ch20 giancoli7e manual

8/15/2019 Ch20 Giancoli7e Manual

1/30

© Copyright 2014 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.

No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

20-1

Responses to Questions

1. The Earth’s magnetic field is not always parallel to the surface of the Earth—it may have a component

perpendicular to the Earth’s surface. The compass will tend to line up with the local direction of the

magnetic field, so one end of the compass will dip downward. The angle that the Earth’s magnetic

field makes with the horizontal is called the dip angle.

2. The pole on a magnetic compass needle that points geographically northward is defined at the north

pole of the compass. This north pole is magnetically attracted to the south pole of other magnets, so the

Earth’s magnetic field must have a south pole at the geographic north pole.

3. The magnetic field lines form clockwise circles centered on the wire.

4. The force is downward. The field lines point from the north pole to the

south pole, or left to right. Use the right-hand rule. Your fingers point in

the direction of the current (away from you). Curl them in the direction of

the field (to the right). Your thumb points in the direction of the force

(downward). See Fig. 20–11a, copied here.

5. A magnet will not attract just any metallic object. For example, while a magnet will attract paper clips

and nails, it will not attract coins or pieces of aluminum foil. This is because magnets will only attract

other ferromagnetic materials (iron, cobalt, nickel, gadolinium, and some of their oxides and alloys).

Iron and its alloys (such as steel) are the only common materials. These ferromagnetic materialscontain magnetic domains that can be made to temporarily align when a strong magnet is brought near.

The alignment occurs in such a way that the north pole of the domain points toward the south pole of

the strong magnet, and vice versa, which creates the attraction.

6. No, they are not both magnets. If they were both magnets, then they would repel one another when

they were placed with like poles facing each other. However, if one is a magnet and the other isn’t,

then they will attract each other no matter which ends are placed together. The magnet will cause an

alignment of the domains of the nonmagnet, causing an attraction.

MAGNETISM 20


2/30

20-2 Chapter 20



7. Typical current in a house circuit is 60 Hz AC. Due to the mass of the compass needle, its reaction to

60 Hz (changing direction back and forth at 60 complete cycles per second) will probably not be

noticeable. A DC current in a single wire could affect a compass, depending on the relative orientation

of the wire and the compass, the magnitude of the current, and the distance from the wire to the

compass. A DC current being carried by two very close wires in opposite directions would not have

much of an effect on the compass needle, since the two currents would cause magnetic fields that

tended to cancel each other.

8. The magnetic force will be exactly perpendicular to the velocity, which means that the force is

perpendicular to the direction of motion. Since there is no component of force in the direction of

motion, the work done by the magnetic force will be zero, and the kinetic energy of the particle will

not change. The particle will change direction, but not change speed.

9. Use the right-hand rule to determine the direction of the force on each particle. In the plane of the

diagram, the magnetic field is coming out of the page for points above the wire and is going into the

page for points below the wire.

a: force down, toward the wire

b: force to the left, opposite of the direction of the currentc: force up, toward the wire

d: force to the left, opposite of the direction of the current

10. Charge a experiences an initial upward force. By the right-hand rule, charge a must be positive.

Charge b experiences no force, so charge b must be uncharged.

Charge c experiences an initial downward force. By the right-hand rule, charge c must be negative.

11. A magnet can attract an iron bar, so by Newton’s third law, an iron bar can attract a magnet. Another

consideration is that the iron has domains that can be made slightly magnetic by an external magnetic

field, as opposed to a substance like plastic or wood. Thus the iron is also a magnet, at least when it is

close to the magnet.

12. The right-hand rule tells us that as the positive particle moves to the right, if it is deflected upward,then the magnetic field must be into the page, and if it is deflected downward, then the magnetic field

must be out of the page. Also, the stronger the magnetic field, the more tightly the charged particle will

turn. The magnetic field at the first bend must be relatively small and pointed into the page. The

magnetic field at the second bend must be medium in size and pointed out of the page. The magnetic

field at the third bend must be relatively very large and pointed into the page. In terms of the indicated

letters:

a: no field

b: relatively weak field, into the page

c: moderate strength field, out of the page

d: no field

e: very strong field, into the page

13. Section 17–11 shows that a CRT television picture is created by shooting a beam of electrons at afluorescent screen. Wherever the electrons hit the screen, the screen glows momentarily. When you

hold a strong magnet too close to the screen, the magnetic field puts a force on the moving electrons

( sin ), F q Bυ θ = which causes them to bend away from their original destination position on the

screen. These errant electrons then cause parts of the screen to light up that aren’t supposed to be lit

and cause other parts to be slightly darker than they are supposed to be. The picture sometimes goes

completely black where the field is strongest because at these points all of the electrons that were

supposed to hit that particular spot have been deflected by the strong field.


3/30


4/30

20-4 Chapter 20



Large current

circuit

Small currentcircuit for

dashboard switch

Starter

Car battery Solenoid

IronRod

SpringyPivot

Piece

of

Iron

22. To design a relay, place the iron

rod inside of a solenoid and then

point one end of the solenoid/rod

combination at the piece of iron on

a pivot. A spring normally holds

the piece of iron away from a

switch, making an open circuit

where current cannot flow. When

the relay is activated with a small

current, a relatively strong

magnetic field is created inside the solenoid, which aligns most of the magnetic domains in the iron

rod and produces a strong magnetic field at the end of the solenoid/rod combination. This magnetic

field attracts the piece of iron on the springy pivot, which causes it to move toward the switch,

connecting it and allowing current to flow through the large current circuit.

23. The two ions will come out of the velocity selector portion of the mass spectrometer at the same speed,

since / . E Bυ = Once the charges reach the area of the mass spectrometer where there is only a

magnetic field, the difference in the ions’ charges will have an effect. Eq. 20–12 for the massspectrometer says that / ,m qBB r E ′= which can be rearranged as / .r mE qBB′= Since every quantity

in the equation is constant except q, the doubly ionized ion will hit the film at a half of the radius as the

singly ionized ion.

24. The magnetic domains in the unmagnetized piece of iron are initially pointing in random directions, as

in Fig. 20–42a (which is why it appears to be unmagnetized). When the south pole of a strong external

magnet is brought close to the random magnetic domains of the iron, many of the domains will rotate

slightly so that their north poles are closer to the external south pole, which causes the unmagnetized

iron to be attracted to the magnet. Similarly, when the north pole of a strong external magnet is

brought close to the random magnetic domains of the iron, many of the domains will rotate slightly so

that their south poles are closer to the external north pole, which causes the unmagnetized iron to now

be attracted to the magnet. Thus, either pole of a magnet will attract an unmagnetized piece of iron.

25. Initially, both the nail and the paper clip have all of their magnetic domains pointing in random

directions (which is why they appear to be unmagnetized). Thus, when you bring them close to each

other, they are not attracted to or repelled from each other. Once the nail is in contact with a magnet

(let’s say the north pole), many of the nail’s domains will align in such a way that the end of the nail

that is touching the magnet becomes a south pole, due to the strong attraction, and the opposite end of

the nail then becomes a north pole. Now, when you bring the nail close to the paper clip, there are

mainly north poles of nail domains close to the paper clip, which causes some of the domains in the

paper clip to align in such a way that the end near the nail becomes a south pole. Since the nail’s

domains are only partially aligned, it will not be a strong magnet and thus the alignment of the paper

clip’s domains will be even weaker. The attraction of the paper clip to the nail will be weaker than the

attraction of the nail to the magnet.

Responsess to MisConceptual Questions

1. (a, b, d , e) A common misconception is that only permanent magnets (such as a magnet and the

Earth) create magnetic fields. However, moving charges and electric currents also produce

magnetic fields. Stationary charges, ordinary pieces of iron, and other pieces of metal do not

create magnetic fields.


5/30

Magnetism 20-5



2. (c) It is common to confuse the directions of the magnetic fields with electric fields and thus

indicate that the magnetic field points toward or away from the current. Since the current is

flowing into the page, the right-hand rule indicates that the magnetic field is in the clockwise

direction around the current. At point A the field points downward.

3. (b) The right-hand rule indicates that the magnetic field is clockwise around the current, so at point

B the current is to the left.

4. (a) The charged particle only experiences a force when it has a component of velocity perpendicular

to the magnetic field. When it moves parallel to the field, it follows a straight line at constant

speed.

5. (c) Electric fields are created by charged objects whether the charges are moving or not. Magnetic

fields are created by moving charged objects. Since the proton is charged and moving, it creates

both an electric field and a magnetic field.

6. (a) A stationary charged particle does not experience a force in a magnetic field. Therefore, the

particle must be moving to experience a force. The force is a maximum when the particle ismoving perpendicular to the field, not parallel to the field. Since the force is perpendicular to the

motion of the particle, it acts as a centripetal force, changing the particle’s direction but not its

kinetic energy. That is, since the force is perpendicular to the motion, it does no work on the

particle. The direction of the force is always perpendicular to the direction of motion and also

perpendicular to the magnetic field.

