Tài liệu Physics exercises_solution: Chapter 32 pptx

32.1: a)
.s28.1
sm1000.3
m1084.3
8
8




c
d
t
b) Light travel time is:
s1072.2
)hour1(
)s3600(
)day1(
)hours24(
)year1(
)days365(
)years61.8(years61.8
8

km.108.16m108.16s)10(2.72s)m100.3(
131688
 ctd
32.2:
.m180)s100.6()sm100.3(
78



tcd
32.3:
jjB
ˆ
2cos
ˆ
cos()
maxmax















t
c
z
fB
ωt)kzB(z,t


jB
ˆ
s)m1000.3(
Hz)1010.6(2cos)T1080.5()(
8
144




















t
z
z,t



.
ˆ
))srad1083.3()m1028.1((cos)T1080.5()(
15174
jB tz,tz 


)
ˆ
()
ˆ
()(
kjE c(z,t)Bz,t
y


.
ˆ
s)rad1083.3()m1028.1((cosm)V1074.1()(
15175
iE t)zz,t 

32.4: a)
.Hz1090.6
m1035.4
sm1000.3
λ
14

7
8





c
f
b)
.T1000.9
sm1000.3
mV1070.2
12
8
3
max
max






c
E
B
c) The electric field is in the
x
-direction, and the wave is propagating in the z-

direction. So the magnetic field is in the y-direction, since
.BES 
Thus:
.
ˆ
))srad1034.4()m1045.1cos(()mV1070.2()(
ˆ
sm1000.3
)Hz1090.6(2cosV/m)1070.2(),(
ˆ
2cos
ˆ
(cos),(
15173
8
143
maxmax
iE
iE
iiE
tzz,t
z
t
πtz
t
c
z
πfEωt)kzEtz






































And
.
ˆ
))srad1034.4()m1045.1cos(()T1000.9(
ˆ
),(
),(
151712
jjB tz
c
tzE
tz




32.5: a)
y

direction.
b)
m.1011.7
)srad1065.2(
)sm1000.3(22
λ
λ

2
2
4
12
8






ω
πcπc
πfω
c) Since the electric field is in the
z

-direction, and the wave is propagating in the
y

-direction, then the magnetic field is in the
x

-direction
).( BES 
So:

iiiB
ˆ
)sin(

ˆ
)sin(
ˆ
),(
),(
00
ty
c
c
E
tky
c
E
c
tyE
ty










iB
ˆ
)srad1065.2(
)sm1000.3(

)srad1065.2(
sin
sm1000.3
mV1010.3
),(
12
8
12
8
5























tyty
.
ˆ
))srad1065.2()m1083.8sin(()T1003.1(),
1233
i(B tyty 

32.6: a)
x

direction.
b)
.Hz1059.6
2
)sm100.3()mrad1038.1(
2
22
11
84



π
π
kc
f
c
fπ

λ
π
k
c) Since the magnetic field is in the
y

-direction, and the wave is propagating in the
x

-direction, then the electric field is in the
z

-direction
).( BES 
So:
.
ˆ
))srad1014.4()mrad1038.1sin(()mV48.2(),(
ˆ
))srad1014.4()mrad1038.1sin(())T1025.3((),(
ˆ
)2sin(
ˆ
),(),(
124
1249
0
kE
kE
kkE

