placed atpoint Y move?

1. Toward charge B correct

2. Toward charge A

3. Along an equipotential plane

Explanation:Electric field lines run from a positive po-

tential to a negative potential, so the chargeB is positive. A negative charge will move to-ward a positive potential, which creates lowerpotential energy and a higher kinetic energy.

011 (part 2 of 3) 10.0 pointsThe electric field at point Y is

1. the same as that the field at point X .

2. stronger than the field at point X . cor-rect

3. weaker than the field at point X .

Explanation:The field at Y is stronger than the field at

X , since the number of field lines per unitvolume at Y is greater than the number offield lines per unit volume at X .

012 (part 3 of 3) 10.0 pointsEstimate the ratio of the magnitude of

charge A to the magnitude of charge B.Your answer must be within ± 5.0%

Correct answer: 1.30769.

Explanation:The number of field lines is proportional to

the magnitude of the charge, so

QA

QB≈

∣∣∣∣−17

13

∣∣∣∣ = 1.30769

E Field Diagrams 03013 (part 1 of 2) 10.0 points


Q

R radius

Consider a solid conducting sphere of ra-dius R and total charge Q. Which diagramdescribes the E(r) vs r (electric field vs radialdistance) function for the sphere?S.

rR0

∝ 1

r2

M.

rR0

∝ 1

r2

∝ 1

r

P.

rR0

∝ 1

r2

L.

rR0

∝ 1

r2

G.

rR0

∝ 1

r2

1. L

2. G correct

3. M

4. S

5. PExplanation:Because the charge distribution is spheri-

cally symmetric, select a spherical Gaussiansurface of radius r and surface area 4 π r2 con-centric with the sphere. The electric field dueto the conducting sphere is directed radiallyoutward by symmetry and is therefore normalto the surface at every point and ~E is parallelto d~A at each point.There is no charge within the Gaussian sur-

face, so E = 0 for r < R .For the region outside the conducting

sphere,

ΦE =

∮~E · d~A =

∮E dA = E

∮dA

= E(4 π r2

)=

qinǫ0

E =qin

4 π ǫ0 r2=

Q

4 π ǫ0 r2for r > R .

G.

rR0

∝ 1

r2

E

014 (part 2 of 2) 10.0 pointsWhich diagram describes the E(r) vs r

(electric field vs radial distance) function if


the sphere is instead a uniformly charged,

non-conducting sphere?

1. P

2. L correct

3. S

4. GExplanation:Select a spherical Gaussian surface of radius

r and volume V ′, where r < R, concentricwith the uniformly charged non-conductingsphere. The charge qin within the Gaussiansurface of the volume V ′ is less than Q; from

the volume charge density ρ ≡ Q

V,

qin = ρ V ′ = ρ

(4

3π r3

).

Applying Gauss’ law, for r < R,

∮E dA = E

∮dA = E

(4 π r2

)=

qenǫ0

E =qen

4 π ǫ0 r2=

ρ4

3π r3

4 π ǫ0 r2=

ρ

3 ǫ0r .

ρ =3Q

4 π R3by definition and k =

1

4 π ǫ0, so

E =4

3k

3Q

4 π R3π r =

kQ

R3r .

In the region outside the uniformly chargednon-conducting sphere, we have the same con-ditions as for the conducting sphere when ap-plying Gauss’ law, so

E =Q

4 π ǫ0 r2.

L.

rR0

∝ 1

r2

E

Charge in Electric Field015 10.0 points

A particle of charge q is placed in a uni-form electric field of magnitude E. Considerthe following statements about the resultingforces on the particle:

I. It has magnitude q E.II. It is perpendicular to the direction of the

field.III. It is parallel to the direction of the field.

Identify the true statement(s).

1. I only

2. None is true.

3. II and III only

4. II only

5. I and II only

6. All are true.

7. III only

8. I and III only correct

Explanation:The electric field at some point in space is

the force per unit charge that the test chargewould feel at that point, so the electric forcethe charge experiences is of magnitude q Eand in the same direction as the direction ofthe electric field.

Electron Deflection016 (part 1 of 3) 10.0 points

An electron traveling at 6× 106 m/s enters a0.06 m region with a uniform electric field of220 N/C , as in the figure.

