Magnetostatic Fields
Biot-Savart Law
According to Coulomb’s law, any distribution of stationary charge
produces a static electric field (electrostatic field). The analogous equation
to Coulomb’s law for electric fields is the Biot-Savart law for magnetic
fields. The Biot-Savart law shows that when charge moves at a constant
rate (direct current - DC), a static magnetic field (magnetostatic field) is
produced. When the rate of charge movement varies with time (for
example, an alternating current - AC), we find that coupled electric and
magnetic fields are produced (electromagnetic field).
The Biot-Savart law defines the magnetostatic field produced by a
steady current. The overall magnetic field produced by an arbitrary vector
line current I (filament, zero cross-section) is expressed as a line integral
of the current. According to the Biot-Savart law, the differential vector
magnetic field (dH) at the field point P produced by a differential element
of current IdlN is
Stationary
Charge
Static
Electric Field
(Electrostatics)
or
Steady
Current
Static Magnetic
Field
(Magnetostatics)
where
Time-varying
Current
Dynamic Electric
and Magnetic Fields
(Electromagnetics)
r - vector locating the field point P
rN - vector locating the source point (IdlN)
Static magnetic fields are also produced by stationary permanent magnets.
When permanent magnets are set in motion such that a time-varying
magnetic field is produced, a time-varying electric field is simultaneously
produced. A time-varying electric field cannot exist without a
corresponding time-varying magnetic field and vice versa.
All of the previously defined equations related to the electric field
have dual equations related to the magnetic field. All of the magnetic field
terms in these dual equations have dual units to those electric field terms
in the electric field equations.
R = r!rN
(vector from the source point to the field point)
R = *R* = *r!rN* (distance from the source point to the field point)
aR = R/R
(unit vector in the direction of R)
l - unit vector in the direction of the current I at dlN
" - angle between l and aR
Note that the direction of the magnetic field is given by the direction
of I×R. For infinite length line currents, the magnetic field direction is
given by the right hand rule.
The scalar and vector forms of the total magnetic field for an arbitrary
surface current K (A/m) are given by
Scalar
Vector
The scalar and vector forms of the total magnetic field for an arbitrary
volume current J (A/m2) are given by
Scalar
The scalar and vector forms of the total magnetic field for an
arbitrary line current I (A) are given by
Scalar
Vector
Vector
Note that the magnetic field is proportional to the product of the
current and the differential element in each case. This product is defined
as the current moment and has units of A-m.
Example (Biot-Savart law / line current)
Transformation of variable:
Determine the magnetic field of a line segment of current lying along
the z-axis extending from z=zA to z =zB.
Given this general result for the magnetic field of a current segment, we
may apply it to several special cases.
(1)
(Given the field point P, the direction
direction of aN does not change as z is
varied along the length of the current)
Line current, symmetric about the x-y plane, zA = !zo, zB = zo
If the field point lies in the x-y plane, then
z = 0, R A = R B = R o = [D2 + zo2 ]1/2
(2)
Semi-infinite line current, zN0 (0, 4 )
zA = 0, zB = 4, RA = [ D2 +z 2 ]1/2, RB = 4
Example (Current loop / straight segments)
Determine the magnetic field at the center of the current loop in the
shape of an equilateral triangle (side length l = 4m) carrying a steady
current of 5A.
If the field point lies in the x-y plane (z = 0),
(3)
Infinite line current, zN0 (!4, 4 )
Wherever the field point is located, the infinite length line
current can be viewed as two semi-infinite line currents. The
resulting magnetic field is twice that of the semi-infinite length
current segment.
The previous formulas are useful when determining the magnetic field of
a closed current loop made up of straight segments. The principle of
superposition may be applied to determine the total magnetic field
produced by the loop. The total magnetic field produced by the loop is the
vector sum of the magnetic field contributions from each current segment.
Magnetic Field Due to a Circular Current Loop
The Biot-Savart law can be used to determine the magnetic field at
the center of a circular loop of steady current.
0
0
The integrals for the x and y components of the magnetic field are zero.
These field components can be shown to be zero by symmetry.