txty
txctx
fkxcBtxcBtx





32.7: a)
m.361
Hz1030.8
sm1000.3
λ
5
8




f
c
b)
1
m0174.0
m361
2
λ
2



ππ
k
c)
s.rad1021.5)Hz1030.8(22
65
 πfω


.mV0145.0)T1082.4()sm1000.3(
118
maxmax


cBE
32.8:
.T1028.1
sm1000.3
mV1085.3
11
8
3
max
max






c

E
B
So
,1056.2
T105
T1028.1
7
5
11
earth
max







B
B
and thus
max
B
is much weaker than
.
earth
B
32.9:
BEBE
KK

cB
KK
BB
vBE

00


.mV779.0
)23.1()74.1(
)T1080.3()sm1000.3(
98




E
32.10: a)
.sm1091.6
)18.5()64.3(
s)m1000.3(
7
8



BE
KK
c
v

b)
.m1006.1
Hz0.65
sm1091.6
λ
6
7



f
v
c)
.T1004.1
sm1091.6
mV1020.7
10
7
3






v
E
B
d)
.mW1075.5

)18.5(2
)T1004.1)(mV1020.7(
2
28
0
103
0






B
K
EB
I
32.11: a)
.m1081.3
Hz1070.5
sm1017.2
λ
7
14
8






f
v
b)
.m1026.5
Hz1070.5
sm1000.3
λ
7
14
8
0





f
c
c)
.38.1
sm1017.2
sm1000.3
8
8




v
c

n
d)
.90.1)38.1(
22
2
2
 n
v
c
K
K
c
v
E
E
32.12: a)
s.m1034.2)m15.6)(Hz1080.3(λ
87
 fv
b)
.64.1
)sm1034.2(
)sm1000.3(
28
28
2
2





v
c
K
E
32.13: a)
25
max
2
max0
mW101.1som,V090.0;
2
1

 IEcEI

b)
T100.3so
10
maxmaxmaxmax

 cEBcBE
c)
W840)m105.2()4()mW10075.1()4(
23252
av



rIP

d) Calculation in part (c) assumes that the transmitter emits uniformly in all directions.
32.14:
The intensity of the electromagnetic wave is given by Eqn. 32.29:
.
2
rms0
2
max0
2
1
cEcEI


Thus the total energy passing through a window of area
A
during
a time
t
is
J
9.15)s0.30)(m500.0()mV0200.0()sm1000.3()mF1085.8(
228122
rms0
μAtcE 


32.15:
J105.2)m100.2()4()mW100.5()4(
25210232
av



rIP
32.16:
a) The average power from the beam is
W104.2)m100.3()mW800.0(
4242 
 IAP
b) We have, using Eq. 32.29,
.
2
rms0
2
max0
2
1
cEcEI


Thus,
mV4.17
s)m1000.3)(mF1085.8(
mW800.0
812
2
0
rms





c
I
E

32.17:
cIp 
rad
so
23
rad
mW1070.2  cpI
Then
W105.8)m0.5()4()mW1070.2()4(
52232


rIP
av
32.18: a)
Hz.1047.8
m354.0
sm1000.3
λ
8
8



c

f
b)
.T1080.1
sm1000.3
mV0540.0
10
8
max
max




c
E
B
c)
.mW1087.3
2
)T1080.1()mV0540.0(
2
26
0
10
0







EB
SI
av
32.19:
2
0
max
2
0
2
max
2
)4(
2
r
Pc
Er
c
E
ASP
av






m.V0.12
)m00.5(2

)sm1000.3()W0.60(
2
0
8
max





E

.T1000.4
sm1000.3
mV0.12
8
8
max
max




c
E
B
32.20: a) The electric field is in the
y

-direction, and the magnetic filed is in the

z

-
direction, so
.
ˆ
ˆ
)
ˆ
(
ˆˆ
ˆ
ikjBES 
That is, the Poynting vector is in the
x

-
direction.
b)
)(cos
),(),(
),(
0
maxmax
0
tkx
BE
txBtxE
txS





))).(2cos(1(
2
0
maxmax
kxt
BE



But over one period, the cosine function averages to zero, so we have:
|
.
2
|
0
maxmax

BE
S
av

32.21: a) The momentum density
s.mkg107.8
s)m100.3(
mW780
215
28

2
2




c
S
dV
dp
av
b) The momentum flow rate
Pa.106.2
sm100.3
mW780
1
6
8
2




c
S
dt
dp
A
av
32.22: a) Absorbed light:

.Pa1033.8
sm100.3
mW2500
1
6
8
2
rad




c
S
dt
dp
A
p
av

.atm1023.8
atmPa10013.1
Pa1033.8
11
5
6
rad







p
b) Reflecting light:
.Pa1067.1
sm100.3
)mW2500(2
2
1
5
8
2
rad




c
S
dt
dp
A
p
av

10
5
5
rad

1065.1
atmPa101.013
Pa1067.1






p
atm. The factor of 2 arises because the
momentum vector totally reverses direction upon reflection. Thus the change in
momentum is twice the original momentum.
c) The momentum density
s.mkg1078.2
s)m100.3(
mW2500
214
28
2
2




c
S
dV
dp
av

32.23:
 EBEBc
c
E
EcEES
0
0
00
0
0
0
0
2
0
0
2
00
0
1












.
2
0
0
2
0
cE
c
EEB



32.24: Recall that
:so,

 BES
a)
.
ˆ
)
ˆ
(
ˆ
ˆ
kjiS 
b)
.
ˆˆˆˆ
kijS 
c)

.j)i()k(S
ˆ
ˆ
ˆ
ˆ

d)
.
ˆ
ˆ
ˆ
ˆ
j)k(iS 
32.25:
T1033.1
8
maxmax

 cEB

 BE
is in the direction of propagation. For
E

in the +
x
-direction,
B
E




is in the +
z
-
direction when

B
is in the +
y
-direction.
32.26: a)
.m00.2
)Hz100.75(2
sm1000.3
22
λ
6
8




f
c
x
b) The distance between the electric and magnetic nodal planes is one-quarter of a
wavelength =
m.00.1
2

m00.2
2
4
λ



x
32.27: a) The node-antinode distance
m.1038.4
Hz)1020.1(4
sm1010.2
44
λ
3
10
8





f
v
b) The distance between the electric and magnetic antinodes is one-quarter of a
wavelength
m.1038.4
Hz)10(1.204
sm102.10
44

λ
3
10
8





f
v
c) The distance between the electric and magnetic nodes is also one-quarter of a
wavelength

m.1038.4
Hz)104(1.20
m/s102.10
44
3
10
8





f
vλ
32.28:
.cm0.20m200.0

)Hz1050.7(2
sm1000.3
22
λ
8
8
nodes




f
c
x
There must be nodes
at the planes, which are 80.0 cm apart, and there are two nodes between the planes, each
20.0 cm from a plane. It is at 20 cm, 40 cm, and 60 cm that a point charge will remain at
rest, since the electric fields there are zero.
32.29: a)
mm.7.10mm)55.3(22λ
2
λ
 xx
b)
.mm55.3




BE

xx
c)
s.m1056.1)m1010.7()Hz1020.2(λ
8310


fv
32.30: a)
)sincos2()sinsin2(
),(
maxmax
2
2
2
2
ωtkxkE
x
ωtkxE
x
x
txE
y











.
),(
sinsin2sinsin2
),(
2
2
00max
2
2
max
2
2
2
t
txE
tkxE
c
tkxEk
x
txE
yy










Similarly:
)cossin2()coscos2(
),(
maxmax
2
2
2
2
tkxkB
x
tkxB
x
x
txB
z










.
),(
coscos2coscos2

),(
2
2
00max
2
2
max
2
2
2
t
txB
tkxB
c
tkxBk
x
txB
zz









b)
tkxkEtkxE
x

x
txE
y

sincos2)sinsin2(
),(
maxmax







.
),(
)coscos2(
),(
.sincos2sincos2sincos2
),(
max
max
max
max
t
txB
tkxB
t
x
txE