−−−−−−−−−0.06 m

+++++++++

ı

6× 106 m/s


Find the magnitude of the acceleration ofthe electron while in the electric field. Themass of an electron is 9.109× 10−31 kg andthe fundamental charge is 1.602× 10−19 C .

Correct answer: 3.86914× 1013 m/s2.

Explanation:

Let : qe = −1.602× 10−19 C ,

me = 9.109× 10−31 kg , and

E = 220 N/C .

F = ma = q E

a =qeE

me

=(−1.602× 10−19 C)(220 N/C)

9.109× 10−31 kg

= (−3.86914× 1013 m/s2) ,

with a magnitude of 3.86914× 1013 m/s2 .

017 (part 2 of 3) 10.0 pointsFind the time it takes the electron to travelthrough the region of the electric field, assum-ing it doesn’t hit the side walls.

Correct answer: 1× 10−8 s.

Explanation:

Let : ℓ = 0.06 m , and

v0 = 6× 106 m/s .

The horizontal distance traveled is

ℓ = v0 t

t =ℓ

v0=

0.06 m

6× 106 m/s

= 1× 10−8 s .

018 (part 3 of 3) 10.0 pointsWhat is the magnitude of the vertical dis-placement ∆y of the electron while it is in theelectric field?

Correct answer: 0.00193457 m.

Explanation:Using the equation for the displacement in

the vertical direction and the results from thefirst two parts of the problem,

∆y =1

2a t2

=−3.86914× 1013 m/s2

2× (1× 10−8 s)2

= −0.00193457 m ,

with a magnitude of 0.00193457 m .

Holt SF 18A 01019 10.0 points

Two alpha particles (helium nuclei), eachconsisting of two protons and two neu-trons, have an electrical potential energy of6.27× 10−19 J .What is the distance between these par-

ticles at this time? The Coulomb constantis 8.98755× 109 N ·m2/C2, the elementalcharge is 1.6021× 10−19 C, and the accel-eration due to gravity is 9.8 m/s2.

Correct answer: 1.47168× 10−9 m.

Explanation:

Let : Ue = 6.27× 10−19 J ,

ke = 8.98755× 109 N ·m2/C2 ,

qn = 0 C ,

qp = 1.6021× 10−19 C , and

qα = 2 qp = 3.2042× 10−19 C .

q1 = q2 = 2 qp + 2 qn

= 2 (1.6021× 10−19 C) + 2 (0 C)

= 3.2042× 10−19 C , so

Ue = keq1 q2r

r = keq1 q2Ue

= keq21

Ue


= (8.99× 109 N ·m2/C2)

× (3.2042× 10−19 C)2

6.27× 10−19 J

= 1.47168× 10−9 m .

Hewitt CP9 22 P06020 10.0 points

The potential difference between a stormcloud and the ground is 2.87× 108 V.If a bolt carrying 2 C falls from a cloud to

Earth, what is the magnitude of the change ofpotential energy of the charge?

Correct answer: 5.74× 108 J.

Explanation:

Let : V = 2.87× 108 V and

q = 2 C .

The potential energy is

U = V q = (2.87× 108 V )(2 C )

= 5.74× 108 J .

AP EM 1993 MC 53 54021 (part 1 of 2) 10.0 points

A battery or batteries connected to twoparallel plates produce the equipotential linesshown between the plates.

2V1V0V−1V−2V

Which of the following configurations ismost likely to produce these equipotentiallines?

1.

+ -2V

+ -2V

2.

- +

2V- +

2V

correct

3.

+ -

2V

4.

- +

2V

5.

+ -

2V- +

2V

Explanation:The potential difference between the two

plates is 4V. With a negative potential onthe left plate, the battery orientation must benegative on the left and positive on the right.

022 (part 2 of 2) 10.0 points

The force ~F on an electron located on the0-volt potential line is

1. directed to the right, but its magnitudecannot be determined without knowing thedistance between the lines. correct

2. directed to the left, but its magnitudecannot be determined without knowing thedistance between the lines.


3. ~F = 0 N.

4. ~F = 1 N, directed to the right.

5. ~F = 1 N, directed to the left.

