Magnetic Force on a Moving Charge
Electric charges moving in a magnetic field experience a force due to
the magnetic field. Given a charge Q moving with velocity u in a magnetic
flux density B, the vector magnetic force Fm on the charge is given by
Note that the force is normal to the plane containing the velocity vector and
the magnetic flux density vector. Also note that the force is zero if the
charge is stationary (u=0).
Example
Determine the vector magnetic force on a point charge +Q moving at
a uniform velocity u=uoay in a uniform magnetic flux density B=Boaz.
Given a charge moving in an electric field and a magnetic field, the
total force on the charge is the superposition of the forces due to the electric
field and the magnetic field. This total force equation is known as the
Lorentz force equation. The vector force component due to the electric
field (Fe) is given by
The total vector force (Lorentz force - F) is
The Lorentz force can also be written in terms of Newton’s law such that
where m is the mass of the charged particle.
Magnetic Force on Current
Given that charge moving in a magnetic field experiences a force, a
current carrying conductor in a magnetic field also experiences a force. The
current carrying conductor (line, surface or volume current) can be
subdivided into current elements (differential lengths, differential surfaces
or differential volumes). The charge-velocity product for a moving point
charge can be related to an equivalent differential length of line current.
The equivalence of the moving point charge and the differential length of
line current yields the equivalent magnetic force equation.
In a similar fashion for surface and volume currents, we find
such that
The overall force on a line, surface and volume current is found by
integrating over the current distribution.
Example (Force between line currents)
Determine the force/unit length on a line current I1 due to the
magnetic flux produced by a parallel line current I2 (separation distance =
d) flowing in the opposite direction.
The magnetic flux produced on the z-axis (I1) due to the current I2 is
The force on a length l of current I1 due to the flux produced by I2 is
The force per unit length on the current I1 is
The force on a length l of current I2 due to the flux produced by I1 is
The force per unit length on the current I2 is
Note that the currents repel each other given the currents flowing in
opposite directions. If the currents flow in the same direction, they attract
each another.
Torque on a Current Loop
Given the change in current directions around a closed current loop,
the magnetic forces on different portions of the loop vary in direction.
Using the Lorentz force equation, we can show that the net force on a
simple circular or rectangular loop is a torque which forces the loop to align
its magnetic moment with the applied magnetic field.
Consider the rectangular current loop shown below. The loop lies in
the x-y plane and carries a DC current I. The loop lies in a uniform
magnetic flux density B given by
The loop consists of four current segments carrying distinct vector current
components.
Given a uniform flux density and a DC current along straight current
segments, the magnetic force on each conductor segment can be simplified
to the following equation.
The forces on the current segments can be determined for each component
of the magnetic flux density.
Forces due to Bz
Forces due to By
The vector torque on the loop is defined in terms of the force magnitude
(IByl2 ), the torque moment arm distance (l1 /2), and the torque direction
(defined by the right hand rule):
where A=l1l2 is the loop area. The vector torque can be written compactly
by defining the magnetic moment (m) of the loop.
where an is the unit normal to the loop (defined by the right hand rule as
applied to the current direction).
The magnitude of the torque in terms of the magnetic moment is
The vector torque is then
Note that the torque on the loop tends to align the loop magnetic moment
with the direction of the applied magnetic field.
Magnetization
Just as dielectric materials are polarized under the influence of an
applied electric field, certain materials can be magnetized under the
influence of an applied magnetic field. Magnetization for magnetic fields
is the dual process to polarization for electric fields. The magnetization
process may be defined using the magnetic moments of the electron orbits
within the atoms of the material. Each orbiting electron can be viewed as
a small current loop with an associated magnetic moment.
An unmagnetized material can be characterized by a random distribution
of the magnetic moments associated with the electron orbits. These
randomly oriented magnetic moments produce magnetic field components
that tend to cancel one another (net H=0). Under the influence of an
applied magnetic field, many of the current loops align their magnetic
moments in the direction of the applied magnetic field.
If most of the magnetic moments stay aligned after the applied magnetic
field is removed, a permanent magnet is formed.
The bar magnet can be viewed as the magnetic analogy to the electric
dipole. The poles of the bar magnet can be represented as equivalent
magnetic charges separated by a distance l (the length of the magnet).
The magnetic flux density produced by the magnetic dipole is equivalent
to the electric field produced by the electric dipole.
The preceding equations assume the dipole is centered at the coordinate
origin and oriented with its dipole moment along the z-axis. A current loop
and a solenoid produce the same B as the bar magnet at large distances (in
the far field) if the magnetic moments of these three devices are equivalent.