ωtkxBtkx
c
E
tkxE
cx
txE
z
y
y
















Similarly:
tkxkBtkxB
x
x
txB

z

cossin2)coscos2(
),(
maxmax








tkxcB
c
tkxB
c
x
txB
z




cossin2cossin2
),(
max
2
max





,(
)sinsin2(cossin2
),(
00max00max00
t
t
xE
tkxE
t
tkxE
x
txB
y
z










32.31: a) Gamma rays:
nm.104.62m1062.4
Hz106.50

sm103.00
λ
514
21
8





f
c
b)
.nm522m105.22
Hz1075.5
sm1000.3
λ:lightGreen
7
14
8





f
c
32.32: a)
Hz.106.0
m5000

sm103.0
λ
4
8



c
f
b)
Hz.106.0
m5.0
sm103.0
λ
7
8



c
f
c)
Hz.106.0
m105.0
sm103.0
λ
13
6
8






c
f
d)
Hz.106.0
m
10
5.0
sm103.0
λ
16
9
8





c
f
32.33:
Using a Gaussian surface such that the front surface is ahead of the wave front (no
electric or magnetic fields) and the back face is behind the wave front (as shown at right),
we have:

 .00
0

encl
xx
E
ε
Q
Ed
AAE



 .00
xx
BABd AB


So the wave must be transverse, since there are no components of the electric or
magnetic field in the direction of propagation.
32.34: Assume
.with),sin(
ˆ
)sin(
ˆ
maxmax



tkxBandtkxE kBjE
Then Eq. (32.12) implies:
.0)cos()cos(
maxmax








tkxBtkxkEx
t
B
x
E
z
y
.λ
λ
/
2
2
maxmaxmaxmaxmaxmaxmax
cBBfB
f
B
k
EBkE 




Similarly for Eq.(32.14)

.0)cos()cos(
max00max00








tkxEtkxkB
t
E
x
B
y
z

.
1
/
2
2
maxmax
2
max
2
max
00

maxmax00max
E
c
E
c
fλ
E
λ
c
f
E
k
BEkB 






32.35: From Eq. (32.12):
2
2
),(),(
),(
t
txB
t
txB
tx
txE

t
zz
y




























But also from Eq. (32.14):

























t
txE
xx
txB

x
y
z
),(
),(
00


2
2
00
),(
t
txB
z





.
),(),(
2
2
00
2
2
t
txB
x

txB
zz







32.36:
)cos(
2
1
2
1
)cos(),(
2
max0
2
0max
tkxEEutkxEtxE
Ey



B
z
E
u
B

tkxBtkx
c
Ec
u








0
2
2
max
0
2
max
2
0
2
)cos(
2
1
)cos(
2






32.37: a) The energy incident on the mirror is
AtcEIAtPt
2
0
2
1


J.105.20s)(1.00)m1000.5()mV028.0()sm1000.3(
2
1
102428
0



E
b) The radiation pressure
.Pa1094.6)mV0280.0(
2
152
0
2
0rad



E

c
I
p
c) Power
2
rad
2
24 RcpRIP


W.101.34m)(3.20Pa)10(6.94s)m1000.3(2
42158 


P
32.38: a)
.Hz1081.7
m0384.0
sm1000.3
λ
9
8



c
f
b)
T.1050.4
sm1000.3

mV1.35
9
8
max
max




c
E
B
c)
.mW102.42m)V(1.35)sm1000.3(
2
1
2
1
2328
0
2
max0



cEI
d)
.N1093.1
s)m1000.3(2
)m(0.240T)10(4.50m)V35.1(

2
12
8
0
29
0







c
EBA
c
IA
pAF
32.39: a) The laser intensity
.mW652
m)10(2.50
W)104(3.204
2
23
3
2







πD
P
A
P
I

But
m.V701
)sm10(3.00
)mW2(652
2
2
1
8
0
2
0
2
0






c
I
EcEI

And
T.102.34
sm103.00
mV701
6
8




c
E
B
b)
.mJ101.09m)V(701
4
1
4
1
362
0
2
max0



Euu
avav
EB
Note the extra factor

of
2
1
since we are averaging.
c) In one meter of the laser beam, the total energy is:
42)(2Vol
2
tottot
LDuALuuE
EE


J.101.07m)/4(1.00m)1050.2()mJ1009.1(2
112336
tot



E
32.40: a) The change in the momentum vector determines
.
rad
p
If
W
is the fraction
absorbed,
.)2()()1(
inout
pWppW 