Explanation:From the definition of the electric field, we

have

F = q E = qV

d,

where d is the distance between the two plates.Thus to know the magnitude of the force onthe electron, we should know the distance d.The force on the negative charge is in the

opposite direction of the electric field; i.e.,from the higher to the lower electric potentialregion, so the force on the electron is directedto the right.

Electron Volt023 10.0 points

The electron volt is a measure of

1. impulse.

2. charge.

3. velocity.

4. energy. correct

5. momentum.

Explanation:Electron volt (eV) is a unit commonly used

in atomic and nuclear physics. It is defined asthe energy that an electron gains or loses bymoving through a potential of 1 V.

Serway CP 16 16024 (part 1 of 2) 10.0 points

Find the speed of an electron that hasa kinetic energy of 2.12 eV. 1 eV=1.602× 10−19 J .

Correct answer: 8.63533× 105 m/s.

Explanation:

Let : Ek = 2.12 eV and

me = 9.109× 10−31 kg .

The kinetic energy is

Ek =1

2me v

2

e

ve =

√2Ke

me

=

√2(2.12 eV)

9.109× 10−31 kg

1.602× 10−19 J

1 eV

= 8.63533× 105 m/s .

025 (part 2 of 2) 10.0 pointsCalculate the speed of a proton with a kineticenergy of 2.12 eV.

Correct answer: 20152 m/s.

Explanation:

Let : Ek = 2.12 eV and

mp = 1.6726× 10−27 kg .

The kinetic energy is

Ek =1

2mp v

2

p

vp =

√2Ek

mp

=

√2(2.12 eV)

1.6726× 10−27 kg

1.602× 10−19 J

1 eV

= 20152 m/s .

AP B 1998 MC 21026 10.0 points

An electron is in a uniform magnetic fieldB that is directed out of the plane of the page,as shown.


ve−

B

B

B

B

When the electron is moving in the planeof the page in the direction indicated by thearrow, the force on the electron is directed

1. into the page.

2. toward the bottom of the page. correct

3. toward the left

4. toward the right

5. out of the page.

6. toward the top of the page.

Explanation:The force on the electron is

~F = q ~v × ~B = −e ~v × ~B.

The direction of the force is thus

F = −v × B ,

pointing toward the bottom of the page , us-

ing right hand rule for v × B, and reversingthe direction due to the negative charge onthe electron.

Conceptual 16 Q16027 10.0 points

The magnetic field at the equator pointsnorth.If you throw a negatively charged object

(for example, a baseball with some electronsadded) to the east, what is the direction ofthe magnetic force on the object?

1. Downward correct

2. Toward the east

3. Toward the west

4. Upward

Explanation:Use the right-hand rule: point your index

finger east and your middle finger north. Yourthumb points upward (representing the forceon a positively charged object).

Holt SF 21A 03028 (part 1 of 2) 10.0 points

An electron in an electron beam experiencesa downward force of 2.7× 10−14 N while trav-eling in a magnetic field of 4.3× 10−2 T west.The charge on a proton is 1.60×10−19.a) What is the magnitude of the velocity?

Correct answer: 3.92442× 106 m/s.

Explanation:

Let : Fm = 2.7× 10−14 N ,

B = 4.3× 10−2 T , and

qe = 1.60× 10−19 C .

The magnetic force is

Fm = q v B

v =Fm

q B

=2.7× 10−14 N

(1.6× 10−19 C) (0.043 T)

= 3.92442× 106 m/s

029 (part 2 of 2) 10.0 pointsb) What is its direction?

1. East

2. None of these

3. West

4. North correct


5. South

Explanation:Apply right-hand rule (for a negative

charge); force directed into the palm of thehand, fingers in the direction of the field,thumb in the direction of the velocity.Palm faces up, fingers point west, so the

thumb points north.

Wire in Magnetic Field 01030 10.0 points

A wire of constant length is moving in aconstant magnetic field, as shown below. Thewire and the velocity vector are perpendicularto each other and also perpendicular to thefield.

v

Magnetic Field

Which graph best represents the potentialdifference E between the ends of the wire as afunction of the speed v of the wire?

1. ε

O v

2. ε

O v

3. ε

O v

4. ε

O v

5. ε

O v

correct

Explanation:By Lorentz’s Law,

~F = q~v × ~B,

the force on the electrons migrating towardone end of the wire increases linearly withthe velocity. This indicates that the potentialdifference between the ends of the wire willalso increase linearly with the velocity.

