Physics 42 Chapter 29 Homework

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Physics 42 Last Homework Fall 2012: Partial Solutions
1. A piece of copper wire with thin insulation, 200 m long and 1.00 mm in diameter, is wound onto a plastic
tube to form a long solenoid. This coil has a circular cross section and consists of tightly wound turns in one
layer. If the current in the solenoid drops linearly from 1.80 A to zero in 0.120 seconds, an emf of 80.0 mV is
induced in the coil. What is the length of the solenoid, measured along its axis?
We can directly find the self inductance of the solenoid:
dI
0 − 1.8 A
+0.08 V =
−L
ε = −L
dt
0.12 s
2
µ0 N A
.
L=
5.33 × 10−3 Vs A =

(
)
Here A = π r 2 , 200 m = N 2π r , and  = N 10−3 m . Eliminating extra unknowns step by step, we have
−3
5.33 × 10
=

µ0 40 000 m 2 10
µ N 2π  200 m 
= 0
=
=
  2π N 
4π 
µ0 N 2π r 2
Vs A =
−7
2

( 40 000 m ) Tm
2

A
−3
4 × 10 WbmA
=
5.33 × 10−3 A Vs
0.750 m
2. A self-induced emf in a solenoid of inductance L changes in time as
ε=εe
. Find the total charge that
0 – kt
passes through the solenoid, assuming the charge is finite.
ε = ε 0 e− kt = − L
dI
dt
dI = −
ε0
L
e− kt dt
=
I
If we require I → 0 as t → ∞ , the solution is
Q=
ε 0 − kt dq
=
e
kL
dt
∫ Idt =
∞
ε0
− kt
∫ kL e
dt = −
0
3. Consider the circuit shown, taking
ε0
2
k L
Q =
ε0
k2L
.
ε = 6.00 V, L = 8.00 mH, and R = 4.00 Ω.
(a) What is the inductive time constant of the circuit?
(b) Calculate the current in the circuit 250 μs after the switch is closed.
(c) What is the value of the final steady-state current?
(d) Find the energy stored in the magnetic field when the current reaches half
its maximum value.
(a)
L
R
τ ==
2.00 × 10−3 s =
2.00 ms
(
)
(
)
(b)
 6.00 V 
1 − e−0.250 2.00 =
I = I max 1 − e− t τ = 
 4.00 Ω 
(c)
I max=
(d)
ε
6.00 V
=
=
R 4.00 Ω
− t 2.00 ms
0.800 =1 − e
0.176 A
1.50 A
FIG. P32.17
→ t =− ( 2.00 ms) ln ( 0.200) = 3.22 ms
4. An LC circuit like the one shown contains an 82.0-mH inductor and a 17.0-μF
capacitor that initially carries a 180-μC charge. The switch is open for t < 0 and then
closed at t = 0. (a) Find the frequency (in hertz) of the resulting oscillations.
At t = 1.00 ms, find (b) the charge on the capacitor and (c) the current in the circuit.
d) If a resistance of 1.00 kΩ is introduced into the circuit, what is the frequency of the
(damped) oscillations?
(a)
=
f
1
=
2π LC 2π
1
( 0.082 0 H ) (17.0 × 10−6 F)
=
135 H z
(180 µ C) cos( 847 × 0.001 00=)
119 µ C
(b)
Q
t
= Qmax cosω=
(c)
dQ
=
−ω Qmax sin ω t =
− ( 847)( 180) sin ( 0.847) =
−114 mA
I=
dt
5. MRI (Magnetic resonance imaging) is a medical technique that produces detailed ‘pictures’ of the interior of
the body. The patient is placed into a solenoid that is 40.0cm in diameter and 1.00 m long. A 100. A current
creates 5.00T magnetic field inside the solenoid. To carry such a large current, the solenoid wires are cooled
with liquid helium unti they become superconducting.
a. How much magnetic energy is stored in the solenoid? Assume the magnetic field is uniform inside the
solenoid and approximately zero outside. b. How many turns of wire does the solenoid have? c. Look online
or wherever and find out the actual imaging works. How do they get an image of the body with magnetic
fields? Just be brief.
Model: Assume the solenoid is long enough that we can approximate the field as being constant inside and zero outside.
Visualize: The energy density depends on the field strength in a particular region of space.
Solve: (a) We can use the given field strength to find the energy density and then determine the volume of a cylinder. We have
1 2
( 5.0 T ) = 9.95 × 106 J/m3
B
=
2µ0
2 ( 4π × 10−7 T m/A )
