PHYSICS 212
CHAPTER 19
MAGNETISM
WORKBOOK
ANSWERS
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made. (Credit will be lost if this is turned in messy.)
3. Complete the chapter outline section as early as possible. Don’t wait for the due
date to be assigned to start.
4. Complete the sections in sequence.
5. Study and learn definitions of terms, physical quantities, units, principles, and
basic equations before attempting problems.
6. You may work on this with other students but do not copy another student’s
workbook or let a student copy this workbook. Do not copy from other sources
either.
7. Wherever possible, include diagrams in your solutions. Diagrams are required.
8. Keep this workbook after it is graded and returned to you.
9. Using the answer key, redo all questions and problems until you can answer
them all correctly by yourself without help.
10. Use the workbook to learn the general problem-solving strategy rather than
how individual problems are solved.
11. Questions marked (Basic) should be answered by students without the need
for additional assistance.
-1-
PHYSICS 212
19-1
CHAPTER 19
MAGNETISM
OUTLINE
Magnets
Complete the following:
1. Magnetic fields affect _moving charges_____ and moving charges produce
__magnetic fields_______ .
2. The ends of a bar magnet are called __poles________. And are identified as the
_north pole___ and the ____south pole__ .
3. Like poles _repel______ and unlike poles __attract______ .
Compare and contrast soft and hard magnetic materials.
Soft materials are easily magnetized but also easily lose their magnetism.
Hard materials are harder to magnetize but hold their magnetism longer.
What symbol is used to designate a magnetic field? ______ B _______________
Draw a sketch of a bar magnet. (fig 19.2)
19.2
EARTH’S MAGNETIC FIELD
If you were sent on a job to Earth’s magnetic south pole, where would you go geographically?
Near the north geographic pole.
Define:
magnetic dip angle: The angle between the magnetic field and horizontal
-2-
magnetic declination: The angle between the magnetic field direction to north and
true north the direction to the geographic north pole
19.3
MAGNETIC FIELDS
What are the SI units of the magnetic field?
Tesla
What are the cgs units of the magnetic field and the conversion factor between cgs and SI units?
Gauss, 1G = 10-4 T
Under what condition will th force on a moving charge in a magnetic field be maximum?
The force is maximum when the velocity is perpendicular to the field.
Under what two different conditions can a charged particle be in a magnetic field and not experience any
force?
When it is stationary or moving parallel to field lines.
What is the name of the rule used to determine the direction of the force on a moving charge in a
magnetic field?
The right hand rule.
19.4
MAGNETIC FORCE ON A CURRENT-CARRYING CONDUCTOR
Write equation 19.6 listing and naming all variables along with their SI units.
F = BIL sin θ
F
Force on wire
N
B
magnetic field
T
I
Current in wire
A
L
Length of wire
m
θ
angle between I and B
deg
A current is directed to the east inside an apparatus in which there is an upward magnetic field. In what
direction is the current displaced?
toward the south
-3-
19.5
TORQUE ON A CURRENT LOOP AND ELECTRIC MOTORS
Does the torque on a current loop depend on the shape of the loop?
No, on area, current and number of turns of wire.
How should a current loop be positioned so it has no torque on it in a magnetic field?
The axis of the loop should be aligned with the magnetic field.
Define the magnetic moment of a coil. What symbol is used and what are the SI units?
m = N µ o IAnˆ
Am2
Describe the following parts of a DC motor:
commutator: rotates with motor, provides electric contact with coil and allows for
changing direction of current flow
brushes:
19.6
make electrical contact with moving commutator
MOTION OF A CHARGED PARTICLE IN A MAGNETIC FIELD
When a charged particle is made to move in a direction perpendicular to a magnetic field what path will
it follow?
circular
Describe the motion of a charged particle that has velocity components both perpendicular and parallel
to a magnetic field?
the particle spirals along the field line
19.7
MAGNETIC FIELD OF A LONG, STRAIGHT WIRE AND AMPÈRE’S LAW
What is the permeability of free space, its symbol, numeric value, and SI units?
µ o = 4π × 10 −7
Tm
A
How does the magnetic field strength of the field produced by a long, current carrying wire vary with
distance from the wire?
it has a 1/r dependance
-4-
State Ampère’s Law.
∑ B cosθ∆l = µ I
o enc
loop
19.8
MAGNETIC FORCE BETWEEN TWO PARALLEL CONDUCTORS
How would you characterize the force between to parallel wires carrying current in the same direction?
attractive
Define the SI units Ampere and Coulomb.
1 A flows through two parallel wires 1 meter apart if the force between them is 2×10-7 N
If 1 A flows for 1 s, then 1 C of charge has passed.
19.9
MAGNETIC FIELDS OF CURRENT LOOPS AND SOLENOIDS
Sketch the magnetic field of a current loop (fig 19.30). Show the direction of the current in the wire.
Sketch the magnetic field of a solenoid (fig19.32). Show the direction of the current in the wire.
