Download Chapter 24-25 Assignment Solutions

Name __________________________________ Date ___________________________ Period _______
Advanced Physics
Chapter 24-25 Assignment Solutions
Contents
Page 664 #32-35, 59-63, 73-80 ..................................................................................................................... 1
Page 690-693 #31, 32, 42, 60, 76 .................................................................................................................. 5
Page 664 #32-35, 59-63, 73-80
32) Complete the following concept map using the following: right-hand rule, 𝐹 = 𝑞𝑣𝐵, and 𝐹 = 𝐼𝐿𝐵.
33) State the rule for magnetic attraction and repulsion. (24.1)
Like poles repel one another; opposite poles attract.
34) Describe how a temporary magnet differs from a permanent magnet. (24.1)
A temporary magnet is like a magnet only while under the influence of another magnet. A permanent
magnet needs no outside influence.
Name __________________________________ Date ___________________________ Period _______
Advanced Physics
35) Name the three most important common magnetic elements. (24.1)
Iron, cobalt, nickel
59) As the magnet below in Figure 24-24 moves toward the suspended magnet, what will the magnet
suspended by the string do?
Since like poles repel, the magnet suspended by the string will either move to the left or begin to rotate.
60) As the magnet in Figure 24-25 moves toward the suspended magnet, what will the magnet that is
suspended by the string do?
Since unlike poles attract, the magnet suspended by the string will move toward the right.
61) Refer to Figure 24-26 to answer the following questions.
a) Where are the poles? 2 and 4
b) Where is the north pole? 2 (magnetic field goes from north to south outside of a magnet)
c) Where is the south pole? 4 (magnetic field goes from north to south outside of a magnet)
Name __________________________________ Date ___________________________ Period _______
Advanced Physics
62) Figure 24-27 shows the response of a compass in two different positions near a magnet. Where is
the south pole of the magnet located?
At the right end. The north end of the compass to the right of the magnet is pointing towards the south
end of the bar magnet, since unlike poles attract.
63) A wire that is 1.50 m long and carrying a current of 10.0 A is at right angles to a uniform magnetic
field. The force acting on the wire is 0.60 N. What is the strength of the magnetic field?
𝐹 = 𝐼𝐿𝐵
𝐵=
𝐹
0.60 N
=
= 0.040 T
𝐼𝐿 (10.0 A)(1.50 m)
73) A wire that is 0.50 m long and carrying a current of 8.0 A is at right angles to a uniform magnetic
field. The force acting on the wire is 0.40 N. What is the strength of the magnetic field?
𝐹 = 𝐼𝐿𝐵
𝐵=
𝐹
0.40 N
=
= 0.10 T
𝐼𝐿 (8.0 A)(0.50 m)
74) The current through a wire that is 0.80 m long is 5.0 A. The wire is perpendicular to a 0.60-T
magnetic field. What is the magnitude of the force on the wire?
𝐹 = 𝐼𝐿𝐵
𝐹 = (5.0 A)(0.80 m)(0.60 T) = 2.4 N
75) A wire that is 25 cm long is at right angles to a 0.30-T uniform magnetic field. The current through
the wire is 6.0 A. What is the magnitude of the force on the wire?
𝐹 = 𝐼𝐿𝐵
𝐹 = (6.0 A)(0.25 m)(0.30 T) = 0.45 N
76) A wire that is 35 cm long is parallel to a 0.53-T uniform magnetic field. The current through the wire
is 4.5 A. What force acts on the wire?
If the wire is parallel to the magnetic field, then it does not cut across any magnetic field lines, so no
force is produced.
Name __________________________________ Date ___________________________ Period _______
Advanced Physics
77) A wire that is 625 m long is perpendicular to a 0.40-T magnetic field. A 1.8-N force acts on the wire.
What current is in the wire?
