In this chapter we investigate forces exerted by magnetic fields. In the next chapter we will study the sources of magnetic fields. The force produced by magnetic fields has played a pivotal role in our production of electrical energy.
Likewise, the transport of this electrical energy (to produce magnetic fields far away from their source) has made it more convenient to live and work at remote locations. We will also introduce the concept of magnetic flux (Φ mag
) and the important role it plays in the production of electrical power.
The magnetic phenomena was first recorded at least 2500 years ago in fragments of magnetised iron ore found near the ancient city Magnesia (Manisa, in western
Turkey). Today we call these fragments permanent fragments .
Magnetic fragments were observed to attract and repel when brought close together.
Figure 1: This figure shows the possible forces of attraction and repulsion that occur with bar magnets.
Before the source of magnetic fields was understood, the interaction of permanent magnets and compass needles were described in terms of magnetic poles .
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The Earth’s Magnetic Field
Figure 2: This figure show the earth’s magnetic field with the internal south-pole in the northern hemisphere. The north-seeking compass needles point northward to the earth’s magnetic south-pole.
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1.1
Magnetic Poles Versus Electric Charge
Figure 3: This figure shows that trying to separate the north pole from the south pole is a futile exercise. It always results in the production of two magnetic dipoles.
To date: magnetic monopoles have not been observed !!
The Danish physicist, Hans Oersted, first noticed the connection between current and the production of magnetic fields in 1820.
Figure 4: This figure shows the effect of a nearby current and its interaction with a compass needle.
Opposite currents result in an opposite change in the direction of the compass needle.
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We would like to introduce the concept of a magnetic field much in the same we that we described the electrostatic force.
1. We said that the distribution of electric charge will create an electric field in the surrounding space.
2. Also, we observed that the electric field exerted a force proportion to the product of the charge and the electric field
~
= E .
In the case of magnetic interactions, we should be able to say the following:
1. A moving charge or a current creates a magnetic field in the surrounding space, and
2. The magnetic field exerts a force are present in the field.
~ on other moving charge or currents that
2.1
Magnetic Forces on Moving Charges
~
= q ~ ×
~
(1) where q is the charge (coulombs), v is the velocity (m/s), and B is the magnetic field (teslas). One tesla is:
1 T = 1 N/ ( A · m )
Figure 5: This figure illustrates the magnetic force resulting from a charge q moving with velocity
~ in a uniform magnetic field
~
.
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2.2
The Right-Hand Rule
Figure 6: This figure illustrates how to apply the right-hand rule to the equation
~
= q~ × ~
.
Figure 7: This figure illustrates how opposite charges result in opposite force vectors
~
.
This figure illustrates how opposite charges result in opposite force vectors
~
.
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2.3
Measuring Magnetic Fields with Test Charges
Figure 8: This figure illustrates how to measure the strength and direction of a magnetic field by tracking the motion of a test charge q .
~
= v ×
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Figure 9: The magnetic fields are stronger at the pole faces (i.e., higher flux) and weaker at field points far away from the pole face (i.e., lower flux).
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3.1
Magnetic Flux and Gauss’s Law for Magnetism
Figure 10: This figure shows the continuity of magnetic field lines for a number of different geometries. Notice that there are no sources and no sinks for the magnetic filed lines.
Φ
B
=
Z
B cos φ dA =
Z
B
⊥ dA =
Z
~
· A (2)
Figure 11: This figure illustrates how to calculate the flux of magnetic field Φ m open surface.
passing through an
If we have a planar surface, then the flux is easily written as:
Φ
B
= B
⊥
A
8
= BA cos φ (3)

The unit of flux is called the weber (1 Wb), in honor of the German physicist
Wilhelm Weber (1804-1891).
1 Wb = 1 T · m
2
= 1 N · m / A
Gauss’s Law for Magnetism
The total magnetic flux through any closed surface :
Φ
B
=
I
S
~
· A = 0
Magnetic Flux Through a Surface
(Always)
Figure 12: This figure illustrates how to calculate the flux of magnetic field Φ m open surface. In this case, the magnetic flux Φ
B
= 0.
passing through an
~
= q~ ×
~
(4)
Motion of a charged particle under the action of a magnetic field alone is always motion with constant speed.
