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 materials contain 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.
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 current c: 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 a fluorescent 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
14. Put one end of one rod close to one end of another rod. The ends will either attract or repel. Continue trying all combinations of rods and ends until two ends repel each other. Then the two rods used in that case are the magnets.
15. No, you cannot set a resting electron into motion with a magnetic field (no matter how big the field is). A magnetic field can only put a force on a moving charge. Thus, with no force (which means no acceleration), the velocity of the electron will not change—it will remain at rest. However, you can set a resting electron into motion with an electric field. An electric field will put a force on any charged particle, moving or not. Thus, the electric force can cause the electron to accelerate from rest to a higher speed. 16. The particle will move in an elongating helical path in the direction of the electric field (for a positive
charge). The radius of the helix will remain constant, but the pitch (see Problem 79) will increase, because the particle will accelerate along the electric field lines.
17. Yes. One possible situation is that the magnetic field is parallel or antiparallel to the velocity of the charged particle. In this case, the magnetic force would be zero, and the particle would continue moving in a straight line. Another possible situation is that there is an electric field with a magnitude and direction (perpendicular to the magnetic field) such that the electric and magnetic forces on the particle cancel each other out. The net force would be zero and the particle would continue moving in a straight line.
18. No. A moving charged particle can be deflected sideways with an electric field that points perpendicular to the direction of the velocity, even when the magnetic field in the region is zero. 19. Consider that the two wires are both horizontal, with the lower one carrying a current pointing east and
with the upper one carrying a current pointing north. The magnetic field from the upper wire points downward on the east half of the lower wire and points upward on the west half of the lower wire. Using the right-hand rule on the east half of the lower wire, where the magnetic field is downward, the force points to the north. Using the right-hand rule on the west half of the lower wire, where the magnetic field is upward, the force points to the south. Thus, the lower wire experiences a
counterclockwise torque about the vertical direction due to the magnetic forces from the upper wire. This torque is attempting to rotate the two wires so that their currents are parallel. Likewise, the upper wire would experience a clockwise torque tending to align the currents in the wires.
20. (a) The current in the lower wire is pointing in the opposite direction of the current in the upper wire. If the current in the lower wire is toward the north, then it creates a magnetic field pointing east at the upper wire. Using the right-hand rule, the current of the upper wire must be pointing toward the south so that the magnetic force created on it is upward.
(b) No, the lower wire is not stable. If the lower wire falls away from the upper wire, then the upward magnetic force on it will weaken, and there will be a net downward force on the wire. It would continue to fall away. Also, if the lower wire were to be moved toward the upper wire, the upward magnetic force on it would increase, and the lower wire would accelerate upward.
21. The equation for the magnetic field strength inside a solenoid is given by B=μ0NI,
A Eq. 20–8. (a) The magnetic field strength is not affected if the diameter of the loops doubles. The equation
shows that the magnetic field is independent of the diameter of the solenoid.
(b) If the spacing between the loops doubles, then the length of the solenoid would increase by a factor of 2, so the magnetic field strength would also decrease by a factor of 2.
Large current circuit
Small current circuit for dashboard switch
Starter Car battery Solenoid
Iron Rod
Springy Pivot
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 υ =E B/ . 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 mass spectrometer 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
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 is moving 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
Solutions to Problems
1. (a) Use Eq. 20–1 to calculate the force with an angle of 90° and a length of 1 meter. F=I BA sinθ → F/A=IBsinθ=(6.40 A)(0.90 T)sin 90° = 5.8 N/m
(b) Change the angle to 35.0 .°
F/A=IBsinθ=(6.40 A)(0.90 T)sin 35.0° = 3.3 N/m 2. Use Eq. 20–2.
max max 0.625 N 2 1.63 A
(4.80 m)(8.00 10 T)
F
F I B I
B −
= → = = =
× A
A
3. Use Eq. 20–1 to calculate the force.
F =I BA sinθ=(120 A)(240 m)(5.0 10× −5T)sin 68° =1.3 N
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.
F =I BA sinθ=(4.5 A)(2.6 m)(5.5 10× −5T)sin 41° = 4.2 10× −4N
5. To have the maximum force, the current must be perpendicular to the magnetic field, as shown in the first diagram. Use F =0.45Fmax
A A to find the angle between the wire and the magnetic field, illustrated in the second diagram. Use Eq. 20–1.