7. (c) Section 20–5 shows that the magnetic field from a current is directed circularly around the wire,

is proportional to the current flowing in the wire, and is inversely proportional to the distance

from the wire. A constant current produces a magnetic field, so the current does not need to be

changing.

8. (e) It is common to confuse the direction of electric fields (which point toward or away from the

charges) with magnetic fields, which always make circles around the current.

9. (c) A common misconception is that a force always does work on the object. Since the magnetic

force is perpendicular to the velocity of the proton, the force acts as a centripetal force, changing

the proton’s direction, but not doing any work and thus not changing its kinetic energy.

10. (a) A common misconception is that a constant magnetic field can change the magnitude of the

particle’s velocity. However, the magnetic force is always perpendicular to the velocity, so it can

do no work on the particle. The magnetic force only serves as a centripetal force to change the

particle’s direction.

11. (e) Equation 20–3 shows that the magnetic force depends upon the particle’s charge, its velocity,

and the strength of the external magnetic field. The direction of the force is always perpendicular

to the magnetic field and the velocity of the particle. Therefore, all four statements are accurate.

12. (c) This question requires a consideration of Newton’s third law. The force that one wire exerts on a

second must be equal in magnitude, but opposite in direction, to the force that the second exerts

on the first.


6/30

20-6 Chapter 20



Solutions to Problems

1. (a) Use Eq. 20–1 to calculate the force with an angle of 90° and a length of 1 meter.

sin / sin (6.40 A)(0.90 T)sin 90 5.8 N/m F I B F IBθ θ = → = = ° =

(b) Change the angle to 35.0 .°

/ sin (6.40 A)(0.90 T)sin 35.0 3.3 N/m F IB θ = = ° =

2. Use Eq. 20–2.

maxmax 2

0.625 N1.63 A

(4.80 m)(8.00 10 T)

F F I B I

B −= → = = =

×

3. Use Eq. 20–1 to calculate the force.

5sin (120 A)(240 m)(5.0 10 T)sin 68 1.3 N F I B θ −= = × ° =

4. The dip angle is the angle between the Earth’s magnetic field and the current in the wire. Use Eq. 20–1

to calculate the force.

5 4sin (4.5 A)(2.6 m)(5.5 10 T)sin 41 4.2 10 N F I B θ − −= = × ° = ×

5. To have the maximum force, the current must be perpendicular to the magnetic field,

as shown in the first diagram. Use max0.45 F F

=

to find the angle between the wire

and the magnetic field, illustrated in the second diagram. Use Eq. 20–1.

1max

0.45 sin 0.45 sin 0.45 27

F F

IB IBθ θ −

= → = → = = °

6. Use Eq. 20–2. The length of wire in the B field is the same as the diameter of the pole faces.

maxmax

1.28 N0.358 T

(6.45 A)(0.555 m)

F F I B B

I = → = = =

7. (a) By the right-hand rule, the magnetic field must be pointing up, so the top pole face must be a

south pole.

(b) Use Eq. 20–2 to relate the maximum force to the current. The length of wire in the magnetic

field is equal to the diameter of the pole faces.

2max

max

(8.50 10 N)

3.864 A 3.86 A(0.100 m)(0.220 T)

F

F I B I B

−×

= → = = = ≈

(c) If the wire is tipped so that it points 10 0. ° downward, then

the angle between the wire and the magnetic field is

changed to 100 0 .. ° But the length of wire now in the field

has increased from to /cos10 0 .. ° The net effect of the

changes is that the force has not changed.

2initial rotated initial; ( / cos10.0 ) sin100.0 8.50 10 N F I B F I B I B F

−= = ° ° = = = ×

B

wire

B

wireθ

initial

rotated


7/30

Magnetism 20-7



8. The magnetic force must be equal in magnitude to the force of gravity on the wire. The maximum

magnetic force is applicable since the wire is perpendicular to the magnetic field. The mass of the wire

is the density of copper times the volume of the wire.

( )

21

B2

2 3 3 3 2 2

5

(8.9 10 kg/m ) (1.00 10 m) (9.80 m/s )1400 A

4 4(5.0 10 T)

F mg I B d g

d g I

B

ρπ

ρπ π −

−

= → = →

× ×= = =

×

This answer does not seem feasible. The current is very large, and the resistive heating in the thin

copper wire would probably melt it.

9. The maximum magnetic force as given in Eq. 20–4 can be used since the velocity is perpendicular to

the magnetic field.

19 5 14max (1.60 10 C)(7.75 10 m/s)(0.45 T) 5.6 10 N F q Bυ

− −= = × × = ×

By the right-hand rule, the force must be directed to the north.

10. The magnetic force will cause centripetal motion, and the electron will move in a clockwise circular

path if viewed in the direction of the magnetic field. The radius of the motion can be determined.

2

max

31 65

19

(9.11 10 kg)(1.70 10 m/s)1.51 10 m

(1.60 10 C)(0.640 T)

F q B mr

mr

qB

υ υ

υ − −−

= = →

× ×= = = ×

×

Assuming the magnetic field existed in a sharply defined region, the electron would make only one-

half of a rotation and then exit the field going in the opposite direction from which it came.

11. In this scenario, the magnetic force is causing centripetal motion, so it must have the form of acentripetal force. The magnetic force is perpendicular to the velocity at all times for circular motion.

2 27 6

max 19

(6.6 10 kg)(1.6 10 m/s)0.24 T

2(1.60 10 C)(0.14 m)

m F q B m B

r qr

υ υ υ

−

−

× ×= = → = = =

×

12. Since the charge is negative, the answer is the OPPOSITE of the result given from the right-hand rule

applied to the velocity and magnetic field.

(a) no force

(b) downward

(c) upward

(d ) inward, into the page

(e) to the left

( f ) to the left

13. The right-hand rule applied to the velocity and magnetic field would give the direction of the force.

Use this to determine the direction of the magnetic field given the velocity and the force.

(a) to the right

(b) downward

(c) into the page


8/30

20-8 Chapter 20



14. The force on the electron due to the electric force must be the same magnitude as the force on the

electron due to the magnetic force.

36 6

E B 3

7.7 10 V/m1.027 10 m/s 1.0 10 m/s

7.5 10 T

E F F qE q B

Bυ υ

−

×= → = → = = = × ≈ ×

×

If the electric field is turned off, then the magnetic force will cause circular motion.

2 31 64

B 19 3

(9.11 10 kg)(1.027 10 m/s)7.8 10 m

(1.60 10 C)(7.5 10 T)

m F q B m r

r qB

υ υ υ

−−

− −

× ×= = → = = = ×

× ×

15. (a) The speed of the ion can be found using energy conservation. The electric potential energy of the

ion becomes kinetic energy as it is accelerated.

21initial final 2

195 5

27

2 2[2(1.60 10 C)](3700 V)5.99 10 m/s 6.0 10 m/s

(6.6 10 kg)

E E qV m

qV

m

υ

υ −

−

= → = →

×= = = × ≈ ×

×

(b) Since the ion is moving perpendicular to the magnetic field, the magnetic force will be a

maximum. That force will cause the ion to move in a circular path.

2

max

27 52 2

9

(6.6 10 kg)(5.99 10 m/s)3.63 10 m 3.6 10 m

2(1.60 10 C)(0.340 T)

F q B mr

mr

qB

υ υ

υ − − −−1

= = →

× ×= = = × ≈ ×

×

(c) The period can be found from the speed and the radius.

27

5

2 2 (3.63 10 m)3.8 10 s

5.99 10 m/s

r π π

υ

−−×

= = = ××

T

16. (a) From Example 20–6, we have ,m

r qB

υ = so .

rqB

mυ = The kinetic energy is given by

2 2 2 221 1

2 2KE ,

2

rqB r q Bm m

m mυ

⎛ ⎞= = =⎜ ⎟

⎝ ⎠ so we see that 2KE .r ∝

(b) The angular momentum of a particle moving in a circular path is . L m r υ = From Example 20–6,

we have ,m

r qB

υ = so .

rqB

mυ = Combining these relationships gives the following.

2rqB L m r m r qBr

mυ = = =

17. The kinetic energy of the proton can be used to find its velocity. The magnetic force produces

centripetal acceleration, and from this the radius can be determined.

221

2

6 19 27

19

KEKE

KE

KE

2

2

2 2(1.5 10 eV)(1.60 10 J/eV)(1.67 10 kg)0.59 m

(1.60 10 C)(0.30 T)

mm q B

m r

mmm m

r qB qB qB

υ υ υ υ

υ − −

−

= → = = →

× × ×= = = = =

×


9/30

Magnetism 20-9



18. The magnetic field can be found from Eq. 20–4, and the direction is found from the right-hand rule.

Remember that the charge is negative.

