453
The Magnetic Field
Conceptual Problems
*1 •
Determine the Concept Because the electrons are initially moving at 90° to the magnetic field, they will be deflected in the direction of the magnetic force acting on them. Use the right-hand rule based on the expression for the magnetic force acting on a moving charge
B
v
F
r
=
q
r
×
r
, remembering that, for a negative charge, the force is in the direction opposite that indicated by the right-hand rule, to convince yourself that the particle will follow the path whose terminal point on the screen is 2. (b)iscorrect.2 •
Determine the Concept One cannot define the direction of the force arbitrarily. By experiment,
F
r
is perpendicular toB
r
.3 •
Determine the ConceptFalse. An object experiences acceleration if either its speed changes or the direction it is moving changes. The magnetic force, acting perpendicular to the direction a charged particle is moving, changes the particle’s velocity by changing the direction it is moving and hence accelerates the particle.
4 •
Determine the Concept Yes; it will be deflected upward. Because the beam passes through undeflected when traveling from left to right, we know that the upward electric force must be balanced by a downward magnetic force. Application of the right-hand rule tells us that the magnetic field must be out of the page. When the beam is reversed, the magnetic force (as well as the electric force) acting on it is upward.
*5 •
Determine the Concept The alternating current running through the filament is changing direction every 1/60 s, so in a magnetic field the filament experiences a force which alternates in direction at that frequency.
6 •
Determine the Concept The magnitude of the torque on a current loop is given by
,
sin
θ
µ
7 •
(a) True. This is an experimental fact and is the basis for the definition of the magnetic force on a moving charged particle being expressed in terms of the cross product of
vrand
B
v
; i.e.F
v
=
q
v
r
×
B
r
.(b) True. This is another experimental fact. The torque on a magnet is a restoring torque, i.e., one that acts in such a direction as to align the magnet with magnetic field.
(c) True. We can use a right-hand rule to relate the direction of the magnetic field around the loop to the direction of the current. Doing so indicates that one side of the loop acts like a north pole and the other like a south pole.
(d) False. The period of a particle moving in a circular path in a magnetic field is given by T =2
π
mr qvBand, hence, is proportional to the square root of the radius of the circle.
(e) True. The drift velocity is related to the Hall voltage according to vd = VH/Bw where w
is the width of the Hall-effect material.
*8 • .
Determine the Concept The direction in which a particle is deflected by a magnetic field will be unchanged by any change in the definition of the direction of the magnetic field. Since we have reversed the direction of the field, we must define the direction in which particles are deflected by a "left-hand" rule instead of a "right-hand" rule.
9 •
Determine the Concept Choose a right-handed coordinate system in which east is the positive x direction and north is the positive y direction. Then the magnetic force acting on the particle is given by
F
r
=
qv
i
ˆ
×
B
ˆ
j
=
qvB
( )
i
ˆ
×
ˆ
j
=
qvB
k
ˆ
. Hence, the magnetic force is upward.10 •
Determine the ConceptApplication of the right-hand rule tells us that this positively charged particle would have to be moving in the northwest direction with the magnetic field upward in order for the magnetic force to be toward the northeast. The situation described cannot exist. (e)iscorrect.
11 •
Apply
∑
Fradial =macto the lithiumnucleus to obtain:
R
v
m
qvB
2
=
Solve for r:
qB
mv
R
=
For the 7Li nucleus this becomes:
eB
p
R
3
Li Li
=
For the proton we have:
eB
p
R
pp
=
Divide equation (2) by equation (1) and simplify to obtain:
Li p
Li p
Li p
3
3
p p
eB p eB
p
R R
= =
Because the momenta are equal:
3
Li p
=
R
R
and (a)iscorrect.
*12 •
Determine the ConceptApplication of the right-hand rule indicates that a positively charged body would experience a downward force and, in the absence of other forces, be deflected downward. Because the direction of the magnetic force on an electron is opposite that of the force on a positively charged object, an electron will be deflected upward. (c)iscorrect.
13 ••
Determine the Concept From relativity; this is equivalent to the electron moving from right to left at velocity v with the magnet stationary. When the electron is directly over the magnet, the field points directly up, so there is a force directed out of the page on the electron.
14 •
Similarities Differences
Magnetic field lines are similar to electric field lines in that their density is a measure of the strength of the field; the lines point in the direction of the field; also, magnetic field lines do not cross.
15 •
Determine the Concept If only Fr and I are known, one can only conclude that the magnetic field Br is in the plane perpendicular toFr. The specific direction of Br is undetermined.
Estimation and Approximation
*16 ••
Picture the Problem If the electron enters the magnetic field in the coil with speed v, it will travel in a circular path under the influence of the magnetic force acting on. We can apply Newton’s 2nd law to the electron in this field to obtain an expression for the
magnetic field. We’ll assume that the deflection of the electron is small over the distance it travels in the magnetic field, but that, once it is through the region of the magnetic field, it travels at an angle θwith respect to the direction it was originally traveling.