F =0.45Fmax → IBsinθ=0.45IB → θ =sin−10.45= 27°
A A
6. Use Eq. 20–2. The length of wire in the B field is the same as the diameter of the pole faces.
max max 1.28 N 0.358 T
(6.45 A)(0.555 m)
F
F I B B
I
= A → = = =
A
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.
2 max
max (0.100 m)(0.220 T)(8.50 10 N) 3.864 A 3.86 A
F
F I B I
B
−
×
= A → = = = ≈
A
(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 A to A/ cos10 0 .. ° The net effect of the changes is that the force has not changed.
Finitial =I BA ; Frotated =I( / cos10.0 ) sin100.0A ° B ° =I B FA = initial= 8.50 10× −2N
BG
wire
BG
wire
θ
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.
( )
2 1
B 2
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 −
−
= → = →
× ×
= = =
×
A A
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.
Fmax =q Bυ =(1.60 10× −19C)(7.75 10 m/s)(0.45 T)× 5 = 5.6 10× −14 N 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 6
5 19
(9.11 10 kg)(1.70 10 m/s)
1.51 10 m (1.60 10 C)(0.640 T)
F q B m
r m
r 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 a
centripetal force. The magnetic force is perpendicular to the velocity at all times for circular motion.
2 27 6
max (6.6 10 kg)(1.6 10 m/s)19 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.
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.
3
6 6
E B 7.7 10 V/m3 1.027 10 m/s 1.0 10 m/s 7.5 10 T
E
F F qE q B
B
υ υ × −
= → = → = = = × ≈ ×
×
If the electric field is turned off, then the magnetic force will cause circular motion.
2 31 6
4 B (9.11 10 19kg)(1.027 10 m/s)3 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.
2 1
initial final 2
19
5 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 5
2 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 m
r m
r qB
υ υ
υ −
− −
−1
= = →
× ×
= = = × ≈ ×
×
(c) The period can be found from the speed and the radius.
2
7 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 r m ,
qB
υ
= so rqB.
m
υ = The kinetic energy is given by
2 2 2 2
2
1 1
2 2
KE ,
2
rqB r q B
m m
m m
υ ⎛ ⎞
= = ⎜ ⎟ =
⎝ ⎠ so we see that
2 KE∝r .
(b) The angular momentum of a particle moving in a circular path is L m r= υ . From Example 20–6, we have r m ,
qB υ
= so rqB.
m
υ = Combining these relationships gives the following.
L m r mrqBr qBr2 m
υ
= = =
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.
2 2
1 2
6 19 27
19 KE
KE
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)
m
m q B
m r
m m
m m
r
qB qB qB
υ
υ υ υ
υ − −
−
= → = = →
× × ×
= = = = =
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.
13 max
max 6.2 1019 N 6 1.4 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.
2 2
1 2
p
27
p p
31
e e e
KE
KE KE
KE
KE
KE
2
2 2
2
1.67 10 kg 42.8
2 9.11 10 kg
m m
m m m
m q B r
m r qB qB qB
m
r qB m
r m m
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.
2 1
initial final 2
2 max
2
2
1 2
qV
E E qV m
m qV m
m m mV
F q B m r
r qB qB B q
υ υ
υ υ
υ
= → = → =
= = → = = =
d d
p d
d
d p
p p d
p p
p
p
p p
p p
2 1
2
2 2
2 1
1
2 1
4 2 2
2 2
1
m m V
m
B q
r
r r
r m V q
q
B q
m m V
m
B q
r
r r
r m V q
q
B q
α α
α α
α α
= = = = → =
= = = = → =
F ma q B a q B m υ υ
⊥ = ⊥ = → ⊥ =
2 2 9 5 3 2
2
1 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 qB
d t a t
m m
υ υ
υ υ
− −
⊥ ⊥ ⊥ −
−
× × ×
⎛ ⎞
= + = ⎜ ⎟⎝ ⎠ = =
×
= ×
A A
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 B V
B B
V B⊥⊥ ⊥ V ⊥
′ ′ ′ ′
= → = = =
(b) When the field is oriented at 60° to the surface, the magnetic field, Bsin 60 ,° is used to create the Hall potential.