13max

max 19 6

6.2 10 N1.4 T

(1.60 10 C)(2.8 10 m/s)

F F q B B

q

υ

υ

−

−

×= → = = =

× ×

The direction would have to be east for the right-hand rule, applied to the velocity and the magnetic

field, to give the proper direction of force.

19. The kinetic energy is used to determine the speed of the particles, and then the speed can be used to

determine the radius of the circular path, since the magnetic force is causing centripetal acceleration.

221

2

p

27 p p

31e ee

KE

KEKEKE

KE

KE

2

22

2

1.67 10 kg42.8

2 9.11 10 kg

mmm m m

m q B r m r qB qB qB

m

r mqB

r mm

qB

υ υ υ υ υ

−

−

= → = = → = = =

×= = = =

×

20. The velocity of each charged particle can be found using energy conservation. The electrical potential

energy of the particle becomes kinetic energy as it is accelerated. Then, since the particle is moving

perpendicularly to the magnetic field, the magnetic force will be a maximum. That force will cause the

ion to move in a circular path, and the radius can be determined in terms of the mass and charge of the

particle.

21initial final 2

2

max

2

2

1 2

qV E E qV m

mqV

mm mV m

F q B m r r qB qB B q

υ υ

υ υ υ

= → = → =

= = → = = =

dd

pddd p

p p d

p p

p

p p p

p p

21

22 2

2 11

21

4

2 22 21

mm V

m B qr r r

r m V q

q B q

mm V

m B qr

r r r m V q

q B q

α α

α α

α α

= = = = → =

= = = = → =

21. The magnetic force produces an acceleration perpendicular to the original motion. If that acceleration

is small, then it will produce a small deflection, and the original velocity can be assumed to always be

perpendicular to the magnetic field. This leads to a constant perpendicular acceleration. The time that

this (approximately) constant acceleration acts can be found from the original velocity υ and thedistance traveled . The starting speed in the perpendicular direction will be zero.


10/30

20-10 Chapter 20



q B F ma q B a

m

υ υ ⊥ ⊥ ⊥= = → =

2 2 9 5 3 221 1

0 2 2 3

6

(18.5 10 C)(5.00 10 T)(1.50 10 m)

2 2(3.40 10 kg)(155 m/s)

1.97 10 m

q B qBd t a t

m m

υ υ

υ υ

− −

⊥ ⊥ ⊥ −

−

× × ×⎛ ⎞= + = = =⎜ ⎟

⎝ ⎠ ×

= ×

This small distance justifies the assumption of constant acceleration.

22. (a) The discussion of the Hall effect in the text states that the Hall emf is proportional to the

magnetic field causing the effect. We can use this proportionality to determine the unknown

resistance. Since the new magnetic field is oriented 90° to the surface, the full magnetic field will

be used to create the Hall potential.

Hall Hall

Hall Hall

63 mV(0.10 T) 0.53 T

12 mV

V V B B B

V B V

⊥⊥ ⊥

⊥

′ ′′′= → = = =

(b) When the field is oriented at 60° to the surface, the magnetic field, sin60 , B ° is used to createthe Hall potential.

Hall

Hall

63 mV (0.10 T)sin 60 0.61 T

12 mV sin60

V B B B

V ⊥ ⊥ ⊥

′′ ′° = → = =

°

23. (a) The sign of the ions will not change the magnitude of the Hall emf but will determine the

polarity of the emf.

(b) The flow velocity corresponds to the drift velocity in the Hall effect relationship.

3Hall

Hall

(0.13 10 V)0.56 m/s

(0.070 T)(0.0033 m)

V V Bd

Bd υ υ

−×= → = = =

24. (a) The drift velocity is found from Hall d .V Bd υ =

65 5Hall

d

(1.02 10 V)4.722 10 m/s 4.7 10 m/s

(1.2 T)(0.018 m)

V

Bd υ

−− −×= = = × ≈ ×

(b) We now find the density by using Eq. 18–10.

19 3 5

29 3

15 A

(1.6 10 C)(1.0 10 m)(0.018 m)(4.722 10 m/s)

1.1 10 electrons/m

d

I n

eAυ − − −= =

× × ×

= ×

25. We assume that the jumper cable is a long straight wire and use Eq. 20–6.7

4 40cable 2

(4 10 T m/A)(65 A)2.889 10 T 2.9 10 T

2 2 (4.5 10 m)

I B

r

µ π

π π

−− −

−

× ⋅= = = × ≈ ×

×

Compare this with the Earth’s field of4

0.5 10 T.−

×

4

cable Earth 5

2.889 10 T/ 5.77,

5.0 10 T B B

−

−

×= =

× so the field of the cable is about 5.8 times that of the Earth.


11/30

Magnetism 20-11



26. We assume that the wire is long and straight and use Eq. 20–6.

40 wire

wire 70

2 2 (0.12 m)(0.50 10 T)30 A

2 4 10 T m/A

I rB B I

r

µ π π

π µ π

−

−

×= → = = =

× ⋅ (2 significant figures)

27. Since the currents are parallel, the force on each wire will be attractive, toward the other wire. Use

Eq. 20–7 to calculate the magnitude of the force.

7 220 1 2

2 2

(4 10 T m/A) (25 A)(25 m) 7.8 10 N, attractive

2 2 (0.040 m)

I I F

d

µ π

π π

−−× ⋅

= = = ×

28. Since the force is attractive, the currents must be in the same direction, so the current in the second

wire must also be upward. Use Eq. 20–7 to calculate the magnitude of the second current.

0 1 22 2

42

2 70 2 1

2

2 2 0.090 m(7.8 10 N/m) 12.54 A 13 A upward

28 A4 10 T m/A

I I F

d

F d I

I

µ

π

π π

µ π

−

−

= →

= = × = ≈

× ⋅

29. To find the direction, draw a radius line from the wire to the field

point. Then at the field point, draw a line perpendicular to the radius,

directed so that the perpendicular line would be part of a

counterclockwise circle. The relative magnitude is given by the

length of the arrow. The farther a point is from the wire, the weaker

the field.

30. For the experiment to be accurate to 3.0%,± the magnetic field due to the current in the cable must be

less than or equal to 3.0% of the Earth’s magnetic field. Use Eq. 20–6 to calculate the magnetic field

due to the current in the cable.

40 Earth

cable Earth 70

2 (0.030 ) 2 (1.00 m)(0.030)(0.50 10 T)0.030 7.5 A

2 4 10 T m/A

I r B B B I

r

µ π π

π µ π

−

−

×= ≤ → ≤ = =

× ⋅

Thus the maximum allowable current is 7.5 A .

31. The magnetic field at the loop due to the long wire is into the page and can be calculated by Eq. 20–6.

The force on the segment of the loop closest to the wire is toward the wire, since the currents are in the

same direction. The force on the segment of the loop farthest from the wire is away from the wire,

since the currents are in the opposite direction.

Because the magnetic field varies with distance, it is difficult to calculate the total force on either the

left or right segments of the loop. Using the right-hand rule, the force on each small piece of the left

segment of wire is to the left, and the force on each small piece of the right segment of wire is to theright. If left and right small pieces are chosen that are equidistant from the long wire, then the net force

on those two small pieces is zero. Thus the total force on the left and right segments of wire is zero, so

only the parallel segments need to be considered in the calculation. Use Eq. 20–7 to find the force.

0 0 01 2 1 2net near far near far 1 2

near far near far

72 6

1 1

2 2 2

4 10 T m/A 1 1(3.5 A) (0.100 m) 5.1 10 N, toward wire

2 0.030 m 0.080 m

I I I I F F F I I

d d d d

µ µ µ

π π π

π

π

−−

⎛ ⎞= − = − = −⎜ ⎟

⎝ ⎠

⎛ ⎞× ⋅= − = ×⎜ ⎟

⎝ ⎠


12/30

20-12 Chapter 20



32. At the location of the compass, the magnetic field caused by the wire will point

to the west, and the part of the Earth’s magnetic field that turns the compass is

pointing due north. The compass needle will point in the direction of the NET

magnetic field.

750

wire(4 10 T m/A)(48 A) 5.33 10 T

2 2 (0.18 m) I Br

µ π π π

−−× ⋅= = = ×

51 1Earth

5wire

4.5 10 Ttan tan 40 N of W (2 significant figures)

5.33 10 T

B

Bθ

−− −

−

×= = = °

×

33. The magnetic field due to the long horizontal wire points straight up at

the point in question, and its magnitude is given by Eq. 20–6. The two

fields are oriented as shown in the diagram. The net field is the vector

sum of the two fields.

