Version 001 – Review 4 Electric Force, Magnetic fields – tubman – (19112)
1
This print-out should have 42 questions.
Multiple-choice questions may continue on
the next column or page – find all choices
before answering.
and Y are separated before the rod moves
away, those charges will remain on X and Y,
X is positively charged and Y is negatively
charged.
AP EM 1993 MC 55 e1
001 10.0 points
Charging Metallic Objects
002 10.0 points
1) Two uncharged metal balls, X and Y,
stand on glass rods and are touching.
Two metal spheres that are initially uncharged are mounted on insulating stands, as
shown.
−
− −−
X
Y
X
Y
2) A third ball, carrying a positive charge, is
brought near the first two.
A negatively charged rubber rod is brought
close to but does not make contact with sphere
X. Sphere Y is then brought close to X on the
side opposite to the rubber rod. Y is allowed
to touch X and then is removed some distance
away. The rubber rod is then moved far away
from 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 charged
electrons will be pushed to sphere Y. If X
+
Y
X
3) Then the first two balls are separated from
each other,
+
Y
X
4) and the third ball is finally removed.
Y
X
What are the resulting charges?
1. Ball X is negative and ball Y 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.
Version 001 – Review 4 Electric Force, Magnetic fields – tubman – (19112)
Explanation:
When a positive ball is moved near a metallic object (X and Y), the positive charge will
attract negative charges, causing X to have
excess positive charge and Y to have excess
negative charge (X and Y are in contact, so
the total net charge on X and Y should be
zero).
+
−
AP EM 1998 MC 39 40
004 (part 1 of 2) 10.0 points
Two charged particles of equal magnitude
(+Q and −Q) are fixed at opposite corners of
a square that lies in a plane (see figure below).
A test charge +q is placed at a third corner.
+Q
+
Later, X and Y are separated, retaining
their charges, so when the third ball is finally
removed, X will have net positive charge and
Y will have net negative charge.
Acceleration of a Particle
003 10.0 points
A particle of mass 40 g and charge 36 µC is
released from rest when it is 19 cm from a
second particle of charge −12 µC.
Determine the magnitude of the initial acceleration of the 40 g particle.
Correct answer: 2688.78 m/s2 .
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 ,
ke = 8.9875 × 109 N · m2 /C2 .
+q
−Q
What is the direction of the force on the
test charge due to the two other charges?
1.
2.
3.
4.
correct
5.
Explanation:
Let :
2
6.
7.
and
8.
The force exerted on the particle is
|q| |Q|
= ma
d2
|q| |Q|
a = ke
m d2
= (8.9875 × 109 N · m2 /C2 )
3.6 × 10−5 C −1.2 × 10−5 C
×
(0.04 kg) (0.19 m2 )
F = ke
= 2688.78 m/s2 .
Explanation:
The force between charges of the same sign
is repulsive and between charges with opposite signs is attractive.
Version 001 – Review 4 Electric Force, Magnetic fields – tubman – (19112)
3
the Earth and moon predominate over electric
forces?
+Q
1. Because the distance between the Earth
and the moon is very large.
+q
−Q
2. Because the masses of the Earth and
moon are very large.
The resultant force is the sum of the two
vectors in the figure.
3. Because there is no electric charge on the
moon.
005 (part 2 of 2) 10.0 points
If F is the magnitude of the force on the test
charge due to only one of the other charges,
what is the magnitude of the net force acting on the test charge due to both of these
charges?
4. Because both the Earth and the moon are
electrically neutral. correct
F
1. Fnet = √
3
2. Fnet = 0
3. Fnet = F
2F
4. Fnet = √
3
2F
5. Fnet =
3
3F
6. Fnet =
2
7. Fnet = 3 F
F
8. Fnet = √
2
√
9. Fnet = 2 F correct
10. Fnet = 2 F
Explanation:
The individual forces form a right angle, so
the magnitude of the net force is
p
√
Fnet = F 2 + F 2 = 2 F .
Hewitt CP9 22 R02
006 10.0 points
Why does the gravitational force between
Explanation:
Since the electrical force between the two
objects varies directly as the product of their
charges and the Earth and the moon are electrically neutral, the electrical force between
them is zero.
AP B 1993 MC 68
007 10.0 points
The diagram shows an isolated, positive
charge Q, where point B is twice as far away
from Q as point A.
+Q
A
B
0
10 cm
20 cm
What is the ratio of the electric field
strength at point A to the electric field
strength at point B?
1.
2.
3.
4.
5.
