31.1. Introduction
The magnetic field at a distance r from a very long straight wire, carrying a
steady current I, has a magnitude equal to
(31.)
and a direction perpendicular to r and I. The path integral along a circle
centered around the wire (see Figure 31.1) is equal to
(31.2)
Here we have used the fact that the magnetic field is tangential at any point
on the circular integration path.
Figure 31.1. Magnetic field generated by current.
Any arbitrary path can be thought of as a collection of radial
segments (r changes and [theta] remains constant) and circular segments
([theta] changes and r remains constant). For the radial segments the magnetic
field will be perpendicular to the displacement and the scaler product between
the magnetic field and the displacement is zero. Consider now a small circular
segment of a trajectory around the wire (see Figure 31.2). The path integral
along this circular segment is equal to
(31.3)
Figure 31.2. Path integral along a small circular path.
Equation (31.3) shows that the contribution of this circular
segment to the total path integral is independent of the distance r and only
depends on the change in the angle [Delta][theta]. For a closed path, the
total change in angle will be 2[pi], and eq.(31.3) can be rewritten as
(31.4)
This expression is
Ampere's Law:
" The integral of B around any closed mathematical path equals u
0
times the current intercepted by the area spanning the path "
Example: Problem 31.5
Six parallel aluminum wires of small, but finite, radius lie in the
same plane. The wires are separated by equal distances d, and they carry equal
currents I in the same direction. Find the magnetic field at the center of the
first wire. Assume that the currents in each wire is uniformly distributed
over its cross section.
A schematic layout of the problem is shown in Figure 31.3. The magnetic field
generated by a single wire is equal to
(31.5)
where r is the distance from the center of the wire. Equation (31.5) is
correct for all points outside the wire, and can therefore be used to determine
the magnetic field generated by wire 2, 3, 4, 5, and 6. The field at the
center of wire 1, due to the current flowing in wire 1, can be determined using
Ampere's law, and is equal to zero. The total magnetic field at the center of
wire 1 can be found by vector addition of the contributions of each of the six
wires. Since the direction of each of these contributions is the same, the
total magnetic field at the center of wire 1 is equal to
(31.6)
Figure 31.3. Problem 31.5
31.2. The solenoid
A solenoid is a device used to generate a homogeneous magnetic field. It can
be made of a thin conducting wire wound in a tight helical coil of many turns.
The magnetic field inside a solenoid can be determined by summing the magnetic
fields generated by N individual rings (where N is the number of turns of the
solenoid). We will limit our discussion of the magnetic field generated by a
solenoid to that generated by an ideal solenoid which is infinitely long, and
has very tightly wound coils.
The ideal solenoid has translational and rotational symmetry. However, since
magnetic field lines must form closed loops, the magnetic field can not be
directed along a radial direction (otherwise field lines would be created or
destroyed on the central axis of the solenoid). Therefore we conclude that the
field lines in a solenoid must be parallel to the solenoid axis. The magnitude
of the magnetic field can be obtained by applying Ampere's law.
Figure 31.4. The ideal solenoid.
Consider the integration path shown in Figure 31.4. The path
integral of the magnetic field around this integration path is equal to
(31.7)
where L is the horizontal length of the integration path. The current enclosed
by the integration path is equal to N
. I
0 where N is the
number of turns enclosed by the integration path and I
0 is the
current in each turn of the solenoid. Using Ampere's law we conclude that
(31.8)
or
(31.9)
where n is the number of turns of the solenoid per unit length. Equation
(31.9) shows that the magnetic field B is independent of the position inside
the solenoid. We conclude that the magnetic field inside an ideal solenoid is
uniform.
Example: Problem 31.14
A long solenoid of n turns per unit length carries a current I, and a
long straight wire lying along the axis of this solenoid carries a current I'.
Find the net magnetic field within the solenoid, at a distance r from the axis.
Describe the shape of the magnetic field lines.
The magnetic field generated by the solenoid is uniform, directed parallel to
the solenoid axis, and has a magnitude equal to
(31.10)
The magnetic field if a long straight wire, carrying a current I' has a
magnitude equal to
(31.11)
and is directed perpendicular to the direction of r and I'. The direction of
B
wire is therefore perpendicular to the direction of
B
sol. The net magnetic field inside the solenoid is equal to the
vector sum of B
wire and B
sol. Its magnitude is equal
to
(31.12)
The angle a between the direction of the magnetic field and the z-axis is given
by
(31.13)
Example: Problem 31.15
A coaxial cable consists of a long cylindrical copper wire of radius
r
1 surrounded by a cylindrical shell of inner radius r
2
and outer radius r
3 (see Figure 31.5). The wire and the shell carry
equal and opposite currents I uniformly distributed over their volumes. Find
formulas for the magnetic field in each of the regions r < r
1,
r
1 < r < r
2, r
2 < r <
r
3, and r > r
3.
The magnetic field lines are circles, centered on the symmetry axis of the
coaxial cable. First consider an integration path with r < r
1.