If the magnetic moments of these three devices are equal,
they produce essentially the same magnetic field at large distances
(equivalent sources). Assuming the same vector magnetic moment, all
three of these devices would experience the same torque when placed in a
given magnetic field.
The parameters associated with the magnetization process are duals
to those of the polarization process. The magnetization vector M is the
dual of the polarization vector P.
Note that the magnetic susceptibility Pm is defined somewhat differently
than the electric susceptibility Pe. However, just as the electric
susceptibility and relative permittivity are a measure of how much
polarization occurs in the material, the magnetic susceptibility and relative
permeability are a measure of how much magnetization occurs in the
material.
Magnetic Materials
Magnetic materials can be classified based on the magnitude of the
relative permeability. Materials with a relative permeability of just under
one (a small negative magnetic susceptibility) are defined as diamagnetic.
In diamagnetic materials, the magnetic moments due to electron orbits and
electron spin are very nearly equal and opposite such that they cancel each
other. Thus, in diamagnetic materials, the response to an applied magnetic
field is a slight magnetic field in the opposite direction. Superconductors
exhibit perfect diamagnetism (Pm = !1) at temperatures near absolute zero
such that magnetic fields cannot exist inside these materials.
Materials with a relative permeability of just greater than one are
defined as paramagnetic. In paramagnetic materials, the magnetic
moments due to electron orbit and spin are unequal, resulting in a small
positive magnetic susceptibility. Magnetization is not significant in
paramagnetic materials. Both diamagnetic and paramagnetic materials are
typically linear media.
Materials with a relative permeability much greater than one are
defined as ferromagnetic. Ferromagnetic materials are always nonlinear.
As such, these materials cannot be described by a single value of relative
permeability. If a single number is given for the relative permeability of
any ferromagnetic material, this number represents an average value of :r.
Diamagnetic
Paramagnetic
Ferromagnetic
:r <1
:r >1
:r o 1
Ferromagnetic materials lose their ferromagnetic properties at very high
temperatures (above a temperature known as the Curie temperature).
The characteristics of ferromagnetic materials are typically presented
using the B-H curve, a plot of the magnetic flux density B in the material
due to a given applied magnetic field H.
The B-H curve shows the initial magnetization curve along with a
curve known as a hysteresis loop. The initial magnetization curve shows
the magnetic flux density that would result when an increasing magnetic
field is applied to an initially unmagnetized material. An unmagnetized
material is defined by the B=H=0 point on the B-H curve (no net magnetic
flux given no applied field). As the magnetic field increases, at some point,
all of the magnetic moments (current loops) within the material align
themselves with the applied field and the magnetic flux density saturates
(Bm). If the magnetic field is then cycled between the saturation magnetic
field value in the forward and reverse directions (±Hm), the hysteresis loop
results. The response of the material to any applied field depends on the
initial state of the material magnetization at that instant.
Magnetic Boundary Conditions
The fundamental boundary conditions involving magnetic fields relate
the tangential components of magnetic field and the normal components of
magnetic flux density on either side of the media interface. The same
techniques used to determine the electric field boundary conditions can be
used to determine the magnetic field boundary conditions.
Tangential Magnetic Field
In order to determine the boundary condition on the tangential
magnetic field at a media interface, Ampere’s law is evaluated around a
closed incremental path that extends into both regions as shown below.
According to Ampere’s law, the line integral of the magnetic field around
the closed loop yields the current enclosed.
If we take the limit of this integral as )y = 0, the integral contributions on
the vertical paths goes to zero.
Assuming the magnetic field and surface current components along the
incremental length )x are uniform, the integrals above reduce to
Dividing this result by )x gives
For this example, if the surface current flows in the opposite direction, we
obtain
or
Thus, the general boundary condition on the tangential magnetic field is
The tangential components of magnetic field are
discontinuous across a media interface by an amount
equal to the surface current density on the interface.
Note that the previous boundary condition relates only scalar
quantities. The vector magnetic field and surface current at the media
interface can be shown to satisfy a vector boundary condition defined by
Vector boundary condition relating the magnetic field
and surface current at a media interface.
where n is a unit normal to the interface pointing into region 1.
Given a surface current on the interface, the tangential magnetic field
components on either side of the interface point in opposite directions.
Normal Magnetic Flux Density
In order to determine the boundary condition on the normal magnetic
flux density at a media interface, we apply Gauss’s law for magnetic fields
to an incremental volume that extends into both regions as shown below.
The application of Gauss’s law for magnetic fields to the closed surface
above gives
If we take the limit as the height of the volume )z approaches 0, the
integral contributions on the four sides of the volume vanish.
The integrals over the upper and lower surfaces on either side of the
interface reduce to
where the magnetic flux density is assumed to be constant over the upper
and lower incremental surfaces. Evaluation of the surface integrals yields
Dividing by )x )y gives
such that the general boundary condition on the normal component of
magnetic flux density becomes.