2
uB
=
Vsol = π rsol2 l = π ( 0.20 m ) (1.0 m ) = 0.126 m3 ⇒ U tot = uBVsol = ( 9.95 × 106 J/m3 )( 0.126 m3 ) = 1.25 × 106 J
2
(b) Using the magnetic field of the solenoid, the number of turns is calculated as follows:
B=
µ0 NI
l
⇒N=
lB
µ0 I
=
(1.0 m )( 5.0 T )
( 4π ×10−7 T m/A ) (100 A )
= 4.0 × 104
Assess: This is a reasonable number for a solenoid that is a meter long.
c. Google: MRI machines make use of the fact that body tissue contains lots of water, and hence protons (1H nuclei),
which get aligned in a large magnetic field.[4] Each water molecule has two hydrogen nuclei or protons. When a person is
inside the powerful magnetic field of the scanner, the average magnetic moment of many protons becomes aligned with the
direction of the field. A radio frequency current is briefly turned on, producing a varying electromagnetic field. This
electromagnetic field has just the right frequency, known as the resonance frequency, to be absorbed and flip the spin of the
protons in the magnetic field. After the electromagnetic field is turned off, the spins of the protons return to thermodynamic
equilibrium and the bulk magnetization becomes re-aligned with the static magnetic field. During this relaxation, a radio
frequency signal (electromagnetic radiation in the RF range) is generated, which can be measured with receiver coils.
6. In a television set the power needed to operate the picture tube is 95W and is derived from the
secondary coil of a transformer. There is a current of 5.3 mA in the secondary coil. The primary coil is
connected to a 120V AC outlet. Find the ration of turns Ns/Np of the transformer
7. A plane electromagnetic sinusoidal wave propagates in the x direction. Suppose that the wavelength is
3.00 m, and the electric field vibrates in the xy plane with an amplitude of 22.0 V/m. Calculate:
(a) the period and frequency of the wave
(b) Write an expression for B with the correct unit vector, with numerical values for Bmax, k, and ω, and with its
magnitude in the form B = Bmax cos(kx − ωt )
(c) What is the value of the magnetic field as t = T/4 (1/4 the period)?
(d) Find the maximum energy density of the wave.
(e) An RLC circuit is used in a radio to tune into the frequency of the EM wave. The resistance is 12.0 Ω, and
the inductance is 1.40µH. What capacitance should be used?
EDIT ANSWER
fλ = c
(a)
f ( 3.00 =
m ) 3.00 × 108 m s
or
so
(b)
(c)
f =
1.00 × 108 Hz =
100 MHz .
E
=c
B
or
22.0
= 3.00 × 108
Bmax
so
Bmax =
k
=
and
2π
=
λ
−73.3kˆ nT .
2π
= 0.126 m −1
50.0
(
)
ω=
2π f =
2π 6.00 × 106 s−1 =
3.77 × 107 rad s
(
)
B=
Bmax cos ( kx − ω t ) =−73.3cos 0.126x − 3.77 × 107 t kˆ nT .
(d)
8. In SI units, an electromagnetic wave traveling in the z x direction has an electric field component described
by
(
E y = 100 sin 1.00 × 10 7 x − ωt
)
Find (a) the amplitude of the corresponding magnetic field oscillations, (b) the wavelength λ, and (c) the
frequency f. (d) Show explicitly that the electric field is a solution of the equation 34.15 in Serway.
100 V m
E
(a)
=
=
B=
3.33 × 10−7 T =
0.333 µT
c 3.00 × 108 m s
(b)
λ
=
2π
2π
=
=
k
1.00 × 107 m −1
(c)
f
=
c 3.00 × 108 m s
=
=
λ 6.28 × 10−7 m
0.628 µm
4.77 × 1014 H z
^
^
^
9. In a region of free space the electric field at an instant of time is E = (80.0 i + 32.0 j – 64.0 k )N/C and the
^
^
^
magnetic field is B = (0.200 i + 0.080 0 j + 0.290 k )μT. (a) Show that the two fields are perpendicular to each
other. (b) Determine the Poynting vector for these fields.
(a)
E=
⋅B
( 16.0 + 2.56 − 18.56)
E⋅=
B
(b)
(80.0iˆ + 32.0ˆj − 64.0kˆ ) ( N C) ⋅ (0.200iˆ + 0.080 0ˆj + 0.290kˆ ) µT
(
N 2 ⋅ s C2 ⋅ m
=
)
0
(
)
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ

 

 80.0i + 32.0j − 64.0k N C  ×  0.200i + 0.080 0j + 0.290k µT 
=
=
S
E× B
µ0
4π × 10−7 T ⋅ m A
1
6.40kˆ − 23.2ˆj − 6.40kˆ + 9.28iˆ − 12.8ˆj + 5.12iˆ ) × 10
(
S=
−6
W m2
4π × 10−7
(
)
S =−
11.5ˆi 28.6ˆj W m 2 =
30.9 W m 2 at −68.2° from the +x axis.
10. A possible means of space flight is to place a perfectly reflecting aluminized sheet into orbit around the
Earth and then use the light from the Sun to push this “solar sail.” Suppose a sail of area 6.00 × 105 m2 and mass
6 000 kg is placed in orbit facing the Sun. (a) What force is exerted on the sail? (b) What is the sail’s
acceleration? (c) How long does it take the sail to reach the Moon, 3.84 × 108 m away? Ignore all gravitational
effects, assume that the acceleration calculated in part (b) remains constant, and assume a solar intensity of
1 340 W/m2.
(a)
(
Multiplying by the total area,
F = 5.36 N .
=
A 6.00 × 10 m gives:
5
)
2 1 340 W m 2
= 8.93 × 10−6 N m 2 .
3.00 × 108 m s2
The radiation pressure is
2
(b)
The acceleration is:
F
5.36 N
a =
=
= 8.93 × 10−4 m s2 .
m 6 000 kg
(c)
It will arrive at time t where
or
(
)
d=
1 2
at
2
2 3.84 × 108 m
2d
t=
=
=9.27 × 105 s = 10.7 days .
a
8.93 × 10−4 m s2
(
)
11. The intensity of solar radiation at the top of the Earth’s atmosphere is 1 340 W/m2. Assuming that 60% of
the incoming solar energy reaches the Earth’s surface and assuming the Earth absorbs 50% of the incident
energy, make an order-of-magnitude estimate of the amount of solar energy absorbed by a square meter in one
hour when the Sun is directly overhead. How does that compare with your own electrical power consumption?
You can look that up to estimate it.
EDIT ANSWER
Suppose you cover a 1.7 m-by-0.3 m section of beach blanket. Suppose the elevation angle of the Sun is 60°. Then the
target area you fill in the Sun’s field of view is
( 1.7 m )( 0.3 m ) cos30° =0.4 m 2 .
Now =
I
=
U IAt
=
P U
=
A At
(1 340 W m ) ( 0.6)( 0.5) ( 0.4 m )  ( 3 600 s)
2
2
~ 106 J .
12. The Earth reflects approximately 38.0% of the incident sunlight from its clouds and surface. (a) Given that
the intensity of solar radiation is 1 340 W/m2, what is the radiation pressure on the Earth, in pascals, at the
location where the Sun is straight overhead? (b) Compare this to normal atmospheric pressure at the Earth’s
surface, which is 101 kPa. You can look up atmospheric pressure in Chapter 14 of Serway.
Of the intensity
S = 1 340 W m 2
the 38.0% that is reflected exerts a pressure
2Sr 2( 0.380) S
.
=
P1 =
c
c
The absorbed light exerts pressure P
=
2
Altogether the pressure at the subsolar point on Earth is
Sa 0.620S
.
=
c
c
(
)
2
1.38S 1.38 1 340 W m
=
= 6.16 × 10−6 Pa
c
3.00 × 108 m s
(a)
Ptotal = P1 + P2 =
(b)
1.01 × 105 N m 2
Pa
=
=
Ptotal 6.16 × 10−6 N m 2
1.64 × 1010 times smaller than atmospheric pressure
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