750

wire

(4 10 T m/A)(24.0 A)2.40 10 T

2 2 (0.200 m)

I B

r

µ π

π π

−−× ⋅= = = ×

5Earth 5.0 10 T B −= ×

5 5net Earth net, wire Earth

2 2 5 2 5 2 5net net net

5net1 1

5net

cos 44 3.60 10 T sin 44 1.07 10 T

(3.60 10 T) ( 1.07 10 T) 3.8 10 T

1.07 10 Ttan tan 17 below the horizontal

3.60 10 T

x y

x y

y

x

B B B B B

B B B

B

Bθ

− −

− − −

−− −

−

= ° = × = − ° = − ×

= + = × + − × = ×

− ×= = = °

×

34. The stream of protons constitutes a current, whose magnitude is found by multiplying the proton rate

by the charge of a proton. Then use Eq. 20–6 to calculate the magnetic field.

7 9 19170

stream

(4 10 T m/A)(2.5 10 protons/s)(1.60 10 C/proton)5.3 10 T

2 2 (1.5 m)

I B

r

µ π

π π

− −−× ⋅ × ×

= = = ×

35. (a) If the currents are in the same direction, then the magnetic fields at the midpoint between the two

currents will oppose each other, so their magnitudes should be subtracted.

750 1 0 2

net1 2

(4 10 T m/A)( 25 A) (2.0 10 T/A)( 25 A)

2 2 2 (0.010 m)

I I B I I

r r

µ µ π

π π π

−−× ⋅= − = − = × −

(b) If the currents are in the opposite direction, then the magnetic fields at the midpoint between the

two currents will reinforce each other, so their magnitudes should be added.

750 1 0 2

net1 2

(4 10 T m/A)( 25 A) (2.0 10 T/A)( 25 A)

2 2 2 (0.010 m)

I I B I I

r r

µ µ π

π π π

−−× ⋅

= + = + = × +

36. Using the right-hand rule, we see that if the currents flow in the same direction, then the magneticfields will oppose each other between the wires and therefore can equal zero at a given point. Set the

sum of the magnetic fields from the two wires equal to zero at the point 2.2 cm from the first wire and

use Eq. 20–6 to solve for the unknown current.

0 1 0 2net

1 2

22 1

1

02 2

7.0 cm 2.2 cm(2.0 A) 4.4 A, same direction as other current

2.2 cm

I I B

r r

r I I

r

µ µ

π π = = − →

⎛ ⎞ ⎛ ⎞−= = =⎜ ⎟ ⎜ ⎟

⎝ ⎠⎝ ⎠

wireB

EarthBnet

B

θ

netB

wireB

EarthB

θ 44°


13/30

Magnetism 20-13



37. Use the right-hand rule to determine the direction of the magnetic field from each

wire. Remembering that the magnetic field is inversely proportional to the distance

from the wire, qualitatively add the magnetic field vectors. The magnetic field at

point 2 is zero.

38. (a) We assume that the power line is long and straight and use Eq. 20–6.

76 60

line

(4 10 T m/A)(95 A)2.235 10 T 2.2 10 T

2 2 (8.5 m)

I B

r

µ π

π π

−− −× ⋅

= = = × ≈ ×

The direction at the ground, from the right-hand rule, is south. Compare this with the Earth’s

field of 40.5 10 T,−× which points approximately north.

6

line Earth 4

2.235 10 T/ 0.0447

0.5 10 T B B

−

−

×= =

×

The field of the cable is about 4% that of the Earth.

(b) We solve for the distance where line Earth . B B=

70 0

line Earth 4Earth

(4 10 T m/A)(95 A)0.38 m 0.4 m

2 2 2 (0.5 10 T)

I I B B r

r B

µ µ π

π π π

−

−

× ⋅= = → = = = ≈

×

So about 0.4 m below the wire, the net B field would be 0, assuming that the Earth’s field points

straight north at this location.

39. The Earth’s magnetic field is present at both locations in the problem, and we

assume that it is the same at both locations. The field east of a vertical wire must

be pointing either due north or due south. The compass shows the direction of the

net magnetic field, and it changes from 17° E of N to 32° E of N when taken

inside. That is a “southerly” change (rather than a “northerly” change”), so the

field due to the wire must be pointing due south. See the diagram, in which17 , 32 ,θ θ φ = ° + = ° and 180 . β θ φ + + = ° Thus 15φ = ° and 148 . β = ° Use the

law of sines to find the magnitude of wireB

and then use Eq. 20–6 to find the

magnitude of the current.

wire Earth 0wire Earth

sin

sin sin sin 2

B B I B B

r

µ φ

φ β β π = → = = →

5Earth 7

0

sin 2 sin15 2(5.0 10 T) (0.120 m) 14.65 A 15 A

sin sin 148 4 10 T m/A I B r

φ π π

β µ π

−

−

°= = × = ≈

° × ⋅

Since the field due to the wire is due south, the current in the wire must be downward.

40. The fields created by the two wires will oppose each other, so the net field is the difference between

the magnitudes of the two fields. The positive direction for the fields is taken to be into the page, so the

closer wire creates a field in the positive direction, and the more distant wire creates a field in the

negative direction. Let d be the separation distance of the wires.

0 0 0 0net 1 1

closer farther closer farther 2 2

1 1 1 1

2 2 2 2

I I I I B

r r r r r d r d

µ µ µ µ

π π π π

⎛ ⎞⎛ ⎞= − = − = −⎜ ⎟⎜ ⎟ ⎜ ⎟− +⎝ ⎠ ⎝ ⎠

θ

φ

β

EarthB

θ wire

B

netB


14/30

20-14 Chapter 20



( ) ( )0

1 12 2

7

6 6

2

(4 10 T m/A)(24.5 A) 0.0028 m

2 (0.10 m 0.0014 m)(0.10 m 0.0014 m)

1.372 10 T 1.4 10 T

I d

r d r d

µ

π

π

π

−

− −

⎛ ⎞⎜ ⎟=⎜ ⎟− +⎝ ⎠

⎛ ⎞× ⋅= ⎜ ⎟

− +⎝ ⎠

= × ≈ ×

Compare this with the Earth’s field of 40.5 10 T.−×

6

net Earth 4

1.372 10 T/ 0.027

0.5 10 T B B

−

−

×= =

×

The field of the wires is about 3% that of the Earth.

41. The center of the third wire is 5.6 mm from the left wire and 2.8 mm from the right wire. The force on

the near (right) wire will attract the near wire, since the currents are in the same direction. The force

on the far (left) wire will repel the far wire, since the currents oppose each other. Use Eq. 20–7 tocalculate the force per unit length.

0 1 2near near

near

near 0 1 2

3near near

0 1 2far far

far

far 0 1 2

far far

72

7

2

25.0 A 24.5 A4.4 10 , attract (to the right)

2 2.8 10 m

2

25.0 A 24.5 A

2 5.6 10

4 10 T m/A ( )( ) N/m

2

4 10 T m/A ( )( )

2

I I F

d

F I I

d

I I F

d

F I I

d

µ

π

µ

π

µ

π

µ

π

π

π

π

π

−

−

−−

−

= →

= = = ××

= →

= =×

× ⋅

× ⋅

3

22.2 10 repel (to the left)

m N/m,

−= ×

42. Since the magnetic field from a current-carrying wire circles the

wire, the individual field at point P from each wire is perpendicular

to the radial line from that wire to point P. We define 1B

as the field

from the top wire and 2B

as the field from the bottom wire. We use

Eq. 20–6 to calculate the magnitude of each individual field.

750

11

750

22

(4 10 T m/A)(28 A)9.333 10 T

2 2 (0.060 m)

(4 10 T m/A)(28 A)5.600 10 T

2 2 (0.100 m)

I B

r

I B

r

µ π

π π

µ π

π π

−−

−−

× ⋅= = = ×

× ⋅= = = ×

We use the law of cosines to determine the angle that the radial line

from each wire to point P makes with the vertical. Since the field is perpendicular to the radial line,

this is the same angle that the magnetic fields make with the horizontal.

2 2 21

1

2 2 21

2

(0.060 m) (0.130 m) (0.100 m)cos 47.7

2(0.060 m)(0.130 m)

(0.100 m) (0.130 m) (0.060 m)cos 26.3

2(0.100 m)(0.130 m)

θ

θ

−

−

⎛ ⎞+ −= = °⎜ ⎟⎜ ⎟

⎝ ⎠

⎛ ⎞+ −= = °⎜ ⎟⎜ ⎟

⎝ ⎠


15/30

Magnetism 20-15



Using the magnitudes and angles of each magnetic field, we calculate the horizontal and vertical

components, add the vectors, and calculate the resultant magnetic field and angle.