EA
EB
EA
EB
EA
EB
EA
EB
EA
EB
=
=
=
=
=
8
1
2
1
1
1
1
2
4
correct
1
Explanation:
Let : rB = 2 rA .
Version 001 – Review 4 Electric Force, Magnetic fields – tubman – (19112)
The electric field strength E ∝
1
r2
EA
= A
1
EB
rB2
=
1
, so
r2
A
−
rB2
(2 r)2
= 4.
=
rA2
r2
AP EM 1993 MC 36
008 10.0 points
From the electric field vector at a point, one
can determine which of the following?
I. the direction of the electrostatic force on
a 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 charge
at 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
~
~ = F , so F
~ = qE
~ and F
~ acts in the same
E
q
~ depending on the
or opposite direction to E
sign of the charge. If we only consider the
magnitude F = q E for a unit charge, the
F
electric field’s magnitude is E = .
q
Four Charges in a Square Short
009 10.0 points
Consider a square with side a. Four charges
−q, +q, +q, and −q are placed at the corners
A, B, C, and D, respectively
+
4
B
a
O
−
+
D
C
What is the magnitude of the electric field
at the center O?
1 kq
√
2 2 a2
√ kq
2. EO = 4 2 2 correct
a
kq
3. EO = 2
a
√ kq
4. EO = 2 2 2
a
kq
5. EO = 3 2
a
1 kq
6. EO = √
2 a2
√ kq
7. EO = 2 2
a
1 kq
8. EO = √
3 2 a2
√ kq
9. EO = 3 2 2
a
1 kq
10. EO = √
4 2 a2
1. EO =
Explanation:
The distance between each corner and the
a
center is √ , so the magnitude of each electric
2
field at D is
E=k q
a
√
2
2 = 2 k
q
a2
The two negative charges yield forces pointing away from them from O and the two positive charges yield forces pointing toward them
from O with the collinear charges adding algebraically:
Version 001 – Review 4 Electric Force, Magnetic fields – tubman – (19112)
~A + E
~ C k = kE
~B +E
~ Dk = 2 E = 4 k
kE
q
.
a2
EA + EC
E
EB + ED
The Cartesian components of the two vectors with the origin at O are
1
q
1
~A + E
~B = 4k
and
E
− √ ı̂ + √ ̂
a2
2
2
q
1
1
~
~
EB + ED = 4 k 2 − √ ı̂ − √ ̂ , so
a
2
2
1
1
−√ − √
ı̂
2
2
1
1
̂
+ √ −√
2
2
√
q
= −4 2 k 2 ı̂ ,
a
√
q
with magnitude −4 2 k 2 .
a
~ = 4k q
E
a2
Two Charge Field
010 (part 1 of 3) 10.0 points
Two point charges at fixed locations produce an electric field as shown.
5
1. Toward charge B correct
2. Toward charge A
3. Along an equipotential plane
Explanation:
Electric field lines run from a positive potential to a negative potential, so the charge
B is positive. A negative charge will move toward a positive potential, which creates lower
potential energy and a higher kinetic energy.
011 (part 2 of 3) 10.0 points
The electric field at point Y is
1. the same as that the field at point X.
2. stronger than the field at point X. correct
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 unit
volume at Y is greater than the number of
field lines per unit volume at X.
012 (part 3 of 3) 10.0 points
Estimate 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
A
B
Y
X
How would a negative charge placed at
point Y move?
QA −17 = 1.30769
≈
QB
13 E Field Diagrams 03
013 (part 1 of 2) 10.0 points
Version 001 – Review 4 Electric Force, Magnetic fields – tubman – (19112)
R radius
G.
∝
6
1
r2
Q
0
Consider a solid conducting sphere of radius R and total charge Q. Which diagram
describes the E(r) vs r (electric field vs radial
distance) function for the sphere?
S.
1
∝ 2
r
0
M.
r
R
∝
1
r
∝
0
1
r2
r
R
1. L
2. G correct
3. M
4. S
5. P
Explanation:
Because the charge distribution is spherically symmetric, select a spherical Gaussian
surface of radius r and surface area 4 π r 2 concentric with the sphere. The electric field due
to the conducting sphere is directed radially
outward by symmetry and is therefore normal
~ is parallel
to the surface at every point and E
~ at each point.
to dA
There is no charge within the Gaussian surface, so E = 0
for r < R .
For the region outside the conducting
sphere,
P.
∝
0
I
I
I
~
~
ΦE = E · dA = E dA = E
dA
qin
= E 4 π r2 =
ǫ0
Q
qin
=
for r > R .
