Magnetic Force on a Moving Charge

Transcription

1 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 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. 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

2 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.

3 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.

4 Example (Force between line currents) 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. 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

5 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 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.

6 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.

7 The forces on the current segments can be determined for each component of the magnetic flux density. Forces due to B z Forces due to B y

8 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. 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.

9 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

10 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.

11 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.

12 Note that the magnetic susceptibility P m is defined somewhat differently than the electric susceptibility P 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.

13 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 (P 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. Diamagnetic : r <1 Paramagnetic : r >1 Ferromagnetic : r o1 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.

14 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.

15 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.

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17 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.

18 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.

19 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.

20 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.

21 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.

22 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 Capacitance Definition Definition

23 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) 8 mn = total flux linkage through circuit m due to current in circuit n R mn = total magnetic flux through circuit m due to current in circuit n

24 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.

25 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=2bd o ) 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

26 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 (bna). If we assume that a current flows in loop â, the magnetic flux density produced by loop â along its axis is

27 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.

28 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

29 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 (L int N) is

30 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

31 The external inductance per unit length (L ext N) 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.

32 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

33 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.

34 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.

35 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.

36 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.

37 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 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.

39 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. The equivalent electric circuit is

40 The reluctance components for the magnetic circuit are The equivalent circuit can be reduced to