Hall
Hall
63 mV (0.10 T)
sin 60 0.61 T
12 mV sin 60
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.
Hall 3
Hall (0.070 T)(0.0033 m)(0.13 10 V) 0.56 m/s
V
V Bd
Bd
υ υ × −
= → = = =
24. (a) The drift velocity is found from VHall =υdBd.
6
5 5
Hall
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 4
0
cable (4 10 T m/A)(65 A)2 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 of 0.5 10× −4T. 4
cable/ Earth 2.889 105 T 5.77, 5.0 10 T
B B = × −− =
26. We assume that the wire is long and straight and use Eq. 20–6.
4
0 wire
wire 7
0
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.
0 1 2 7 2 2
2 2 I I 2 (4 102 T m/A) (25 A)(0.040 m)(25 m) 7.8 10 N, attractive
F
d
μ π
π π
−
−
× ⋅
= A = = ×
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 2
2 2
4 2
2 7
0 2 1 2
2 2 0.090 m
(7.8 10 N/m) 12.54 A 13 A upward 28 A
4 10 T m/A
I I F
d F d I
I μ
π
π π
μ π
− −
= →
= = × = ≈
× ⋅
A
A
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.
4
0 Earth
cable Earth 7
0
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 the right. 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 1 2 0 1 2 0
net near far near far 1 2
near far near far
7
2 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
−
−
⎛ ⎞
= − = − = ⎜ − ⎟
⎝ ⎠
⎛ ⎞
× ⋅
= ⎜ − ⎟= ×
⎝ ⎠
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.
7
5 0
wire (4 10 T m/A)(48 A) 5.33 10 T 2 2 (0.18 m)
I B r μ π π π − − × ⋅ = = = × 5
1 Earth 1
5 wire
4.5 10 T
tan tan 40 N of W (2 significant figures) 5.33 10 T
B B
θ − − −
−
×
= = = °
×
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.
0 7 5
wire 2 (4 102 (0.200 m)T m/A)(24.0 A) 2.40 10 T
I B r μ π π π − − × ⋅ = = = ×
BEarth =5.0 10× −5T
5 5
net Earth net, wire Earth
2 2 5 2 5 2 5
net net net
5 net
1 1
5 net
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 T
tan tan 17 below the horizontal 3.60 10 T
x y
x y
y x
B B B B B
B B B
B B θ − − − − − − − − − = ° = × = − ° = − × = + = × + − × = × − × = = = ° ×
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 19
17 0
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.
7
5 0 1 0 2
net
1 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.
7
5 0 1 0 2
net
1 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 magnetic fields 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 2 net 1 2 2 2 1 1 0 2 2
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.
7
6 6
0
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 0.5 10 T,× −4 which points approximately north.
6
line/ Earth 2.235 104 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 Bline=BEarth.
7
0 0
line Earth 4
Earth
(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 which
17 , 32 ,
θ= ° + = °θ φ and β θ φ+ + =180 .° Thus φ= °15 and β =148 .° Use the law of sines to find the magnitude of BGwire and then use Eq. 20–6 to find the magnitude of the current.
Bsinwireφ = BsinEarthβ → Bwire =BEarthsinsinφβ = 2μπ0 rI →
Earth 5 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 φ πr 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.
net 0 0 0 0 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
Earth
BG
θ
wire
BG
net
(
)(
)
0 1 1 2 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 0.5 10× −4T.
6
net/ Earth 1.372 104 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 to calculate the force per unit length.
0 1 2
near near
near near 0 1 2
3
near near
0 1 2
far far
far far 0 1 2
far far
7
2
7
2
25.0 A 24.5 A
4.4 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 − − − − − = → = = = × × = → = = × × ⋅ × ⋅ A A A A 3 2
2.2 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 BG1 as the field from the top wire and B2
G
as the field from the bottom wire. We use Eq. 20–6 to calculate the magnitude of each individual field.
7 5 0 1 1 7 5 0 2 2
(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 2
1 1
2 2 2
1 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
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.