E=
2
4 π ǫ0 r
4 π ǫ0 r 2
1
r2
∝
r
R
1
r2
E
L.
r
R
∝
1
r2
G.
0
0
R
r
R
r
014 (part 2 of 2) 10.0 points
Which diagram describes the E(r) vs r
(electric field vs radial distance) function if
Version 001 – Review 4 Electric Force, Magnetic fields – tubman – (19112)
the sphere is instead a uniformly charged,
non-conducting sphere?
1. P
2. L correct
3. S
4. G
Explanation:
Select a spherical Gaussian surface of radius
r and volume V ′ , where r < R, concentric
with the uniformly charged non-conducting
sphere. The charge qin within the Gaussian
surface of the volume V ′ is less than Q; from
Q
the volume charge density ρ ≡ ,
V
4
′
3
qin = ρ V = ρ
.
πr
3
Applying Gauss’ law, for r < R,
I
I
3Q
1
ρ=
by definition and k =
, so
3
4πR
4 π ǫ0
E=
3Q
kQ
4
k
πr = 3 r.
3
3 4πR
R
In the region outside the uniformly charged
non-conducting sphere, we have the same conditions as for the conducting sphere when applying Gauss’ law, so
E=
Charge in Electric Field
015 10.0 points
A particle of charge q is placed in a uniform electric field of magnitude E. Consider
the following statements about the resulting
forces 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
qen
dA = E 4 π r 2 =
ǫ0
4
ρ π r3
ρ
qen
3
r.
=
=
E=
2
2
4 π ǫ0 r
4 π ǫ0 r
3 ǫ0
E dA = E
Q
.
4 π ǫ0 r 2
1
r2
E
∝
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 charge
would feel at that point, so the electric force
the charge experiences is of magnitude q E
and in the same direction as the direction of
the electric field.
Electron Deflection
016 (part 1 of 3) 10.0 points
An electron traveling at 6 × 106 m/s enters a
0.06 m region with a uniform electric field of
220 N/C , as in the figure.
0.06 m
̂
−−−−−−−−−
6 × 106 m/s
L.
0
R
7
r
+++++++++
ı̂
Version 001 – Review 4 Electric Force, Magnetic fields – tubman – (19112)
Find the magnitude of the acceleration of
the electron while in the electric field. The
mass of an electron is 9.109 × 10−31 kg and
the fundamental charge is 1.602 × 10−19 C .
Correct answer: 3.86914 × 1013 m/s2 .
8
Correct answer: 0.00193457 m.
Explanation:
Using the equation for the displacement in
the vertical direction and the results from the
first two parts of the problem,
Explanation:
1 2
at
2
−3.86914 × 1013 m/s2
=
2
−8 2
× (1 × 10 s)
= −0.00193457 m ,
∆y =
Let : qe = −1.602 × 10−19 C ,
me = 9.109 × 10−31 kg , and
E = 220 N/C .
F = ma = qE
qe E
a=
̂
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 points
Find the time it takes the electron to travel
through the region of the electric field, assuming it doesn’t hit the side walls.
Correct answer: 1 × 10
−8
s.
Explanation:
with a magnitude of 0.00193457 m .
Holt SF 18A 01
019 10.0 points
Two alpha particles (helium nuclei), each
consisting of two protons and two neutrons, have an electrical potential energy of
6.27 × 10−19 J .
What is the distance between these particles at this time? The Coulomb constant
is 8.98755 × 109 N · m2 /C2 , the elemental
charge is 1.6021 × 10−19 C, and the acceleration due to gravity is 9.8 m/s2 .
Correct answer: 1.47168 × 10−9 m.
Explanation:
Let :
Let :
ℓ = 0.06 m , and
v0 = 6 × 106 m/s .
The horizontal distance traveled is
ℓ = v0 t
ℓ
0.06 m
t=
=
v0
6 × 106 m/s
= 1 × 10
−8
s .
018 (part 3 of 3) 10.0 points
What is the magnitude of the vertical displacement ∆y of the electron while it is in the
electric field?
Ue
ke
qn
qp
= 6.27 × 10−19 J ,
= 8.98755 × 109 N · m2 /C2 ,
= 0 C,
= 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
q1 q2
r
q2
q1 q2
= ke 1
r = ke
Ue
Ue
Ue = k e
Version 001 – Review 4 Electric Force, Magnetic fields – tubman – (19112)
= (8.99 × 109 N · m2 /C2 )
(3.2042 × 10−19 C)2
×
6.27 × 10−19 J
= 1.47168 × 10
−9
1.
m .