5 5 5net 1 1 2 2cos ( ) cos (9.333 10 T)cos 47.7 (5.600 10 T)cos 26.3 1.261 10 T x B B Bθ θ

− − −= − = × ° − × ° = ×

4 5 5net 1 1 2 1sin ( ) sin (9.333 10 T)sin 47.7 (5.600 10 T)sin 26.3 9.384 10 T y B B Bθ θ

− − −= + = × ° + × ° = ×

2 2 5 2 5 2 5net, net,

5net1 1

5net,

5 5

(1.261 10 T) (9.384 10 T) 9.468 10 T

8.379 10 Ttan tan 82.3

1.126 10 T

9.468 10 T at 82.3 9.5 10 T at 82

x y

y

y

B B B

B

Bθ

− − −

−,− −

−

− −

= + = × + × = ×

×= = = °

×

= × ° ≈ × °B

43. The magnetic fields created by the individual currents will be at right angles to each other. The field

due to the top wire will be to the right, and the field due to the bottom wire will be out of the page.

Since they are at right angles, the net field is the hypotenuse of the two individual fields.

2 2 70 top 2 2 2 20 bottom 0

net top bottomtop bottom

5

4 10 T m/A(20.0 A) (12.0 A)

2 2 2 2 (0.100 m)

4.66 10 T

I I B I I

r r r

µ µ µ π

π π π π

−

−

⎛ ⎞ ⎛ ⎞ × ⋅= + = + = +⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠

= ×

44. Use Eq. 20–8 for the field inside a solenoid.

730 (4 10 T m/A)(460)(2.0 A) 9.6 10 T

0.12 m

NI B

µ π − −× ⋅= = = ×

45. Use Eq. 20–8 for the field inside a solenoid.

30

70

(4.65 10 T)(0.300 m)1.19 A

(4 10 T m/A)(935)

NI B B I

N

µ

µ π

−

−

×= → = = =

× ⋅

46. The field inside a solenoid is given by Eq. 20–8.

30

70

(0.030 T)(0.42 m)2.2 10 turns

(4 10 T m/A)(4.5 A)

NI B B N

I

µ

µ π −= → = = = ×

× ⋅

47. The field due to the solenoid is given by Eq. 20–8. Since the field due to the solenoid is perpendicular

to the current in the wire, Eq. 20–2 can be used to find the force on the wire segment.

70 solenoid

wire wire solenoid wire wiresolenoid

(4 10 T m/A)(38 A)(550)(22 A)(0.030 m)

(0.15 m)

0.1156 N 0.12 N to the south

NI F I B I

µ π −× ⋅= = =

= ≈


16/30

20-16 Chapter 20



48. Since the mass of copper is fixed and the density is fixed, the volume of copper is fixed, and we

designate it as Cu Cu Cu Cu Cu .V m A ρ = = We call the fixed voltage 0.V The magnetic field in the

solenoid is given by Eq. 20–8. For the resistance, we used the resistivity-based definition from Eq. 18–3,

with resistivity represented by RCu . ρ

0 0 0 0 0 Cu 0 0 Cu Cu0 0 2

Cusol sol Cu sol RCu sol Cu RCu sol CuRCu

Cu

0 0 Cu Cu

2RCu sol Cu

NI V V V A V m N N N N B

R

A

V m N

µ µ µ ρ µ µ

ρ ρ ρ

µ ρ

ρ

= = = = =

=

The number of turns of wire is the length of wire divided by the circumference of the solenoid.

Cu

Cu 0 0 Cu Cu 0 0 Cu Cu sol 0 0 Cu Cu

2 2sol RCu RCu RCu sol sol Cusol Cu sol Cu

2 1

2 2

V m V m r V m N N B

r r

µ ρ µ ρ π µ ρ

π ρ ρ πρ = → = = =

The first factor in the expression for B is made of constants, so we havesol sol Cu

1. B

r ∝

Thus we

want the wire to be short and fat. Also, the radius of the solenoid should be small and the length of the

solenoid should be small.

49. (a) Each loop of wire produces a field along its axis, similar to Fig. 20–9. For path 1, with all the

loops taken together, that symmetry leads to a magnetic field that is the same anywhere along the

path and parallel to the path. One side of every turn of the wire is enclosed by path 1, so the

enclosed current is . NI Apply Eq. 20–9.

0 0 0(2 ) /2 B NI B R NI B NI R µ π µ µ π ∆ = → = → =∑

(b) For path 2, each loop of wire pierces the enclosed area twice—once going up and once goingdown. Ampere’s law takes the direction of the current into account, so the net current through

the area enclosed by path 2 is 0.

0 (2 ) 0 0 B NI B R B µ π ∆ = → = → =∑ (c) The field inside a toroid is not uniform. As seen in the result to part (a), the field varies inversely

as the radius of the toroid, so1

. B R

∝ The field is strongest at the inside wall of the toroid, and

weakest at the outside wall.

50. (a) Ampere’s law (Eq. 20–9) says that along a closed path, 0 encl . B I µ ∆ =∑ For the path, choosea circle of radius r , centered on the center of the wire, greater than the radius of the inner wireand less than the radius of the outer cylindrical braid. Because the wire is long and straight, the

magnetic field is tangent to the chosen path, so . B B= The current enclosed is I .

00 encl (2 )

2

I I B B B B r B

r

µ µ π

π = ∆ = ∆ = ∆ = → =∑ ∑ ∑


17/30

Magnetism 20-17



(b) We make a similar argument, but now choose the path to be a circle of radius r , greater than the

radius of the outer cylindrical braid. Because the wire and the braid are long and straight, the

magnetic field is tangent to the chosen path, so . B B= The current enclosed by the path is zero,

since there are two equal but oppositely directed currents.

0 encl0 encl (2 ) 0

2

I I B B B B r B

r

µ µ π

π = ∆ = ∆ = ∆ = → = =∑ ∑ ∑

51. If the face of the loop of wire is parallel to the magnetic field, then the angle between the perpendicular

to the loop and the magnetic field is 90 .° Use Eq. 20–10 to calculate the magnetic field strength.

2

0.325 m Nsin 1.18 T

sin (1)(5.70 A)(0.220 m) sin 90 NIAB B

NIA

τ τ θ

θ

⋅= → = = =

°

52. From Eq. 20–10, we see that the torque is proportional to the current, so if the current drops by 12%,

then the output torque will also drop by 12%. Thus the final torque is 0.88 times the initial torque.

53. In Section 20–10, it is shown that the angular deflection of the galvanometer needle is proportional to

the product of the current and the magnetic field. Thus if the magnetic field is decreased to 0.760 times

its original value, then the current must be increased by dividing the original value by 0.760 to obtain

the same deflection.

initial initial initialinitial final final

final initial

(53.0 A)( ) ( ) 69.7 A

0.760

I B B IB IB I

B B

µ µ = → = = =

54. (a) The torque is given by Eq. 20–10. The angle is the angle between the B field and the

perpendicular to the coil face.

2 5 50.120 msin 9(7.20 A) (5.50 10 T)sin 34.0 2.25 10 m N2

NIABτ θ π − −⎡ ⎤⎛ ⎞= = × ° = × ⋅⎢ ⎥⎜ ⎟⎝ ⎠⎢ ⎥⎣ ⎦

(b) In Example 20–13 it is stated that if the coil is free to turn, then it will rotate toward the

orientation so that the angle is 0°. In this case, that means the north edge of the coil will rise, so

that a perpendicular to its face will be parallel with the Earth’s magnetic field.

55. The radius and magnetic field values can be used to find the speed of the protons. The electric field is

then found from the fact that the magnetic force must be the same magnitude as the electric force for

the protons to have straight paths.

2E B

19 2 22 6

27

/ /

(1.60 10 C)(0.566 T) (6.10 10 m)/ 1.87 10 V/m1.67 10 kg

q B m r qBr m F F qE q B

E B qB r m

υ υ υ υ

υ − −

−

= → = = → = →

× ×= = = = ××

The direction of the electric field must be perpendicular to both the velocity and the magnetic field and

must be in the opposite direction to the magnetic force on the protons.


18/30

20-18 Chapter 20



56. The magnetic force on the ions causes them to move in a circular path, so the magnetic force is a

centripetal force. This results in the ion mass being proportional to the path’s radius of curvature.

2

21.0 21.621.0 21.6

21.9 22.221.9 22.2

/ / / / constant 76 u/22.8 cm

76 u 76 u

70 u 72 u21.0 cm 22.8 cm 21.6 cm 22.8 cm

76 u 76 u73 u 74 u

21.9 cm 22.8 cm 22.2 cm 22.8 cm

q B m r m qBr m r qB

m m

m m

m mm m

υ υ υ υ = → = → = = =

= → = = → =

= → = = → =

The other masses are 70 u, 72 u, 73 u, and 74 u.

57. The location of each line on the film is twice the radius of curvature of the ion. The radius of curvature

can be found from Eq. 20–12.

27 42

12 19 2

2 213 14

22

2(12)(1.67 10 kg)(2.88 10 V/m)

2 1.560 10 m(1.60 10 C)(0.68 T)

2 1.690 10 m 2 1.820 10 m

qBB r mE mE m r r

E qBB qBB

r

r r

−−

−

− −

′= → = → =

′ ′

× ×

= = ××

= × = ×

The distances between the lines are the following.