At the loop center (h=0),
Symmetry of the loop magnetic
field when the field point lies
on the loop axis
Solenoid
A solenoid is a cylindrically shaped current carrying coil. The
solenoid is the magnetic field equivalent to the parallel plate capacitor for
electric fields. Just as the parallel plate capacitor concentrates the electric
field between the plates, the solenoid concentrates the magnetic field
within the coil. For the purpose of determining the solenoid magnetic
field, the solenoid of length l and radius a which is tightly wound with N
turns can be modeled as an equivalent uniform surface current on the
cylinder surface.
The Biot-Savart law integral to determine the magnetic field of the
solenoid equivalent surface current is
To determine the characteristics of the magnetic field inside the solenoid,
we choose the field point P on the z-axis.
The cross product in the numerator of the Biot-Savart law integral is
The equivalent uniform surface current density (Ko) for the solenoid is
found by spreading the total current of NI over the length l.
The aD unit vector can be transformed into rectangular components as
The Biot-Savart law integral for the solenoid becomes
0
0
For a long solenoid (l o a), the approximate magnetic field values at the
center and at the ends of the solenoid are
Thus, the magnetic field at the ends of a long solenoid is approximately
half that at the center of the solenoid. It can be shown that the magnetic
field over the length is a long solenoid is relatively constant. At the ends
of the long solenoid, the magnetic field falls rapidly to about one-half of
the peak value.
The remaining zN integration is the same form as that found for the current
segment on the z-axis. The result of the integration is
Example (Solenoid magnetic field)
Plot the magnetic field along the axis of the following three
solenoids. All three solenoids are characterized by N =100, l =10 cm, and
I =1 mA. The radii of the three solenoids are 2 cm, 1 cm, and 0.5
cm. These three radii yield solenoid length to radius ratios of 5, 10,
and1 20.
Ç
É
È
É
Ç È
0.9
0.8
The magnetic field at the center of the solenoid (z = 0) is
0.7
Ç
H (A/m)
0.6
È
É
0.5
Note that as the length to radius
ratio grows large, the assumption
of a uniform field along the axis
of the solenoid becomes more
accurate.
0.4
0.3
0.2
At either end of the solenoid (z = +l/2,!l/2),
0.1
0
-5
-4
-3
-2
-1
0
z (cm)
1
2
3
4
5
Ampere’s Law
Gauss’s law is the Maxwell equation that relates the electrostatic
field (flux) to the source of the electrostatic field (charge). Ampere’s law
is the Maxwell equation that relates the magnetostatic field (flux) to the
source of the magnetostatic field (current).
Example (Ampere’s law / magnetic field of a surface current)
Determine the vector magnetic field produced by a uniform axdirected surface current covering the x-y plane
Ampere’s Law - The line integral of the magnetic field around a
closed path equals the net current enclosed (the current direction is
implied by the direction of the path according to the right hand rule).
Example (Ampere’s law / infinite-length line current)
Given a infinite-length line current I lying along the z-axis, use
Ampere’s law to determine the magnetic field by integrating the magnetic
field around a circular path of radius D lying in the x-y plane.
Ampere’s law may be applied on the path shown below. Note that the path
dimensions are defined in terms of arbitrary distances (!y,+y) and (!z,+z)
such that the result is valid for any value of y and z.
From Ampere’s law,
By symmetry, the magnetic field is
uniform on the given path so that
or
This result agrees with that found using the Biot-Savart law.
To evaluate the Ampere’s law integral, we must first determine the vector
characteristics of the magnetic field. By symmetry, the magnetic field on
the horizontal segments of the path ( and ) must be uniform.
È
Ê
Also by symmetry, we can show that the magnetic field above the surface
current is everywhere !ay directed while the magnetic field everywhere
below the surface current is +ay directed. The overall surface current can
be subdivided into differential lengths (Ko dy) with each differential length
equivalent to a line current. For any given field point, and any given line
current, there is always another line current in the opposite direction that
produces a magnetic field component that when added to the magnetic
field component of the original line current, produces only a !ay
component (above) or +ay component (below) of magnetic field.