+ - + 2V 2V
Hewitt CP9 22 P06
020 10.0 points
The potential difference between a storm
cloud 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 of
potential energy of the charge?
Correct answer: 5.74 × 108 J.
2.
correct
- + - +
2V 2V
3.
Explanation:
Let :
9
+ 2V
V = 2.87 × 108 V
q = 2 C.
and
4.
- +
2V
The potential energy is
U = V q = (2.87 × 108 V )(2 C )
= 5.74 × 108 J .
5.
AP EM 1993 MC 53 54
021 (part 1 of 2) 10.0 points
A battery or batteries connected to two
parallel plates produce the equipotential lines
shown between the plates.
+ - - +
2V 2V
Explanation:
The potential difference between the two
plates is 4 V. With a negative potential on
the left plate, the battery orientation must be
negative on the left and positive on the right.
022 (part 2 of 2) 10.0 points
~ on an electron located on the
The force F
0-volt potential line is
−2V −1V 0V
1V
2V
Which of the following configurations is
most likely to produce these equipotential
lines?
1. directed to the right, but its magnitude
cannot be determined without knowing the
distance between the lines. correct
2. directed to the left, but its magnitude
cannot be determined without knowing the
distance between the lines.
Version 001 – Review 4 Electric Force, Magnetic fields – tubman – (19112)
10
~ = 0 N.
3. F
Let :
~ = 1 N, directed to the right.
4. F
~ = 1 N, directed to the left.
5. F
Explanation:
From the definition of the electric field, we
have
V
F = qE = q ,
d
where d is the distance between the two plates.
Thus to know the magnitude of the force on
the 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 potential
region, so the force on the electron is directed
to the right.
Electron Volt
023 10.0 points
The electron volt is a measure of
Ek = 2.12 eV and
me = 9.109 × 10−31 kg .
The kinetic energy is
1
me ve2
2
r
2 Ke
ve =
m
s e
2(2.12 eV)
1.602 × 10−19 J
=
9.109 × 10−31 kg
1 eV
Ek =
= 8.63533 × 105 m/s .
025 (part 2 of 2) 10.0 points
Calculate the speed of a proton with a kinetic
energy of 2.12 eV.
Correct answer: 20152 m/s.
Explanation:
1. impulse.
2. charge.
Let :
3. velocity.
Ek = 2.12 eV and
mp = 1.6726 × 10−27 kg .
The kinetic energy is
4. energy. correct
5. momentum.
Explanation:
Electron volt (eV) is a unit commonly used
in atomic and nuclear physics. It is defined as
the energy that an electron gains or loses by
moving through a potential of 1 V.
Serway CP 16 16
024 (part 1 of 2) 10.0 points
Find the speed of an electron that has
a kinetic energy of 2.12 eV.
1 eV=
1.602 × 10−19 J .
Correct answer: 8.63533 × 105 m/s.
Explanation:
1
mp vp2
2
s
2 Ek
vp =
mp
s
2(2.12 eV)
1.602 × 10−19 J
=
1.6726 × 10−27 kg
1 eV
Ek =
= 20152 m/s .
AP B 1998 MC 21
026 10.0 points
An electron is in a uniform magnetic field
B that is directed out of the plane of the page,
as shown.
Version 001 – Review 4 Electric Force, Magnetic fields – tubman – (19112)
B
B
v
3. Toward the west
4. Upward
e−
B
11
B
When the electron is moving in the plane
of the page in the direction indicated by the
arrow, the force on the electron is directed
Explanation:
Use the right-hand rule: point your index
finger east and your middle finger north. Your
thumb points upward (representing the force
on a positively charged object).
3. toward the left
Holt SF 21A 03
028 (part 1 of 2) 10.0 points
An electron in an electron beam experiences
a downward force of 2.7 × 10−14 N while traveling 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?
4. toward the right
Correct answer: 3.92442 × 106 m/s.
1. into the page.
2. toward the bottom of the page. correct
Explanation:
5. out of the page.
6. toward the top of the page.
Explanation:
The force on the electron is
~ = q ~v × B
~ = −e ~v × B.
~
F
Let : Fm = 2.7 × 10−14 N ,
B = 4.3 × 10−2 T , and
qe = 1.60 × 10−19 C .
The magnetic force is
The direction of the force is thus
b = −b
b,
F
v×B
pointing toward the bottom of the page , usb and reversing
ing right hand rule for b
v × B,
the direction due to the negative charge on
the electron.