19.10 MAGNETIC DOMAINS
What causes magnetism in materials?
unpaired spinning electrons
-5-
What are magnetic domains?
Groups of atoms whose electron spins are aligned.
Describe ferromagnetic materials.
materials with large magnetic domains
What determines if a magnetic material retains its magnetism?
Thermal energy can easily randomize the orientation of aligned domains in soft
materials.
-6-
PHYSICS 212
CHAPTER 19 MULTIPLE CHOICE QUESTIONS
MAGNETISM
Multiple Choice
Identify the letter of the choice that best completes the statement or answers the question.
C____ 1.
Electrical charges and magnetic poles have many similarities, but one difference is:
A. opposite magnetic poles repel.
B. one magnetic pole cannot create magnetic poles in other materials.
C. a magnetic pole cannot be isolated.
D. magnetic poles do not produce magnetic fields.
E. magnetic poles produce only alternating fields.
(Basic)
A____ 2.
Which of the following is not a hard magnetic material?
A. iron
D. neodymium
B. cobalt
E. both b and c
C. nickel
(Basic)
A____ 3.
Geophysicists today generally attribute the existence of the Earth's magnetic field to
which of the following?
A. convection currents within the liquid interior
B. iron ore deposits in the crust
C. nickel-iron deposits in the crust
D. solar flares
E. iron-cobalt deposits in the crust
(Basic)
D____ 4.
The term magnetic declination refers to which of the following?
A. angle between Earth's magnetic field and Earth's surface
B. Earth's magnetic field strength at the equator
C. tendency for Earth's field to reverse itself
D. angle between directions to true north and magnetic north
E. angle between Earth's magnetic field and Earth's rotational axis
(Basic)
B____ 5.
The magnetic field of the Earth is believed responsible for which of the following?
A. deflection of both charged and uncharged cosmic rays
B. deflection of charged cosmic rays
C. ozone in the upper atmosphere
D. solar flares
E. deflection of uncharged cosmic rays
(Basic)
B____ 6.
The magnetic pole of the Earth nearest the geographic North Pole corresponds to which
of the following?
A. a magnetic north pole
C. a magnetic arctic pole
B. a magnetic south pole
D. a magnetic antarctic pole
(Basic)
-7-
C____ 7.
The force on a charged particle created by its motion in a magnetic field is maximum at
what angle between the particle velocity and field?
A. zero
D. 45°
B. 180°
E. 135°
C. 90°
(Basic)
B____ 8.
Assume that a uniform magnetic field is directed into this page. If an electron is released
with an initial velocity directed from the bottom edge to the top edge of the page, which
of the following describes the direction of the resultant force acting on the electron?
A. out of the page
D. into the page
B. to the right
E. from top edge to bottom edge of the
page
C. to the left
B____ 9.
A proton moves across the Earth's equator in a northeasterly direction. At this point the
Earth's magnetic field has a direction due north and is parallel to the surface. What is the
direction of the force acting on the proton at this instant?
A. toward the northwest
D. toward the northeast
B. out of the Earth's surface
E. toward the southwest
C. into the Earth's surface
B____ 10.
Different units can be used to measure the same physical quantity, differing only by some
multiplicative factor. The cgs unit for magnetic field, the gauss, is equal to ____ tesla.
A. 104
B. 10-4
C. 0.5
D. 4π
E. These units do not measure the same physical quantity.
C____ 11.
If a proton is released at the equator and falls toward the Earth under the influence of
gravity, the magnetic force on the proton will be toward the:
A. north.
C. east.
B. south.
D. west.
E____ 12.
A stationary positive charge +Q is located in a magnetic field B, which is directed toward
the right as indicated. The direction of the magnetic force on Q is:
A. toward the right.
B. up.
C. down.
D. toward the left.
E. There is no magnetic force.
-8-
D____ 13.
There is a magnetic force on a particle. It is possible that the particle is:
A. uncharged.
B. stationary.
C. moving in the direction of the magnetic field.
D. not part of a wire.
E. moving opposite the direction of the magnetic field.
(Basic)
D____ 14.
Which of the following devices makes use of an electromagnet?
A. loudspeaker
D. both A and B
B. galvanometer
E. none of the above
C. gyrocompass
D____ 15.
The direction of the force on a current carrying wire located in an external magnetic field
is which of the following?
A. perpendicular to the current
D. Both choices A and B are valid.
B. perpendicular to the field
E. None of the above are valid.
C. parallel to the wire
(Basic)
C____ 16.
A circular current loop is placed in an external magnetic field. How is the torque related
to the radius of the loop?
A. directly proportional to radius
B. inversely proportional to radius
C. directly proportional to radius squared
D. inversely proportional to radius squared
E. directly proportional to square root of radius
(Basic)
C____ 17.
Magnetism had been a known phenomenon for some time before its relation to electric
currents was found. That a current in a wire produces a magnetic field was discovered by:
A. Maxwell.
D. Tesla.
