Lecture 3 Dipoles & Electric Field for Continuous

Total Page:16

File Type:pdf, Size:1020Kb

Lecture 3 Dipoles & Electric Field for Continuous LECTURE 3 DIPOLES & ELECTRIC FIELD FOR CONTINUOUS CHARGE DISTRIBUTION Lecture 3 2 ¨ Reading chapter 21-6 to 22-1. ¤ Dipoles ¤ Coulomb’s law for continuous charge distributions Electric dipoles 3 ¨ An electric dipole is a system of two equal and opposite charges separated by some distance. ¨ For an electric dipole with a negative charge, -q, and a positive charge, q, separated by a distance L, its dipole moment is defined by ¨ The electric field on the dipole axis at a field point, P, a long way away from the dipole is in the same direction as the dipole, and its magnitude is P |x| Torque on a dipole in an electric field 4 ¨ The torque on a dipole is given by ¨ The magnitude of the torque is given by Quiz: 1 5 Microwave cooking 6 ¨ A microwave oven produces a rapidly oscillating electric field. ¨ Water molecules have permanent dipole moment. ¨ The oscillating field exerts a torque on water molecules in the microwave. ¨ Water molecules often form groups of molecules, the energy from the rotation can causes these groups to break apart, and the broken apart molecules have increased thermal energy. ¨ The water molecules will then form more groups, releasing their electric potential energy as thermal energy, and the process continues. Nonpolar molecules in an electric field 7 ¨ A nonpolar molecule has no permanent dipole moment. ¨ In an external electric field, the positive and negative charges can separate creating a dipole, or polarized. ¨ This is how a charged object can attract a neutral one. ¨ In a nonuniform electric field this can cause the molecule to accelerate. Example: 1 8 ¨ A dipole of moment p = 50. e·nm is placed in a uniform electric field that has a magnitude of E = 4.0×104 NC-1. What is the magnitude of the torque on the dipole when the direction of dipole makes an angle of θ = 30.° with the direction of electric field? Demo: 1 9 ¨ Wooden “needle” ¤ Demonstration of polarization Quiz: 2 10 Continuous charge distribution 11 ¨ If we have a charge Q uniformly distributed over a volume, V, the volume charge density is given by ¨ Similarly, for Q uniformly distributed over an area, A, the area charge density is given by ¨ If we have a charge Q uniformly distributed over a length, L, the linear charge density is given by Example: 2 12 ¨ Calculate the total charge on a thin rod starting at x = 0 and finishing at x = L that has a linear charge density of � � = �� %. Symmetries 13 ¨ An object has: ¤ Rotational symmetry if it does not change under an arbitrary rotation ¤ Translational symmetry if it does not change under an arbitrary translation ¤ Reflection symmetry if it does not change under a reflection ¨ This circle has rotational symmetry about the center of the circle. ¨ This is the same circle rotated 45 degrees. Can you tell? ¨ Does this object have rotational symmetry about the center? ¨ The electric field for an object that has one of these symmetries must have the same symmetry. Quiz: 3 14 E due to continuous charge distribution 15 ¨ The electric field from a small charge element, dq, is given by ¨ The total electric field E due to a continuous charge distribution is calculated by where the integral is over the entire charge distribution. E field from an infinite line of charge 16 ¨ The E field from an infinite line of charge with linear charge density λ: ¨ From rotational symmetry of the charge, the direction of E is radial. Direction: radial dE1+dE2 P r1 R r2 + + + + + + + + + ++ + + dq1 dq2 E field from a ring of charge 17 ¨ The E field along the axis of a circular charge ring with radius a and linear charge density λ or total charge Q: ¨ From rotational symmetry of the charge, E cannot have component perpendicular to the axis. Direction: along axis E field from a disk of charge 18 ¨ The E field along the axis of a circular charge disk with radius b and area charge density σ: ¤ Think of a disk of charge as a collection of thin rings of charge, and integrate from a = 0 to a = b. Direction: normal to disk E field from an infinite plane of charge 19 ¨ The magnitude of the E field from an infinite plane of charge with area charge density σ: ¤ Think of an infinite plane of charge as a disk of charge with an infinite radius. Direction: normal to plane ¤ Note that the magnitude of E does not depend on the distance from the plane. Quiz: 4 20 Example: 3 21 ¨ A non-conducting rod of length L has charge -q uniformly distributed along its length. a) What is the linear charge density of the rod? b) What is the electric field at point P, a distance a from the end of the rod? c) If P were very far from the rod compared to L, the rod would look like a point charge. Show that your answer to b) reduces to the electric field of a point charge for a >> L. -q P - - - - - - - - - - - - - - - - L a.