According to Ampere’s law, this integral is equal to the total current
enclosed by the path. The total current enclosed is the surface current
located between !y and +y. Given a uniform surface current, the total
current is the product of the surface current density and the distance (2y)
such that
Given the direction of the integration path, the direction of the enclosed
current is the +ax direction. From Ampere’s law,
With only horizontal components of magnetic field, the Ampere’s law
integrals on the vertical paths ( and ) are zero. The magnetic field on
the horizontal paths ( and ) may be written as
È
Ê
Ç
É
The magnetic field above and below the current sheet is
where Ho is a constant (the magnetic field is uniform). Given the magnetic
field characteristics on the horizontal and vertical paths, the Ampere’s law
integral can be evaluated to determine the magnetic field of the surface
current.
Note that the magnetic field is independent of the distance x so that the
magnetic field is uniform above and below the surface current.
Example (Ampere’s law / Magnetic field in a coaxial transmission line)
The coaxial transmission line shown below carries a total current I in
the +az direction through the inner conductor and in the !az direction
through the inner conductor. Assume uniform current densities in both
conductors of the coaxial transmission line. Use Ampere’s law to
determine the magnetic field everywhere.
By symmetry, on all four of the integration paths, the magnetic field is
uniform and has only an aN component. Thus, for each path, Ampere’s law
reduces to
or
The magnetic field in each region is proportional to the net current
enclosed by the path.
Within the inner conductor (D < a) [L1]
The uniform vector current densities in the inner and outer conductors of
the coaxial transmission line (Ji and Jo, respectively) are
To determine the magnetic field everywhere, Ampere’s law is applied on
circular paths in the four distinct regions for the coaxial transmission line.
Between the conductors (a < D < b) [L2]
I
2π a
Within the outer conductor (b < D < c) [L3]
I
2π b
According to the equations obtained from Ampere’s law, the magnetic
field in the four regions of the coaxial transmission line is
Outside the outer conductor (D < c) [L4]
The magnetic field of a single conductor carrying a uniform current density
can be determined from the results of the coaxial transmission line. With
no outer conductor, the curve for the region between the coaxial
conductors would continue for the single conductor.
Toroid
Another commonly encountered magnetic energy storage geometry
is the toroid. A toroid is formed by wrapping a conductor around a ring
of uniform cross-section (typically circular cross-section).
Solving for the toroid magnetic field yields
where l = 2BDo is the equivalent length of the toroid. The magnetic field
at any point within the toroid is the same as that found at the center of the
long solenoid. The primary advantage of the toroid over the solenoid is the
confinement of the magnetic field within the toroid as opposed to the
solenoid which produces magnetic fields external to the coil. Also, the
toroid does not suffer from the end effects (fringing) seen in the solenoid.
Differential Form of Ampere’s Law
(Curl Operator)
The differential form of Ampere’s law may be determined by
applying Ampere’s law to a differential surface. Given the differential
surface shown below, integration of the magnetic field around the path L
gives
The distance from the center of the ring to the center of the ring crosssection is defined as the mean radius Do. Given a circular cross-section of
radius a, if the mean radius is large relative to the radius of the cross
section (Do o a), then the toroid may be viewed as a long solenoid bent into
the shape of a circle (the magnetic field within the toroid may be assumed
to be uniform). Application of Ampere’s law on the mean radius path
gives
Dividing the Ampere’s law integral by the differential surface )S and
taking the limit as )S approaches zero defines the x-component of the curl
operator.
To evaluate the Ampere’s law integral needed to define the x-component
of the curl operator, the magnetic field is first defined at the point P.
Expanding the y and z components of the magnetic field in a Taylor series
about the point P gives
For the differential surface )S, the distances (y!yo) and (z!zo) are very
small and the higher order terms may be neglected. The approximations
for the magnetic field terms are thus
The results of the integrals on the four segments are
or
The x-component of the curl operator becomes
The magnetic field must be evaluated on each of the four segments that
make up the closed path L. These magnetic field terms are
The y- and z-components of the curl operator may be determined in a
similar fashion by applying Ampere’s law on differential surfaces that are
normal to the y and z directions, respectively.
The results for the y- and z-components of the curl operator are
Rectangular Coordinates
Cylindrical Coordinates
The overall curl operator in rectangular coordinates is
Thus, the differential form of Ampere’s law is
The curl operator can be written in terms of the gradient operator as
which can also be written in determinant form as
The same technique is used to determine the curl operator in
cylindrical and spherical coordinates. In these cases, we must use
differential surfaces that match the given coordinate system.