Conceptual 16 Q16
027 10.0 points
The magnetic field at the equator points
north.
If you throw a negatively charged object
(for example, a baseball with some electrons
added) to the east, what is the direction of
the magnetic force on the object?
Fm = q v B
v=
=
Fm
qB
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 points
b) What is its direction?
1. East
2. None of these
1. Downward correct
3. West
2. Toward the east
4. North correct
Version 001 – Review 4 Electric Force, Magnetic fields – tubman – (19112)
ε
5.
5. South
Explanation:
Apply right-hand rule (for a negative
charge); force directed into the palm of the
hand, 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 01
030 10.0 points
A wire of constant length is moving in a
constant magnetic field, as shown below. The
wire and the velocity vector are perpendicular
to each other and also perpendicular to the
field.
O
Which graph best represents the potential
difference E between the ends of the wire as a
function of the speed v of the wire?
1.
ε
v
correct
Explanation:
By Lorentz’s Law,
~ = q ~v × B,
~
F
the force on the electrons migrating toward
one end of the wire increases linearly with
the velocity. This indicates that the potential
difference between the ends of the wire will
also increase linearly with the velocity.
AP B 1993 MC 19
031 10.0 points
Magnetic Field
v
12
Two long, parallel wires are separated by a
distance 2 d, as shown below. Wire #1 carries
a steady current I out of the plane of the page
while wire #2 carries a steady current I out
of the page.
S
I
I
d
wire #1
P
d
wire #2
S′
O
2.
v
ε
At what points in the plane of the page
(besides points at infinity), is the magnetic
field due to the currents zero?
1. At only point P . correct
O
3.
2. At all points on the line SS ′ , a perpendicular bisector of a line connecting the two
wires.
ε
O
4.
v
v
ε
O
3. At no points.
4. At all points on a circle of radius d centered at either wire.
v
5. At all points on the line connecting the
two wires.
Version 001 – Review 4 Electric Force, Magnetic fields – tubman – (19112)
Explanation:
The only way that the total magnetic field
would be zero is if the magnetic fields due to
the two wires have the same magnitude but
opposite directions at the same point.
Only at points on the line SS ′ do the magnetic fields have the same magnitude. Only
at point P are the magnetic fields parallel
(aligned with the vertical axis). Using the
right hand rule, they are in opposite direction.
Thus, at only point P (besides points at infinity) is the magnetic field due to the currents
zero.
Drummond HW2 02
032 10.0 points
A rectangular loop of wire hangs vertically
as shown in the figure. A magnetic field is directed horizontally, perpendicular to the wire,
and points out of the page at all points as represented by the symbol ⊙. The magnetic field
is very nearly uniform along the horizontal
portion of the wire ab (length is 0.1 m) which
is near the center of a large magnet producing the field. The top portion of the wire loop
is free of the field. The loop hangs from a
balance which measures a downward force (in
addition to the gravitational force) of 3×10−2
N when the wire carries a current 0.2 A.
Explanation:
Let :
I
ℓ
B
a
b
F
What is the magnitude of the magnet field
B at the center of the magnet?
Correct answer: 1.5 T.
F = 3 × 10−2 ,
ℓ = 0.1 m , and
I = 0.2 A .
The magnetic forces on the two vertical sections of the wire loop point to the left and
right, respectively. They are equal but in
opposite directions, so they add up to zero.
Thus the net magnetic force on the loop is that
on the horizontal section ab whose length is
ℓ = 0.1 m (and θ = 90◦ so sin θ = 1), and
F
3 × 10−2 N
B=
=
= 1.5 T .
Iℓ
(0.2 A) (0.1 m)
Holt SF 21Rev 41
033 (part 1 of 2) 10.0 points
In the figure, a 33 cm length of conducting
wire that is free to move is held in place
between two thin conducting wires. All of the
wires are in a magnetic field. When a 6.0 A
current is in the wire, as shown in the figure,
the wire segment moves upward at a constant
velocity.
The acceleration of gravity is 9.81 m/s2 .
33 cm
6A
I
13
6A
6A
a) Assuming the wire slides without friction
on the two vertical conductors and has a mass
of 0.13 kg, find the magnitude of the minimum
magnetic field that is required to move the
wire.
Correct answer: 0.644091 T.
Explanation:
Let : ℓ = 33 cm ,
I = 6.0 A ,
m = 0.13 kg , and
g = 9.81 m/s2 .