Recommended publications
  • Chapter 2 Introduction to Electrostatics
    Chapter 2 Introduction to electrostatics 2.1 Coulomb and Gauss’ Laws We will restrict our discussion to the case of static electric and magnetic fields in a homogeneous, isotropic medium. In this case the electric field satisfies the two equations, Eq. 1.59a with a time independent charge density and Eq. 1.77 with a time independent magnetic flux density, D (r)= ρ (r) , (1.59a) ∇ · 0 E (r)=0. (1.77) ∇ × Because we are working with static fields in a homogeneous, isotropic medium the constituent equation is D (r)=εE (r) . (1.78) Note : D is sometimes written : (1.78b) D = ²oE + P .... SI units D = E +4πP in Gaussian units in these cases ε = [1+4πP/E] Gaussian The solution of Eq. 1.59 is 1 ρ0 (r0)(r r0) 3 D (r)= − d r0 + D0 (r) , SI units (1.79) 4π r r 3 ZZZ | − 0| with D0 (r)=0 ∇ · If we are seeking the contribution of the charge density, ρ0 (r) , to the electric displacement vector then D0 (r)=0. The given charge density generates the electric field 1 ρ0 (r0)(r r0) 3 E (r)= − d r0 SI units (1.80) 4πε r r 3 ZZZ | − 0| 18 Section 2.2 The electric or scalar potential 2.2 TheelectricorscalarpotentialFaraday’s law with static fields, Eq. 1.77, is automatically satisfied by any electric field E(r) which is given by E (r)= φ (r) (1.81) −∇ The function φ (r) is the scalar potential for the electric field. It is also possible to obtain the difference in the values of the scalar potential at two points by integrating the tangent component of the electric field along any path connecting the two points E (r) d` = φ (r) d` (1.82) − path · path ∇ · ra rb ra rb Z → Z → ∂φ(r) ∂φ(r) ∂φ(r) = dx + dy + dz path ∂x ∂y ∂z ra rb Z → · ¸ = dφ (r)=φ (rb) φ (ra) path − ra rb Z → The result obtained in Eq.
    [Show full text]
  • 6.007 Lecture 5: Electrostatics (Gauss's Law and Boundary
    Electrostatics (Free Space With Charges & Conductors) Reading - Shen and Kong – Ch. 9 Outline Maxwell’s Equations (In Free Space) Gauss’ Law & Faraday’s Law Applications of Gauss’ Law Electrostatic Boundary Conditions Electrostatic Energy Storage 1 Maxwell’s Equations (in Free Space with Electric Charges present) DIFFERENTIAL FORM INTEGRAL FORM E-Gauss: Faraday: H-Gauss: Ampere: Static arise when , and Maxwell’s Equations split into decoupled electrostatic and magnetostatic eqns. Electro-quasistatic and magneto-quasitatic systems arise when one (but not both) time derivative becomes important. Note that the Differential and Integral forms of Maxwell’s Equations are related through ’ ’ Stoke s Theorem and2 Gauss Theorem Charges and Currents Charge conservation and KCL for ideal nodes There can be a nonzero charge density in the absence of a current density . There can be a nonzero current density in the absence of a charge density . 3 Gauss’ Law Flux of through closed surface S = net charge inside V 4 Point Charge Example Apply Gauss’ Law in integral form making use of symmetry to find • Assume that the image charge is uniformly distributed at . Why is this important ? • Symmetry 5 Gauss’ Law Tells Us … … the electric charge can reside only on the surface of the conductor. [If charge was present inside a conductor, we can draw a Gaussian surface around that charge and the electric field in vicinity of that charge would be non-zero ! A non-zero field implies current flow through the conductor, which will transport the charge to the surface.] … there is no charge at all on the inner surface of a hollow conductor.
