Chapter 20
Magnetism
Units of Chapter 20
• Magnets and Magnetic Fields
• Electric Currents Produce Magnetic Fields
• Force on an Electric Current in a Magnetic
Field; Definition of B
• Force on Electric Charge Moving in a
Magnetic Field
• Magnetic Field Due to a Long Straight Wire
• Force between Two Parallel Wires
Units of Chapter 20
• Solenoids and Electromagnets
• Ampère’s Law
• Torque on a Current Loop; Magnetic
Moment
• Applications: Galvanometers, Motors,
Loudspeakers
• Mass Spectrometer
• Ferromagnetism: Domains and Hysteresis
20.1 Magnets and Magnetic Fields
Magnets have two ends – poles – called
north and south.
Like poles repel; unlike poles attract.
20.1 Magnets and Magnetic Fields
However, if you cut a magnet in half, you don’t
get a north pole and a south pole – you get two
smaller magnets.
20.1 Magnets and Magnetic Fields
Magnetic fields can be visualized using
magnetic field lines, which are always closed
loops.
20.1 Magnets and Magnetic Fields
The Earth’s magnetic field is similar to that of a
bar magnet.
Note that the Earth’s
“North Pole” is really
a south magnetic
pole, as the north
ends of magnets are
attracted to it.
20.1 Magnets and Magnetic Fields
A uniform magnetic field is constant in
magnitude and direction.
The field between
these two wide poles
is nearly uniform.
20.2 Electric Currents Produce Magnetic
Fields
Experiment shows that an electric current
produces a magnetic field.
20.2 Electric Currents Produce Magnetic
Fields
The direction of the
field is given by a
right-hand rule.
20.3 Force on an Electric Current in a
Magnetic Field; Definition of B
A magnet exerts a force on a currentcarrying wire. The direction of the force is
given by a right-hand rule.
20.3 Force on an Electric Current in a
Magnetic Field; Definition of B
The force on the wire depends on the
current, the length of the wire, the magnetic
field, and its orientation.
(20-1)
This equation defines the magnetic field B.
20.3 Force on an Electric Current in a
Magnetic Field; Definition of B
Unit of B: the tesla, T.
1 T = 1 N/A·m.
Another unit sometimes used: the gauss (G).
1 G = 10-4 T.
Example 20-1
A wire carrying a 30 A current has a length of 12 cm between the pole
faces of a magnet at an angle of 60 degrees. The magnetic field is
approximately uniform at 0.90 T. We ignore the field beyond the pole
pieces. What is the magnitude of the force on the wire?
l =12 cm, I = 30 A, B = 0.90 T,  = 60
F = IlBsin 
F = (30 A)(0.12 m)(0.90 T)sin60
F = 2.8 N

Example 20-1
A rectangular loop of wire hangs vertically as shown. A magnetic field B is
directed horizontally, perpendicular to the wire, and points out of the page at
all points. The magnetic field is very nearly uniform along the horizontal
portion of wire ab (length = 10.0 cm) which is near the center of the gap of a
large magnet producing the field. The top portion of the wire loop is free of
the field. The loop hangs from a balance which measures a downward force
(in addition to the gravitational force) of F=3.48x10-2 N when the wire carries
a current of 0.245 A. What is the magnitude of the magnetic field B?
F = IlBsin ,  = 90
F
3.48x10 -2 N
B= =
Il (0.245 A)(0.100 m)
B = 1.42 T
20.4 Force on Electric Charge Moving in a
Magnetic Field
The force on a moving charge is related to
the force on a current:
(20-3)
Once again, the
direction is given by
a right-hand rule.
20.4 Force on Electric Charge Moving in a
Magnetic Field
If a charged particle is
moving perpendicular
to a uniform magnetic
field, its path will be a
circle.
20.4 Force on Electric Charge Moving in a
Magnetic Field
Problem solving: Magnetic fields – things to
remember
1. The magnetic force is perpendicular to the
magnetic field direction.
2. The right-hand rule is useful for determining
directions.
3. Equations in this chapter give magnitudes
only. The right-hand rule gives the direction.
20.4 Force on Electric Charge Moving in a
Magnetic Field
Example 20-4
A proton having a speed of 5.0x106 m/s in a magnetic field feels a force of
8.0x10-14 N toward the west when it moves vertically upward. When moving
horizontally in a northern direction, it feels zero force. Determine the
magnitude and direction of the magnetic field in this region. (The charge on
a proton is q=+e=1.6x10-19 C.)
F = qvBsin ,  = 90
F
8.0x10 -14 N
B=
=
qv (1.6x10 -19 C)(5.0x10 6 s)
B = 0.10 T
No force when moving north means the field must

be in a north-south direction. In order for the force
to pint west in (a), the field must point north (which
is into the page in this figure) from the right hand
rule.
Example 20-5
An electron travels at 2.0x107 m/s in a plane perpendicular to a uniform
0.010 T magnetic field. Describe its path qualitatively.
F = ma
v2
qvB = m
r
mv (9.1x10 -31 kg)(2.0x10 7 m/s)
r=
=
qB
(1.6x10 -19 C)(0.010 T)
So the electron moves in a circle of radius
1.1x10 -2 m, or 1.1 cm.

20.5 Magnetic Field Due to a Long Straight
Wire
The field is inversely proportional to the
distance from the wire:
(20-6)
The constant μ0 is called the permeability of
free space, and has the value:
Example 20-7
An electric wire in the wall of a building carries a dc current of 25 A vertically
upward. What is the magnetic field due to this current at a point P 10 cm due
north of the wire?
I
B= 0
2r
(4 x10 -7 Tm/A)(25 A)
B=
(2 )(0.10 m)
B = 5.0x10 -5 T
Note: The wire’s field has about the same magnitude as
Earth’s so a compass would not point north but in a

northwesterly direction.
Note: Most electrical wiring in buildings consists of
cables with two wires in each cable. Since the two wires
carry current in opposite direction, their magnetic fields
will cancel to a large extent.
Example 20-8
Two parallel straight wires 10.0 cm apart carry currents in opposite directions.
Current I1=5.0 A is out of the page, and I2=7.0 A is into the page. Determine
the magnitude and direction of the magnetic field halfway between the two
wires.
From right - hand - rule 1, both fields point
upwards at the midpoint, which is 0.050 m
from each wire.
0I1 (4 x10 -7 Tm/A)(5.0 A)
B1 =
=
= 2.0x10 -5 T
2r
(2 )(0.050 m)
0I 2 (4 x10 -7 Tm/A)(7.0 A)
B2 =
=
= 2.8x10 -5 T
2r
(2 )(0.050 m)
B = B1 + B2 = 4.8x10 -5 T