PPP
Here,
)1( W

is the fraction
reflected. The positive direction was chosen in the direction of reflection. p is the
magnitude of the incoming momentum. With Eq. 32.31, and taking the average, we
get
.)2(
rad
C
I
Wp 
Be careful not to confuse p, the momentum of the incoming wave,
with
,
rad
p
the radiation pressure.
b) (i) totally absorbing
C
I
pW

rad
so1
(ii) totally reflecting
C
pW

2
so0
rad

These are just equations 32.32 and 32.33.
c)
Pa1013.5
1000.3
)mW1040.1()9.02(
W/m1040.1,9.0
6
s
m
8
23
rad
22





pIW
Pa1087.8
1000.3
)mW1040.1()1.02(
mW1040.1,1.0
6
s
m

8
22
rad
23





pIW
32.41:
a) At the sun’s surface:
.Pa21.0
sm1000.3
mW104.6
mW104.6
m)1096.6(4
W109.3
4
8
27
rad
27
28
26
2









c
I
p
R
P
A
P
IIAP

Halfway out from the sun’s center, the intensity is 4 times more intense, and so is the
radiation pressure:
Pa.85.0)2/(
sunrad
Rp
At the top of the earth’s atmosphere, the measured sunlight intensity is

2
mW1400
,Pa105
6

which is about 100,000 times less than the values above.
b) The gas pressure at the sun’s surface is 50,000 times greater than the radiation
pressure, and halfway out of the sun the gas pressure is believed to be about 6
13
10

times greater than the radiation, pressure. Therefore it is reasonable to ignore radiation
pressure when modeling the sun’s interior structure.
32.42: a)
,1)(2cos0),(
ˆ
))(2cos1(
2
),(
0
maxmax
 tkxtxStkx
BE
tx


iS
which never happens. So the Poynting vector is always positive, which makes sense since
the direction of wave propagation by definition is the direction of energy flow.
b)
32.43: a)
.
000
dt
di
nAA
dt
dB
dt
d
dt

di
n
dt
dB
niB
B




So,




dt
di
rn
dt
di
nArE
dt
d
d
B
2
00
2

lE


.
2
0
dt
di
nr
E


b) The direction of the Poynting vector is radially inward, since the magnetic field is
along the solenoid’s axis and the electric filed is circumferential. It’s magnitude
.
2
2
0
0
dt
di
rin
EB
S



c)
.
2
)(
22

)(
2
222
0
2
22
0
0
2
0
0
2
lain
aullAuU
inni
B
u







But also
,
2
2
22
0

222
0
2
ilan
i
lain
i
U
Li
Li
U



and so the rate of
energy increase due to the increasing current is given by
.
22
0
dt
di
ilan
dt
di
LiP


d) The in-flow of electromagnetic energy through a cylindrical surface located at the
solenoid coils is


 .2
2
2
22
0
2
0
dt
di
ilanal
dt
di
ain
πalSd


AS


e) The values from parts (c) and (d) are identical for the flow of energy, and hence we
can consider the energy stored in a current carrying solenoid as having entered through its
cylindrical walls while the current was attaining its steady-state value.
32.44:
a) The energy density, as a function of x, for the equations for the electrical and
magnetic fields of Eqs. (32.34) and (32.35) is given by:
tkxEEu

sinsin4
22
max0

2
0

b) At
.
2
1
4
sinsinand
2
1
4
coscos,
4






ωttt
For
.
ˆˆˆ
ˆˆ
ˆ
0cos,0sin,
2
0 ikjBES  kxkx
k

x

And for
.
ˆˆˆ
ˆˆ
ˆ
0cos,0sin,
2
ikjBES  kxkx
k
x
k



2
1
4
3
sinsinand
2
1
4
3
coscos,
4
3
At








ttt
.
For 0
.
ˆˆˆ
ˆˆ
ˆ
0cos,0sin,
2
ikjBES  kxkx
k
x

And for
.
ˆˆˆ
ˆˆ
ˆ
0cos,0sin,
2
ikjBES  kxkx
k
x
k



c) the plots from part (a) can be interpreted as two waves passing through each other
in opposite directions, adding constructively at
certain times, and destructively at others.
32.45: a)
2
a
I
A
I
JE





