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Physics 12 Assignment KEY
Chapters 20 & 21
Define the following terms:
 Magnetic dipole - another name for a magnet that always has two poles such as a
bar magnet
 Domain - small region under a particular magnetic influence so that all its magnets
point a particular direction
 Electromagnetism - phenomena associated with moving electrons producing a
magnetic field and a changing magnetic field causing electrons to move
 Solenoid - a closely wound helix of wire that acts as a magnet when current runs
through the wire
 Right-Hand Rules #1, #2, & #3 -the rules which help to visualize the directions of
vectors by using the fingers and thumb of your right hand.
#1. used to determine the direction of the magnetic lines of force from a currentcarrying conductor: imagine grasping a current-carrying conductor so that your
thumb lies along the wire in the direction of the current and your fingers curl
around it; then the magnetic lines of force encircle the wire in the direction of
your fingers
#2. used to indicate the direction of the magnetic lines of force in a solenoid:
imagine grasping a solenoid so that your fingers point in the direction of the
current, then your thumb points in the direction of the magnetic lines of force
inside the coil
#3. used to find the direction of the force exerted on a conductor by a magnet:
with your hand flat, point your thumb along a conductor in the direction of the
current and your fingers in the direction of the magnetic field (from the magnet),
then your palm faces in the direction of the force that the magnet’s field exerts
on the conductor
 Rotor - the rotating part of an electric motor, which consists of a coil with an iron
core, also called armature
 Armature – see rotor
 Split ring commutator - a device which allows continuous connection of the rotating
rotor (or armature) to the rest of the circuit in a motor or generator; used in DC
motors and generators to reverse the current direction
 Slip ring commutator - a device that allows the continuous connection of the
rotating rotor (or armature) to the rest of the circuit in a generator; used in an AC
generator to transmit alternating current
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
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AC generator - a device which converts mechanical energy into electrical energy.
Here, the induced current alternates or changes its direction every half-cycle; the
current from an AC generator has a sinusoidal form
DC generator - a device which converts mechanical energy into electrical energy.
Here, the current always leaves in the same direction
AC motor - a device which converts electrical energy into mechanical energy. The
AC motor runs on alternating current through the use of a slips-ring commutator
DC motor – a device which converts electrical energy into mechanical energy. The
DC motor runs on direct current through the use of a split-ring commutator
Electromagnetic induction - the generation of a current in a wire (circuit) due to the
relative motion of the wire and a magnetic field
Lenz’s Law – a law which states that when a conductor interacts with a magnetic
field, there must be an induced current that opposes the interaction, because of the
law of conservation of energy
Chapter 20
Questions – Pages 614-5: #3, 9, 13, 16, 17, 18, 21, 24
3. The magnetic field lines form counter-clockwise circles centered on the
wire (see figure to right).
9. Put one end of one rod close to one end of another rod. The ends will
either attract or repel. Continue trying all combinations of rods and ends
until two ends repel each other. Then the two rods used in that case are the
magnets.
13. The charged particle will now travel in a helix. It will continue in its original circular path
while it is traveling in the direction of the magnetic and electric fields, which is perpendicular to
the plane of the original circle. For as long as the electric field stays on, the particle will move in
a constantly elongating helix (the pitch of the helix will continue to increase), since the electric
field will cause the charged particle to accelerate along the electric field lines.
16. No, a moving charged particle can be deflected sideways with an electric field that points
perpendicular to the direction of the velocity, even when the magnetic field in the region is zero.
17. Assuming no electric field, the magnetic force will be exactly perpendicular to the velocity,
and so also perpendicular to the direction of motion. Since there is no component of force in the
direction of motion, the work done by the magnetic force will be zero, and the kinetic energy of
the particle will not change. The particle will change direction, but not change speed.
18. Use the right hand rule #3 to determine the direction of the force on each particle. In the
plane of the diagram, the magnetic field is coming out of the paper for points above the wire, and
is going into the paper for points below the wire.
Upper left hand particle: force down, toward the wire
Upper right hand particle: force to the left, opposite of the direction of the current
Lower right hand particle: force to the left, opposite of the direction of the current
Lower left hand particle: force up, toward the wire
21. If the moving electrons are changing speed as they are being deflected, then an electric field
is present. This will be the case whether or not a magnetic field exists in the same region of
space. If the moving electrons are being deflected but they are not changing speed, then only a
magnetic field is present.
24. (a) The current in the lower wire is pointing in the opposite direction of the current in the
upper wire. If the current in the lower wire is toward the north, then it creates a magnetic field
pointing east at the upper wire. Using the Right Hand Rule #3, the current of the upper wire must
be toward the south so that the magnetic force created on it is upward.
(b) Yes, the upper wire is vertically stable. If the upper wire falls toward the lower wire slightly,
then the magnetic force increases to push it back up to its stable position. If the upper wire rises
away from the lower wire slightly, although the magnetic field decreases, gravity will pull it
back into its stable position.
Problems – Pages 615-6: #1, 2, 3, 5, 7, 8, 10, 13
1. (a) The maximum force will be produced when the wire and the magnetic field are
perpendicular, so we have
Fmax = ILB, or Fmax/L = IB = (9.80 A)(0.80 T) = 7.8 N/m.
(b) We find the force per unit length from
F/L = IB sin 45.0° = (Fmax/L ) sin 45.0° = (7.8 N/m) sin 45.0° = 5.5 N/m.
2. The force on the wire is produced by the component of the magnetic field perpendicular to the
wire:
F = ILB sin 40°
F = (6.5 A)(1.5 m)(5.5 × 10–5 T) sin 40°
F = 3.45 × 10–4 N perpendicular to the wire and to B.