Apply
∑
F =macto the electronin the magnetic field to obtain:
r
v
m
evB
2
=
Solve for B:
er
mv
B
=
The kinetic energy of the electron is: 2 2 1
mv
eV
K
=
=
Solve for v to obtain:m eV v= 2
Substitute for v in the expression for r:
e mV r
m eV er
m
B= 2 = 1 2
Because θ << 1:
θ
sin r
d ≈ ⇒
θ
sin
d
r
≈
Substitute for r in the expression for B to obtain:
e mV d
B= sin
θ
2For maximum deflection,
θ ≈ 45°. Substitute numerical values and evaluate B:
(
)
(
)
mT 84 . 5
C 10 60 . 1
kV 15 kg 10 11 . 9 2 m 05 . 0
45 sin
19 31
=
× × °
= − −
B
17 ••
Picture the Problem Let h be the height of the orbit above the surface of the earth, m the mass of the micrometeorite, and v its speed. We can apply Newton’s 2nd law to the orbiting micrometeorite with Fmag = qvB to derive an expression for the charge-to-mass
(a) Apply
∑
F =macto themicrometeorite orbiting under the influence of the magnetic force:
earth 2
R
h
v
m
qvB
+
=
Solve for q/m to obtain:
(
h
R
earth)
B
v
m
q
+
=
Substitute numerical values and evaluate q/m:
(
)
(
)
88
.
6
C/kg
km
6370
km
400
T
10
5
km/s
30
5
+
=
×
=
−m
q
(b) Solve the result for q/m obtained in (a) for q to obtain:
(
)
mq= 88.6C/kg
Substitute numerical values and evaluate q:
(
)
(
)
nC
6
.
26
kg
10
3
C/kg
6
.
88
10=
×
=
−q
Force Exerted by a Magnetic Field
18 •
Picture the ProblemThe magnetic force acting on a charge is given by
F
r
=
q
v
r
×
B
r
. We can express vrandB
r
, form their vector (a.k.a. ″cross″) product, and multiply by the scalar q to findF
r
.Express the force acting on the proton:
F
r
=
q
v
r
×
B
r
Express vr:v
r
=
(
4
.
46
Mm/s
)
i
ˆ
ExpressB
r
:B
r
=
(
1
.
75
T
)
k
ˆ
Substitute numerical values and evaluate
F
r
:(
)
[
(
) (
i
)
k
]
(
)
j
F
r
=
1
.
60
×
10
−19C
4
.
46
Mm/s
ˆ
×
1
.
75
T
ˆ
=
−
1
.
25
pN
ˆ
19 •
scalar q to find
F
r
.Express the force acting on the charge:
B
v
F
r
=
q
r
×
r
Substitute numerical values to obtain:
(
)
[
(
)
i
B
]
F
r
=
−
3
.
64
nC
2
.
75
×
10
6m/s
ˆ
×
r
(a) Evaluate
F
r
forB
r
= 0.38 Tj
ˆ
:(
)
[
(
)
i
(
)
j
]
(
)
k
F
r
=
−
3
.
64
nC
2
.
75
×
10
6m/s
ˆ
×
0
.
38
T
ˆ
=
−
3
.
80
mN
ˆ
(b) Evaluate
F
r
forB
r
= 0.75 T iˆ + 0.75 Tj
ˆ
:(
)
[
(
)
i
{
(
) (
i
)
j
}
]
(
)
k
F
r
=
−
3
.
64
nC
2
.
75
×
10
6m/s
ˆ
×
0
.
75
T
ˆ
+
0
.
75
T
ˆ
=
−
7
.
51
mN
ˆ
(c) Evaluate
F
r
forB
r
= 0.65 T iˆ:(
−3.64nC)
[
(
2.75×106m/s)
ˆ×(
0.65T)
ˆ]
= 0= i i
Fr
(d) Evaluate
F
r
for Br = 0.75 T iˆ + 0.75 Tkˆ:(
)
[
(
)
i
(
) (
i
)
k
]
(
)
j
F
r
=
−
3
.
64
nC
2
.
75
×
10
6m/s
ˆ
×
0
.
75
T
ˆ
+
0
.
75
T
ˆ
=
7
.
51
mN
ˆ
20 •
Picture the Problem The magnetic force acting on the proton is given by
F
r
=
q
v
r
×
B
r
. We can express vrandB
r
, form their vector (a.k.a. ″cross″) product, and multiply by the scalar q to findF
r
.Express the force acting on the proton:
F
r
=
q
v
r
×
B
r
(a) EvaluateF
r
for vr= 2.7 Mm/s iˆ:(
)(
[
)
i
(
)
k
]
(
)
j
F
r
=
1
.
60
×
10
−19C
2
.
7
×
10
6m/s
ˆ
×
1
.
48
T
ˆ
=
−
0
.