B. Ampere.
E. Faraday.
C. Oersted.
(Basic)
D____ 18.
A current in a long, straight wire produces a magnetic field. The magnetic field lines:
A. go out from the wire to infinity.
D. form circles that go around the wire.
B. come in from infinity to the wire.
E. are parallel to the wire.
C. form circles that pass through the wire.
(Basic)
C____ 19.
Two parallel wires are separated by 0.25 m. Wire A carries 5.0 A and Wire B carries 10
A, both currents in the same direction. The force on 0.80 m of Wire A is:
A. half that on 0.80 m of wire B.
D. away from Wire B.
B. one-fourth that on 0.80 m of wire B.
E. one-eighth that on 0.80 m of wire B.
C. toward Wire B.
(Basic)
-9-
B____ 20.
A current in a solenoid coil creates a magnetic field inside that coil. The field strength is
directly proportional to:
A. the solenoid area.
D. Both A and B are valid choices.
B. the current.
E. None of the above choices are valid.
C. the solenoid diameter.
(Basic)
D____ 21.
A current in a coil with N turns creates a magnetic field at the center of that loop. The
field strength is directly proportional to:
A. number of turns in the loop.
D. Both choices A and B are valid.
B. current strength.
E. None of the above are valid.
C. length of the coil.
(Basic)
C____ 22.
The magnetic domains in a non-magnetized piece of iron are characterized by which
orientation?
A. parallel to the magnetic axis
B. anti-parallel (opposite direction) to the magnetic axis
C. random
D. perpendicular to the magnetic axis
E. any of the above is possible.
(Basic)
A____ 23.
When an electromagnet has an iron core inserted, what happens to the strength of the
magnet?
A. It increases.
B. It remains the same.
C. It decreases.
D. Since it depends on the metal used in the wires of the electromagnet, any of the
above.
(Basic)
-10-
PHYSICS 212
1.
CHAPTER 19 ADDITIONAL QUESTIONS
MAGNETISM
Calculate the net force on a dust particle moving in Earth’s magnetic field. The magnetic field is
horizontal and directed southward with a magnitude of 0.5 G. The dust particle has a mass of 0.05
grams, a charge of -4e and is traveling horizontally eastward at 15 m/s. (Basic)
B=0.5G=0.5×10-4T
F = qvB sin θ = ( −4e)(15 ms )(0.5 × 10−4 T )sin(90)
F = 4.80 × 10−22 N
According to the right-hand-rule, the direction of the force on the moving charge is initially directed
upward.
2.
What is the force on a 50m length of wire carrying 15 A in a vertically upward magnetic field of 2 T?
(Basic)
Assuming that the wire is horizontal.
F = ILB sin θ = (15 A)(50m)(2T )sin(90)
F = 1500T
If the current in the wire is directed to the north, the force on the wire will be to the east.
-11-
3.
A coil is made of 15 loops of wire with a radius of 30 cm. If it carries a current of 5 A, what is its
magnetic dipole moment? Add the magnetic dipole moment to the diagram. (Basic)
m
m = NIA = NI π r 2 = (15)(5 A)π (.3m) 2
m = 21.21Am 2
4.
Wire 1 lies along the y-axis with 2.5 A of current flowing in the +y direction. Wire 2 lies along the
x-axis with 3.75 A of current flowing in the -x direction. What is the net magnetic field due to these
currents at the point P located at coordinates (5 cm, 6 cm)?
Both B1 and B2 are directed inward, according to the right hand rule.
BP = BP ,1 + BP ,2
Bp =
µo I1 µo I 2
+
2π r1 2π r2
(4π × 10−7 Tm
(4π × 10 −7 Tm
A )(2.5 A)
A )(3.75 A)
Bp =
+
2π (.05m)
2π (.06m)
BP = 2.25 × 10 −5 T inward
-12-
5.
Calculate the magnetic field inside a 40 cm long solenoid if the solenoid current is 3.0 A and the
there are 300 turns of wire per cm. Indicate the direction of the magnetic field also. (Basic)
B=
µ 0 IN
L
=
(4π × 10−7
Tm
A
B = 1.13 × 10−1 T upward
6.
)(3 A)(300cm −1 )(40cm)
.4m
In some region of space a uniform magnetic field of 3.5 T points to the right. A microsphere of
charge -4nC and mass 1.0 mg moves in this field with a speed of 1.5 × 106 m/s initially directed
upward and perpendicular to the field. Draw this magnetic field and the path the microsphere will
follow. What is the radius of this path?
FB = FC
mv 2
qvB sin θ =
r
(1.0 × 10−6 kg )(1.5 × 106 ms )
mv
=
r=
qB sin 90o
(4.0 × 10 −9 C )(3.5T )(1)
r = 1.07 × 108 m
The path the particle will follow is circular and perpendicular to the page. The initial force is outward.
-13-
7.
Suppose the coil (loop) of problem 3 is placed in the magnetic field of problem 5. The plane of the
coil makes a 35 degree angle with the magnetic field. Calculate the magnitude of the torque acting
on the loop and specify how the loop will respond to this torque.
τ = mB sin θ = (21.21Am 2 )(.113T )sin(550 )
τ = 1.96 Nm
The coil rotates so that its magnetic dipole moment aligns with the solenoid’s magnetic field.
8.
A long wire carrying 7.0 A in the +y direction lies along the y-axis. Next to the wire, in the x-y
plane is a square loop whose sides are 5 cm long and is made with 15 turns of wire. The left edge of
the loop is 7 cm from the long wire. Compute the net force on the loop. Current in the loop is 1A
going around the loop clockwise.
FNET = FL − FR = NI 2 aBL − NI 2 aBR = NI 2 a ( BL − BR )
µ I
µ I  NI aµ I  1 1 
FNET = NI 2 a  o 1 − o 1  = 2 o 1  − 
2π
 2π rL 2π rR 
 rL rR 
(15)(1A)(.05m)(4π × 10 −7 Tm
1
1 
A )(7 A) 
FNET =
−