    [Show full text]
  • A Problem-Solving Approach – Chapter 2: the Electric Field
    chapter 2 the electric field 50 The Ekelric Field The ancient Greeks observed that when the fossil resin amber was rubbed, small light-weight objects were attracted. Yet, upon contact with the amber, they were then repelled. No further significant advances in the understanding of this mysterious phenomenon were made until the eighteenth century when more quantitative electrification experiments showed that these effects were due to electric charges, the source of all effects we will study in this text. 2·1 ELECTRIC CHARGE 2·1·1 Charginl by Contact We now know that all matter is held together by the aurae· tive force between equal numbers of negatively charged elec· trons and positively charged protons. The early researchers in the 1700s discovered the existence of these two species of charges by performing experiments like those in Figures 2·1 to 2·4. When a glass rod is rubbed by a dry doth, as in Figure 2-1, some of the electrons in the glass are rubbed off onto the doth. The doth then becomes negatively charged because it now has more electrons than protons. The glass rod becomes • • • • • • • • • • • • • • • • • • • • • , • • , ~ ,., ,» Figure 2·1 A glass rod rubbed with a dry doth loses some of iu electrons to the doth. The glau rod then has a net positive charge while the doth has acquired an equal amount of negative charge. The total charge in the system remains zero. £kctric Charge 51 positively charged as it has lost electrons leaving behind a surplus number of protons. If the positively charged glass rod is brought near a metal ball that is free to move as in Figure 2-2a, the electrons in the ball nt~ar the rod are attracted to the surface leaving uncovered positive charge on the other side of the ball.
    [Show full text]
  • Chapter 5 Capacitance and Dielectrics
    Chapter 5 Capacitance and Dielectrics 5.1 Introduction...........................................................................................................5-3 5.2 Calculation of Capacitance ...................................................................................5-4 Example 5.1: Parallel-Plate Capacitor ....................................................................5-4 Interactive Simulation 5.1: Parallel-Plate Capacitor ...........................................5-6 Example 5.2: Cylindrical Capacitor........................................................................5-6 Example 5.3: Spherical Capacitor...........................................................................5-8 5.3 Capacitors in Electric Circuits ..............................................................................5-9 5.3.1 Parallel Connection......................................................................................5-10 5.3.2 Series Connection ........................................................................................5-11 Example 5.4: Equivalent Capacitance ..................................................................5-12 5.4 Storing Energy in a Capacitor.............................................................................5-13 5.4.1 Energy Density of the Electric Field............................................................5-14 Interactive Simulation 5.2: Charge Placed between Capacitor Plates..............5-14 Example 5.5: Electric Energy Density of Dry Air................................................5-15
    [Show full text]
  • Chapter 4. Electric Fields in Matter 4.4 Linear Dielectrics 4.4.1 Susceptibility, Permittivity, Dielectric Constant
    Chapter 4. Electric Fields in Matter 4.4 Linear Dielectrics 4.4.1 Susceptibility, Permittivity, Dielectric Constant PE 0 e In linear dielectrics, P is proportional to E, provided E is not too strong. e : Electric susceptibility (It would be a tensor in general cases) Note that E is the total field from free charges and the polarization itself. If, for instance, we put a piece of dielectric into an external field E0, we cannot compute P directly from PE 0 e ; EE 0 E0 produces P, P will produce its own field, this in turn modifies P, which ... Breaking where? To calculate P, the simplest approach is to begin with the displacement D, at least in those cases where D can be deduced directly from the free charge distribution. DEP001 e E DE 0 1 e : Permittivity re1 : Relative permittivity 0 (Dielectric constant) Susceptibility, Permittivity, Dielectric Constant Problem 4.41 In a linear dielectric, the polarization is proportional to the field: P = 0e E. If the material consists of atoms (or nonpolar molecules), the induced dipole moment of each one is likewise proportional to the field p = E. What is the relation between atomic polarizability and susceptibility e? Note that, the atomic polarizability was defined for an isolated atom subject to an external field coming from somewhere else, Eelse p = Eelse For N atoms in unit volume, the polarization can be set There is another electric field, Eself , produced by the polarization P: Therefore, the total field is The total field E finally produce the polarization P: or Clausius-Mossotti formula Susceptibility, Permittivity, Dielectric Constant Example 4.5 A metal sphere of radius a carries a charge Q.