2 2 3 313 12

2 2 3 314 13

2 2 1.690 10 m 1.560 10 m 1.11 10 m 1.3 10 m

2 2 1.820 10 m 1.690 10 m 1.12 10 m 1.3 10 m

r r

r r

− − − −

− − − −

− = × − × = × ≈ ×

− = × − × = × ≈ ×

If the ions are doubly charged, then the value of q in the denominator of the expression would double,

so the actual distances on the film would be halved. Thus the distances between the lines would also be

halved.

3 3 413 12

3 3 414 13

2 2 8.451 10 m 7.801 10 m 6.5 10 m

2 2 9.101 10 m 8.451 10 m 6.5 10 m

r r

r r

− − −

− − −

− = × − × = ×

− = × − × = ×

58. The velocity of the ions is found using energy conservation. The electrical potential energy of the ions

becomes kinetic energy as they are accelerated. Then, since the ions move perpendicularly to the

magnetic field, the magnetic force will be a maximum. That force will cause the ions to move in a

circular path.

22 2 2 2 2 221 1

2 2 2 2

m qBR qBR q B R qR Bq B qV m m m

R m m m V

υ υ υ υ

⎛ ⎞= → = = = = → =⎜ ⎟

⎝ ⎠

59. Since the particle is undeflected in the crossed fields, its speed is given by / , E Bυ = as stated inSection 20–11. Without the electric field, the particle will travel in a circle due to the magnetic force.

Using the centripetal acceleration, we can calculate the mass of the particle. Also, the charge must be

an integer multiple of the fundamental charge.

2

2 19 227

3

(1.60 10 C)(0.034 T) (0.027 m)(3.3 10 kg) (2.0 u)

( / ) 1.5 10 V/m

mq B

r

qBr qBr neB r nm n n

E B E

υ υ

υ

−−

= →

×= = = = = × ≈

×


19/30

Magnetism 20-19



The particle has an atomic mass of a multiple of 2.0 u. The simplest two cases are that it could be a

hydrogen-2 nucleus (called a deuteron) or a helium-4 nucleus (called an alpha particle): 2 41 2H, He .

60. The particles in the mass spectrometer follow a semicircular path as shown in Fig. 20–41. A particle

has a displacement of 2r from the point of entering the semicircular region to where it strikes the film.So if the separation of the two molecules on the film is 0.50 mm, then the difference in radii of the two

molecules is 0.25 mm. The mass to radius ratio is the same for the two molecules.

2

2

4CO N

/ / / constant

28.0106 u 28.0134 u2.5 m

2.5 10 m

q B m r m qBr m r

m mr

r r r r

υ υ υ

−

= → = → =

⎛ ⎞ ⎛ ⎞= → = → =⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠ + ×

61. The field inside the solenoid is given by Eq. 20–8 with 0 µ replaced by the permeability of the iron.

73000(4 10 T m/A)(380)(0.35 A)0.5 T

(1.0 m)

NI B

µ π −× ⋅= = =

We assume that the factor of 3000 only has 1 significant figure.

62. The field inside the solenoid is given by Eq. 20–8 with 0 µ replaced by the permeability of the iron.

5 50

(2.2 T)(0.38 m)2.239 10 T m/A 2.2 10 T m/A 18

(780)(48 A)

NI B B

NI

µ µ µ − −= → = = = × ⋅ ≈ × ⋅ ≈

63. The magnetic permeability is found from the two fields.

0 0 00 0 0

; ; B B

B nI B nI B B

µ µ µ µ µ

µ = = = → =

Here is a data table with the given values as well as the calculated values of µ .

For the graph, we have not plotted the last three data

points so that the structure for low fields is seen. It

would appear from the data that the value of µ isasymptotically approaching 0 for large fields.

40 (10 T) B

− B (T) 4 (10 )(T-m/A) µ −

0 0

0.13 0.0042 4.06

0.25 0.01 5.03

0.50 0.028 7.04

0.63 0.043 8.58

0.78 0.095 15.3

1.0 0.45 56.5

1.3 0.67 64.81.9 1.01 66.8

2.5 1.18 59.3

6.3 1.44 28.7

13.0 1.58 15.3

130 1.72 1.66

1300 2.26 0.218

10000 3.15 0.040

0

10

20

30

40

50

60

70

0 2 4 6 8 10 12 14

B 0(10-4

T)

µ ( 1 0 - 4

T - m / A )


20/30

20-20 Chapter 20



64. (a) Use Eq. 20–6 to calculate the field due to a long straight wire.

76 60 A

A at BA to B

(4 10 T m/A)(2.0 A)2.667 10 T 2.7 10 T

2 2 (0.15 m)

I B

r

µ π

π π

−− −× ⋅= = = × ≈ ×

(b)7

6 60 BB at A

B to A

(4 10 T m/A)(4.0 A) 5.333 10 T 5.3 10 T2 2 (0.15 m)

I Br

µ π π π

−− −× ⋅= = = × ≈ ×

(c) The two fields are not equal and opposite. Each individual field is due to a single wire and has no

dependence on the other wire. The magnitude of current in the second wire has nothing to do

with the value of the field caused by the first wire.

(d ) Use Eq. 20–7 to calculate the force per unit length on one wire due to the other wire. The forces

are attractive since the currents are in the same direction.

7on A due to B on B due to A 0 A B

A B A to B

5 5

(4 10 T m/A) (2.0 A)(4.0 A)

2 2 (0.15 m)

1.067 10 N/m 1.1 10 N/m

F F I I

d

µ π

π π

−

− −

× ⋅= = =

= × ≈ ×

These two forces per unit length are equal and opposite because they are a Newton’s third law

pair of forces.

65. The magnetic force produces centripetal acceleration.

212 2

19

4.8 10 kg m/s/ 2.7 10 T

(1.60 10 C)(1.1 m)

pq B r m p qBr B

qr υ υ υ

−−

−

× ⋅= → = = → = = = ×

×m

The magnetic field must point upward to cause an inward-pointing (centripetal) force that steers the

protons clockwise.

66. There will be no force on either the top or bottom part of the wire, because the current is either parallel

to or opposite to the magnetic field. So the only force is on the left branch, which we define to be of

length . Since the current is perpendicular to the magnetic field, use Eq. 20–2. The magnetic field can

be calculated by Eq. 20–8 for the magnetic field inside a solenoid. By the right-hand rule, the force on

the left branch is up out of the page.

70 solenoid

wire wire wire wire wiresolenoid

wire wire

(4 10 T m/A)(700)(7.0 A)(5.0 A)

0.15 m

0.2 N (assuming given in m)

NI F I B I

µ π −⎡ ⎤⎛ ⎞ × ⋅= = = ⎢ ⎥⎜ ⎟

⎢ ⎥⎝ ⎠ ⎣ ⎦

=

67. We assume that the horizontal component of the

Earth’s magnetic field is pointing due north. The

Earth’s magnetic field also has the dip angle of 22°.

The angle between the magnetic field and the eastwardcurrent is 90°. Use Eq. 20–1 to calculate the magnitude

of the force.

5sin (330 A)(18 m)(5.0 10 T) sin 90

0.297 N 0.30 N

F I B θ −= = × °

= ≈

Using the right-hand rule with the eastward current and the Earth’s magnetic field, the force on the

wire is northerly and 68° above the horizontal.

EarthB

E

N

BF

o22

o68


21/30

Magnetism 20-21



68. From Example 20–6, we have .m

r qB

υ = The quantity mυ is the momentum, , p so .

pr

qB= Thus

. p qBr =

69. The airplane is a charge moving in a magnetic field. Since it is flying perpendicular to the magnetic

field, Eq. 20–4 applies.

6 5 6max (1280 10 C)(120 m/s)(5.0 10 T) 7.7 10 N F q Bυ

− − −= = × × = ×

70. The field inside a solenoid is given by Eq. 20–8.

30

70

(0.050 T)(0.32 m)1989 2.0 10 turns

(4 10 T m/A)(6.4 A)

NI B B N

I

µ

µ π −= → = = = ≈ ×

× ⋅

71. The magnetic force must be equal in magnitude to the weight of the electron.

31 2

619 4

(9.11 10 kg)(9.80 m/s ) 1.1 10 m/s(1.60 10 C)(0.50 10 T)

mg mg q BqB

υ υ −

−− −

×= → = = = ×× ×

The magnetic force must point upward, so by the right-hand rule and the negative charge of the

electron, the electron must be moving west.

72. (a) The velocity of the ions is found using energy conservation. The electrical potential energy of

the ions becomes kinetic energy as they are accelerated. Then, assuming the ions move

perpendicularly to the magnetic field, the magnetic force will be a maximum. That force will

cause the ion to move in a circular path.

