Solutions of HW7
P29 .1
Ó Ó
FB = q v ‰ B. Consider a three dimensional coordinate system with the xy - plane in the plane of this page,
the + x - direction toward the right edge of the page and the + y direction toward the top of the page. Then,
the z - axis is perpendicular to the page with the + z - direction being upward,
out of the page. The magnitude field is directed in the + x - direction, toward the right.
HaL When a proton Hpositively chargedL moves in the + y - direction, the right hand rule gives the direction of the magnitude force as into the page or in the - z - direction.
HbL With velocity in the - y - direction,
the right - hand rule gives the direction of the force on the proton as out of the page,
in the + z - direction.
HcL When the proton moves in the + x - direction, parallel to he magnetic field,
the magnitude of the magnetic force it experiences is F =
qvB sin H0L = 0. The magnitude force is zero in this case
P29 .7
First find the speed of the electron.
1
HDK + DULi = HDK + DULf
”
mv2 = eDV :
2
2 eDV
v=
=
m
2 I1.60 ´ 10-19 CM H2400 J  CL
9.11 ´ 10-31 kg
= 2.90 ´ 107 m  s
HaL FB,max = qvB = I1.60 ´ 10-19 CM I2.90 ´ 107 m  sM H1.70 TL = 7.90 ´ 10-12 N
Ó
Ó
HbL FB,min = 0 occurs when v is either parallel to or anti - parallel to B
P29 .8
Ó Ó
FB = q v ‰ B
` `
i j
Ó Ó
v‰B =
2 -4
1 2
Ó Ó
v‰B
FB
=
= q
`
k
1
-3
`
`
`
`
`
`
= H12 - 2L i + H1 + 6L j + H4 + 4L k = 10 i + 7 j + 8 k
102 + 72 + 82 = 14.6 Tm  s
Ó Ó
v ‰ B = 2.34 ´ 10-18 N
P29 .12
HaL The magnitude force acting on the electron provides the centripetal acceleration ,
holding the electron in the circular path. Therefore, F = q vB sin90 o = me v2 ‘ r,
2
HW7.nb
me v
r=
=
I9.11 ´ 10-31 kgM I1.50 ´ 107 m  sM
eB
I1.60 ´ 10-19 CM I2.00 ´ 10-3 TM
= 0.0427 m = 4.27 cm
HbL The time to complete one revolution around the orbit Hi.e., the periodL is
2 Πr
distance traveled
T=
=
constant speed
=
v
2 Π H0.0427 mL
1.50 ´ 107 m  s
= 1.79 ´ 10-8 s
P29 .18
HaL The magnetic force provides the centripetal force to keep the particle moving on a circle,
S F = ma ” qvB sin90 o =
mv2
R
and the kinetic energy of the particle is
1
K=
mv2
2
Both equations have the same term mv2 in common :
From the first equation, mv2 = qvBR, and from the second one, mv2 = 2 K.
Setting these equal to each other gives
mv2 = qvBR = 2 K
2K
v=
qBR
HbL From the above , we have m =
qBR
m=
qBR
= qBR
v
=
2K
qBR
v
. Using our result from HaL, we get
HqBRL2
2K
P29 .22
HaL The boundary between a region of strong magnetic field and a
region of zero field cannot be perfectly sharp, but we ignore the thickness
of the transition zone. In the field the electron moves on an arc of a circle :
S F = ma ”
vB sin90 o =
q
mv2
:
r
v
q
B
=Ω=
r
=
m
I1.60 ´ 10-19 CM I10-3 Ns  CmM
9.11 ´ 10-31 kg
= 1.76 ´ 108 rad  s
The time for one half revolution is, from DΘ = ΩDt,
DΘ
Π rad
= 1.79 ´ 10-8 s
1.76 ´ 108 rad  s
HbL The maximum depth of penetration is the radius of the
magnetic force cannnot alter the kinetic energy of the
Dt =
=
Ω
HW7.nb
path. The magnetic force cannnot alter the kinetic energy of the electron.
Then v = Ωr = I1.76 ´ 108 s-1 M H0.02 mL = 3.51 ´ 106 m  s
1
and K =
mv2 =
2
I9.11 ´ 10-31 kgM I3.51 ´ 106 m  sM = 5.62 ´ 10-18 J =
1
2
2
5.62 ´ 10-18 J × e
1.60 ´ 10-19 C
= 35.1 eV
P29 .24
HaL FB = qvB =
v
Ω=
qBR
=
R
HbL v =
mv2
R
qB
=
mR
qBR
m
m
= 4.31 ´ 107 rad  s
= 5.17 ´ 107 m  s
P29 .25
FB = Fe
2K
so qvB = qE, where v =
and K is kinetic energy of the electron.
m
2K
E = vB =
B=
m
2 H750L I1.60 ´ 10-19 M
H0.015L = 244 kV  m
9.11 ´ 10-31
P29 .31
HaL From F = BIL sinΘ, the magnetic field is
HF  LL
0.120 N  m
B=
=
= 8 ´ 10-3 T
IsinΘ
H15.0 AL sin90 o
HbL The magnetic field must be in the + z - direction to produce a force in the y - direction when the current is in the + x - direction
P29 .35
HaL The magnetic force must be upward to lift the wire. For current in the south direction,
the magnetic field must be east to produce an upward force,
which is given by the right - hand rule
HbL FB = ILB sinΘ with FB = Fg = mg
mg = ILB sinΘ ” mg  L = IB sin Θ ” B =
m
g
×
L
I sin Θ
3
4
HW7.nb
0.500 ´ 10-3 kg
B=
1.00 ´ 10-2 m
9.80 m ‘ s2
H2.00 AL sin90 o
= 0.245 T
P29 .44
HaL 2 Πr = 2.00 m
so r = 0.318 m
Μ = IA = I IΠr2 M = I17.0 ´ 10-3 AM IΠ H0.318L2 m2 M = 5.41 mA × m2
Ó Ó Ó
HbL Τ = Μ ‰ B
so Τ = I5.41 ´ 10-3 A × m2 M H0.800 TL = 4.33 mN × m
P29 .47
Τ = NBAI sinΦ
Τ = H100 NL H0.8 TL I0.4 ´ 0.3 m2 M H1.2 AL sin 60o = 9.98 N × m
Ó
Note that Φ is the angle between the magnetic moment and the B field. The
Ó
loop will rotate so as to align the magnetic moment with the B field,
clockwise as seen looking down from a position on the positive y axis.