    [Show full text]
  • Physics 2102 Lecture 2
    Physics 2102 Jonathan Dowling PPhhyyssicicss 22110022 LLeeccttuurree 22 Charles-Augustin de Coulomb EElleeccttrriicc FFiieellddss (1736-1806) January 17, 07 Version: 1/17/07 WWhhaatt aarree wwee ggooiinngg ttoo lleeaarrnn?? AA rrooaadd mmaapp • Electric charge Electric force on other electric charges Electric field, and electric potential • Moving electric charges : current • Electronic circuit components: batteries, resistors, capacitors • Electric currents Magnetic field Magnetic force on moving charges • Time-varying magnetic field Electric Field • More circuit components: inductors. • Electromagnetic waves light waves • Geometrical Optics (light rays). • Physical optics (light waves) CoulombCoulomb’’ss lawlaw +q1 F12 F21 !q2 r12 For charges in a k | q || q | VACUUM | F | 1 2 12 = 2 2 N m r k = 8.99 !109 12 C 2 Often, we write k as: 2 1 !12 C k = with #0 = 8.85"10 2 4$#0 N m EEleleccttrricic FFieieldldss • Electric field E at some point in space is defined as the force experienced by an imaginary point charge of +1 C, divided by Electric field of a point charge 1 C. • Note that E is a VECTOR. +1 C • Since E is the force per unit q charge, it is measured in units of E N/C. • We measure the electric field R using very small “test charges”, and dividing the measured force k | q | by the magnitude of the charge. | E |= R2 SSuuppeerrppoossititioionn • Question: How do we figure out the field due to several point charges? • Answer: consider one charge at a time, calculate the field (a vector!) produced by each charge, and then add all the vectors! (“superposition”) • Useful to look out for SYMMETRY to simplify calculations! Example Total electric field +q -2q • 4 charges are placed at the corners of a square as shown.
    [Show full text]
  • Ee334lect37summaryelectroma
    EE334 Electromagnetic Theory I Todd Kaiser Maxwell’s Equations: Maxwell’s equations were developed on experimental evidence and have been found to govern all classical electromagnetic phenomena. They can be written in differential or integral form. r r r Gauss'sLaw ∇ ⋅ D = ρ D ⋅ dS = ρ dv = Q ∫∫ enclosed SV r r r Nomagneticmonopoles ∇ ⋅ B = 0 ∫ B ⋅ dS = 0 S r r ∂B r r ∂ r r Faraday'sLaw ∇× E = − E ⋅ dl = − B ⋅ dS ∫∫S ∂t C ∂t r r r ∂D r r r r ∂ r r Modified Ampere'sLaw ∇× H = J + H ⋅ dl = J ⋅ dS + D ⋅ dS ∫ ∫∫SS ∂t C ∂t where: E = Electric Field Intensity (V/m) D = Electric Flux Density (C/m2) H = Magnetic Field Intensity (A/m) B = Magnetic Flux Density (T) J = Electric Current Density (A/m2) ρ = Electric Charge Density (C/m3) The Continuity Equation for current is consistent with Maxwell’s Equations and the conservation of charge. It can be used to derive Kirchhoff’s Current Law: r ∂ρ ∂ρ r ∇ ⋅ J + = 0 if = 0 ∇ ⋅ J = 0 implies KCL ∂t ∂t Constitutive Relationships: The field intensities and flux densities are related by using the constitutive equations. In general, the permittivity (ε) and the permeability (µ) are tensors (different values in different directions) and are functions of the material. In simple materials they are scalars. r r r r D = ε E ⇒ D = ε rε 0 E r r r r B = µ H ⇒ B = µ r µ0 H where: εr = Relative permittivity ε0 = Vacuum permittivity µr = Relative permeability µ0 = Vacuum permeability Boundary Conditions: At abrupt interfaces between different materials the following conditions hold: r r r r nˆ × (E1 − E2 )= 0 nˆ ⋅(D1 − D2 )= ρ S r r r r r nˆ × ()H1 − H 2 = J S nˆ ⋅ ()B1 − B2 = 0 where: n is the normal vector from region-2 to region-1 Js is the surface current density (A/m) 2 ρs is the surface charge density (C/m ) 1 Electrostatic Fields: When there are no time dependent fields, electric and magnetic fields can exist as independent fields.
