Download PHYS 102 Practice Problems Chapters 18-22

PHYS 102
Practice Problems
Chapters 18-22
Chapter 18.
45.
The peak value of an alternating current in a 1500-W device is 5.4 A. What is the rms voltage
across?
The power and current can be used to find the peak voltage, and then the rms voltage can be found
from the peak voltage.
I peak
2 (1500 W )
2P
2
P = I rmsVrms =
Vrms → Vrms =
=
= 3.9 × 10 V
I peak
5.4 A
2
Chapter 19.
15.
Eight 7.0-W Christmas tree lights are connected in series to each other and to a 110-V
source. What is the resistance of each bulb?
Each bulb will get one-eighth of the total voltage, and so Vbulb =
Vtot
8
. Use that voltage and the
power dissipated by each bulb to calculate the resistance of a bulb.
2
Pbulb =
20.
Vbulb
R
2
→ R=
Vbulb
P
2
=
Vtot
64 P
=
(110 V )2
=
64 ( 7.0 W )
27 Ω
What is the net resistance of the circuit connected to the battery in Fig.
19-41? Each resistance has R = 2.8 kΩ.
The resistors have been numbered in the accompanying diagram to help in
the analysis. R1 and R2 are in series with an equivalent resistance of
R12 = R + R = 2 R . This combination is in parallel with R3 , with an
R2
R1
R3
R4
−1
1 1 
equivalent resistance of R123 =  +  = 23 R . This combination is in
 R 2R 
series with R4 , with an equivalent resistance of R1234 = 23 R + R = 35 R . This
combination is in parallel with R5 , with an equivalent resistance of
−1
R5
1 3 
5
+
 = 8 R . Finally, this combination is in series with R6 , and we calculate the
 R 5R 
final equivalent resistance.
R12345 = 
Req = 58 R + R = 138 R = 138 ( 2.8 kΩ ) = 4.55 kΩ
R6
27.
Determine the magnitude and directions of the currents
through R1 and R2 in Fig. 19-47.
Because there are no resistors in the bottom branch, it is
possible to write Kirchhoff loop equations that only have one
current term, making them easier to solve. To find the current
through R1 , go around the outer loop counterclockwise, starting
at the lower left corner.
V3 − I1 R1 + V1 = 0 → I1 =
V3 + V1
R1
=
6.0 V + 9.0 V
22 Ω
= 0.68 A , left
To find the current through R2 , go around the lower loop counterclockwise, starting at the
lower left corner.
V3 − I 2 R2 = 0 → I 2 =
V3
R2
=
6.0 V
15 Ω
= 0.40 A , left
50. In Fig. 19–56 (same as Fig. 19–20a), the total resistance is 15.0 kΩ, and the battery’s emf is
24.0 V. If the time constant is measured to be 35.0 µs, calculate (a) the total capacitance of
the circuit and (b) the time it takes for the voltage across the resistor to reach 16.0 V after
the switch is closed.
(a) From Eq. 19-7, the product RC is equal to the time constant.
τ 35.0 ×10−6 s
τ = RC → C = =
= 2.33 ×10−9 F
3
R
15.0 ×10 Ω
(b) Since the battery has an EMF of 24.0 V, if the voltage across the resistor is 16.0 V, the
voltage across the capacitor will be 8.0 V as it charges. Use the expression for the
voltage across a charging capacitor.
t
 V 
 V 
− t /τ
−t / τ
VC = E (1 − e ) → e
=  1 − C  → − = ln  1 − C  →
τ
 E 
 E 