The path integral of B along this path is equal to
(31.14)
The current enclosed by this integration path is equal to
(31.15)
Applying Faraday's law we can relate the current enclosed to the path integral
of B
(31.16)
Therefore, the magnetic field is B is equal to
(31.17)
Figure 31.5. Problem 31.15.
In the region between the wire and the shell, the enclosed
current is equal to I and the path integral of the magnetic field is given by
eq.(31.14). Ampere's law states then that
(31.18)
and the magnetic field is given by
(31.19)
In the third region (r
2 < r < r
3) the path integral
of the magnetic field along a circular path with radius r is given by
eq.(31.14). The enclosed current is equal to
(31.20)
The magnetic field is therefore equal to
(31.21)
The current enclosed by an integration path with a radius r > r
3
is equal to zero (since the current in the wire and in the shell are flowing in
opposite directions). The magnetic field in this region is therefore also
equal to zero.
31.3. Motion of charges in electric and magnetic fields
The magnetic force acting on particle with charge q moving with velocity v is
equal to
(31.22)
This force is always perpendicular to the direction of motion of the particle,
and will therefore only change the direction of motion, and not the magnitude
of the velocity. If the charged particle is moving in a uniform magnetic
field, with strength B, that is perpendicular to the velocity v, then the
magnitude of the magnetic force is given by
(31.23)
and its direction is perpendicular to v. As a result of this force, the
particle will carry out uniform circular motion. The radius of the circle is
determined by the requirement that the strength magnetic force is equal to the
centripetal force. Thus
(31.24)
The radius r of the orbit is equal to
(31.25)
where p is the momentum of the charged particle. The distance traveled by the
particle in one revolution is equal to
(31.26)
The time T required to complete one revolution is equal to
(31.27)
The frequency of this motion is equal to
(31.28)
and is called the
cyclotron frequency. Equation (31.28) shows
that the cyclotron frequency is independent of the energy of the particle, and
depends only on its mass m and charge q.
The effect of a magnetic field on the motion of a charged particle can be used
to determine some of its properties. One example is a measurement of the
charge of the electron. An electron moving in a uniform magnetic field will
described a circular motion with a radius given by eq.(31.25). Suppose the
electron is accelerated by a potential V
0. The final kinetic energy
of the electron is given by
(31.29)
The momentum p of the electron is determined by its kinetic energy
(31.30)
The radius of curvature of the trajectory of the electron is thus equal to
(31.31)
Equation (31.31) shows that a measurement of r can be used to determine the
mass over charge ratio of the electron.
Another application of the effect of a magnetic field on the motion of a
charged particle is the cyclotron. A cyclotron consists of an evacuated cavity
placed between the poles of a large electromagnet. The cavity is cut into two
D-shaped pieces (called dees) with a gap between them. An oscillating high
voltage is connected to the plates, generating an oscillating electric field in
the region between the two dees. A charged particle, injected in the center of
the cyclotron, will carry out a uniform circular motion for the first half of
one turn. The frequency of the motion of the particle depends on its mass, its
charge and the magnetic field strength. The frequency of the oscillator is
chosen such that each time the particle crosses the gap between the dees, it
will be accelerated by the electric field. As the energy of the ion increases,
its radius of curvature will increase until it reaches the edge of the
cyclotron and is extracted. During its motion in the cyclotron, the ion will
cross the gap between the dees many times, and it will be accelerated to high
energies.
Up to now we have assumed that the direction of the motion of the charged
particle is perpendicular to the direction of the magnetic field. If this is
the case, uniform circular motion will result. If the direction of motion of
the ion is not perpendicular to the magnetic field, spiral motion will result.
The velocity of the charged particle can be decomposed into two components: one
parallel and one perpendicular to the magnetic field. The magnetic force
acting on the particle will be determined by the component of its velocity
perpendicular to the magnetic field. The projection of the motion of the
particle on the x-y plane (assumed to be perpendicular to the magnetic field)
will be circular. The magnetic field will not effect the component of the
motion parallel to the field, and this component of the velocity will remain
constant. The net result will be spiral motion.
31.4. Crossed electric and magnetic fields
A charged particle moving in a region with an electric and magnetic field will
experience a total force equal to
(31.32)
This force is called the
Lorentz force.
Figure 31.6. Charged particle moving in crossed E and B
fields.
Consider a special case in which the electric field is
perpendicular to the magnetic field. The motion of a charged particle in such
a region can be quit complicated. A charged particle with a positive charge q
and velocity v is moving in this field (see Figure 31.6). The direction of the
particle shown in Figure 31.6 is perpendicular to both the electric field and
the magnetic field. The electric force acting on the particle is directed
along the direction of the electric field and has a magnitude equal to
(31.33)
The magnetic force acting on the charge particle is directed perpendicular to
both v and B and has a magnitude equal to
(31.34)
The net force acting on the particle is the sum of these two components and has
a magnitude equal to
(31.35)
If the charged particle has a velocity equal to
(31.36)
then the net force will be equal to zero, and the motion of the particle will
be uniform linear motion. A device with crossed electric and magnetic fields
is called a velocity selector. If slit are placed in the appropriate
positions, it will transport only those particles that have a velocity defined
by the magnitudes of the electric and magnetic fields.