    [Show full text]
  • Electron Charge Density: a Clue from Quantum Chemistry for Quantum Foundations
    Electron Charge Density: A Clue from Quantum Chemistry for Quantum Foundations Charles T. Sebens California Institute of Technology arXiv v.2 June 24, 2021 Forthcoming in Foundations of Physics Abstract Within quantum chemistry, the electron clouds that surround nuclei in atoms and molecules are sometimes treated as clouds of probability and sometimes as clouds of charge. These two roles, tracing back to Schr¨odingerand Born, are in tension with one another but are not incompatible. Schr¨odinger'sidea that the nucleus of an atom is surrounded by a spread-out electron charge density is supported by a variety of evidence from quantum chemistry, including two methods that are used to determine atomic and molecular structure: the Hartree-Fock method and density functional theory. Taking this evidence as a clue to the foundations of quantum physics, Schr¨odinger'selectron charge density can be incorporated into many different interpretations of quantum mechanics (and extensions of such interpretations to quantum field theory). Contents 1 Introduction2 2 Probability Density and Charge Density3 3 Charge Density in Quantum Chemistry9 3.1 The Hartree-Fock Method . 10 arXiv:2105.11988v2 [quant-ph] 24 Jun 2021 3.2 Density Functional Theory . 20 3.3 Further Evidence . 25 4 Charge Density in Quantum Foundations 26 4.1 GRW Theory . 26 4.2 The Many-Worlds Interpretation . 29 4.3 Bohmian Mechanics and Other Particle Interpretations . 31 4.4 Quantum Field Theory . 33 5 Conclusion 35 1 1 Introduction Despite the massive progress that has been made in physics, the composition of the atom remains unsettled. J. J. Thomson [1] famously advocated a \plum pudding" model where electrons are seen as tiny negative charges inside a sphere of uniformly distributed positive charge (like the raisins|once called \plums"|suspended in a plum pudding).
    [Show full text]
  • Charge Density Empty 1/1
    9/30/2004 sec 3_2 charge density empty 1/1 3-2 Charge and Charge Density (pp.61-63) HO: Electric Charge Q: A: HO: Charge Density Q: A: HO: Total Charge Jim Stiles The Univ. of Kansas Dept. of EECS 9/30/2004 Electric Charge 1/2 Electric Charge Most of classical physics can be described in terms of three fundamental units, which define our physical “reality” 1. Mass (e.g., kg) 2. Distance (e.g., meters) 3. Time (e.g., seconds) From these fundamental units, we can define other important physical parameters. For example, energy can always be described in units of kg m2/s2. But, these three fundamental units alone are insufficient for describing all of classic physics—we require one more to completely describe physical reality! This fourth fundamental unit is Coulomb, the unit of electric charge. All electromagnetic phenomena can be attributed to electric charge! Jim Stiles The Univ. of Kansas Dept. of EECS 9/30/2004 Electric Charge 2/2 We shall find that electric charge is somewhat analogous to mass. However, one important difference between mass and charge is that charge can be either positive or negative! Essentially, charge (like mass) is a property of atomic particles. Specifically, we find that: The charge “on” a proton is +1.6 x 10-19 C The charge “on” a neutron is 0.0 C The charge “on” an electron is -1.6 x 10-19 C Charged particles (of all types) can be distributed (unevenly) across a volume, surface, or contour. Jim Stiles The Univ.