8.0 V 
 V 
−6
−5
t = −τ ln  1 − C  = − ( 35.0 × 10 s ) ln  1 −
= 1.42 × 10 s

 E 
 24.0 V 
Chapter 20.
5.
The force on a wire carrying 8.75 A is a maximum of 1.28 N when placed between the pole
faces of a magnet. If the pole faces are 55.5 cm in diameter, what is the approximate
strength of the magnetic field?
Use Eq. 20-2. The length of wire in the B-field is the same as the diameter of the pole faces.
Fmax = IlB → B =
Fmax
Il
=
1.28 N
( 8.75A )( 0.555m )
= 0.264 T
12.
r
r
Determine the direction of B for each case in Fig. 20–52, where F represents the
r
maximum magnetic force on a positively charged particle moving with velocity v .
The right hand rule applied to the velocity and magnetic field would give the direction of
the force. Use this to determine the direction of the magnetic field given the velocity and
the force.
(a)
downward
(b)
inward into the paper
(c)
right
15.
An electron experiences the greatest force as it travels 2.9 x 106 m/s in a magnetic field
when it is moving northward. The force is upward and of magnitude 7.2 x 10-13 N What
are the magnitude and direction of the magnetic field?
The magnetic field can be found from Eq. 20-4, and the direction is found from the right
hand rule. Remember that the charge is negative.
Fmax = qvB → B =
Fmax
qv
=
7.2 × 10−13 N
(1.60 ×10
−19
)(
C 2.9 × 106 m s
)
= 1.55 T
The direction would have to be East for the right hand rule, applied to the velocity and the
magnetic field, to give the proper direction of force.
30. Determine the magnitude and direction of the force between two parallel wires 35 m long
and 6.0 cm apart, each carrying 25 A in the same direction.
Since the currents are parallel, the force on each wire will be attractive, toward the other
wire. Use Eq. 20-7 to calculate the magnitude of the force.
F2 =
36.
( 4π ×10−7 T ⋅ m A ) ( 25 A )2 ( 35 m ) = 0.0729 N, attractive
µ0 I1 I 2
l2 =
2π d
2π
( 6.0 ×10−2 m )
A straight stream of protons passes a given point in space at a rate of 1.5 x 109 protons/s.
What magnetic field do they produce 2.0 m from the beam?
The stream of protons constitutes a current, whose magnitude is found by multiplying the
proton rate times the charge of a proton. Then use Eq. 20-6 to calculate the magnetic field.
Bstream
−7
9
−19
µ0 I ( 4π ×10 T ⋅ m A )(1.5 ×10 protons s )(1.60 ×10 C proton )
−17
=
=
= 2.40 ×10 T
2π r
2π ( 2.0 m)
Chapter 21.
5.
A 12.0-cm diameter loop of wire is initially oriented perpendicular to a 1.5-T magnetic
field. The loop is rotated so that its plane is parallel to the field direction in 0.20 s. What
is the average induced emf in the loop?
The flux changes because the loop rotates. The angle changes from 0o to 90o. The
magnitude of the average induced emf is given by Eq. 21-2a.
E =−
∆Φ B
∆t
=−
AB∆ cos θ
∆t
π ( 0.060 m ) 1.5 T ( cos 90o − cos 0o )
2
=−
0.20 s
2
=−
17.
π ( 0.060 m ) 1.5 T ( 0 − 1)
0.20 s
= 84.8 mV
In Fig. 21-12, the rod moves with a speed of 1.6 m/s, is 30.0 cm long, and has a
resistance of 2.5 Ω. The magnetic field is 0.35 T, and the resistance of the U-shaped
conductor is 25.0 Ω at a given instant. Calculate (a) the induced emf, (b) the current in
the U-shaped conductor, and (c) the external force needed to keep the rod’s velocity
constant at that instant.
(a)
Because the velocity is perpendicular to the magnetic field and the rod, we find
the induced emf from Eq. 21-3.
E = Blv = ( 0.35 T )( 0.300 m )(1.6 m s ) = 0.168 V
(b)
Find the induced current from Ohm’s law.
E 0.168 V
−3
I=
(c)
R
=
27.5 Ω
= 6.11 × 10 A
The induced current in the rod will be down. Because this current is in an upward
magnetic field, there will be a magnetic force to the left. To keep the rod moving,
there must be an equal external force to the right, given by Eq. 20-2.
(
−3
)
−4
F = IlB = 6.11 × 10 A ( 0.300 m )( 0.35 T ) = 6.41 × 10 N
20.
A simple generator is used to generate a peak output voltage of 24.0 V. The square
armature consists of windings that are 6.0 cm on a side and rotates in a field of 0.420 T at
a rate of 60.0 rev/s. How many loops of wire should be wound on the square armature?
We find the number of turns from Eq. 21-5. The factor multiplying the sine term is the peak
output voltage.
Epeak = NBω A → N =
Epeak
Bω A
=
24.0 V
( 0.420 T ) 2π rad
rev ( 60 rev s )( 0.060 m )
2
= 42.1 ≈ 42 loops
32.
A step-up transformer increases 25 V to 120 V. What is the current in the secondary coil
as compared to the primary coil?
Use Eqs. 21-6 and 21-7 to relate the voltage and current ratios.
VS
=
VP
42.
NS
;
NP
IS
IP
=
NP
VS
→
NS
=
VP
IP
→
IS
IS
IP
VP
=
=
VS
25 V
120 V
= 0.208
How many turns of wire would be required to make a 130-mH inductance out of a 30.0cm-long air-filled coil with a diameter of 5.2 cm?
Use the relationship for the inductance of a solenoid, as given in Example 21-14.
L=
55.
µ0 N 2 A
→ N=
l
Ll
=
µ0 A
( 0.130 H )( 0.300 m )
( 4π × 10
−7
) (
T ⋅ m A π 2.6 × 10−2 m
)
2
= 3823 turns
At what frequency will a 2.40-µF capacitor have a reactance of 6.70 kΩ?
We find the frequency from Eq. 21-12b for the reactance of a capacitor.
XC =
66.
1
→ f =
2π fC
1
2π X C C
=
1
(
3
)(
−6
2π 6.70 × 10 Ω 2.40 × 10 F
)
= 9.90 Hz
A 2.5-kΩ resistor in series with a 420-mH inductor is driven by an ac power supply. At
what frequency is the impedance double that of the impedance of 60 Hz?
The impedance is given by Eq. 21-15 with no capacitive reactance.
2
2
2
2
Z = R + X L = R + ( 2π fL ) .
2
2
R + 4π f L = 4  R + 4π
2
2
2
2
2
f =
2