3. For the maximum force the wire is perpendicular to the field, so we find the current from F =
ILB;
I = F/(LB)
I = 0.900 N/(4.20 m)(0.0800 T), which gives I = 2.68 A.
5. To find the direction of the force on the electron, we point our fingers west and curl them
upward into the magnetic field. Our thumb points north, which would be the direction of the
force on a positive charge. Thus the force on the electron is south.
F = qvB = (1.60 × 10–19 C)(3.58 × 106 m/s)(1.30 T) = 7.45 × 10–13 N south.
7. To find the direction of the force on the electron, we point our fingers in the direction of v and
curl them into the magnetic field B. Our thumb points in the direction of the force on a positive
charge. Thus the force on the electron is opposite to our thumb.
(a) Fingers out, curl down, thumb right, force is left.
(b) Fingers down, curl back, thumb right, force is left.
(c) Fingers in, curl right, thumb down, force is up.
(d) Fingers right, curl up, thumb out, force is into the page.
(e) Fingers left, but cannot curl into B, so force is zero.
(f) Fingers left, curl out, thumb up, force is down.
8. We assume that we want the direction of B that produces the maximum force, i.e.,
perpendicular to v. Because the charge is positive, we point our thumb in the direction of F and
our fingers in the direction of v. To find the direction of B, we note which way we should curl
our fingers, which will be the direction of the magnetic field B.
(a) Thumb out, fingers left, curl down.
(b) Thumb up, fingers right, curl into the page.
(c) Thumb down, fingers in, curl right.
10. The greatest force will be produced when the velocity and the magnetic field are
perpendicular. We point our thumb down (a negative charge!), and our fingers south. We must
curl our fingers to the west, which will be the direction of the magnetic field. We find the
magnitude from F = qvB;
B = F/(qv)
–12
B = 2.2 × 10 N/(1.60 × 10–19 C)(1.8 × 106 m/s) = 7.6 T west.
13. The magnetic force provides the centripetal acceleration:
mv 2
qvB 
r
 mv  qBr
The kinetic energy of the electron is:
mv  qBr
1
(qBr ) 2
1 2 2
E k  mv 
2
m
2
1
(1.60  10 19 C)(1.15 T)(8.40  10 3 m)
1
E k  mv 2  2
2
(1.67  10  27 kg)

E k  7.15  10 -16 J  4.47  10 3 eV

Chapter 21
Questions – Pages 652-3: #1, 3, 8, 11
1. The advantage of using many turns (N = large number) in Faraday’s experiments is that the
emf and induced current are proportional to N, which makes it easier to experimentally measure
those quantities.
3. Yes, a current is induced in the ring when you bring the south pole toward the ring. An emf and current
are induced in the ring due to the changing magnetic flux (as the magnet gets closer to the ring, more
magnetic field lines are going through the ring).
No, a current is not induced in the ring when the magnet is held steady within the ring. An emf and
current are not induced in the ring since the magnetic flux through the ring is not changing while the
magnet is held steady.
Yes, a current is induced in the ring when you withdraw the magnet. An emf and current are induced in
the ring due to the changing magnetic flux (as you pull the magnet out of the ring toward you, fewer
magnetic field lines are going through the ring). Using Lenz’s law and the Right Hand Rule, the direction
of the induced current when you bring the south pole toward the ring is clockwise. In this case, the
number of magnetic field lines coming through the loop and pointing toward you is increasing
(remember, magnetic field lines point toward the south pole of the magnet). The induced current in the
loop is going to try to oppose this change in flux and will attempt to create magnetic field lines through
the loop that point away from you. A clockwise induced current will provide this opposing magnetic
field. Using Lenz’s law and the Right Hand Rule again, the direction of the induced current when you
withdraw the south pole from the ring is counter-clockwise. In this case, the number of magnetic field
lines coming through the loop and pointing toward you is decreasing. The induced current in the loop is
going to try to oppose this change in flux and will attempt to create more magnetic field lines through the
loop that point toward you. A counter-clockwise induced current will provide this opposing magnetic
field.
8. The starter motor is used to convert electrical energy to mechanical energy to start the engine. In early
automobiles, it was possible to connect the lead wires to the armature or rotor and then to the battery. As
the rotor moved in a magnetic field, electric current would be generated and transferred to the battery,
thus charging it.
11. Figure 21-12 shows that the induced current in the upper armature segment points into the page. This
can be shown using a Right Hand Rule: The charges in the top metal armature segment are moving in the
direction of the velocity shown with the green arrow (up and to the right) and these moving charges are in
a magnetic field shown with the blue arrows (to the right), and then the Right Hand Rule says that the
charges experience a force into the page producing the induced current. This induced current is also in the
magnetic field. Using another Right Hand Rule, a current-carrying wire, with the current going into the
page (as in the upper armature segment), in a magnetic field pointed to the right, will experience a force
in a downward direction. Thus, there is a counter-clockwise torque on the armature while it is rotation in a
clockwise direction.
Problems – Page 655-6: #30, 32, 34
30. We find the number of turns in the secondary from
VS/VP = NS/NP ;
(10,000V)/(120 V) = NS/(5000 turns), which gives NS = 4.17 x 105 turns.
32. With 100% efficiency, the power on each side of the transformer is the same:
IPVP = ISVS , so we have
IS/IP = VP/VS = (16 V)/(120) = 0.13.
34. (a) We assume 100% efficiency, so we have
IPVP = ISVS ;
(15 A)/(0.75 A) = (120 V)/VS , which gives VS = 6.0 V.
(b) Because VS < VP , this is a step-down transformer.