    [Show full text]
  • Ampere's Law Was Still Pretty Easy for Me
    Ampere’s Law Phys 122 Lecture 20 G. Rybka Some good questions from the Prelecture Electric fields do not exist inside conductors. Do magnetic fields exist? Could we spend more time talking about the properties of magnetic fields inside and outside hollow tubes? How does that differ from magnetic fields inside and outside solid cylinders? So, I can't go to your office hours because I have class right after Physics on Friday, what is the best source of getting down the right hand rule down in all of the different cases? This is still holding me back, even though this Ampere's Law was still pretty easy for me. We learned that charges are distributed on the surface of a conductor, and the current is the result of flowing charges. If the charges are all distributed on the surface of a wire, does that mean the current only flows on the surface of the wire? Right-Hand-Rules: Get ready to Click ! ! ! ! ! ! F = qv × B F = IL× B ! ! ! ! τ = r × F τ = µ × B ! ! µ I d s ×rˆ dB = 0 4π r 2 Clicker L1 rail A conducting rail of length L1 rests on the top of the circuit loop as L2 C S shown. It is free to move. A B uniform magnetic field exists in the box of dimension L by L . V 2 3 When switch s is closed, which way R does the rail move (if at all) ? L3 A) Left B) Right C) Rotates clockwise D) Rotates counterclockwise E) Does not move • Current through rail is down • B is pointed toward you • IL x B is to the LEFT Clicker L1 rail A conducting rail of length L1 rests on the top of the circuit loop as L2 C S shown.
    [Show full text]
  • Dielectric Polarization, Bound Charges, and the Electric Displacement Field
    Dielectric Polarization, Bound Charges, and the Electric Displacement Field Any kind of matter is full of positive and negative electric charges. In a dielectric, these charges are bound — they cannot move separately from each other through any macroscopic distance, — so when an electric field is applied there is no net electric current. However, the field does push the positive charges just a tiny bit in the direction of E while the negative charges are pushed in the opposite directions. Consequently, the atoms and the molecules comprising the dielectric acquire tiny electric dipole moments in the direction of E. −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+ −+
    [Show full text]
  • Electromagnetic Fields and Energy
    MIT OpenCourseWare http://ocw.mit.edu Haus, Hermann A., and James R. Melcher. Electromagnetic Fields and Energy. Englewood Cliffs, NJ: Prentice-Hall, 1989. ISBN: 9780132490207. Please use the following citation format: Haus, Hermann A., and James R. Melcher, Electromagnetic Fields and Energy. (Massachusetts Institute of Technology: MIT OpenCourseWare). http://ocw.mit.edu (accessed [Date]). License: Creative Commons Attribution-NonCommercial-Share Alike. Also available from Prentice-Hall: Englewood Cliffs, NJ, 1989. ISBN: 9780132490207. Note: Please use the actual date you accessed this material in your citation. For more information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms 9 MAGNETIZATION 9.0 INTRODUCTION The sources of the magnetic fields considered in Chap. 8 were conduction currents associated with the motion of unpaired charge carriers through materials. Typically, the current was in a metal and the carriers were conduction electrons. In this chapter, we recognize that materials provide still other magnetic field sources. These account for the fields of permanent magnets and for the increase in inductance produced in a coil by insertion of a magnetizable material. Magnetization effects are due to the propensity of the atomic constituents of matter to behave as magnetic dipoles. It is natural to think of electrons circulating around a nucleus as comprising a circulating current, and hence giving rise to a magnetic moment similar to that for a current loop, as discussed in Example 8.3.2. More surprising is the magnetic dipole moment found for individual electrons. This moment, associated with the electronic property of spin, is defined as the Bohr magneton e 1 m = ± ¯h (1) e m 2 11 where e/m is the electronic charge­to­mass ratio, 1.76 × 10 coulomb/kg, and 2π¯h −34 2 is Planck’s constant, ¯h = 1.05 × 10 joule­sec so that me has the units A − m .
    [Show full text]