Figure 31.7. Current in a magnetic field.
A technique used to determine the density and sign of charge
carriers in a metal is based on the forces exerted by crossed E and B fields on
the charge carriers. The diagram shown in Figure 31.7 shows a metallic strip
carrying a current in the direction shown and placed in a uniform magnetic
field with the direction of the magnetic field being perpendicular to the
electric field (which generates the current I). Suppose the charge carriers in
the material are electrons, than the electrons will move in a direction
opposite to that of the current (see Figure 31.7). Since the magnetic field is
perpendicular to the electric field, it is also perpendicular to the direction
of motion of the electrons. As a result of the magnetic force, the electrons
are deflected downwards, and an excess of negative charge will be created on
the bottom of the strip. At the same time, a deficit of negative charge will
be created at the top of the strip. This charge distribution will generate an
electric field that is perpendicular to the external electric field and, under
equilibrium conditions, the electric force produced by this field will cancel
the magnetic force acting on the electrons. When this occurs, the internal
electric field, E
in, is equal to the product of the electron
velocity, v
d, and the strength of the magnetic field, B. As a
result of the internal electric field, a potential difference will be created
between the top and bottom of the strip. If the metallic strip has a width w,
then the potential difference [Delta]V will be equal to
(31.37)
This effect is called the Hall effect.
The drift velocity of the electrons depend on the current I in the wire, its
cross sectional area A and the density n of electrons (see Chapter 28):
(31.38)
Combining eq.(31.38) and eq.(31.37) we obtain the following expression for
[Delta]V
(31.39)
A measurement of [Delta]V can therefore be used to determine n.
31.5. Forces on a wire
A current I flowing through a wire is equivalent to a collection of charges
moving with a certain velocity v
d along the wire. The amount of
charge dq present in a segment dL of the wire is equal to
(31.40)
If the wire is placed in a magnetic field, a magnetic force will be exerted on
each of the charge carriers, and as a result, a force will be exerted on the
wire. Suppose the angle between the direction of the current and the direction
of the field is equal to [theta] (see Figure 31.8). The magnetic force acting
on the segment dL of the wire is equal to
(31.41)
The total force exerted by the magnetic field on the wire can be found by
integrating eq.(31.41) along the entire wire.
Figure 31.8. Magnetic force on wire.
Example: Problem 31.33
A balance can be used to measure the strength of the magnetic field.
Consider a loop of wire, carrying a precisely known current, shown in Figure
31.9 which is partially immersed in the magnetic field. The force that the
magnetic field exerts on the loop can be measured with the balance, and this
permits the calculation of the strength of the magnetic field. Suppose that
the short side of the loop measured 10.0 cm, the current in the wire is 0.225
A, and the magnetic force is 5.35 x 10
-2 N. What is the strength of
the magnetic field ?
Consider the three segments of the current loop shown in Figure 31.9 which are
immersed in the magnetic field. The magnetic force acting on segment 1 and 3
have equal magnitude, but are directed in an opposite direction, and therefore
cancel. The magnitude of the magnetic force acting on segment 2 can be
calculated using eq.(31.41) and is equal to
(31.42)
This force is measured using a balance and is equal to 5.35 x 10
-2
N. The strength of the magnetic field is thus equal to
(31.43)
Figure 31.9. Current loop in immersed in magnetic field.
31.6. Torque on a current loop
If a current loop is immersed in a magnetic field, the net magnetic force will
be equal to zero. However, the torque on this loop will in general not be
equal to zero. Suppose a rectangular current loop is placed in a uniform
magnetic field (see Figure 31.10). The angle between the normal of the current
loop and the magnetic field is equal to [theta]. The magnetic forces acting on
the top and the bottom sections of the current loop are equal to
(31.44)
where L is the length of the top and bottom edge. The torque exerted on the
current loop, with respect to its axis, is equal to
(31.45)
Figure 31.10. Current loop placed in uniform magnetic field.
Using the definition of the magnetic dipole moment u, discussed
in Chapter 30, eq.(31.45) can be rewritten as
(31.46)
where
(31.47)
Using vector notation, eq.(31.45) can be rewritten as
(31.48)
where the direction of the magnetic moment is defined using the right-hand
rule.
The work that must be done against the magnetic field to rotate the current
loop by an angle d[theta] is equal to - [tau] d[theta]. The change in
potential energy of the current loop when it rotates between
[theta]
0 and [theta]
1 is given by
(31.49)
A common choice for the reference point is [theta]
0 = 90deg. and
U([theta]
0) = 0 J. If this choice is made we can rewrite eq.(31.50)
as
(31.50)
In vector notation:
(31.51)
The potential energy of the current loop has a minimum when u and B are
parallel, and a maximum when u and B are anti-parallel.