3R + 16π
= 1.64 kHz
2
2
2
R + 4π f L = 2 R + 4π
Z f = 2 Z 60 →
2
( 60 Hz )
2
2
4π L
2
2
( 60 Hz )
2
L
=
2
2
( 60 Hz )
2
2
→
L
2 2
2
2
2
L  = 4 R + 16π ( 60 Hz ) L →

3R 2
2
2
2
4π L
+ 4 ( 60 Hz ) =
3 ( 2500 Ω )
4π
2
2
( 0.42 H )
2
+ 4 ( 60 Hz )
2
Chapter 22.
4.
In an EM wave traveling west, the B field oscillates vertically and has a frequency of 80.0
kHz and an rms strength of 6.75 x 10-9 T. What are the frequency and rms strength of the
electric field, and what is its direction? (See Fig. 22-7)
The frequency of the two fields must be the same: 80.0 kHz.
The rms strength of the electric field is
Erms = cBrms = ( 3.00 × 108 m s )( 6.75 × 10−9 T ) = 2.03V m.
The electric field is perpendicular to both the direction of travel and the magnetic field, so the
electric field oscillates along the horizontal north-south line.
18.
r
The E field in an EM wave in free space has a peak of 21.8 mV/m. What is the average
rate at which this wave carries energy across unit area per unit time?
The energy per unit area per unit time is
S = 12 cε 0 E0 2
=
32.
1
2
( 3.00 × 10
8
2
m s )(8.85 × 10−12 C 2 N ⋅ m 2 )( 21.8 × 10−3 V m ) = 6.31 × 10−7 W m2 .
The oscillator of a 96.1-MHz FM station has an inductance of 1.8 µH. What value must
the capacitance be?
We find the capacitance from the resonant frequency:
1
 1  1  2
f0 = 

 ;
 2π  LC 
1
2
1
 1 

 , which gives C = 1.53× 10−12 F = 1.53pF.
96.1 × 106 Hz = 

−6
2
π
1.8
×
10
H
C

  (
) 