The normal components of magnetic flux density
are continuous across a media interface.
Example (Magnetic field boundary conditions, ferromagnetic materials)
Region 1 is air (z>0, : = :o) and region 2 is iron (z<0,
: =200 :o). The magnetic field in the air region is given by
Determine the vector magnetic field and vector magnetic flux density
in the iron.
Given no current on the interface, both the tangential magnetic field
and the normal magnetic flux density are continuous across the media
interface.
Inductors and Inductance
An inductor is an energy storage device that stores energy in an
magnetic field. A inductor typically consists of some configuration of
conductor coils (an efficient way of concentrating the magnetic field). Yet,
even straight conductors contain inductance. The parameters that define
inductors and inductance can be defined as parallel quantities to those of
capacitors and capacitance.
Inductance
Definition
Capacitance
Definition
The flux linkage of an inductor defines the total magnetic flux that
links the current. If the magnetic flux produced by a given current links
that same current, the resulting inductance is defined as a self inductance.
If the magnetic flux produced by a given current links the current in another
circuit, the resulting inductance is defined as a mutual inductance.
Example (Transformer / self and mutual inductance)
8mn = total flux linkage through circuit m due to current in circuit n
Rmn = total magnetic flux through circuit m due to current in circuit n
Example (Self inductance / long solenoid)
Given the long solenoid (loa), the magnetic field throughout the
solenoid can be assumed to be constant.
According to the definition of inductance,
the inductance of the long solenoid is
The total magnetic flux through the
solenoid is
Inserting the total magnetic flux expression into the inductance equation
yields
Note that the inductance of the long solenoid is directly proportional to the
permeability of the medium inside the core of the solenoid. By using a
ferromagnetic material such as iron as the solenoid core, the inductance can
be increased significantly given the large relative permeability of a
ferromagnetic material.
Example (Self inductance / toroid)
The equation for the self inductance of the long solenoid can be used
to determine the self inductance of the toroid assuming that the toroid mean
length (l=2BDo) is large relative to the cross-sectional radius a.
The length l in the long solenoid equation is simply replaced by the mean
length of the toroid to give
Mutual Inductance Calculations
The mutual inductance between two distinct circuits can be
determined by assuming a current in one circuit and determining the flux
linkage to the opposite circuit.
Example (Mutual inductance between coaxial loops)
Determine the mutual inductance between two coaxial loops as
shown below. Assume that the loop separation distance h is large
relative to the radii of both loops (a, b). Also assume that loop ã is
much smaller than loop â (b n a).
If we assume that a current flows in loop â, the magnetic flux density
produced by loop â along its axis is
Given that loop ã is much smaller than loop â, we may assume that the
flux produced by loop â is nearly uniform over the area of loop ã and is
approximately equal to that at the loop center. The flux produced by loop
â can also be assumed to be approximately normal to loop ã over its area.
Thus, the total flux produced by loop â linking loop ã is approximately
From the definition of inductance, the mutual inductance between the loops
is
Note that the calculations required by assuming the current in loop â are
much easier than those required by assuming the current in loop ã. The
calculations are easier based on the approximations employed. If the
current is assumed in loop ã, the resulting magnetic flux density over loop
â is a rather complicated function of position which requires a complex
integration to arrive at the same approximate answer.
Internal and External Inductance
In general, a current carrying conductor has magnetic flux internal
and external to the conductor. Thus, the magnetic flux inside the conductor
can link portions of the conductor current which produces a component of
inductance designated as internal inductance. The magnetic flux outside
the conductor that links the conductor current is designated as external
inductance.
The most efficient technique in determining the internal and external
components of inductance is the energy method. The energy method for
determining inductance is based on the total magnetic energy expression
for an inductor given by
Solving this equation for the inductance yields
The internal and external components of the inductance are found by
integrating the internal and external magnetic flux density expressions,
respectively.
Example (Internal inductance of a cylindrical conductor)
Determine the internal inductance for a cylindrical conductor of
radius a carrying a uniform current density J.
The magnetic flux density internal to the conductor is given by
The internal inductance of a length l of conductor is found by integrating
the internal magnetic flux density throughout the internal volume of the
conductor which gives
Note that the internal inductance of the conductor is independent of the
radius. The internal inductance per unit length of a non-magnetic
conductor (LintN) is
If the same process is attempted for the external inductance, the
integral of the external magnetic flux density yields an infinite value. The
infinite result is obtained due to the fact that an infinite length conductor
with no return current is a nonphysical situation. Realistic current
configurations such as the two-wire transmission line or the coaxial
transmission line have a return current yielding a finite result.