2π
 .07 m .13m 
FNET = 6.92 × 10−6 N to the left
The magnetic field produced by current 1 is directed into the page where the coil is. The forces on the top
and bottom of the coil are equal and opposite. The force on the left side is attractive and stronger than the
force on the right side which is repulsive.
-14-
PHYSICS 212
PRACTICE PROBLEMS:
HOMEWORK PROBLEMS:
CHAPTER 19
MAGNETISM
PROBLEMS
3, 4, 10, 11, 12, 15, 19, 20, 23, 27, 34, 39, 41, 44, 47, 61
1, 7, 13, 17, 29, 32, 40, 46, 49, 58
[NOTE: Problem numbers in italic are included in the student study guide; Problem numbers underlined are
online at www.cp7e.com.]
1.
An electron gun fires electrons into a magnetic field directed straight downward. Find the direction
of the force exerted by the field on an electron for each of the following directions of the electron’s
velocity: (a) horizontal and due north; (b) horizontal and 30/ west of north; (c) due north, but at 30/
below the horizontal; (d) straight upward. (Remember that an electron has a negative charge.)
(Basic)
The direction in parts (a) through (d) is found by use of the right hand rule. You must remember that the
electron is negatively charged and thus experiences a force in the direction exactly opposite that predicted by
the right hand rule for a positively charged particle.
(a) horizontal and due east
(b) horizontal and 30° N of E
(c) horizontal and due east
(d) zero force , F = qvB sin θ = qvB sin (180° ) = 0
7.
What velocity would a proton need to circle Earth 1 000 km above the magnetic equator, where
Earth’s magnetic field is directed horizontally north and has a magnitude of 4.00 × 10!8 T?
The gravitational force is small enough to be ignored, so the magnetic force must supply the needed
centripetal acceleration. Thus,
v2
qBr
where r = RE + 1000 km=7.38 × 106 m
m = qvB sin 90° , or v =
r
m
(1.60 × 10
v=
−19
)(
)(
C 4.00 × 10−8 T 7.38 × 106 m
1.67 × 10
−27
kg
)=
2.83 × 107 m s
If v is toward the west and B is northward, F will be directed downward as required.
-15-
13.
In Figure P19.3, assume that in each case the velocity vector shown is replaced with a wire carrying a
current in the direction of the velocity vector. For each case, find the direction of the magnetic field
that will produce the magnetic force shown. (Basic)
Figure P19.3
Use the right hand rule, holding your right hand with the fingers in the direction of the current and the
thumb pointing in the direction of the force. As you close your hand, the fingers will move toward the
direction of the magnetic field. The results are
(a) into the page
17.
(b) toward the right
(c) toward the bottom of the page
A wire with a mass of 1.00 g/cm is placed on a horizontal surface with a coefficient of friction of 0.200. The wire carries
a current of 1.50 A eastward and moves horizontally to the north. What are the magnitude and the direction of the
smallest vertical magnetic field that enables the wire to move in this fashion?
For minimum field, B should be perpendicular to the wire. If the force is to be northward, the field must be
directed downward .
To keep the wire moving, the magnitude of the magnetic force must equal that of the kinetic friction force.
Thus, B I L sin 90° = µ k ( mg ) , or
B=
µk ( m L ) g
I sin 90°
( 0.200 )(1.00 g cm ) ( 9.80
=
(1.50 A )(1.00 )
-16-
)
m s 2  1 kg   10 2 cm 
 = 0.131 T
 3 
 10 g  1 m 
29.
Figure P19.29a is a diagram of a device called a velocity selector, in which particles of a specific velocity pass through
undeflected while those with greater or lesser velocities are deflected either upwards or downwards. An electric field is
directed perpendicular to a magnetic field, producing an electric force and a magnetic force on the charged particle that
can be equal in magnitude and opposite in direction (Fig. P19.29b) and hence cancel. Show that particles with a speed of
v = E/B will pass through the velocity selector undeflected.
Figure P19.29
For the particle to pass through with no deflection, the net force acting on it must be zero. Thus, the
magnetic force and the electric force must be in opposite directions and have equal magnitudes. This gives
Fm = Fe , or qvB = qE which reduces to v = E B
-17-
32.
A mass spectrometer is used to examine the isotopes of uranium. Ions in the beam emerge from the velocity selector at a
speed of 3.00 × 105 m/s and enter a uniform magnetic field of 0.600 T directed perpendicularly to the velocity of the ions.
What is the distance between the impact points formed on the photographic plate by singly charged ions of 235U and 238U?
Since the centripetal acceleration is furnished by the magnetic force acting on the ions, qvB =
mv
mv 2
or the
r
radius of the path is r =
. Thus, the distance between the impact points (that is, the difference in the
diameters of the paths qB
followed by the
U 238 and the U 235 isotopes) is
∆d = 2 ( r238 − r235 ) =
=
or
(
2v
( m238 − m235 )
qB
2 3.00 × 105 m s
(1.60 × 10
−19
)


