1 Magnetostatic Fields 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). Biot-Savart Law 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) a R = 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 a R
2 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/m 2 ) are given by The scalar and vector forms of the total magnetic field for an arbitrary line current I (A) are given by Scalar 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.
3 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=z A to z =z B. Given this general result for the magnetic field of a current segment, we may apply it to several special cases. (1) Line current, symmetric about the x-y plane, z A = z o, z B = z o (Given the field point P, the direction direction of a N does not change as z is varied along the length of the current) If the field point lies in the x-y plane, then z = 0, R A = R B = R o = [ 2 + z o 2 ] 1/2
4 (2) Semi-infinite line current, zn0(0,4) z A = 0, z B = 4, R A = [D 2 +z 2 ] 1/2, R B = 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.
5 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
6 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 (K o ) for the solenoid is found by spreading the total current of NI over the length l. The a D unit vector can be transformed into rectangular components as
7 The Biot-Savart law integral for the solenoid becomes 0 0 For a long solenoid (loa), the approximate magnetic field values at the center and at the ends of the solenoid are 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 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. 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 Ç 2 cm, È 1 cm, and É 0.5 cm. These three radii yield solenoid length to radius ratios of Ç 5, È 10, and1é 20. The magnetic field at the center of the solenoid (z =0) is È Ç É At either end of the solenoid (z =+l/2,!l/2), H (A/m) 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 z (cm)
8 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 a x - directed 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 (È and Ê) must be uniform.
9 Also by symmetry, we can show that the magnetic field above the surface current is everywhere!a y directed while the magnetic field everywhere below the surface current is +a y directed. The overall surface current can be subdivided into differential lengths (K o 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!a y component (above) or +a y 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 +a x direction. From Ampere s law, With only horizontal components of magnetic field, the Ampere s law integrals on the vertical paths (Ç and É) are zero. The magnetic field on the horizontal paths (È and Ê) may be written as The magnetic field above and below the current sheet is where H o 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.
10 Example (Ampere s law / Magnetic field in a coaxial transmission line) The coaxial transmission line shown below carries a total current I in the +a z direction through the inner conductor and in the!a z 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 a N 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) [L 1 ] The uniform vector current densities in the inner and outer conductors of the coaxial transmission line (J i and J o, 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.
11 Between the conductors (a < D < b) [L 2 ] I 2πa Within the outer conductor (b < D < c) [L 3 ] I 2πb 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) [L 4 ] 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.
12 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 = 2BD o 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 D o. Given a circular cross-section of radius a, if the mean radius is large relative to the radius of the cross section (D o oa), 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
13 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 The results of the integrals on the four segments are For the differential surface )S, the distances (y!y o ) and (z!z o ) are very small and the higher order terms may be neglected. The approximations for the magnetic field terms are thus 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.
14 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 Spherical Coordinates 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.
15 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 a z 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 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. Since the magnetic fields for r < a and r > a have only a N components that are functions of D only, all terms in the curl expression are zero except the
16 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 )S k bounded by paths L k. 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 )S k approaches zero yields In the limit as )S k 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 a n is the unit normal to the surface S. The Ampere s law integral above becomes The integrals are related by or 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. Using Stoke s theorem, the integral form of Ampere s law can be directly transformed into the differential form.
17 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 Magnetostatic 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 Magnetic flux lines form charge and end on negative charge. closed loops (Discontinuous) (Continuous) Gauss s Law for Magnetic 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. Faraday s Law for Electrostatic Fields 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.
18 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: 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. Integral form Differential form Static E 9 Static H 9 Ç È É Ê Gauss s law (electric fields) Faraday s law Gauss s law (magnetic fields) Ampere s law
19 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. 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. 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 (D v = 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 Fú0) vary in a direction z to the direction of the field. Vectors with zero curl ( L F= 0) do not vary in a direction z to the direction of the field.
20 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 F m on the charge 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 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=u o a y in a uniform magnetic flux density B=B o a z. 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 (F e ) is given by The equivalence of the moving point charge and the differential length of line current yields the equivalent magnetic force equation.
21 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 I 1 due to the magnetic flux produced by a parallel line current I 2 (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 (I 1 ) due to the current I 2 is The force on a length l of current I 1 due to the flux produced by I 2 is
22 Torque on a Current Loop The force per unit length on the current I 1 is The force on a length l of current I 2 due to the flux produced by I 1 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 I 2 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.
23 The forces on the current segments can be determined for each component of the magnetic flux density. Forces due to B z The vector torque on the loop is defined in terms of the force magnitude (IB y l 2 ), the torque moment arm distance (l 1 /2), and the torque direction (defined by the right hand rule): where A=l 1 l 2 is the loop area. The vector torque can be written compactly by defining the magnetic moment (m) of the loop. Forces due to B y where a n 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.
24 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. dipole. The poles of the bar magnet can be represented as equivalent magnetic charges separated by a distance l (the length of the magnet). 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. The magnetic flux density produced by the magnetic dipole is equivalent to the electric field produced by the electric dipole. 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 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.
25 If the magnetic moments of these three devices are equal, Note that the magnetic susceptibility m is defined somewhat differently than the electric susceptibility e. 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. 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.
26 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 ( m = 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. 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 (B m ). If the magnetic field is then cycled between the saturation magnetic field value in the forward and reverse directions (±H m ), 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. Diamagnetic r <1 Paramagnetic r >1 Ferromagnetic 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.
27 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 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 the closed loop yields the current enclosed. 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.
28 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.
29 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 Dividing by x y gives 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. 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.
30 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. 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) Inductance Capacitance Definition Definition mn = total flux linkage through circuit m due to current in circuit n mn = total magnetic flux through circuit m due to current in circuit n
31 Example (Self inductance / long solenoid) Given the long solenoid (l a), the magnetic field throughout the solenoid can be assumed to be constant. 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=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
32 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 is much smaller than loop, we may assume that the flux produced by loop is nearly uniform over the area of loop and is approximately equal to that at the loop center. The flux produced by loop can also be assumed to be approximately normal to loop over its area. Thus, the total flux produced by loop linking loop 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 is much smaller than loop (b a). From the definition of inductance, the mutual inductance between the loops is Note that the calculations required by assuming the current in loop are much easier than those required by assuming the current in loop. The calculations are easier based on the approximations employed. If the current is assumed in loop, the resulting magnetic flux density over loop 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, the magnetic flux density produced by loop along its axis is
33 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. Note that the internal inductance of the conductor is independent of the radius. The internal inductance per unit length of a non-magnetic conductor (L int ) is The magnetic flux density internal to the conductor is given by
34 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 (L ext ) for the coaxial transmission line is If we compare the coaxial transmission line external inductance per unit length with the capacitance per unit length (C ) given by The magnetic flux density between the conductors of the coaxial line is we find that 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.
35 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.
36 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. 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. 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.
37 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.
38 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 cm 2 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 The equivalent electric circuit is which is the same equation found using the magnetic circuit.
39 The reluctance components for the magnetic circuit are The equivalent circuit can be reduced to