, in the direction of the current.
b)

 ,
2
0
0
a
I
BId




lB
counterclockwise when looking into the current.
c) The direction of the Poynting vector
,
ˆ
ˆ
ˆ
ˆˆ
ˆ
ρkBES 

where we have used
cylindrical coordinates, with the current in the z-direction.
Its magnitude is
32
2
0
2
00
22
1
a
ρI
a
I
a
ρIEB
S





.
d) Over a length l, the rate of energy flowing in is
.2
2
2
2
32
2
a
lI
al
a
I
SA






The thermal power loss is
,
2
2
22
a
lI
A

l
IRI



which exactly equals the flow of
electromagnetic energy.
32.46:
,and,
2
2
00
0
r
q
E
q
EAd
r
i
B
S




AE


so the magnitude of the

Poynting vector is
.
22
32
0
32
0
0
dt
dq
r
q
r
qiEB
S



Now, the rate of energy flow into the region between the plates is:





















.
22
1)(
2
1
)2(
2
2
0
2
2
0
2
0
dt
dU
C
q
dt
d
q

A
l
dt
d
dt
qd
r
l
dt
dq
r
lq
rlSd


AS


This is just rate of increase in electrostatic energy U stored in the capacitor.
32.47: The power from the antenna is
.4
2
2
0
2
max
r
cB
IAP




So
T1042.2
)sm1000.3()m2500(4
)W1050.5(2
4
2
9
82
4
0
2
0
max









cr
P
B

sT44.1)T10(2.42Hz)1050.9(22
97

maxmax



fBB
dt
dB
V.0366.0
4
)sT44.1()m180.0(
4
22





dt
dBD
dt
dB
A
dt
d
32.48:
.mV242
)sm1000.3(
)m36W1080.2(22
2
1

8
0
23
0
2
0






c
I
EcE
A
P
I
32.49:
a) Find the force on you due to the momentum carried off by the light:
ApFcIp
radrad
and 
gives
cPcAIF /
av

298
av
sm1044.4)]sm10kg)(3.00150[()W200()(


 mcPmFa
x
Then


)sm1044.4()m0.16(2)(2gives
2
9
0
21
200
xxx
axxttatvxx
h6.23s1049.8
4

The radiation force is very small. In the calculation we have ignored any other forces on
you.
b) You could throw the flashlight in the direction away from the ship. By conservation
of linear momentum you would move toward the ship with the same magnitude of
momentum as you gave the flashlight.
32.50:
cA
Vi
cA
P
EcE
A
P

IIAP
00
2
0
22
2
1




.mV106.14
)sm10(3.00)m(100
A)(1000V)102(5.002
4
8
0
2
5
0




εcA
Vi
E

And
.T102.05

sm103.00
mV106.14
4
8
4




c
E
B
32.51: a)
.
3
4
3
4
.
2
3
3
22
r
ρRGM
ρR
r
GM
r
mGM

F
SSS
G



b) Assuming that the sun’s radiation is intercepted by the particle’s cross-section, we
can write the force on the particle as:
.
4
.
4
2
22
2
cr
LR
c
R
r
L
c
IA
F



c) So if the force of gravity and the force from the radiat
ion pressure on a particle from
the sun are equal, we can solve for the particle’s radius:

m.109.1
)sm10(3.0)mkg(3000kg)10(2.0)kgmN10(6.716
)W109.3(3
.
16
3
43
4
7
83302211
26
2
2
2
3