221

initial final 2PE KE

2qV mqV m q B

m r

υ υ υ υ = → = → = = →

272 2

2 19 2

2

2 2(6.6 10 kg)(3200 V)4.787 10 m 4.8 10 m

2(1.60 10 C)(0.240 T)

qV m

m mV mr

qB qB qB

υ − − −−

×= = = = = × ≈ ×

×

(b) The period is the circumference of the circular path divided by the speed.

2 277

19

22

2 2 2 (6.6 10 kg)5.5 10 s

2 2(1.60 10 C)(0.240 T)

mV

qBr mT

qBqV

m

π π π π

υ

−−

−

×= = = = = ×

×

73. The magnetic field at the center of the square is the vector sum of themagnetic field created by each current. Since the magnitudes of the

currents are equal and the distance from each corner to the center is the

same, the magnitude of the magnetic field from each wire is the same

and is given by Eq. 20–6. The direction of the magnetic field is directed

by the right-hand rule and is shown in the diagram. By symmetry, we

see that the vertical components of the magnetic field cancel and the

horizontal components add.


22/30

20-22 Chapter 20



01 2 3 4

0 0

4 cos 45 to the left2

22

4 to the left to the left222

2

I

r

I I

µ

π

µ µ

π π

⎛ ⎞= + + + = °⎜ ⎟

⎝ ⎠

⎛ ⎞⎜ ⎟⎜ ⎟= =⎜ ⎟⎜ ⎟⎝ ⎠

B B B B B

74. (a) For the beam of electrons to be undeflected, the magnitude of the magnetic force must equal the

magnitude of the electric force. We assume that the magnetic field will be perpendicular to the

velocity of the electrons so that the maximum magnetic force is obtained.

3B E 6

12,000 V/m2.5 10 T

4.8 10 m/s

E F F q B qE Bυ

υ

−= → = → = = = ××

(b) Since the electric field is pointing south, the electric force is to the north. Thus the magnetic

force must be to the south. Using the right-hand rule with the negative electrons, the magnetic

field must be vertically upward.

(c) If the electric field is turned off, then the magnetic field will cause a centripetal force, moving

the electrons in a circular path. The frequency is the cyclotron frequency, Eq. 20–5.

197

31 6

(1.60 10 C)(12,000 V/m)7.0 10 Hz

2 2 2 (9.11 10 kg)(4.8 10 m/s)

qB qE f

m mπ π υ π

−

−

× = = = = ×

× ×

75. The protons will follow a circular path as they move through the

region of magnetic field, with a radius of curvature given in

Example 20–6 as .m

r qB

υ = Fast-moving protons will have a radius

of curvature that is too large, so they will exit above the second

tube. Likewise, slow-moving protons will have a radius of curvature

that is too small, so they will exit below the second tube. Since the

exit velocity is perpendicular to the radius line from the center of

curvature, the bending angle can be calculated.

2 191 1 1 1

27 6

sin

(5.0 10 m)(1.60 10 C)(0.41 T)sin sin sin sin 0.7856 52

(1.67 10 kg)(2.5 10 m/s)

r

qB

r mv

θ

θ − −

− − − −

−

= →

× ×= = = = = °

× ×

76. With one turn of wire, we have 0 .2

I B

r

µ = Use the radius of the Earth for r .

6 49 90

70

2 2(6.38 10 m)(1 10 T)1.015 10 A 1 10 A

2 4 10 T m/A

I rB B I

r

µ

µ π

−

−

× ×= → = = = × ≈ ×

× ⋅

r r

θ


23/30

Magnetism 20-23



77. The speed of the proton can be calculated based on the radius of curvature of the (almost) circular

motion. From that the kinetic energy can be calculated.

22 2 2 221 1

2 2

2 2 19 2 22 2 3 3

2 1 27

20

2 2

KE

KE

2

(1.60 10 C) (0.010 T)( ) [(8.5 10 m) (10.0 10 m) ]

2 2(1.67 10 kg)

2.1 10 J or 0.13 eV

m qBr qBr q B r q B m m

r m m m

q Br r

m

υ υ υ υ

−− −

−

−

= = = =

= =

⎛ ⎞→ = ⎜ ⎟

⎝ ⎠

×∆ − × − ×

×

= − × −

78. There are three forces on each wire—its weight, the magnetic force of repulsion

from the other wire, and the tension in the attached string. See the diagram. The

magnetic force is given by Eq. 20–7. The mass of the wire is its density times its

volume. The length of the current-carrying wires is . The net force in both the

vertical and horizontal directions is zero.

B T T T

2B Al Al

sin ; coscos

sin tan tancos

mg F F F mg F

mg F mg r g

θ θ θ

θ θ ρ π θ θ

= = → = →

= = =

2 220 0

B Al Al tan2 2

I I F r g

S S

µ µ ρ π θ

π π = → = →

2 3 4 2 2Al Al

70

2 tan 2 [2(0.50 m) sin 3 ](2700 kg/m ) (2.1 10 m) (9.8 m/s ) tan 3

(4 10 T m/A)

7.1 A

S r g I

π ρ π θ π π

µ π

−

−

° × °= =

× ⋅

=

79. The centripetal force is caused by the magnetic field and is given by Eq. 20–3.

2

31 65 5

19

sin

(9.11 10 kg)(3.0 10 m/s) sin 455.251 10 m 5.3 10 m

(1.60 10 C)(0.23 T)

F q B q B mr

mr

qB

υ υ θ υ

υ

⊥⊥

−− −⊥

−

= = = →

× × °= = = × ≈ ×

×

The component of the velocity that is parallel to the magnetic field is unchanged, so the pitch is that

velocity component times the period of the circular motion.

316 4

19

22 2

2 2 (9.11 10 kg)cos 45 (3.0 10 m/s)cos 45 3.3 10 m(1.60 10 C)(0.23 T)

m

r mqBT

qB

m p T qB

υ π

π π

υ υ

π π υ υ

⊥

⊥ ⊥

−−

−

= = =

⎛ ⎞ ×= = ° = × ° = ×⎜ ⎟×⎝ ⎠

80. The maximum torque is found using Eq. 20–10 with sin 1.θ = Set the current equal to the voltage

divided by the resistance and the area as the square of the side length.

2 2 49.0 V20 (0.050 m) (0.020 T) 3.2 10 m N28

V NIAB N B

Rτ −

⎛ ⎞⎛ ⎞= = = = × ⋅⎜ ⎟⎜ ⎟

Ω⎝ ⎠ ⎝ ⎠

mg

TF

BF θ


24/30

20-24 Chapter 20



81. We find the speed of the electron using conservation of

energy. The accelerating potential energy becomes the

kinetic energy of the electron.

212

19

31

7

2 2(2200 eV)(1.60 10 J/eV)

(9.11 10 kg)

2.78 10 m/s

eV m

eV

m

υ

υ −

−

= →

×= =

×

= ×

Upon entering the magnetic field, the electron is

traveling horizontally. The magnetic field will cause the

path of the electron to be an arc of a circle of radius r

and deflect an angle θ from the horizontal. While in the field, the electron will travel a horizontaldistance d and a vertical distance 0.h Our approximation will be to ignore the distance 0.h We have

the following relationships.

1 1

31 73 3

19

tan ; sin ;

11 cm 0.038 mtan tan 26.6 ; 0.0849 m

22 cm sin sin 26.6

(9.11 10 kg)(2.78 10 m/s)1.86 10 T 1.9 10 T

(1.60 10 C)(0.0849 m)

h d mr r eB

h d r

m B

er

υ θ θ

θ θ

υ

− −

−− −

−

= = = →

= = = ° = = =°

× ×= = = × ≈ ×

×

82. The magnetic field from the wire at the location of the plane is perpendicular to the velocity of the

plane since the plane is flying parallel to the wire. We calculate the force on the plane, and thus the

acceleration, using Eq. 20–4, with the magnetic field of the wire given by Eq. 20–6.

0

3 70

5 2 6

2

2

(18.0 10 C)(3.4 m/s)(4 10 T m/A)(25 A)

2 2 (0.175 kg)(0.086 m)

12.033 10 m/s 2.1 10 's

9.80 m/s

I F q B q

r

I F qa

m m r

g g

µ υ υ

π µ υ π

π π

− −

− −

= =

× × ⋅= = =

⎛ ⎞= × = ×⎜ ⎟⎜ ⎟

⎝ ⎠

83. Since the magnetic and gravitational force along the entire rod is uniform, we consider the two forces

acting at the center of mass of the rod. To be balanced, the net torque about the fulcrum must be zero.

Using the usual sign convention for torques and Eq. 8–10a, we solve for the magnetic force on the rod.

( ) ( ) ( )1 1 14 4 40 ( ) M M g mg F F M m g τ = = − − → = −∑

We solve for the current using Eq. 20–2.