The angular velocity of the charged-particle’s circular motion can be found by using
Newton’s 2nd law:
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Figure 13: This figure illustrates how to calculate the flux of magnetic field Φ m open surface.
passing through an
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X
F ext
= ma c
→ qvB = mv 2
R p = mv = qBR Also v = Rω
So,
ω = v
R
=
| q | B m
Magnetic Bottle in Plasma Physics
(cyclotron frequency) (5)
Figure 14: This figure demonstrates how a magnetic field can be used to contain high-temperature ions in a magnetic bottle .
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Magnetic Fields in High Energy Physics
Figure 15: This figure shows a gamma-ray conversion into an electron-positron pair, along with a δ -ray (electron) kicked out of a neutral hydrogen atom during the conversion process. At first glance, it appears that charge is not conserved; however, if you include the δ -ray electron in the initial part of the “balanced charge” equation, charge is conserved .
( γ + e
− → e
+ e
−
+ e
−
)
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Velocity Selector
If electric and magnetic fields are simultaneously applied to the motion of a charged particle (as shown in the figure below), it is possible to build a velocity selector:
~
= q
~
+ ~ ×
~
Figure 16: This figure illustrates how to calculate the flux of magnetic field Φ m open surface.
passing through an
The ions that successfully make it through the collimator at the bottom of the velocity selector satisfy the following condition (ignore gravity): v =
E
B
(ions that move in a straight line)
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(6)

Mass Spectrometers
A mass spectrometer is shown in the following figure. This device is used to separate nuclear isotopes with slightly different atomic masses (e.g., 235 U and 238 U .)
Figure 17: This figure shows ions with the same velocity but different masses being separated in the second magnetic field at the bottom of the figure. In this figure, the magnetic field
~ is presumed to be the same inside and outside the velocity selector.
R = mE
| q | B 2 v =
E
B
R = mv
| q | B
(Radius as a function of mass) (7)
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We can compute the force on a current-carrying conductor starting with the magnetic force
~
= q~ ×
~ on a single moving charge. Let’s assume that all the positive charges are moving with a uniform velocity v d
.
Figure 18: This figure shows the current moving “upwards” in a uniform magnetic field into the page. The resulting force
~
= v d
× ~ points to the left.
~ pointing
F = ( nA` )( qv d
B ) where n is the number density of positive charges, A is the cross-sectional area, and ` is the “short” segment length of wire. If we identify the current density as
J = nqv d and J A = I , we can rewrite the above equation as:
F = I`B
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If the magnetic field
~ is not perpendicular to the wire but makes an angle φ it, then F = I`B sin φ . This equation can be written using vector notation as: with
~
= I ~ ×
~
(8) where
~ points in the current direction. If the conductor is not straight, we can divide it into infinitesimal segments d~ . The differential force due to this differential line segment will be: d ~ = I d~ ×
~ where, once again, d~ points in the direction of the current .
(9)
Figure 19: This figure shows the force on a current-carrying wire of length ` subtending an angle φ with respect to the magnetic field
~
.
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5.1
Magnetic Force on a Curved Conductor
Figure 20: This figure shows ions with the same velocity but different masses being separated in the second magnetic field at the bottom of the figure. In this figure, the magnetic field
~ is presumed to be the same inside and outside the velocity selector.
dF x
= IR dθ B cos θ dF y
= IR dθ B
F x
F y
=
=
Z
Z dF x dF y
= IRB
Z
π
= IRB
Z
π
0 cos θ dθ = 0 sin θ dθ = 2 IRB
0 sin θ
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Figure 21: This figure shows how to calculate the forces on each of the four segments of the current loop. From this, the resulting torques can be calculated.
The force on each segment is going to be:
~
= I ~ ×
~
While the sum of the forces is zero , the sum of the torques is generally not .
X
F i
= 0 however, in general
X
~ = 0
τ = 2 F ( b/ 2) sin φ = ( IBa )( b sin φ )
The magnitude of the magnetic torque on a current loop is:
τ = IBA sin φ (10) where A is the area of the loop and φ is the angle between the area-normal and the magnetic field
~
= A n
~
. The produce IA is called the magnetic dipole moment
~ . So, we can write the torque as
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Torque on a Magnetic Dipole
~ = µ × (Torque on a magnetic dipole)
Potential Energy for a Magnetic Dipole
U = − ~ · B
(11)
(12)
Figure 22: This figure shows the principle design of a dc motor (direct current). As the loop is rotating, the torque varies but always in a direction to cause the loop to continuously turn in the counter-clockwise direction.
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