R
π
R
cGM
L
R
cr
LR

r
ρRGM
FF
S
S
G


d) If the particle has a radius smaller than that found in part (c), then the radiation
pressure overcomes the gravitational force and results in an acceleration away from the
sun, thus removing all such particles from the solar system.
32.52:
a) The momentum transfer is always greatest when reflecting surfaces are used
(consider a ball colliding with a wallthe wall exerts a greater force if the ball rebounds
rather than sticks). So in solar sailing one would want to use a reflecting sail.
b) The equation for repulsion comes from balancing the gravitational force and the
force from the radiation pressure. As seen in Problem 32.51, the latter is:
2
2
2
226
26
8302211
22
rad
2
rad
mi2.53
)milekm1.6(
km6.48

km6.48m1048.6
W10(2)3.9
)sm10(3.0kg)(10000kg)10(2.0)kgmN10(6.74
2
4
4
2
:Thus.
4
2






A
A
L
mcGM
A
cr
LA
r
mGM
FF
cr
LA
F
SS

G



c) This answer is independent of the distance from the sun since both the gravitational
force and the radiation pressure go down like one over the distance squared, and thus the
distance cancels out of the problem.
32.53: a)
.W
s
J
s
Nm
)sm()mN(
)sm(
6
322
2
2
2
3
0
22
















dt
dE
C
C
c
aq

b) For a proton moving in a circle, the acceleration can be rewritten:
.sm101.53
m)(0.75kg)10(1.67
)eVJ10(1.6eV)102(6.00
2
15
27
196
2
1
2
2
1
2







mR
mv
R
v
a
The rate at which it emits energy because of its acceleration is:
s.eV1032.8
sJ1033.1
s)m10(3.06
)sm10(1.53C)10(1.6
6
5
23
38
0
2215219
3
0
22










c
aq
dt
dE
So the fraction of its energy that it radiates every second is:
.1039.1
eV106.00
eV108.32s)1)((
11
6
5






E
dtdE
c)
Carrying out the same calculations as in part (b), but now for an electron at the same
speed and radius. That means the electron’s acceleration is the same as the proton, and
thus so is the rate at which it emits energy, since they also have the same charge.
However, the electron’s initial energy differs from the proton’s by the ratio of their
masses:
eV.3273
kg)10(1.67

kg)10(9.11
eV)1000.6(
27
31
6






p
e
pe
m
m
EE
So the fraction of its energy that it radiates every second is:
.1054.2
eV3273
eV108.32s)(1)(
8
5





E
dtdE

32.54:
For the electron in the classical hydrogen atom, its acceleration is:

.s/m109.03
m)10(5.29kg)10(9.11
)eVJ10(1.60eV)2(13.6
222
1131
19
2
1
2
2
1
2






mR
mv
R
v
a
Then using the formula for the rate of energy emission given in Pr. (33-49):
38
0
2222219

3
0
22
s)m10(3.00
)sm10(9.03C)10(1.60
6





6πc
aq
dt
dE
s,eV102.89sJ1064.4
118


dt
dE
which means that the electron would almost
immediately lose all its energy!
32.55: a)
).(sine),(
max
txkEtxE
c
xk
y

c




)cos(e)()sin(e)(
).cos(e)()sin(e)(
2
max
2
max
2
2
maxmax
txkkEtxkkE
x
E
txkkEtxkkE
x
E
c
xk
cc
xk
c
y
c
xk
cc
xk

c
y
cc
cc











).cos(e
).cos(e2
)sin(e)()cos(e)(
max
2
max
2
max
2
max
txkE
t
E
txkkE
txkkEtxkkE

c
xk
y
c
xk
c
c
xk
cc
xk
c
c
c
cc












Set
).cos(e)cos(e2
max
2

max
2
2
txkωEtxkkE
t
E
x
E
c
xk
c
xk
c
yy
cc










This will
only be true if
.
2
or

2
2






c
c
k
k
b) The hint basically answers the question.
c)
m.1060.6
Hz)10(1.02
m)1072.1(221
,1
e
5
0
6
8
0









c
c
y
y
k
xxk
E
E