𝐹 = 𝐼𝐿𝐵
𝐼=
𝐹
1.8 N
=
= 7.2 mA
𝐿𝐵 (625 m)(0.40 T)
78) The force on a 0.80-m wire that is perpendicular to Earth’s magnetic field is 0.12 N. What is the
current in the wire? Use 5.0 × 10−5 T for Earth’s magnetic field.
𝐹 = 𝐼𝐿𝐵
𝐼=
𝐹
0.12 N
=
= 3.0 × 103 A
𝐿𝐵 (0.80 m)(5.0 × 10−5 T)
79) The force acting on a wire that is at right angles to a 0.80-T magnetic field is 3.6 N. The current in
the wire is 7.5 A. How long is the wire?
𝐹 = 𝐼𝐿𝐵
𝐿=
𝐹
3.6 N
=
= 0.60 m
𝐼𝐵 (7.5 A)(0.80 T)
80) A power line carries a 225-A current from east to west, parallel to the surface of Earth.
a) What is the magnitude of the force resulting from Earth’s magnetic field acting on each meter of
the wire. Use 𝐵𝐸𝑎𝑟𝑡ℎ = 5.0 × 10−5 T.
𝐹 = 𝐼𝐿𝐵
𝐹
= 𝐼𝐵 = (225 A)(5.0 × 10−5 T) = 0.011 N/m
𝐿
b) What is the direction of the force?
The force must be perpendicular to the direction of the current (east to west) and perpendicular
to the direction of the magnetic field (south to north). This means the force must be up or
down.
Using the right hand rule for cross-products (currents producing magnetic fields), the force is
directed down. (Note: In this class, you need to be able to determine that the force is “either up
or down” but you do not need to determine that it is down.)
c) In your judgment, would this force be important in designing towers to hold this power line?
Explain.
No. The force is so much smaller than the weight of the wires (or any bird that might rest on the
wire) that it is negligible.
Name __________________________________ Date ___________________________ Period _______
Advanced Physics
Page 690-693 #31, 32, 42, 60, 76
31) You have a coil of wire and a bar magnet. Describe how you could use them to generate an electric
current. (25.1)
Either move the magnet repeatedly into and out of the coil, or move the coil up and down over the end
of the magnet.
32) What does 𝐸𝑀𝐹 stand for? Why is the name inaccurate? (25.1)
Electromotive force; it is not a force (it is not measured in Newtons), but an electric potential (energy
per unit of charge).
42) Upon what does the ratio of the 𝐸𝑀𝐹 in the primary circuit of a transformer to the 𝐸𝑀𝐹 in the
secondary circuit of the transformer depend? (25.2)
The 𝐸𝑀𝐹 ratio is equal to the ratio of number of turns of wire in the primary coil to the number of turns
of wire in the secondary coil.
𝑉𝑃 𝑁𝑃
=
𝑉𝑆 𝑁𝑆
60) A wire, 20.0-m long, moves at 4.0 m/s perpendicularly through a magnetic field. An 𝐸𝑀𝐹 of 40 V is
induced in the wire. What is the strength of the magnetic field?
𝜀 = 𝐵𝐿𝑉
𝐵=
𝜀
40 V
=
= 0.5 T
𝐿𝑣 (20.0 m)(4.0 m/s)
76) Laptop Computers: The power supply in a laptop computer requires an effective voltage of 9.0 V
from a 120-V line.
a) If the primary coil has 475 turns, how many does the secondary have?
𝑉𝑃 𝑁𝑃
=
𝑉𝑆 𝑁𝑆
𝑉𝑆 𝑁𝑝 (9.0 V)(475)
=
= 36 turns
𝑉𝑃
120 V
b) A 125-mA current is in the computer. What current is in the primary circuit?
𝑁𝑆 =
𝑃𝑃 = 𝑃𝑆
𝐼𝑃 𝑉𝑃 = 𝐼𝑆 𝑉𝑆
𝐼𝑃 =
𝐼𝑆 𝑉𝑆 (9.0 V)(125 mA)
=
= 9.4 mA
𝑉𝑃
120 V