Version 001 – Review 4 Electric Force, Magnetic fields – tubman – (19112)
The magnetic and gravitational forces are
equal:
Fm = Fg
B I ℓ = mg
mg
B=
Iℓ
(0.13 kg) (9.81 m/s2 )
=
(6 A) (0.33 m)
= 0.644091 T
034 (part 2 of 2) 10.0 points
b) What is its direction?
1. Toward the right edge of the page
2. Toward the bottom edge of the page
14
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 = qp B v
= qp B v
= (1.60218 × 10−19 C)
× (0.177 T) (3.22 × 107 m/s)
= 9.13145 × 10−13 N .
4. Toward the top edge of the page
036 (part 2 of 2) 10.0 points
If the proton of mass 1.67262 × 10−27 kg continues to move in a direction that is consistently perpendicular to the field, what is the
radius of curvature of its path?.
5. None of these
Correct answer: 1.8992 m.
6. Toward the left edge of the page
Explanation:
3. Out of the page correct
7. Into the page
Explanation:
Apply right-hand rule; force directed out of
the palm of the hand, fingers in the direction
of the field, thumb in the direction of the
current.
Thumb points to the left, palm faces toward
the top of the page, and the fingers point out
of the page.
Accelerated Proton
035 (part 1 of 2) 10.0 points
In a nuclear research laboratory, a proton
moves in a particle accelerator through a magnetic field of intensity 0.177 T at a speed of
3.22 × 107 m/s.
The
charge
of
a
proton
is
−19
1.60218 × 10
C.
If the proton is moving perpendicular to the
field, what force acts on it?
Correct answer: 9.13145 × 10−13 N.
Let : m = 1.67262 × 10−27 kg .
The centripetal force for the circular path is
mp v 2
, so
r
mp v 2
r=
F
= (1.67262 × 10−27 kg)
(3.22 × 107 m/s)2
×
(9.13145 × 10−13 N)
F =
= 1.8992 m .
AP B 1993 FR 3
037 (part 1 of 6) 10.0 points
A particle of mass 1.3114 × 10−25 kg and
charge of magnitude 4.8 × 10−19 C is accelerated from rest in the plane of the page through
a potential difference of 189 V between two
Version 001 – Review 4 Electric Force, Magnetic fields – tubman – (19112)
parallel plates as shown. The particle is injected through a hole in the right-hand plate
into a region of space containing a uniform
magnetic field of magnitude 0.491 T oriented
perpendicular to the plane of the page. The
particle curves in a semicircular path and
strikes a detector.
q
Region of
Magnetic
m
Field B
hole
E
What is the sign of the charge of the particle? 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 the
charge is positive.
038 (part 2 of 6) 10.0 points
Which 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:
+q
m
+
+
+
15
−
hole
−
−
E
+
−
B
−
Because the particle curves down, the di~ is down. By the right-hand
rection of ~v × B
~ points out of the page.
rule, B
039 (part 3 of 6) 10.0 points
What is the speed of the charged particle as
it 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 charged
particle is equal to the work done on it by the
potential difference:
1
m v2 = q V
2
r
2qV
v=
m
s
2 (4.8 × 10−19 C) (189 V)
=
1.3114 × 10−25 kg
= 37196.2 m/s .
040 (part 4 of 6) 10.0 points
What is the magnitude of the force exerted on
the charged particle as it enters the region of
~?
the magnetic field B
Correct answer: 8.76641 × 10−15 N.
Explanation:
Let : B = 0.491 T .
Version 001 – Review 4 Electric Force, Magnetic fields – tubman – (19112)
The force on the particle is given by the
Lorentz force law
~ = q ~v × B
~
F
~k = qvB
kF
= (4.8 × 10−19 C) (37196.2 m/s) (0.491 T)
= 8.76641 × 10−15 N .
041 (part 5 of 6) 10.0 points
What is the distance from the point of injection to the detector?
Correct answer: 0.0413944.
Explanation:
The acceleration of the particle will be centripetal:
m v2
= qvB
r
mv
r=
qB
(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 injection
to the detector is
2 r = 2 (0.0206972 m) = 0.0413944 m .
042 (part 6 of 6) 10.0 points
What is the work done by the magnetic field
on the charged particle during the semicircular 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
16
7. W = −4.536 × 10−17 J
8. W = −5.77541 × 10−17 J
9. W = 9.072 × 10−17 J
Explanation:
~ · ~d .
W =F
The magnetic field causes a force which is
perpendicular to the displacement, so no work
is done.