−27 kg  

( 238 u − 235 u ) 1.66 × 10
u  
C ( 0.600 T ) 

)
-18-
40.
The two wires in Figure P19.40 carry currents of 3.00 A and 5.00 A in the direction indicated. (a) Find the direction and
magnitude of the magnetic field at a point midway between the wires. (b) Find the magnitude and direction of the
magnetic field at point P, located 20.0 cm above the wire carrying the 5.00-A current.
Figure P19.40
Call the wire carrying a current of 3.00 A wire 1 and the other wire 2. Also, choose the line running from
wire 1 to wire 2 as the positive x direction.
(a)
At the point midway between the wires, the field due to
each wire is parallel to the y axis and the net field is
Bnet = + B1 y − B2 y = µ 0 ( I1 − I 2 ) 2π r
Thus,
Bnet
or
( 4π × 10
=
−7
T⋅m A
2π ( 0.100 m )
) ( 3.00 A − 5.00 A ) = − 4.00 × 10
Bnet = 4.00 µ T toward the bottom of the page
-19-
−6
T
(b)
At point P,
r1 = ( 0.200 m ) 2 and B1 is directed at θ1 = +135° .
The magnitude of B1 is
−7
µ0 I1 ( 4π × 10 T ⋅ m A ) ( 3.00 A )
B1 =
=
= 2.12 µ T
2π r1
2π 0.200 2 m
(
)
The contribution from wire 2 is in the –x direction and has magnitude
( 4π × 10 T ⋅ m A ) ( 5.00 A ) = 5.00 µ T
µI
B2 = 0 2 =
2π r2
2π ( 0.200 m )
−7
Therefore, the components of the net field at point P are:
Bx = B1 cos135° + B2 cos180°
= ( 2.12 µ T ) cos135° + ( 5.00 µ T ) cos180° = −6.50 µ T
and
B y = B1 sin135° + B2 sin180° = ( 2.12 µ T ) sin135° + 0 = +1.50 µ T
Therefore, at
Bnet = Bx2 + B y2 = 6.67 µ T
 Bx 
 6.50 µ T 
= tan −1 