Spherical Coordinates
Example (Differential form of Ampere’s law)
Given the magnetic field inside and outside a conductor of radius a
carrying a uniform current density (total current = I out ), show that the
differential form of Ampere’s law yields the current density in both
regions.
first term of the az component.
Application of the curl operator yields
From the differential form of Ampere’s law, evaluating the curl (in
cylindrical coordinates) of the magnetic field inside and outside the
conductor should yield the current density inside and outside the
conductor. The current density inside and outside the conductor is
The curl of the magnetic field in cylindrical coordinates is
Since the magnetic fields for r < a and r > a have only aN components that
are functions of D only, all terms in the curl expression are zero except the
Thus, given the magnetic field everywhere for a particular current
distribution, application of Ampere’s law in differential form (curl of H)
gives the current density that produces the magnetic field.
Stoke’s Theorem
Stoke’s theorem is a vector identity that defines the transformation
of a line integral of a vector around a closed path into a surface integral
over the surface bounded by that path. The integrand of the resulting
surface integral is the curl of the vector.
Given a surface S bounded by a path L, the surface can be subdivided
into cells of surface area )Sk bounded by paths Lk. If we apply Ampere’s
law to each cell and sum the results, the contributions from the internal
paths on adjacent cells cancel. The net result is the integral of the
magnetic field around the outer path L.
Taking the limit as )Sk approaches zero yields
In the limit as )Sk approaches zero, the discrete sum becomes a continuous
sum (an integral).
The term in brackets above is the definition of the curl of H in the
direction normal to the surface S such that
where an is the unit normal to the surface S. The Ampere’s law integral
above becomes
The integrals are related by
This integral relationship, shown here in terms of Ampere’s law, is actually
a vector identity which is valid for any vector F and any surface S.
or
Using Stoke’s theorem, the integral form of Ampere’s law can be directly
transformed into the differential form.
Applying Stoke’s theorem gives
The charge terms on the right hand side of Gauss’s law for magnetic
fields (integral and differential form) are zero since the dual parameter to
electric charge (magnetic charge) does not exist. The characteristics of
electrostatic and magnetostatic fields are fundamentally different based on
the existence or nonexistence of charge.
Electrostatic Fields
Since the surface integrals in this equation are valid for any surface S, the
integrands of the two integrals must be equal. This yields Ampere’s law
in differential form:
.
Electric flux lines begin on positive
charge and end on negative charge.
(Discontinuous)
Magnetostatic Fields
Magnetic flux lines form
closed loops
(Continuous)
Gauss’s Law for Magnetic Fields
Faraday’s Law for Electrostatic Fields
Two of the four Maxwell’s equations have been defined thus far:
Gauss’s law (for electric fields) and Ampere’s law. The third Maxwell’s
equation is Gauss’s law applied to magnetic fields. The integral and
differential forms of Gauss’s law for magnetic fields can be determined
from the corresponding electric field equations.
The last of the four Maxwell’s equations is Faraday’s law. The
general time-varying form of Faraday’s law will be discussed when timevarying fields are considered. The electrostatic form of Faraday’s law is
simply a statement of the conservative nature of the electrostatic field (the
closed line integral of the electrostatic field is zero).
Using Stoke’s theorem, this integral can also be written as
Since the surface integral above is valid for any surface S, the electric field
must satisfy
which is the differential form of Faraday’s law for electrostatic fields.
Maxwell’s Equations for Static Fields
All four of Maxwell’s equations for static fields have been defined in
both integral form and differential form. Maxwell’s equations for timevarying fields contain additional terms which form a complete set of
coupled equations (all four equations must be satisfied simultaneously).
For static fields, Maxwell’s equations de-couple into two sets of two
equations: two for electrostatic fields and two for magnetostatic fields.
Maxwell’s equations for static fields are:
Integral form
Static E
Static H
Differential form
9
9
Ç
È
É
Ê
Gauss’s law (electric fields)
Faraday’s law
Gauss’s law (magnetic fields)
Ampere’s law
The divergence of a vector F at a point P can be visualized by
enclosing the point with an infinitesimally small differential volume and
examining the flux of the vector in and out of the volume. If there is a net
flux out of the volume (more flux out of the volume than into the volume),
the divergence of F is positive at the point P. If there is a net flux into the
volume (more flux into the volume than out the volume), the divergence
of F is negative at the point P.