AP B 1993 MC 19031 10.0 points

Two long, parallel wires are separated by adistance 2 d, as shown below. Wire #1 carriesa steady current I out of the plane of the pagewhile wire #2 carries a steady current I outof the page.

d dP

I I

S

S′

wire #2wire #1

At what points in the plane of the page(besides points at infinity), is the magneticfield due to the currents zero?

1. At only point P . correct

2. At all points on the line SS′, a perpen-dicular bisector of a line connecting the twowires.

3. At no points.

4. At all points on a circle of radius d cen-tered at either wire.

5. At all points on the line connecting thetwo wires.


Explanation:The only way that the total magnetic field

would be zero is if the magnetic fields due tothe two wires have the same magnitude butopposite directions at the same point.Only at points on the line SS′ do the mag-

netic fields have the same magnitude. Onlyat point P are the magnetic fields parallel(aligned with the vertical axis). Using theright hand rule, they are in opposite direc-tion.Thus, at only point P (besides points at in-

finity) is the magnetic field due to the currentszero.

Drummond HW2 02032 10.0 points

A rectangular loop of wire hangs verticallyas shown in the figure. A magnetic field is di-rected horizontally, perpendicular to the wire,and points out of the page at all points as rep-resented by the symbol ⊙. The magnetic fieldis very nearly uniform along the horizontalportion of the wire ab (length is 0.1 m) whichis near the center of a large magnet produc-ing the field. The top portion of the wire loopis free of the field. The loop hangs from abalance which measures a downward force (inaddition to the gravitational force) of 3×10−2

N when the wire carries a current 0.2 A.

a b

I I

F

ℓ

B

What is the magnitude of the magnet fieldB at the center of the magnet?

Correct answer: 1.5 T.

Explanation:

Let : F = 3× 10−2 ,

ℓ = 0.1 m , and

I = 0.2 A .

The magnetic forces on the two vertical sec-tions of the wire loop point to the left andright, respectively. They are equal but inopposite directions, so they add up to zero.Thus the net magnetic force on the loop is thaton the horizontal section ab whose length isℓ = 0.1 m (and θ = 90◦ so sin θ = 1), and

B =F

I ℓ=

3× 10−2 N

(0.2 A) (0.1 m)= 1.5 T .

Holt SF 21Rev 41033 (part 1 of 2) 10.0 points

In the figure, a 33 cm length of conductingwire that is free to move is held in placebetween two thin conducting wires. All of thewires are in a magnetic field. When a 6.0 Acurrent is in the wire, as shown in the figure,the wire segment moves upward at a constantvelocity.The acceleration of gravity is 9.81 m/s2 .

33 cm

6 A6 A6 A

a) Assuming the wire slides without frictionon the two vertical conductors and has a massof 0.13 kg, find the magnitude of the minimummagnetic field that is required to move thewire.

Correct answer: 0.644091 T.

Explanation:

Let : ℓ = 33 cm ,

I = 6.0 A ,

m = 0.13 kg , and

g = 9.81 m/s2 .


The magnetic and gravitational forces areequal:

Fm = Fg

B I ℓ = mg

B =mg

I ℓ

=(0.13 kg) (9.81 m/s2)

(6 A) (0.33 m)

= 0.644091 T

034 (part 2 of 2) 10.0 pointsb) What is its direction?

1. Toward the right edge of the page

2. Toward the bottom edge of the page

3. Out of the page correct

4. Toward the top edge of the page

5. None of these

6. Toward the left edge of the page

7. Into the page

Explanation:Apply right-hand rule; force directed out of

the palm of the hand, fingers in the directionof the field, thumb in the direction of thecurrent.Thumb points to the left, palm faces toward

the top of the page, and the fingers point outof the page.

Accelerated Proton035 (part 1 of 2) 10.0 points

In a nuclear research laboratory, a protonmoves in a particle accelerator through a mag-netic field of intensity 0.177 T at a speed of3.22× 107 m/s.The charge of a proton is

1.60218× 10−19 C.If the proton is moving perpendicular to the

field, what force acts on it?

Correct answer: 9.13145× 10−13 N.

Explanation:

Let : v = 3.22× 107 m/s ,

B = 0.177 T , and

qp = 1.60218× 10−19 C .