Bnetx=B1cos ( )θ1 −B2cosθ2=(9.333 10× −5T) cos 47.7° −(5.600 10× −5T) cos 26.3° =1.261 10× −5T Bnety =B1sin ( )θ1 +B2sinθ1=(9.333 10× −4T)sin 47.7° +(5.600 10× −5T)sin 26.3° =9.384 10× −5T
2 2 5 2 5 2 5
net, net,
5 net
1 1
5 net,
5 5
(1.261 10 T) (9.384 10 T) 9.468 10 T
8.379 10 T
tan 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 B θ
− − −
− ,
− −
−
− −
= + = × + × = ×
×
= = = °
×
= × ° ≈ × °
BG
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 7
0 top 0 bottom 0 2 2 2 2
net top bottom
top 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.
7
3 0 (4 10 T m/A)(460)(2.0 A) 9.6 10 T
0.12 m
NI
B=μ = π× − ⋅ = × −
A
45. Use Eq. 20–8 for the field inside a solenoid.
3 0
7 0
(4.65 10 T)(0.300 m)
1.19 A (4 10 T m/A)(935)
NI B
B I
N μ
μ π
− −
×
= → = = =
× ⋅
A A
46. The field inside a solenoid is given by Eq. 20–8.
0 3
7 0
(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 −
= → = = = ×
× ⋅
A A
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.
7 0 solenoid
wire wire solenoid wire wire
solenoid
(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× − ⋅
= ≈
A A
48. Since the mass of copper is fixed and the density is fixed, the volume of copper is fixed, and we designate it as VCu =mCu Cuρ =ACu CuA . We call the fixed voltage V0. 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 ρRCu.
0 0 0 0 0 Cu 0 0 Cu Cu
0 0 2
Cu
sol sol Cu sol RCu RCu sol Cu RCu sol Cu
Cu 0 0 Cu Cu
2 RCu sol Cu
NI N V N V V N A V N m
B
R
A
V m N
μ μ μ μ μ ρ
ρ ρ
ρ
μ ρ
ρ
= = = = =
=
A
A A A A A A A
A A
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 2
sol RCu sol Cu RCu sol Cu RCu sol sol Cu
2 1
2 2
V m N V m r V m
N B
r r
μ ρ μ ρ π μ ρ
π ρ ρ πρ
= → = = =
A A
A A
A A A A
The first factor in the expression for B is made of constants, so we have
sol sol Cu 1 .
B r
∝
A A 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.
∑
B&Δ =A μ0NI → B(2πR)=μ0NI → B=μ0NI/2πR(b) For path 2, each loop of wire pierces the enclosed area twice—once going up and once going down. Ampere’s law takes the direction of the current into account, so the net current through the area enclosed by path 2 is 0.
∑
B&Δ =A μ0NI → B(2πR) 0= → B=0(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, so B 1.
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,
∑
B&Δ =A μ0 enclI . For the path, choose a circle of radius r, centered on the center of the wire, greater than the radius of the inner wire and 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.0 encl (2 ) 0
2
I
I B B B B r B
r μ
μ π
π
(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 encl (2 ) 0 encl 0
2
I
I B B B B r B
r
μ
μ π
π
=
∑
&Δ =A∑
Δ =A∑
Δ =A → = =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.
sin 0.325 m N2 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 final final initial initial initial
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 5
0.120 m
sin 9(7.20 A) (5.50 10 T)sin 34.0 2.25 10 m N 2
NIAB
τ= 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.
2
E B
19 2 2
2 6
27
/ /
(1.60 10 C)(0.566 T) (6.10 10 m)
/ 1.87 10 V/m
1.67 10 kg
q B m r qBr m F F qE q B
E B qB r m
υ υ υ υ
υ − − −
= → = = → = →
× ×
= = = = ×
×
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 21.6
21.0 21.6
21.9 22.2
21.9 22.2
/ / / / constant 76 u/22.8 cm
76 u 76 u
70 u 72 u
21.0 cm 22.8 cm 21.6 cm 22.8 cm
76 u 76 u
73 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 m
m 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 4
2
12 19 2
2 2
13 14
2 2
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 3
13 12
2 2 3 3
14 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 4
13 12
3 3 4
14 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.