640
pN
ˆ
(b) Evaluate
F
r
for vr = 3.7 Mm/sˆ
j
:(
)(
[
)
j
(
)
k
]
(
)
i
(c) Evaluate
F
r
for vr = 6.8 Mm/s kˆ:(
1.60×10 19C)(
[
6.8×106m/s)
ˆ×(
1.48T)
ˆ]
= 0= − k k
Fr
(d) Evaluate
F
r
forv
r
=
4
.
0
Mm/s
i
ˆ
+
3
.
0
Mm/s
ˆ
j
:(
)
[
{
(
) (
)
}
(
)
]
(
) (
i
)
j
k
j
i
F
ˆ
pN
947
.
0
ˆ
pN
710
.
0
ˆ
T
48
.
1
ˆ
Mm/s
0
.
3
ˆ
Mm/s
0
.
4
C
10
60
.
1
19−
=
×
+
×
=
−r
21 •
Picture the ProblemThe magnitude of the magnetic force acting on a segment of wire is given by F =IlBsin
θ
where lis the length of the segment of wire, B is the magnetic field, and θ is the angle between the segment of wire and the direction of the magnetic field.Express the magnitude of the magnetic force acting on the segment of wire:
θ
sin B I F = lSubstitute numerical values and evaluate F:
(
)( )(
)
N
962
.
0
30
sin
T
37
.
0
m
2
A
6
.
2
=
°
=
F
*22 •
Picture the Problem We can use
F
r
=
I
L
r
×
B
r
to find the force acting on the wire segment.Express the force acting on the wire segment:
B
L
F
r
=
I
r
×
r
Substitute numerical values and evaluate
F
r
:(
) (
[
) (
)
]
(
)
(
)
k
i
j
i
F
ˆ
N
140
.
0
ˆ
T
3
.
1
ˆ
cm
4
ˆ
cm
3
A
7
.
2
−
=
×
+
=
r
23 •
Express the force acting on the proton:
F
r
=
q
v
r
×
B
r
Express the magnitude ofF
r
in termsof its components:
2 2 2
z y x
F
F
F
F
=
+
+
(1)Substitute numerical values and evaluate
F
r
:(
)
[
{
(
)
(
)
} (
)
]
(
) (
) (
) (
)
(
) (
i
) (
j
)
k
i
k
j
k
k
j
i
j
i
F
ˆ
pN
576
.
0
ˆ
pN
128
.
0
ˆ
pN
192
.
0
ˆ
pN
192
.
0
ˆ
pN
384
.
0
ˆ
pN
128
.
0
ˆ
pN
192
.
0
T
ˆ
4
.
0
ˆ
6
.
0
ˆ
8
.
0
ˆ
Mm/s
3
ˆ
Mm/s
2
C
10
60
.
1
19−
−
−
=
−
+
−
+
−
+
−
=
−
+
×
−
×
−
=
−r
Substitute in equation (1) to obtain:
(
−0.192pN) (
2+ −0.128pN) (
2+ −0.576pN)
2 = 0.621pN=
F
Express and evaluate the angle
F
r
makes with the x axis:° = ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ − = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = − − 108 pN 0.621 pN 192 . 0 cos
cos1 1
F Fx
x
θ
Express and evaluate the angle
F
r
makes with the y axis:° = ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ − = ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ = − − 102 pN 0.621 pN 128 . 0 cos
cos 1 1
F Fy
y
θ
Express and evaluate the angle
F
r
makes with the z axis:° = ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ − = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = − − 158 pN 0.621 pN 576 . 0 cos
cos 1 1
F Fz
z
θ
24 ••
Picture the Problem We can use Fr =Ilr×Brto find the force acting on the segments of the wire as well as the magnetic force acting on the wire if it were a straight segment from a to b.
Express the magnetic force acting on the wire:
cm 4 cm
3 F
Evaluate Fr3cm:
(
) (
[
) (
)
]
(
)
( )
(
)
jj k i F ˆ N 0648 . 0 ˆ N 0648 . 0 ˆ T 2 . 1 ˆ cm 3 A 8 . 1 cm 3 − = − = × = r
Evaluate F4cm
r
:
(
) (
[
) (
)
]
(
)
i
k
j
F
ˆ
N
0864
.
0
ˆ
T
2
.
1
ˆ
cm
4
A
8
.
1
cm 4=
×
=
r
Substitute to obtain:
(
) (
)
(
) (
i
)
j
i
j
F
ˆ
N
0648
.
0
ˆ
N
0864
.
0
ˆ
N
0864
.
0
ˆ
N
0648
.
0
−
=
+
−
=
r
If the wire were straight from a to b:
r
l
=
(
3
cm
) (
i
ˆ
+
4
cm
)
ˆ
j
The magnetic force acting on the wire is:
(
) (
[
) (
)
]
(
)
(
) (
)
(
) (
i
)
j
i
j
k
j
i
F
ˆ
N
0648
.
0
ˆ
N
0864
.
0
ˆ
N
0864
.
