The minimum capacitance of a variable capacitor in a radio is
4.80 pF.
(a) [10 pts] What is the inductance of a coil connected to this
capacitor if the oscillatory frequency is 1600×103 Hz,
corresponding to one end of the AM radio broadcast band, when the capacitor is
set to its minimum capacitance?
w
=
1
Ö
LC
L
=
1
4p2f2C
=
1
4p2(1.60×106 Hz)(4.80× 10-12 F)
=
2.06 mH
(b) [10 pts] The frequency at the other end of the broadcast band is
540×103 Hz. What is the maximum capacitance of the capacitor
if the oscillation frequency is adjustable over the range of the broadcast
band?
C
=
1
w2L
=
1
4p2(5.40×105 Hz)2(2.06 ×10-3 H)
=
42.1 pF
(c) [5 pts] In part (b), if the maximum current in the circuit is
120. mA, what will be the maximum voltage drop across the capacitor when
the capacitor is momentarily fully charged?
Io
=
wQo
=
wCVo
Vo
=
Io
wC
=
0.120 A
2p(5.40×105 Hz) (42.1×10-12 F)
=
839 V
An electromagnetic wave travels through a medium of uniform density and
index of refraction n. The wavefunction for the magnetic field in the wave
is
B(x,t)=(4.80×10-6 T)
^
k
sin[ (1.083×107 rad/m)x+wt]
(a) [4 pts] State the axis along which the wave is traveling and give
the direction of travel.
ANS: The wave travels in the NEGATIVE x-direction.
(b) [6 pts] If the wavelength in air is 900 nm, calculate the index of
refraction of the medium.
nairlair
=
nmedlmed
=
nmed(2p/kmed)
nmed
=
nairlairkmed
2p
=
(1)(9.00×10-7 m)(1.083× 107 rad/m)
2p rad
=
1.55
(c) [4 pts] Calculate the speed of the wave in the medium.
v
=
c
nmed
=
3.00×108 m/s
1.55
=
1.93×108 m/s
(d) [5 pts] Calculate the value of the angular frequency, w.
w
=
vk
=
(1.93×108 m/s)(1.083×107 rad/m )
=
2.09×1015 rad/s
(e) [6 pts] Write the equation for the equivalent wavefunction for
the electric field, E(x,t). Provide numerical values for all of
the constants.
E(x,t)
=
Eo(-
^
j
)sin(kx+wt)
The constants are
Eo
=
vBo
=
(2.09×108 m/s)(4.80×10-6 T )
=
928 V/m
k
=
1.083×107 rad/m
w
=
2.09×1015 rad/s
An intense point source radiates isotropically (uniformly in all
directions). An observer, 10.0 m from the source, measures the amplitude of
the electric field in the radiation to be 1240 V/m.
(a) [7 pts] Calculate the intensity of the radiation at the observer.
I
=
1
2
eocE2o
=
1
2
(8.85×10-12 C/V·m)( 3.00×108 m/s)(1240 V/m)2
=
2040 W/m2
(b) [7 pts] Calculate the total average power output of the source.
P
=
IA
=
4p(2040 W/m2)(10.0 m)2
=
2.57×106 W
(c) [11 pts] If the observer moves to a new position 20.0 m from
the source, calculate electric field amplitude measured by the observer?
The power through a surface stays the same regardless of the distance from the
source.
P1
=
P2
1
2
eocE2o1(4pr21)
=
1
2
eocE2o2(4pr22)
This equation reduces to
Eo2
=
æ è
r1
r2
ö ø
Eo1
=
æ è
10.0 m
20.0 m
ö ø
(1240 V/m)
=
620 V/m
Very short pulses of high-intensity laser beams are used to repair
detached portions of the retina of the eye. The brief pulses of energy
absorbed by the retina weld the detached portions back into place. In one
such procedure, a laser beam has a wavelength of 810 nm and delivers
250 mW of power spread over a circular spot 510 mm in diameter.
The vitreous humor (the transparent fluid that fills most of the eye) has
an index of refraction of 1.34.
(a) [4 pts] If the laser pulses are 1.50 ms long, how much energy is
delivered to the retina with each pulse?
U
=
Pt
=
(0.250 W)(0.00150 s)
=
3.75×10-4 J
(b) [6 pts] What average pressure does the pulse of the laser beam
exert on the retina as it is fully absorbed by the circular spot?
prad
=
I
c
=
P
cA
=
4P
pcd2
=
4(0.250 W)
p(510×10-6 m)2 (3.00×108 m/s)
=
4.08×10-3 Pa
I used c in the intensity equation instead of v. The elecrric field also
changes as the signal goes into the eye. Index is involved. A compensation
occurs. We didn't cover that aspect.
(c) [8 pts] What are the wavelength and frequency of the laser light
inside the vitreous humor of the eye?
lmed
=
nairlair
nmed
=
(1)(810 nm)
1.34
=
604 nm
fmed
=
fair
=
c
lair
=
3.00×108 m/s
810×10-9 m
=
3.70×1014 Hz
(d) [7 pts] What is the electric field amplitude in the laser beam?
I
=
P
A
=
1
2
eocE2o
Eo
=
æ è
8P
eocpd2
ö ø
1/2
=
é ë
8(0.250 W)
(8.85×10-12 C/V·m )(3.00×108 m/s)p(510 ×10-6 m)2
ù û
1/2
=
3.04×104 V/m
CONSTANTS
c=3.00×108 m/s
mo=4p×10-7 T·m/A
eo = 8.85×10-12 C/V·m
m=10-3
m = 10-6
n = 10-9
p = 10-12
FORMULAS
E=vB
I=P/A
I=eocEo2/2
prad=I/c
prad=2I/c
A=pr2
Fav=pradA
E=cB
k=w/c
U=Pt
k=w/v
nl = constant
w = 2pf
A=4pr2
k=2p/l
nv=constant
v=lf
E(x,t)=Eocos(kx-wt)
w = 1/
Ö
LC
Io=wQo
Q=CV
File translated from
TEX
by
TTH,
version 3.85. On 01 May 2016, 17:16.