 = 77.0°
 By 
µ
1.50
T




θ = tan −1 
or
B net = 6.67 µ T at 77.0° to the left of vertical
-20-
46.
In Figure P19.46, the current in the long, straight wire is I1 = 5.00 A, and the wire lies in the plane of the rectangular
loop, which carries 10.0 A. The dimensions shown are c = 0.100 m, a = 0.150 m, and R = 0.450 m. Find the magnitude
and direction of the net force exerted by the magnetic field due to the straight wire on the loop.
Fnet = +
or Fnet
( 4π × 10
=
−7
µ 0 I1 I 2 ℓ
µIIℓ
µ I I ℓ1
1 
− 0 1 2 = 0 1 2  −

2π c
2π ( c + a )
2π  c c + a 
T ⋅ m A ) ( 5.00 A )(10.0 A )( 0.450 m ) 
1
1

−


2π
 0.100 m 0.250 m 
= + 2.70 × 10 −5 N = 2.70 × 10 −5 N to the left
-21-
49.
A single-turn square loop of wire 2.00 cm on a side carries a counterclockwise current of 0.200 A. The loop is inside a
solenoid, with the plane of the loop perpendicular to the magnetic field of the solenoid. The solenoid has 30 turns per
centimeter and carries a counterclockwise current of 15.0 A. Find the force on each side of the loop and the torque acting
on the loop.
The magnetic field inside the solenoid is

turns  100 cm  
B = µ 0 nI1 = 4π × 10 −7 T ⋅ m A  30
(15.0 A ) = 5.65 × 10−2 T



cm  1 m  

(
)
Therefore, the magnitude of the magnetic force on any one of the sides of the square loop is
(
)
(
)
F = BI 2 L sin 90.0° = 5.65 × 10 −2 T ( 0.200 A ) 2.00 × 10 −2 m = 2.26 × 10 −4 N
The forces acting on the sides of the loop lie in the plane of the loop, are perpendicular to the sides,
and are directed away from the interior of the loop. Thus, they tend to stretch the loop but do not tend
to rotate it. The torque acting on the loop is
τ =0
-22-
58.
Two circular loops are parallel, coaxial, and almost in contact 1.00 mm apart (Fig. P19.58). Each loop is 10.0 cm in
radius. The top loop carries a clockwise current of 140 A. The bottom loop carries a counterclockwise current of 140 A.
(a) Calculate the magnetic force that the bottom loop exerts on the top loop. (b) The upper loop has a mass of 0.021 0 kg.
Calculate its acceleration, assuming that the only forces acting on it are the force in part (a) and its weight. [Hint: The
distance between the loops is small in comparison to their radius of curvature, so the loops may be treated as long,
straight parallel wires.]
Figure P19.58
(a)
Since the distance between them is so small in comparison to the radius of curvature, the hoops
may be treated as long, straight, parallel wires. Because the currents are in opposite directions, the
hoops repel each other. The magnetic force on the top loop is
µ I I
Fm =  0 1 2
 2π d
µ0 I 2 ( 2π r ) µ0 I 2 r

=
L =
2π d
d

( 4π × 10
=
(b)
or
−7
)
T ⋅ m A (140 A ) ( 0.100 m )
1.00 × 10 −3 m
2
= 2.46 N upward
ΣFy = ma y = Fm − mg
ay =
Fm
2.46 N
−g=
− 9.80 m s 2 = 107 m s 2 upward
m
0.021 kg
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