Download Magnets Review Sheet

Magnets Review Sheet Answers
AP Physics
1. This picture shows a solenoid acting as an electromagnet (though
an electromagnet will usually have an iron core to increase the
magnetic field).
a) Which end of the solenoid is acting as the north pole of the
electromagnet?
The right-hand side because that is where the fields lines come
out of the electromagnet and field lines go from the north pole
of a magnet to the south pole of a magnet.
b) Which way is the conventional current flowing through the
solenoid?
It loops over the top coming toward us and away from us on the bottom. Overall it is traveling from left to
right. Test this using the second right hand rule (the one showing the direction of the field around a current
carrying wire). Run your thumb along the wire in the direction carrying the current and your fingers show
the direction of the magnetic field loops around the wire. Look to see that they are pointing the same way
as the overall field (from all the loops) on the inside and outside of the solenoid.
c) Where is the magnetic field of the solenoid strongest? How can you tell?
The magnetic field is strongest inside the coils because that is where the field lines are closest.
2. The aurora borealis is produced by particles from space (mostly from the sun) hitting and exciting
atmospheric molecules. Where the aurora happens is determined by the earth’s magnetic field. What do you
know about the particles from space if they are affected by a magnetic field?
They must be charged particles because only moving charged particles are affected by magnetic fields.
3. What are two ways that you can destroy a permanent magnet (i.e erase a magnetic floppy disk)? Why does
that work?
The reason that permanent magnets are magnetic is because they are made up from materials whose
atoms are themselves tiny magnets. (Atomic magnets are essentially tiny electromagnets. Their
magnetic fields are created by the way that the electrons “orbit” around the nucleus. A simple, but
revealing, model is to consider an electron orbit to be a loop of wire creating a magnetic field).
Because opposite poles of magnets attract and like poles repel, there is a tendency for these atomic
magnets to line up with the other atomic magnets around them, though it is uncommon for all of the
atoms to align exactly the same way. These localized clusters of aligned magnets are called domains
and are indicated by the arrows in the pictures below.
In an unmagnetized object, the domains are aligned at random and produce fields that cancel out on a
large scale. In a magnet, the
domains are aligned so that their
magnetic fields add up to
produce a large scale magnetic
field. These two states are
shown in the drawings here:
This makes it clear that the only difference between a permanent magnet and an unmagnetized piece
of the same material is how well the magnetic domains are aligned. In order to destroy a permanent
magnet, all you have to do is get the domains out of alignment. Two ways to that are: 1) physical
violence – hitting the magnet can shake the domains out of alignment; 2) heating it – because a higher
temperature means greater movement of the atoms in the materials, raising the temperature will induce
the atoms in the domains to knock themselves and each other out of alignment.
4. Find the size and direction of the magnetic force on the following charged particles:
a) B = 0.3 T
b) B = 1.7 T
c) B = 0.09 T
q = 1.8 x 10-15 C; v = 104 m/s
q = - 3.05 x 10-9 C; v = 3 x 105 m/s
q = - 5.3 x 10-8 C; v = 150 m/s
a) F = qvB = 1.8 x 10-15 C (104 m/s) 0.3 T = 5.4 x 10-12 N up
b) F = qvBsin = -3.05 x 10-9 C (3 x 105 m/s) 1.7 T sin 60 = -0.00135 N out of page or 0.00135 N into page
c) F = qvB = -5.3 x 10-8 C (150 m/s) 0.09 T = -7.2 x 10-7 N left or 7.2 x 10-7 N right
5. A space heater has a resistance of 420  when running
at 120 V. When the space heater is running the two wires
in the 3 m long cord carry current in opposite direction.
The wires in the cord are 4 mm apart. Presume that the
cord is laying flat on the ground, so the two wires are next
to each other and running straight east and west, with the
more northerly carrying current to the west.
a) How much current flows through the wires between the outlet to the space heater?
I = V/R = 120 V/420  = 0.286 A
b) What will the total magnetic field from the two wires be at a point 2 mm to the south of the more southerly
wire? (include magnitude and direction).
At that point both wires create fields:
0 ·I 4 x 10-7 · 0.286 A
B1 =
=
= 2.86 x 10-5 T down
2d1
2 · 0.002 m
0 ·I 4 x 10-7 · 0.286 A
B2 =
=
= 9.53 x 10-6 T up
2d2
2 · 0.006 m
BT = B1 + B2 = 2.86 x 10-5 T – 0.953 x 10-5 T = 1.907 x 10-5 T down
c) What will the total magnetic field from the two wires be at a point 5 cm to the south of the southern wire?
Must take field created by each wire into consideration:
0 ·I 4 x 10-7 · 0.286 A
B1 =
=
= 1.144 x 10-6 T down
2d1
2 · 0.050 m
0 ·I 4 x 10-7 · 0.286 A
B2 =
=
= 1.059 x 10-6 T up
2d2
2 · 0.054 m
BT = B1 + B2 = 1.144 x 10-6 T – 1.059 x 10-6 T = 0.085 x 10-6 T down = 8.5 x 10-8 T down
d) What is the force from each wire within the cord on the other wire within the cord?