0
ˆ
N
0648
.
0
ˆ
T
2
.
1
ˆ
cm
4
ˆ
cm
3
A
8
.
1
−
=
+
−
=
×
+
=
r
in agreement with the result obtained above when we treated the two straight segments of the wire separately.
25 ••
Picture the Problem Because the magnetic field is horizontal and perpendicular to the wire, the force it exerts on the current-carrying wire will be vertical. Under equilibrium conditions, this upward magnetic force will be equal to the downward gravitational force acting on the wire.
Apply
∑
Fvertical=0to the wire:F
mag−
w
=
0
Express Fmag:
F
mag=
I
l
B
because θ = 90°.
Substitute to obtain:
I
l
B
−
mg
=
0
Solve for I:B
mg
I
l
=
Substitute numerical values and evaluate I:
( )
(
)
*26 ••
Picture the Problem The diagram shows the gaussmeter displaced from equilibrium under the influence of the gravitational and magnetic forces acting on it. We can apply the condition for translational equilibrium in the x direction to find the equilibrium angular displacement of the wire from the vertical. In part (b) we can solve the equation derived in part (a) for B and evaluate this expression for the given data to find the horizontal magnetic field sensitivity of this gaussmeter.
(a) Apply
∑
Fx =0to the wire to obtain:0
cos
sin
θ
−
F
θ
=
mg
Substitute for F and solve for θ to obtain:
0
cos
sin
θ
−
I
B
θ
=
mg
l
(1) and⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
=
−mg
B
I
l
1
tan
θ
Substitute numerical values and evaluate θ:
(
)(
)(
)
(
)
(
)
° =
⎥ ⎦ ⎤ ⎢
⎣ ⎡
= −
66 . 4
m/s 81 . 9 kg 005 . 0
T 04 . 0 m 5 . 0 A 2 . 0
tan 1 2
θ
(b) Solve equation (1) for B to
obtain:
I
l
mg
B
=
tan
θ
For a displacement from vertical of
0.5 mm:
0.5
m
0
.
001
mm
5
.
0
sin
tan
θ
≈
θ
=
=
and
rad
001
.
0
=
θ
Substitute numerical values and evaluate B:
(
)
(
)
(
)
(
)(
)
T
91
.
4
m
5
.
0
A
20
rad
001
.
0
m/s
81
.
9
kg
005
.
0
2µ
=
=
B
Picture the Problem With the current in the direction indicated and the magnetic field in the z direction, pointing out of the plane of the page, the force is in the radial direction and we can integrate the element of force dF acting on an element of length dℓ between θ = 0 and π to find the force acting on the semicircular portion of the loop and use the expression for the force on a current-carrying wire in a uniform
magnetic field to find the force on the straight segment of the loop.
Express the net force acting on the semicircular loop of wire:
segment straight loop
ar semicircul
F
F
F
=
+
(1)Express the force acting on the straight segment of the loop:
RIB
I
2
segment
straight
=
×
B
=
−
F
r
r
l
r
Express the force dF acting on the element of the wire of length dℓ:
θ
IRBd B
Id dF = l =
Express the x and y components of dF:
θ
cos
dF
dF
x=
andθ
sin
dF
dF
y=
Because, by symmetry, the x component of the force is zero, we can integrate the y component to find the force on the wire:
θ
θ
d
IRB
dF
y=
sin
andRIB d
RIB
Fy sin 2
0
=
= π
∫
θ
θ
28 ••
Picture the Problem We can use the information given in the 1st and 2nd sentences to obtain an expression containing the components of the magnetic fieldBr . We can then use the information in the 1st and 3rd sentences to obtain a second equation in these components that we can solve simultaneously for the components ofBr .
Express the magnetic field Br in terms of its components:
k
j
i
B
ˆ
ˆ
ˆ
z y
x
B
B
B
+
+
=
r
(1)
Express Frin terms of Br:
( )(
[
)
]
(
)
(
)
(
)
(
)
j(
)
ik j i k k j i k B F ˆ m A 4 . 0 ˆ m A 4 . 0 ˆ ˆ ˆ ˆ m A 4 . 0 ˆ ˆ ˆ ˆ m 1 . 0 A 4 y y z y x z y x B B B B B B B B I ⋅ − ⋅ = + + × ⋅ = + + × = ×
= rl r
r
Equate the components of this expression for Frwith those given in the second sentence of the statement of the problem to obtain:
(
0
.
4
A
⋅
m
)
B
y=
0
.
2
N
and(
0
.
4
A
⋅
m
)
B
x=
0
.
2
N
Noting that Bz is undetermined, solve for Bx and By:
T
5
.
0
=
xB
andB
y=
0
.