If the net flux into the differential volume is zero (the flux into the volume
equals the flux out of the volume), the divergence of F is zero at P.
According to Gauss’s law for electric fields in differential form,
the divergence of the electric flux density is zero in a charge-free region
(Dv = 0) and non-zero in a region where charge is present. Thus, the
divergence of the electric flux density locates the source of the electrostatic
field (net positive charge = net flux out, net negative charge = net flux in).
According to Gauss’s law for magnetic fields in differential form,
the divergence of the magnetic flux density is always zero since there is no
magnetic charge (net flux = 0).
Characteristics of F based on L ×F
Vectors with nonzero curl ( L ×F ú 0) vary in a direction
the direction of the field.
z to
Vectors with zero curl ( L ×F= 0) do not vary in a direction z
to the direction of the field.
The curl of a vector F at a point P can be visualized by inserting a
small paddle wheel into the field (interpreting the vector F as a force field)
and noting if the paddle wheel rotates or not. If there is an imbalance of
force on the sides of the paddle wheel, the wheel will rotate and the curl of
F is in the direction of the wheel axis (according to the right hand rule).
If the forces on both sides are equal, there is no rotation, and the curl is
zero. The magnitude of the rotation velocity represents the magnitude of
the curl of F at P. The curl of the vector field F is therefore a measure of
the circulation of F about the point P.
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.
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.
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 equivalence of the moving point charge and the differential length of
line current yields the equivalent magnetic force equation.
Example (Force between line currents)
In a similar fashion for surface and volume currents, we find
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.
such that
The overall force on a line, surface and volume current is found by
integrating over the current distribution.
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
Torque on a Current Loop
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
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 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.
The forces on the current segments can be determined for each component
of the magnetic flux density.
Forces due to Bz
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.
Forces due to By
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 3m is defined somewhat differently
than the electric susceptibility 3e. 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 (3m = 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 R 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
If we take the limit of this integral as y = 0, the integral contributions on
the vertical paths goes to zero.
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
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 closed loop yields the current enclosed.
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.
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.
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.
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
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
where the magnetic flux density is assumed to be constant over the upper
and lower incremental surfaces. Evaluation of the surface integrals yields
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.
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.
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
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)
Capacitance
Definition
mn = total flux linkage through circuit m due to current in circuit n
5mn = total magnetic flux through circuit m due to current in circuit n
Example (Self inductance / long solenoid)
Example (Self inductance / toroid)
Given the long solenoid (lRa), the magnetic field throughout the
solenoid can be assumed to be constant.
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=2%'o) is large relative to the cross-sectional radius a.
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.
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.
Given that loop Æ is much smaller than loop Å, we may assume that the
flux produced by loop Å is nearly uniform over the area of loop Æ and is
approximately equal to that at the loop center. The flux produced by loop
Å can also be assumed to be approximately normal to loop Æ over its area.
Thus, the total flux produced by loop Å linking loop Æ is approximately
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 Æ is
much smaller than loop Å (b Q a).
From the definition of inductance, the mutual inductance between the loops
is
Note that the calculations required by assuming the current in loop Å are
much easier than those required by assuming the current in loop Æ. The
calculations are easier based on the approximations employed. If the
current is assumed in loop Æ, the resulting magnetic flux density over loop
Å is a rather complicated function of position which requires a complex
integration to arrive at the same approximate answer.
If we assume that a current flows in loop Å, the magnetic flux density
produced by loop Å along its axis is
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
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
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
Note that the internal inductance of the conductor is independent of the
radius. The internal inductance per unit length of a non-magnetic
conductor (Lint1) 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 external inductance per unit length (Lext1) for the coaxial transmission
line is
If we compare the coaxial transmission line external inductance per unit
length with the capacitance per unit length (C1) given by
we find that
The magnetic flux density between the conductors of the coaxial line is
This relationship is true in general for any conductor geometry just as
The external inductance is found by integrating the magnetic flux density
external to the conductors (between the conductors) according to
Thus, the external inductance can be found quickly if the capacitance of the
conductor configuration is already known.
Magnetic Forces on Magnetic Materials
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 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
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.
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.
The equivalent electric circuit is
The reluctance components for the magnetic circuit are
The equivalent circuit can be reduced to