0 ·I 4 x 10-7 · 0.286 A
On southern wire (wire 1): B2 =
=
= 1.43 x 10-5 T up
2d
2 · 0.004 m
F1 = ILB sin = (0.286 A)(3 m)(1.43 x 10-5 T) sin90 = 1.23 x 10-5 N south
0 ·I 4 x 10-7 · 0.286 A
=
= 1.43 x 10-5 T down
2d
2 · 0.004 m
F2 = ILB sin = (0.286 A)(3 m)(1.43 x 10-5 T) sin90 = 1.23 x 10-5 N north
On northern wire (wire 2): B1 =
6. The voice coil on a speaker is made of several coils of current carrying wire. Take a voice coil that is made
from 20 loops of wire that are 7 cm in diameter. How large is the magnetic field at the center of the voice coil if
it is carrying 15 A of current?
0·I·N 4 x 10-7 · 15 A · 20
B = 2r =
= 0.00539 T
2 · 0.035 m
7. Several years ago, Dutch and British researchers used magnetic forces in a 32 mm wide and 20 cm long
solenoid made with 214 turns of wire to levitate a frog (and other unusual objects). They needed a 16 Tesla
magnetic field to levitate the frog. How big was the current in the solenoid?
B = 0·n·I
N
214
B
16 T
n = L = 0.20 m = 1070 turns/m
I=
=
= 1.190 x 104 A
-7
0·n 4 x 10 ·1070 turns/m
8. The following problems occur within the velocity selector section of mass spectrometer machine.
a) If 3.2 x 10-18 C particles
b) What magnetic field
c) What electric field would
are shot eastward through an
would be required to balance
be required to balance out a
overlapping electric field of
out a 1.5 V/m electric field to
450 T magnetic field
1.55 x 104 N/C down and
the east so that a –1.6 x 10-19
pointing down so that an ion
magnetic field of 7.4 mT
C muon going north at 3.8 x
with 1.7 x 10-17 C of charge
north and are not deflected
105 m/s will pass straight
moving at 1.95 x 103 m/s
by the fields, how fast must
through the fields?
would not be deflected?
the particles be moving?
E
E
v = so
v
=
E 1.55 x 104 N/C
B
B so E = vB
v = B = 7.4 x 10-3 T
E
1.5 V/m
E= 1.95 x 103 m/s · 450 x 10-6 T
B
=
=
6
5
v = 2.09 x 10 m/s
v 3.8 x 10 m/s
E = 0.8775 V/m = 0.8775 N/C
B = 3.95 x 10-6 T
9. The following problems take place in the deflector section of a mass spectrometer where the
magnetic field is 0.8 mT pointing downward into the page.
a) A particle with 2.4 x 10-17 C of charge and
b) A 3.5 x 10-23 kg ion moving at 2.17 x 105 m/s
enters the chamber at 3000 m/s and makes an
is deflected in a semicircle with a diameter of
arc with a radius of 38.9 cm. What is the mass
23.2 cm. What is the charge on the particle?
of the particle?
qBr 2.4 x 10-17 C · 0.8 x 10-3 T · 0.389 m
m= v =
3000 m/s
mv 3.5 x 10-23 kg·2.17 x 105 m/s
q = Br =
0.8 x 10-3 T·0.116 m
m = 2.490 x 10-24 kg
q = 8.184 x 10-14 C
10. A wire inside an MRI device runs through a magnetic field of 3.8 T to the east. The wire carries a
0.73 A current and runs for 1 m to the north, then runs 50 cm down, then runs 30 cm to the south and,
finally, runs 50 cm up. What is the net force acting on the wire?
We need to calculate the force on each section of the wire separately and then add up the total force.
Section 1: 1 m north
F = BILsin = 3.8 T · 0.73 A · 1 m · sin90
F1 = 2.774 N down
Section 2: 0.5 m down
F = BILsin = 3.8 T · 0.73 A · 0.5 m · sin90
F1 = 1.387 N north
Total:
Hor: 1.387 N, N + 1.387 N, S = 0
Section 3: 0.3 m south
F = BILsin = 3.8 T · 0.73 A · 0.3 m · sin90
F1 = 0.832 N up
Section 4: 0.5 m up
F = BILsin = 3.8 T · 0.73 A · 0.5 m · sin90
F1 = 1.387 N south
Vert: 2.774 N, D + 0.832 N, U = 1.942 N, U
Fnet = 1.942 N, U
11. The rotor of a motor is made from 5 overlapping rectangular coils of wire, 10 cm x 15 cm, in a 0.4
mT magnetic field pointing due east. If there are 5 A of current running through the wire, how large
will the torque on it be in each of the following situations?
a) when the coils are oriented up and down
b) when the coils are oriented at a 60° angle to the
ground
 = IABsin
 = 5A·0.015m2·0.4 x 10-3 T·sin0
 = 0 m·N
 = IABsin
 = 5A·0.015m2·0.4 x 10-3 T·sin30
 = 1.5 x 10-5 m·N