5
T
When the wire is rotated so that the current flows in the positive x
direction:
( )(
[
)
]
(
)
(
)
(
)
(
)
j(
)
kk j i i k j i i B F ˆ m A 4 . 0 ˆ m A 4 . 0 ˆ ˆ ˆ ˆ m A 4 . 0 ˆ ˆ ˆ ˆ m 1 . 0 A 4 y z z y x z y x B B B B B B B B I ⋅ + ⋅ − = + + × ⋅ = + + × = ×
= rl r
r
Equate the components of this expression for Frwith those given in the third sentence of the problem statement to obtain:
(
0.4A⋅m)
=0− Bz
and
(
0
.
4
A
⋅
m
)
B
y=
0
.
2
N
Solve for Bz and By to obtain: Bz =0
and, in agreement with our results above,
T
5
.
0
=
yB
29 ••
Picture the Problem We can use the information given in the 1st and 2nd sentences to obtain an expression containing the components of the magnetic fieldBr . We can then use the information in the 1st and 3rd sentences to obtain a second equation in these components that we can solve simultaneously for the components ofBr .
Express the magnetic field Br in terms of its components:
k
j
i
B
ˆ
ˆ
ˆ
z y
x
B
B
B
+
+
=
r
(1)
Express Frin terms of Br:
( )(
[
)
]
(
)
(
)
(
)
(
)
j(
)
kk j i i k j i i B F ˆ m A 2 . 0 ˆ m A 2 . 0 ˆ ˆ ˆ ˆ m A 2 . 0 ˆ ˆ ˆ ˆ m 1 . 0 A 2 y z z y x z y x B B B B B B B B I ⋅ + ⋅ − = + + × ⋅ = + + × = ×
= rl r
r
Equate the components of this expression for Frwith those given in the second sentence of the statement of the problem to obtain:
(
0.2A⋅m)
=3N− Bz
and
(
0
.
2
A
⋅
m
)
B
y=
2
N
Noting that Bx is undetermined, solve for Bz and By:
T 15 − = z B and
T
10
=
yB
When the wire is rotated so that the current flows in the positive y
direction:
( )(
[
)
]
(
)
(
)
(
)
(
)
i
(
)
k
k
j
i
j
k
j
i
j
B
F
ˆ
m
A
2
.
0
ˆ
m
A
2
.
0
ˆ
ˆ
ˆ
ˆ
m
A
2
.
0
ˆ
ˆ
ˆ
ˆ
m
1
.
0
A
2
x z z y x z y xB
B
B
B
B
B
B
B
I
⋅
−
⋅
=
+
+
×
⋅
=
+
+
×
=
×
=
r
l
r
r
Equate the components of this expression for Frwith those given in the third sentence of the problem statement to obtain:
(
0
.
2
A
⋅
m
)
B
x=
−
3
N
and(
0.2A⋅m)
=−2N− Bz
Solve for Bx and Bz to obtain:
B
x=
−
15
T
and, in agreement with our results above, T
10
=
z
Substitute in equation (1) to obtain:
B
r
=
(
10
T
) (
i
ˆ
+
10
T
) ( )
ˆ
j
−
15
T
k
ˆ
30 •••
Picture the Problem We can integrate the expression for the force dFracting on an element of the wire of length dLrfrom a to b to show thatFr =ILr×Br.
Express the force dFr acting on the element of the wire of length dLr:
B L Fr =Idr× r
d
Integrate this expression to obtain:
∫
×= b
a
IdL B
Fr r r
Because
B
r
and I are constant:B
L
B
L
F
r
r
⎟⎟
×
r
=
r
×
r
⎠
⎞
⎜⎜
⎝
⎛
=
I
∫
d
I
ba
where
L
r
is the vector from a to b.Motion of a Point Charge in a Magnetic Field
*31 •
Picture the Problem We can apply Newton’s 2nd law to the orbiting proton to relate its speed to its radius. We can then use T = 2πr/v to find its period. In Part (b) we can use the relationship between T and v to determine v. In Part (c) we can use its definition to find the kinetic energy of the proton.
(a) Relate the period T of the motion of the proton to its orbital speed v:
v
r
T
=
2
π
(1)Apply
∑
Fradial =macto the protonto obtain:
r
v
m
qvB
2
=
Solve for v/r to obtain:
m
qB
r
v
=
Substitute to obtain:
qB
m
T
=
2
π
Substitute numerical values and evaluate T:
(
)
(
)
(
)
87
.
4
ns
T
75
.
0
C
10
60
.
1
kg
10
67
.
1
2
19 27
=
×
×
=
π
− −(b) From equation (1) we have:
(
)
m/s
10
67
.
4
ns
4
.
87
m
65
.
0
2
2
7
×
=
=
=
π
π
T
r
v
(c) Using its definition, express and evaluate the kinetic energy of the proton:
(
)(
)
MeV
11.4
J
10
1.60
eV
1
J
10
82
.
1
m/s
10
67
.
4
kg
10
67
.