( ) (6.0 ) 5.0 F M m g m m g mg I

B B B B

− −= = = =

The right-hand rule indicates that the current must flow toward the left since the magnetic field is into

the page and the magnetic force is downward.


25/30

Magnetism 20-25



84. The magnetic force will produce centripetal acceleration. Use that relationship to calculate the speed.

The radius of the Earth is 66.38 10 km,× and the altitude is added to that.

2 19 6 48

B27

(1.60 10 C)(6.385 10 m)(0.50 10 T)1.3 10 m/s

238(1.66 10 kg)

qrB F q B m

r m

υ υ υ

− −

−

× × ×= = → = = = ×

×

Compare the size of the magnetic force to the force of gravity on the ion.

19 8 48B

27 2g

(1.60 10 C)(1.3 10 m/s)(0.50 10 T)2.7 10

238(1.66 10 kg)(9.80 m/s )

F q B

F mg

υ − −

−

× × ×= = = ×

×

It is fine to ignore gravity—the magnetic force is almost 300 million times larger than gravity.

85. (a) For the particle to move upward, the magnetic force must point upward; by the right-hand rule

we see that the force on a positively charged particle would be downward. Therefore, the charge

on the particle must be negative.

(b) In the figure we have created a right triangle to relate the

horizontal distance , the displacement d , and the radius of

curvature, r . Using the Pythagorean theorem we can write

an expression for the radius in terms of the other two

distances.

2 22 2 2

( )2

d r r d r

d

+= − + → =

Since the momentum is perpendicular to the magnetic field,

we can solve for the momentum by relating the maximum

force (Eq. 20–4) to the centripetal force on the particle.

2 22

0max 0 0

( )

2

qB d m F q B p m qB r

r d υ υ υ

+= = → = = =

86. Example 18–10 estimates the current in a lightning bolt at 100 A. Use Eq. 20–2, with the Earth’s

magnetic field. We estimate the flag pole as being 65 m tall.

4(100 A)(6 m)(0.5 10 T) 0.03 N F I B −

= = × =

87. The accelerating force on the bar is due to the magnetic force on the current. If the current is constant,

then the magnetic force will be constant, so constant-acceleration kinematics (Eq. 2–11c) can be used.

2 2 22 2 0

02

2 2a x a

x x

υ υ υ υ υ

−= + ∆ → = =

∆ ∆

2

2 3 2

net

2 (1.5 10 kg)(28 m/s)1.2 A

2 2(0.28 m)(1.7 T)(1.0 m)

m xma mv

F ma I B I B B B x

υ −

⎛ ⎞⎜ ⎟⎜ ⎟∆ ×⎝ ⎠

= = → = = = = =∆

Using the right-hand rule, for the force on the bar to be in the direction of the acceleration shown in

Fig. 20–71, the magnetic field must be down.


26/30

20-26 Chapter 20



88. (a) The frequency of the voltage must match the frequency of circular motion of the particles so that

the electric field is synchronized with the circular motion. The radius of each circular orbit is

given in Example 20–6 as .m

r qB

υ = For an object moving in circular motion, the period is given

by 2 ,r T π υ

= and the frequency is the reciprocal of the period.

2

22

r BqT f

mT m

qB

π υ

υ υ π π

= → = = =

In particular, we note that this frequency is independent of the radius, so the same frequency can

be used throughout the acceleration. This frequency is also given as Eq. 20–5.

(b) For a small gap, the electric field across the gap will be approximately constant and uniform as

the particles cross the gap. If the motion and the voltage are synchronized so that the maximum

voltage occurs when the particles are at the gap, then the particles receive an energy increase of

0KE qV = as they pass each gap. The energy gain from one revolution will include the passing of

two gaps, so the total KE increase is 02 .qV

(c) The maximum kinetic energy will occur at the outside of the cyclotron.

2 2 2 2 2 19 2 22 max max1 1 1 1

max max2 2 2 2 27

12

19 6

KE(2.0 m) (1.60 10 C) (0.50 T)

1.67 10 kg

1 eV 1 MeV7.66 10 J 48 MeV

1.60 10 J 10 eV

rqB

m

r qB r q Bm m

m m

υ

υ −

−

−

−

=

×⎛ ⎞= = = =⎜ ⎟

×⎝ ⎠

⎛ ⎞⎛ ⎞= × =⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟×⎝ ⎠⎝ ⎠

89. (a) The forces on wire M due to the other wires are repelling forces, awayfrom the other wires. Use Eq. 20–7 to calculate the force per unit length

on wire M due to each of the other wires and then add the force vectors

together. The horizontal parts of the forces cancel, and the sum is

vertical.

M net MN MP

M M M

cos30° cos30 y F F F

= + °

0 M N 0 M P

MN MP

7 24

cos30 cos302 2

(4 10 T m/A) (8.00 A)2 cos30 5.8 10 N/m, upward

2 (0.038 m)

I I I I

d d

µ µ

π π

π

π

−−

= ° + °

× ⋅= ° = ×

The forces on wire N due to the other wires are an attractive force

toward wire P and a repelling force away from wire M. Use Eq. 20–7

to calculate the force per unit length on wire N due to each of the other

wires and then add the force vectors together. From symmetry, we

expect the net force to lie exactly between the two individual force

vectors, which is 60° below the horizontal.

MNF

MPF

o30o30

NMF

NPF

o90o30


27/30

Magnetism 20-27



0 M N N net NM

MN

7 24

N net NM NP

7 2 7 2

5

cos30 cos302

(4 10 T m/A) (8.00 A)cos30 2.917 10 N/m

2 (0.038 m)

sin 30

(4 10 T m/A) (8.00 A) (4 10 T m/A) (8.00 A)sin 30

2 (0.038 m) 2 (0.038 m)

1.684 10 N/m

y

x

I I F F

d

F F F

µ

π

π

π

π π

π π

−−

− −

−

= − ° = − °

× ⋅= − ° = − ×

= − ° +

× ⋅ × ⋅= − ° +

= ×

N net2 2 4 1 1 N net N net N net

N net

2.9173.4 10 N/m tan tan 300

1.684

y x y

x

F F F F

F θ − − −

−= + = × = = = °

The forces on wire P due to the other wires are an attractive force toward

wire N and a repelling force away from wire M. Use Eq. 20–7 to calculate

the force per unit length on wire P due to each of the other wires and then

add the force vectors together. From symmetry, this is just a mirror image

of the previous solution, so the net force is as follows.

4P net 3.4 10 N/m 240 F θ

−= × = °

(b) There will be three magnetic fields to sum—one from each wire. Each

field will point perpendicularly to the line connecting the wire to the

midpoint. The two fields due to M and N are drawn slightly separated

from each other, but should be collinear. The magnitude of each field

is given by Eq. 20–6.

net M N P

750

M NM

750

PP

(4 10 Τ m/A) 8.00 A8.421 10 T

2 2 0.019 m

(4 10 T m/A) 8.00 A4.862 10 T

2 2 3(0.019 m)

I B B

r

I B

r

µ π

π π

µ π

π π

−−

−−

= + +

× ⋅= = = = ×

× ⋅= = = ×

B B B B

5net M P

5 4

2 cos30 cos60 2(8.421 10 T)cos30

(4.862 10 T)cos 60 1.702 10 T

x B B B −

− −

= ° + ° = × °

+ × ° = ×

5net M P

5 5

2 sin 30 sin 60 2(8.421 10 T)sin 30

(4.862 10 T)sin 60 4.210 10 T

y B B B −

− −

= − ° + ° = − × °

+ × ° = − ×

2 2 4 2 5 4net net net (1.702 10 T) ( 4.210 10 T) 1.75 10 T x y B B B

− − −= + = × + − × = ×

5net1 1

net 4net

4.210 10 T

tan tan 141.702 10 T

y

x

B

Bθ

−− −

−

− ×

= = = − °×

The net field points slightly below the horizontal direction.

90. Each of the bottom wires will repel the top wire since each bottom current is

opposite to the top current. The repelling forces will be along the line directly from

the bottom wires to the top wires. Only the vertical components of those forces will

be counteracting gravity. Use Eq. 20–7 to calculate the magnetic forces. The mass of

the wire is its density times its volume. The length of the wire is represented by .

PMF

PNF

o90

o30

M

N P

MB

NB

PB

o60o30

mg

BF

BF

o30o30


28/30

20-28 Chapter 20



( )2 0 bottom Cu1

B Cu Cu22 cos30 2 cos30

2

I I mg F d g

d

µ ρ π

π = ° → = ° →

( )22 1 3 2 4 2 2

Cu Cu2Cu 7

0 bottom

2

(8900 kg/m ) (5.0 10 m) (0.038 m)(9.80 m/s )

cos30 (4 10 T m/A)(75 A)(cos30 )

100.2 A 1.0 10 A

d dg I

I

ρ π π

µ π

−

−

ch20 giancoli7e manual

Documents