The force acting on the proton is

F = qpB v

= qpB v

= (1.60218× 10−19 C)

× (0.177 T) (3.22× 107 m/s)

= 9.13145× 10−13 N .

036 (part 2 of 2) 10.0 pointsIf the proton of mass 1.67262× 10−27 kg con-tinues to move in a direction that is consis-tently perpendicular to the field, what is theradius of curvature of its path?.

Correct answer: 1.8992 m.

Explanation:

Let : m = 1.67262× 10−27 kg .

The centripetal force for the circular path is

F =mp v

2

r, so

r =mp v

2

F= (1.67262× 10−27 kg)

× (3.22× 107 m/s)2

(9.13145× 10−13 N)

= 1.8992 m .

AP B 1993 FR 3037 (part 1 of 6) 10.0 points

A particle of mass 1.3114× 10−25 kg andcharge of magnitude 4.8× 10−19 C is acceler-ated from rest in the plane of the page througha potential difference of 189 V between two


parallel plates as shown. The particle is in-jected through a hole in the right-hand plateinto a region of space containing a uniformmagnetic field of magnitude 0.491 T orientedperpendicular to the plane of the page. Theparticle curves in a semicircular path andstrikes a detector.

q

m

Region ofMagnetic

Field B

E

hole

What is the sign of the charge of the parti-cle? Neglect relativistic effects.

1. The charge q cannot be determined.

2. The charge q is positive (+) . correct

3. The charge q is negative (−) .

Explanation:The charge accelerates toward the negative

plate and away from the positive plate, so thecharge is positive.

038 (part 2 of 6) 10.0 pointsWhich way does the magnetic field point?

1. Into the page

2. Cannot be determined

3. To the left

4. Toward the bottom of page

5. Toward the top of page

6. Out of the page correct

7. To the right

Explanation:

B

+q

m

E

hole

+−

−−

−

−

+

+

+

Because the particle curves down, the di-rection of ~v × ~B is down. By the right-handrule, ~B points out of the page.

039 (part 3 of 6) 10.0 pointsWhat is the speed of the charged particle asit enters the region of the magnetic field?

Correct answer: 37196.2 m/s.

Explanation:

Let : m = 1.3114× 10−25 kg ,

V = 189 V , and

|q| = 4.8× 10−19 C .

The change in kinetic energy of the chargedparticle is equal to the work done on it by thepotential difference:

1

2mv2 = q V

v =

√2 q V

m

=

√2 (4.8× 10−19 C) (189 V)

1.3114× 10−25 kg

= 37196.2 m/s .

040 (part 4 of 6) 10.0 pointsWhat is the magnitude of the force exerted onthe charged particle as it enters the region ofthe magnetic field ~B ?

Correct answer: 8.76641× 10−15 N.

Explanation:

Let : B = 0.491 T .


The force on the particle is given by theLorentz force law

~F = q~v × ~B

‖~F‖ = q v B

= (4.8× 10−19 C) (37196.2 m/s) (0.491 T)

= 8.76641× 10−15 N .

041 (part 5 of 6) 10.0 pointsWhat is the distance from the point of injec-tion to the detector?

Correct answer: 0.0413944.

Explanation:The acceleration of the particle will be cen-

tripetal:

mv2

r= q v B

r =mv

q B

=(1.3114× 10−25 kg) (37196.2 m/s)

(4.8× 10−19 C) (0.491 T)= 0.0206972 m ,

and the distance from the point of injectionto the detector is

2 r = 2 (0.0206972 m) = 0.0413944 m .

042 (part 6 of 6) 10.0 pointsWhat is the work done by the magnetic fieldon the charged particle during the semicircu-lar trip?

1. W = 1.8144× 10−16 J

2. W = 4.536× 10−17 J

3. W = −1.8144× 10−16 J

4. W = 0 J correct

5. W = 5.77541× 10−17 J

6. W = −9.072× 10−17 J

7. W = −4.536× 10−17 J

8. W = −5.77541× 10−17 J

9. W = 9.072× 10−17 J

Explanation:

W = ~F · ~d .The magnetic field causes a force which isperpendicular to the displacement, so no workis done.

Documents

Quest Review 4 Electric Force, Magnetic Fields KEY