1
27 7 2 12 192 1 2 2 1
=
×
×
×
=
×
×
=
=
mv
− − −K
32 •
Picture the Problem We can apply Newton’s 2nd law to the orbiting electron to obtain an expression for the radius of its orbit as a function of its mass m, charge q, speed v, and the magnetic field B. Using the definition of its kinetic energy will allow us to express r in terms of m, q, B, and its kinetic energy K. We can use
T = 2πr/v to find the period of the motion and calculate the frequency from the reciprocal of the period of the motion.
(a) Apply
∑
Fradial =macto theproton to obtain:
r
v
m
qvB
2
=
Solve for r:
qB
mv
r
=
(1)Express the kinetic energy of the electron:
2 2 1
mv
K
=
Solve for v to obtain:
m K
v= 2 (2)
Substitute in equation (1) to obtain:
Km
qB
m
K
qB
m
r
=
2
=
1
2
Find the frequency from the
reciprocal of the period:
0
.
110
ns
9
.
10
GHz
1
1
=
=
=
T
f
(
)
(
)
(
)
2
.
20
mm
T
325
.
0
C
10
.6
1
eV
J
10
1.6
kg
10
11
.
9
keV
45
2
19
19 31
=
×
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
×
×
×
=
−− −
r
(b) Relate the period of the
electron’s motion to the radius of its orbit and its orbital speed:
v
r
T
=
2
π
Substitute equation (2) to obtain:
K
m
r
m
K
r
T
2
2
2
π
π
=
=
Substitute numerical values and evaluate T:
(
)
(
)
0
.
110
ns
eV
J
10
1.6
keV
45
kg
10
11
.
9
2
mm
20
.
2
1931
=
×
×
×
=
π
− −T
33 •
Picture the Problem We can apply Newton’s 2nd law to the orbiting electron to obtain an expression for the radius of its orbit as a function of its mass m, charge q, speed v, and the magnetic field B.
(a) Apply
∑
Fradial =macto theproton to obtain:
r
v
m
qvB
2
=
Solve for r:
qB
mv
r
=
Substitute numerical values and evaluate r:
(
)(
)
(
)(
)
m
142
T
10
4
C
10
60
.
1
m/s
10
kg
10
11
.
9
7 19
7 31
=
×
×
×
=
−− −r
(b) For B = 2×10−5 T:
(
)(
)
(
)(
)
m
84
.
2
T
10
2
C
10
60
.
1
m/s
10
kg
10
11
.
9
5 19
7 31
=
×
×
×
=
−− −r
34 ••
Picture the Problem We can apply Newton’s 2nd law to an orbiting particle to obtain an expression for the radius of its orbit R as a function of its mass m, charge q, speed v, and the magnetic field B.
Apply
∑
Fradial =macto an orbitingparticle to obtain:
r
v
m
qvB
2
=
Solve for r:
qB
mv
r
=
Express the kinetic energy of the particle:
2 2 1
mv
K
=
Solve for v to obtain:
m K
v= 2
Substitute in equation (1) to obtain:
Km
qB
m
K
qB
m
r
=
2
=
1
2
(1)Using equation (1), express the ratio Rd/Rp:
2
2
2
1
2
1
p p
p d
d p
p p
d d
p d
=
=
=
=
m
m
e
e
m
m
q
q
Km
B
q
Km
B
q
R
R
Using equation (1), express the ratio Rα /Rp:
1
4
2
2
1
2
1
p p
p p
p p
p
=
=
=
=
m
m
e
e
m
m
q
q
Km
B
q
Km
B
q
R
R
αα α α
α
35 ••
with the definitions of kinetic energy and angular momentum.
(a) Apply
∑
Fradial =macto anorbiting particle to obtain:
r
v
m
qvB
2
=
Solve for v:
m qBr
v =
Express the velocities of the
particles: p
p p
m
Br
q
v
=
andα α α
m
Br
q
v
=
Divide the second of these equations by the first to obtain:
( )
2
1
4
2
p p p p p p p=
=
=
=
m
e
em
m
q
m
q
m
Br
q
m
Br
q
v
v
α α α α α(b) Express the kinetic energy of an
orbiting particle:
m
r
B
q
m
qBr
m
mv
K
2 2 2 2 1 2 2 1 2 21
⎟
=
⎠
⎞
⎜
⎝
⎛
=
=
Using this relationship, express the ratio of Kα to Kp:
( )
( )
1
4
2
p 2 p 2 2 p p 2 p 2 2 2 p 2 1 2 2 2 2 1 p=
=
=
=
m
e
m
e
m
q
m
q
m
r
B
q
m
r
B
q
K
K
α α α α α(c) Express the angular momenta of the particles:
r
v
m
L
α=
α α andL
p=
m
pv
pr
Express the ratio of Lα to Lp:
( )( )
2
4
p p p 2 1 p p p p=
=
=
v
m
v
m
r
v
m
r
v
m
L
L
α α α36 ••
(a) Express the momentum of the particle:
mv
p
=
(1)Apply
∑
Fradial =macto theorbiting particle to obtain:
R
v
m
qvB
2
=
Solve for v:
m
qBR
v
=
Substitute in equation (1) to obtain:
qBR
m
qBR
m
p
⎟
=
⎠
⎞
⎜
⎝
⎛
=
(b) Express the kinetic energy of the orbiting particle as a function of its momentum:
m
p
K
2
2
=
Substitute our result from part (a) to obtain:
(
)
m
R
B
q
m
qBR
K
2
2
2 2 2 2
=
=
*37 ••
Picture the Problem The particle’s velocity has a component v1 parallel to B
r
and a component v2 normal toB
r
. v1 = v cosθ and is constant, whereas v2 = v sinθ , being
normal toBr , will result in a magnetic force acting on the beam of particles and circular motion perpendicular toBr . We can use the relationship between distance, rate, and time and Newton’s 2nd law to express the distance the particle moves in the direction of the field during one period of the motion.