Example (External inductance of a coaxial transmission line)
Determine the external inductance for a coaxial transmission
line assuming uniform current densities in both the inner conductor
(radius=a) and the outer conductor (inner radius=b).
The magnetic flux density between the conductors of the coaxial line is
The external inductance is found by integrating the magnetic flux density
external to the conductors (between the conductors) according to
The external inductance per unit length (LextN) for the coaxial transmission
line is
If we compare the coaxial transmission line external inductance per unit
length with the capacitance per unit length (CN) given by
we find that
This relationship is true in general for any conductor geometry just as
Thus, the external inductance can be found quickly if the capacitance of the
conductor configuration is already known.
Example (External inductance of a two-wire transmission line)
The capacitance per unit length of the two wire transmission line is
The corresponding external inductance per unit length for the two wire line
is
Magnetic Forces on Magnetic Materials
Magnetic materials experience a force when placed in an applied
magnetic field. This type of force is seen when a bar magnet is placed in
a magnetic field, such as that of another bar
magnet. The like poles repel each other while
unlike poles attract each other. Note that the bar
magnets behave in the same way as current
loops in an applied magnetic field as each
magnet tends to align its magnetic moment in
the direction of the applied field of the opposite
magnet.
The magnetized needle on a compass
obeys the same principle as the compass needle
aligns itself with the Earth’s magnetic field. The
magnetic field on the Earth’s surface has a
horizontal component that points toward the
magnetic south pole (near the geographic north pole). The horizontal
component of magnetic field on the Earth’s surface is maximum near the
equator (approximately 35 :T) and falls to zero at the magnetic poles.
An electromagnet can be used to lift ferromagnetic materials (such as
scrap iron). The lifting force F of the electromagnet can be determined by
considering how much energy is stored in the air gaps when the
ferromagnetic materials are
separated. The magnitude of the
force necessary to separate the
two pieces of ferromagnetic
material equals the total amount
of magnetic energy stored in the
air gaps after separation.
The total energy required
to move the lower magnetic
piece a distance l from the
electromagnet is the product of
the force magnitude F and the
distance l. The total magnetic
energy in the two air gaps is found by integrating the magnetic energy
densities in the air gaps. Assuming short air gaps, the magnetic field in the
air gaps can be assumed the be uniform and confined to the volume below
the electromagnet poles (the fringing effect of the magnetic field is
negligible for small air gaps). The uniform magnetic field in the air gap
produces a uniform energy density so that the total magnetic energy stored
in each air gap is a simply the product of the energy density and the volume
of the air gap.
Solving this equation for the force magnitude gives
The air gap magnetic field can be determined according to the boundary
condition for the normal component of magnetic flux across the air gap.
The magnetic flux density is normal to the interfaces between the air
and the ferromagnetic core. According to the boundary condition on the
normal component of magnetic flux density, the magnetic flux density must
be continuous across the air gap.
Rewriting this boundary condition in terms of
the magnetic fields gives
or
Given either the magnetic field or the magnetic
flux density in the core, the corresponding
quantity in the air gap can be found according
to the previous equations. Note that the
magnetic field in the air gap is much larger than that in the core given the
large relative permeability of the ferromagnetic core.
Magnetic Circuits
Magnetic field problems involving components such as current coils,
ferromagnetic cores and air gaps can be solved as magnetic circuits
according to the analogous behavior of the magnetic quantities to the
corresponding electric quantities in an electric circuit.
Given that reluctance in a magnetic circuit is analogous to resistance
in an electric circuit, and permeability in a magnetic circuit is analogous to
conductivity in an electric circuit, we may interpret the permeability of a
medium as a measure of the resistance of the material to magnetic flux.
Just as current in an electric circuit follows the path of least resistance, the
magnetic flux in a magnetic circuit follows the path of least reluctance.
Example (Magnetic circuit)
For the magnetic circuit shown, determine the current I required
to produce a total magnetic flux of 0.5mWb in the iron core. Assume
the relative permeability of the iron is 1000.
This problem could also be solved using Ampere’s law by integrating
the magnetic field along the centerline of the iron core. The magnetic field
inside the iron core can be assumed to be uniform. Integrating along the
core centerline in the direction of H gives
or
Given a uniform magnetic field in the iron core, the total magnetic flux in
the core is
Inserting the magnetic field expression into the current expression above
gives
which is the same equation found using the magnetic circuit.
Example (Series/parallel magnetic circuits)
Determine the magnetic field in the air gap of the magnetic
circuit shown below. The cross sectional area of all branches is 10
cm2 and :r =50.
The equivalent electric circuit is
The reluctance components for the magnetic circuit are
The equivalent circuit can be reduced to