2
2 2 2 2 2 2
2
1 1
2 2 2 2
m qBR qBR q B R qR B
q 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 υ=E B/ , as stated in Section 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 2
27 3
(1.60 10 C)(0.034 T) (0.027 m) (3.3 10 kg) (2.0 u)
( / ) 1.5 10 V/m
m q B
r
qBr qBr neB r n
m n n
E B E
υ υ
υ
−
−
= →
×
= = = = = × ≈
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): 12H, He .42 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
4
CO N
/ / / constant
28.0106 u 28.0134 u 2.5 m 2.5 10 m
q B m r m qBr m r
m m r
r r r r
υ υ υ
−
= → = → =
⎛ ⎞ =⎛ ⎞ → = → =
⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠ + ×
61. The field inside the solenoid is given by Eq. 20–8 with μ0 replaced by the permeability of the iron.
7
3000(4 10 T m/A)(380)(0.35 A) 0.5 T (1.0 m)
NI
B=μ = π× − ⋅ =
A
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 μ0 replaced by the permeability of the iron.
(2.2 T)(0.38 m) 2.239 10 5T m/A 2.2 10 5T m/A 18 0 (780)(48 A)
NI B
B
NI
μ μ − − μ
= → = A = = × ⋅ ≈ × ⋅ ≈
A
63. The magnetic permeability is found from the two fields.
0 0 0
0 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 is asymptotically approaching 0 for large fields.
4 0(10 T)
B − B (T) μ (10 )(T-m/A)−4
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 64.8 1.9 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
B0(10-4 T)
μ
(1
0
-4
T-m
/A
64. (a) Use Eq. 20–6 to calculate the field due to a long straight wire.
7
6 6
0 A A at B
A 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 6
0 B B at A
B to A
(4 10 T m/A)(4.0 A) 5.333 10 T 5.3 10 T
2 2 (0.15 m)
I B
r
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.
7 on 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
−
− −
× ⋅
= = =
= × ≈ ×
A A
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.
2/ 4.8 10 2119kg m/s 2.7 10 2T
(1.60 10 C)(1.1 m)
p
q 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 .A 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.
7 0 solenoid
wire wire wire wire wire
solenoid
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× − ⋅ ⎤⎥
⎢ ⎥
⎝ ⎠ ⎣ ⎦
=
A A A
A
A A
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 eastward current is 90°. Use Eq. 20–1 to calculate the magnitude of the force.
5
sin (330 A)(18 m)(5.0 10 T)sin 90 0.297 N 0.30 N
F =I B θ= × − °
= ≈
A
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.
Earth
BG
E
N
B
FG
o
22
o
68. From Example 20–6, we have r m .
qB
υ
= The quantity mυ is the momentum, ,p so r p .
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.
Fmax =q Bυ =(1280 10× −6C)(120 m/s)(5.0 10× −5T)= 7.7 10× −6N 70. The field inside a solenoid is given by Eq. 20–8.
0 7 3
0
(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 −
= → = = = ≈ ×
× ⋅
A A
71. The magnetic force must be equal in magnitude to the weight of the electron.
31 2
6
19 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 B
qB
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.
2 2
1
initial final 2
PE KE qV m 2qV q B m
m r
υ
υ υ υ
= → = → = = →
27
2 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 m mV
r
qB qB qB
υ −
− −
−
×
= = = = = × ≈ ×
×
(b) The period is the circumference of the circular path divided by the speed.
2 27
7 19
2 2
2 2 2 (6.6 10 kg)
5.5 10 s 2 2(1.60 10 C)(0.240 T)
mV qB
r m
T
qB qV
m
π
π π π
υ
−
− −
×
= = = = = ×
×
73. The magnetic field at the center of the square is the vector sum of the magnetic 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.
A
A
A
0
1 2 3 4
0 0
4 cos 45 to the left 2
2 2
4 to the left to the left 2
2 2
2
I r
I I
μ π
μ μ
π π
⎛ ⎞
= + + + = ⎜ ⎟ °
⎝ ⎠
⎛ ⎞
⎜ ⎟
⎜ ⎟
= =
⎜ ⎟
⎜ ⎟
⎝ ⎠
B BG G BG BG BG
A A
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.
B E 12,000 V/m6 2.5 10 3T
4.8 10 m/s
E
F F q B qEυ B
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.
19
7
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 m
π 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 r m .
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 19
1 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 − − − − − − −
= →
× ×
= = = = = °
× ×
A
A A
76. With one turn of wire, we have 0 . 2
I B
r μ
= Use the radius of the Earth for r.
6 4
9 9
0
7 0
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