Express the distance moved in the direction of Br by the particle during one period:
T v
x= 1 (1)
Express the period of the circular
motion of the particles in the beam: 2
2
v
r
T
=
π
Apply
∑
Fradial =macto a particlein the beam to obtain:
r
v
m
B
qv
2 2
2
=
Solve for v2:
m
qBr
Substitute to obtain:
qB
m
m
qBr
r
T
=
2
π
=
2
π
Because v1 = v cosθ, equation (1)
becomes:
(
)
π
θ
π
θ
2
2
cos
cos
v
qB
m
qB
m
v
x
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
=
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
=
38 ••
Picture the Problem The trajectory of the proton is shown to the right. We know that, because the proton enters the field
perpendicularly to the field, its trajectory while in the field will be circular. We can use symmetry considerations to determine
φ. The application of Newton’s 2nd law to the proton while it is in the magnetic field and of trigonometry will allow us to conclude that r = d and to determine their value.
From symmetry, it is evident that the angle θ in Figure 26-35 equals the angle φ:
° = 60.0
φ
Use trigonometry to obtain:
(
)
r
d
2
2
1
30
sin
90
sin
°
−
θ
=
°
=
=
or r = d.
Apply
∑
Fradial =macto the proton while it is in the magnetic field to obtain:r
v
m
qvB
2
=
Solve for r:
qB
mv
r
=
Substitute numerical values and evaluate r = d:
(
)(
)
(
)
(
)
m
130
.
0
T
8
.
0
C
10
60
.
1
m/s
10
kg
10
67
.
1
19 7 27
=
×
×
=
=
r
−−39 ••
Picture the Problem The trajectory of the proton is shown to the right. We know that, because the proton enters the field
perpendicularly to the field, its trajectory while in the field will be circular. We can use symmetry considerations to determine
φ. The application of Newton’s 2nd law to the proton while it is in the magnetic field and of trigonometry will allow us to conclude that r = d and to determine their value.
(a) From symmetry, it is evident that the angle θ in Figure 26-33 equals the angle φ:
° = 24.0
φ
Use trigonometry to obtain:
(
)
p
2 24
sin 90
sin
r d
= ° =
−
°
θ
or
m
492
.
0
24
sin
2
m
4
.
0
24
sin
2
p
=
°
=
°
=
d
r
Apply
∑
Fradial =macto the proton while it is in the magnetic field to obtain:p 2 p p p p
r v m B v
q =
Solve for and evaluate vp:
p p p p
m
B
r
q
v
=
Substitute numerical values and evaluate vp:
(
)
(
)(
)
m/s
10
83
.
2
kg
10
1.67
T
6
.
0
m
492
.
0
C
10
60
.
1
7
27 19
p
×
=
×
×
=
− −v
(b) Express vd:
p p p
d d d d
2
m
B
r
q
m
B
r
q
Substitute numerical values and evaluate vd:
(
)
(
)(
)
(
)
m/s
10
41
.
1
kg
10
1.67
2
T
6
.
0
m
492
.
0
C
10
60
.
1
7
27 19
d
×
=
×
×
=
− −v
40 ••
Picture the Problem We can apply Newton’s 2nd law of motion to express the orbital speed of the particle and then find the period of the dust particle from this orbital speed.
The period of the dust particle’s
motion is given by:
v
r
T
=
2
π
Apply
∑
F =macto the particle:r
v
m
qvB
2
=
Solve for v to obtain:
m
qBr
v
=
Substitute for v in the expression for
T and simplify:
qB
m
qBr
rm
T
=
2
π
=
2
π
Substitute numerical values and evaluate T:
(
)
(
)
(
)
y
10
64
.
6
Ms
31.56
y
1
s
10
094
.
2
T
10
nC
3
.
0
kg/g
10
g
10
10
2
3 11
9 3 6
×
=
×
×
=
×
×
=
π
− − −T
The Velocity Selector
*41 •
Picture the ProblemSuppose that, for positively charged particles, their motion is from left to right through the velocity selector and the electric field is upward. Then the magnetic force must be downward and the magnetic field out of the page. We can apply the condition for translational equilibrium to relate v to E and B. In (b) and (c) we can use the definition of kinetic energy to find the energies of protons and electrons that pass through the velocity selector undeflected.
(a) Apply
∑
Fy =0to the particle to obtain:0
mag elec
−
F
=
F
or
Solve for v to obtain:
B
E
v
=
Substitute numerical values and
evaluate v:
0
.
28
T
1
.
64
10
m/s
MV/m
46
.
0
=
×
6=
v
(b) Express and evaluate the kinetic energy of protons passing through
the velocity selector undeflected:
(
)(
)
keV
14.0
J
10
1.60
eV
1
J
10
26
.
2
m/s
10
64
.
1
10
67
.
1
19 15 2 6 kg 27 2 1 2 p 2 1 p=
×
×
×
=
×
×
=
=
− − −v
m
K
(c) The kinetic energy of electrons passing through the velocity selector
undeflected is given by:
(
)(
)
eV
66
.
7
J
10
1.60
eV
1
J
10
23
.
1
m/s
10
64
.
1
10
11
.
9
19 18 2 6 kg 31 2 1 2 e 2 1 e=
×
×
×
=
×
×
=
=
− − −v
m
K
42 •
Picture the ProblemBecause the beam of protons is not deflected; we can conclude that the electric force acting on them is balanced by the magnetic force. Hence, we can find the magnetic force from the given data and use its definition to express the electric field.
(a) Use the definition of electric field to relate it to the electric force acting on the beam of protons:
q elec elec F E r r =
Express the magnetic force acting on the beam of protons:
k
j
i
F
ˆ
ˆ
ˆ
mag
=
qv
×
B
=
qvB
r
Because the electric force must be equal in magnitude but opposite in direction:
(
)
(
)(
)
k(
)
kk
F ˆ 1.60 10 19C 12.4km/s 0.85T ˆ 1.69 10 15N ˆ
elec − − =− × × − = − = qvB r
Substitute in the equation for the electric field to obtain:
(b) Becauseboth FmagandFelecarereversed,electronsarenot deflected.
r r
Thomson’s Measurement of
q
/
m
for Electrons and the Mass
Spectrometer
*43 ••
Picture the Problem Figure 26-18 is reproduced below. We can express the total
deflection of the electron beam as the sum of the deflections while the beam is in the field between the plates and its deflection while it is in the field-free space. We can, in turn, use constant-acceleration equations to express each of these deflections. The resulting equation is in terms of v0 and E. We can find v0 from the kinetic energy of the beam and
E from the potential difference across the plates and their separation. In part (b) we can equate the electric and magnetic forces acting on an electron to express B in terms of E and v0.
(a) Express the total deflection ∆y of the electrons:
2
1 y
y y=∆ +∆
∆ (1) where
∆y1 is the deflection of the beam while it is
in the electric field and ∆y2 is the deflection
of the beam while it travels along a straight-line path outside the electric field.
Use a constant-acceleration equation to express ∆y1:
( )
2 21
1
a
t
y
=
y∆
∆
(2) where ∆t = x1/v0 is the time an electron is inthe electric field between the plates.
Apply Newton’s 2nd law to an electron between the plates to obtain:
y
ma
qE
=
Solve for ay and substitute into
equation (2) to obtain:
m
2 0 1 2 1 1
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
⎟
⎠
⎞
⎜
⎝
⎛
=
∆
v
x
m
qE
y
(3)Express the vertical deflection ∆y2 of
the electrons once they are out of the electric field:
2
2
v
t
y
=
y∆
∆
(4)Use a constant-acceleration equation to find the vertical speed of an electron as it leaves the electric field:
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
+
=
∆
+
=
0 1 1 00
v
x
m
qE
t
a
v
v
y y ySubstitute in equation (4) to obtain:
2 0 2 1 0 2 0 1 2 mv x qEx v x v x m qE
y ⎟⎟=
⎠ ⎞ ⎜⎜ ⎝ ⎛ ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ =
∆ (5)
Substitute equations (3) and (5) in
equation (1) to obtain: 2
0 2 1 2 0 1 2 1
mv
x
qEx
v
x
m
qE
y
⎟⎟
+
⎠
⎞
⎜⎜
⎝
⎛
⎟
⎠
⎞
⎜
⎝
⎛
=
∆
or ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + =∆ 1 2
2 0 1 2 x x mv qEx
y (6)
Use the definition of kinetic energy to find the speed of the electrons:
2 0 2 1
mv
K
=
and(
)
m/s 10 14 . 3 kg 10 9.11 keV 8 . 2 2 2 7 31 0 × = × = = − m K vExpress the electric field between the plates in terms of their potential difference:
d
V
E
=
Substitute numerical values and
evaluate E:
1.2
cm
2
.
08
kV/m
V
25
=
=
=
d
V
E
Substitute numerical values in equation (6) and evaluate ∆y:
