Single Particle Motion


 Isabella York
 7 months ago
 Views:
Transcription
1 Single Particle Motion C ontents Uniform E and B E =  guiding centers Definition of guiding center E gravitation Non Uniform B 'grad B' drift, B B Curvature drift Grad B drift, B B invariance of µ. Magnetic mirrors, loss cone Bounce Period in a dipole Drift period in a dipole The Ring Current around the earth Non Uniform E (finite Larmor radius) Time varying E (polarization drift) Time Varying B (magnetic moment) Adiabatic Invariants Appendix (Geometry) p 3.1
2 Plasmas somewhere between fluids and single particles. Here consider single particles. Uniform E and B E =  guiding centers m dv dt = qv B Take B = B ˆ z. The m v x = qbv y ; m v y = qbv x ; m v z = v x = qb m v = qb y v m x v y = qb v m x = qb m v y Harmonic oscillator at cyclotron frequency i.e. ω c = q B m v x,y = v e ± iω c t +iδ x,y ± denotes sign of q. Choose phase δ so that v x = v e iω c t = x where v is a positive constant, speed in plane perpendicular to vector B. Then v y = m qb v x = ± 1 ω c v x = ±v e iω c t = y Integrate x  x = i v ω c e iω c t ; y y = ± v ω c e iω c t Define Larmor radius p 3.
3 r L = v = mv ω c q B Taking real parts (fn + complex conj)/) gives x x = r L sin( ω c t); y y = ±r L cos( ω c t) i.e. a circular orbit around B. Ions and electrons circulate in opposite directions. The sense is such that the B field generated by the particle always tends to reduce the external B, i.e. plasmas are diamagnetic. Electrons have smaller Larmor radii than ions. Definition of guiding center Defined as r gc = r r is position vector of particle and is radius of curvature, which is a vector from the position of the particle to the center of gyration. In the plane perpendicular to B vector write in terms of momentum vector p: mω c = qv B Now ω c = q B / m, so and = p B qb r gc = r p B qb Now consider a collision in which a force f is applied to the particle in a direction perpendicular to the vector B, and this force is f >> the Lorentz force. Let the field be homogeneous. Let the impact time be very short, << the Larmor frequency. At the collision the momentum p changes a lot but the particle position vector r does not. The momentum p changes from p to p' = p + p, where p = t + t t f dt p 3.3
4 Now the guiding center must move by an amount r gc p B qb Generalizing to a continuous force d dt r gc = d dr r + now use dp dt = qv B + F d dt p B qb where F is a non magnetic force. Then dr gc dt ( = v + F + qv B ) B qb ( B ) = v + F B + q v B qb = v + F B qb (expand triple vector product and use v = v + v i.e. if the force is continuous, the guiding center motion can be viewed as a continuous series of small impacts. p 3.4
5 E y E Ḅ z E x ion electron B Will find motion is sum of usual circular Larmor orbit plus a drift of the 'guiding center'. Choose E in the xz plane so that E y =. As before, z component of velocity is unrelated to the transverse components and can be treated separately. z component m dv dt = q ( E + v B ) m dv z dt = q( E z ); v z = qe z m t + v z i.e. acceleration along the vector B. Transverse components give v x = q m E x ±ω c v y ; v y = mω c v x Differentiate i.e. v x = ω c v x q v y = mω c m E ± ω v E x c y = ω x c B + v y d dt v = ω x c v x d v dt y + E x = ω B c v y + E x B i.e. just like E = case except replace v y by v y + E x /B. Therefore solution is p 3.5
6 v x = v e iω c t v y =±iv e iω ct E x B i.e. there is a drift in the y direction. Electrons and ions drift in the same direction. More generally, can obtain an equation for the 'guiding center' drift v gc. Omit dv/dt terms, as we know this just gives the Larmor orbit and frequency. Then ( E + v B ) = E B = v B B = B ( v B ) = vb B v B (Use a ( b c ) = b a c a = B;b = v;c = B : B v B c a b ) = v ( B B ) B ( B v ) transverse component gives the electric field drift of the guiding center" v gc = E B B, independent of q, m, ω c. This is because on first half of orbit a particle gains energy from E field, so velocity and Larmor radius increase. But on second half of orbit the particle loses energy and Larmor radius decreases. The difference in Larmor radius causes the drift. gravitation Generalize by replacing electrostatic by a general force. Then v gc = 1 q e.g. for gravity F B B v gc = m q g B B p 3.6
7 Note this is charge dependent, but very small. Therefore under gravity the charges drift and produce a net current j = n i q i v i Non Uniform B = n( m + M) g B B Need to expand in a small parameter; orbit theory 'grad B' drift B B y gradient of B ion x electron z B Take straight field lines in y direction, but with a gradient in their density,. Anticipate result: gradient in B causes the Larmor radius to be different, If gradient in y direction then smaller radius at top than bottom,, which will give a drift opposite for e and i, perpendicular to both the field and its gradient. Average the Lorentz force over a gyration. Note F x = as equal time moving up and down. To calculate average F y use undisturbed orbit. Expand the field vector around the point x, y so that B = B + ( r )B +.. B z = B + y B z y +.. p 3.7
8 F y = qv x B z ( y) = qv cos ω c t qv cos ω c t B ± r L cos( ω c t) B B + y B y +... y +... This assumes r L /L << 1, where L is scale of B/ y. Then F y = m 1 qv r B L y and the guiding center drift is v B = ± 1 v r L B B B with ± standing for the sign of the charge. Curvature drift Assume lines of force have constant radius of curvature. Take B constant. This does not satisfy Maxwell's equations, so we will always add the grad B drift to the answer we are about to get. Let v be the average of the square of the (random) velocity along the vector B. then the average centrifugal force is Therefore F cf = mv R r ˆ = mv c v R = 1 q F cf B = mv B qb Now compute the associated gradb drift. Use cylindrical coordinates. B = in vacuum.. B has only a θ component, and B only a radial component, so and ( B ) z = 1 r r rb θ = ; B θ 1 r B 1 ; B B = p 3.8
9 so that v B = m 1 v r L B B B = ± 1 v ω c Adding the curvature term to this gives B = 1 m B q v B B v R + v B = m q B B v + v Now for a Maxwellian plasma v = 1 v = k b T (because the perpendicular component has degrees of freedom), so that v R + B = ± v th ω c where y ˆ = R B c B y ˆ = ± r L,atv th i.e. dependent on q but not m. ˆ y Grad  B drift, B B Now consider B field along z, with magnitude dependent on z. Consider symmetric case; / θ =. Then lines of force diverge and converge, and B r. From B = 1 r r rb r + B z z = Suppose B z / z is given at r =, and is not strongly dependent on r, then r rb r = r B z dr 1 r r B z r B r 1 r B z r r = r = Variation of B with r causes a gradb drift of the guiding centers about the axis of symmetry. However there is no radial grad  B drift because B/ θ =. The component of the Lorentz force are p 3.9
10 The terms F r = q( v θ B z v z B θ ) = q( v θ B z ) F θ = q( v r B z + v z B r ) F z = q( v r B θ v θ B r ) = q( v θ B r ) F θ part b = q v z B r F r = q( v θ B z ); F θ part a = q( v r B z ) give the Larmor gyration. The term vanishes on axis, and off axis it causes a drift in the radial direction. This is written as makes the guiding centers follow the 'lines of force'. The final term, F z = q v θ B r F z = 1 qv r B z θ z Average over a gyration period. Consider a particle with guiding center on axis. The v θ = v ; r = r L F z = m 1 qv r B z L = m 1 z q v ω c B z z = 1 mv B B z z Define magnetic moment µ = 1 mv B Then in general the parallel force on a particle is given in terms of the element ds along the vector B: F = µ B s = µ B Note for a current loop area A current I then µ = IA = eω c π πr L = eω c πv = 1 ev π ω c ω c = mv B Invariance of µ. Consider parallel equation of motion m dv dt = µ B s mv dv dt = d 1 mv dt = µ B ds s dt = µ db dt p 3.1
11 where db/dt is the variation of B seen by the particle. (B itself is or can be constant). Now conserve energy Then d 1 dt mv + 1 mv = d 1 dt mv + µb = d dt ( µb) µ db dt = ; dµ dt Magnetic mirrors, loss cone = Finally we have the magnetic mirror. A particle moves from weak to strong field. B increases, so v must increase to conserve µ. Then v must decrease to conserve energy. If B is high enough at some place along the trajectory, then v =, and the particle is reflected. A particle with small v /v at the mid plane where B = B is a minimum can escape if the maximum field B m is not sufficiently large. Let B = B, v = v and v = v ' turning point B = B', v =, and v = v. Conserve µ 1 mv = 1 ' mv B B ' Conserve energy ' v = v + v v at the mid plane. At the B = v B = v ' ' v v = sin ( θ) θ is pitch angle of particle in weak field. Particles with smaller pitch θ will mirror in regions of higher B. If θ is too small B' > B m and the particle escapes. The smallest θ of a confined particle is given by the mirror ratio R m : sin ( θ) = B = 1 B m R m Particles which escape are said to be in the loss cone, independent of q and m. Collisions can scatter particles into loss cones. Electrons are lost easily because collision frequency is higher. p 3.11
12 v α dφ v  B Consider the particles at the low field point where B = B. Let there be a uniform distribution of pitch angles α. The probability of getting lost is P = Φ /(π) = α = 1 1 sin ( α ) sin( α )dα = 1 cos α 1 / = 1 1 B B M 1 / = Planetary Loss Cones exist. Take account of atmosphere. e.g. define equatorial loss cone for Earth as that pitch angle α such that its mirror point is at 1 km from the surface. This is arbitrary, but used because at 1 km the density is high enough for scattering of electrons to occur. Therefore at 1 km and below a mirroring electron will probably get absorbed by the atmosphere and lost from the radiation belt. The equatorial loss cone for a dipole line of force crossing the equator at 6 R E is about 3. All electrons within 3 equatorial pitch angle cone are precipitated because they are mirroring below 1 km. Electrons outside of the loss cone mirror at heights above 1 km are trapped radiation belt electrons. R M 1 / p 3.1
13 Bounce Period in a dipole Consider trapped particles in the Earth's dipole field. Period of north south motion of guiding center is T b = ds v ds along path, v along path vector B, integral over complete period. Note v = v v = 1 B sin α B = v 1 sin ( α) where conservation of the first invariant µ has been applied, v = v and subscript means at the equator. We need an expression for ds. Work in spherical coordinates M Q r+dr O dθ θ r φ P (r,θ) is polar coordinate of point P. PM perpendicular to OQ. sin δθ sin( OQP) = r PQ sin δθ = r δθ δθ δs δs PQ Let Q approach P, then angle OQP becomes the angle between the tangent and the radius vector, denoted as φ. Also sin δθ /δθ 1; δs / PQ 1; δθ /δs dθ / ds and Similarly sin( OQP) sin( φ) = r dθ ds p 3.13
14 i.e. Then cos( OQP) = MQ OQ OM = PQ PQ r +δr rcos ( δθ ) = and since then δs ( ) r 1 cos δθ = δs cos φ = + dr tan( φ) = r dθ dr δs PQ + δr δs δs PQ x1 = dr ds ds sec ( φ) =1 + tan ( φ); cosec ( φ) = 1+ cot ( φ) ds dr 1 r i.e ds dθ = 1+ r dθ dr ds dθ =1 + 1 dr r dθ = r + dr dθ For a dipole field r = r cos ( λ ), where λ = π/θ, and finally ds = r cos λ 1 / dλ 1+ 3sin ( λ) Also for a dipole B B ( ) 1 / = 1+ 3sin λ cos 6 λ, so that T b = λ max 1/ 1 + 3sin ( λ) r cos λ v 1 sin ( α ) 1 + dλ = 4r ( ( 3sin λ )) 1/ v I 1 cos 6 ( λ) p 3.14
15 I 1 = λ max 1 + 3sin ( λ) 1 + 3sin ( λ) cos 6 ( λ) cos λ 1 sin α 1/ 1/ dλ λ 1 is the mirror point of the particle in the northern hemisphere. Here v = and λ m is given by the solution of cos 6 ( λ m ) sin α 1 / = 1+ 3sin ( λ m ) Now I 1 is dimensionless, and solved numerically to be I sin( α ) Note 4r /v depends on the particle energy. Now particles are precipitated (lost) if the distance of closest approach R min r < 1 L i.e. particles are trapped if the distance of closest approach is larger than R min. Now the field line equation is r = r cos ( λ ), so that for particles mirroring at the plane surface we have λ min given by cos ( λ m ) = 1 L For Earth, r = L R E (L is a parameter introduced by McIlwain). Figure shows bounce period as a function of electron energy for the marginally trapped electrons, for L = 1 to 8. For Auroral lines of force (L = 6), and typical electron energies of 1 kev to 5 kev, the bounce period is a few seconds. Drift period in a dipole field To calculate the guiding center drift period in a dipole field, remember that there are two components, the gradient drift and the curvature drift. They can be combined together as v R + v B = m q B B v + v = 1 B v ω c B + v p 3.15
16 where ω c is the cyclotron frequency, and is the radius of curvature. This drift is perpendicular to vector B and vector ρ, that is around the word (equivalent to along the equator). The expression for the radius of curvature of a line in polar coordinates is r + dr dθ ρ = r + dr d r dθ dθ For the dipole field r = r cos ( λ ); λ = 9 θ so that the radius of curvature vector is sin ( λ) cos λ ρ = r sin λ The cyclotron frequency for the dipole is ( 1 + 3sin ( λ) ) 1 ω c = ω cos 6 λ 3 where ω is the value at the equator. T hen the drift velocity at a latitude λ is written as v D = 3v ( 1 +sin ( λ ) cos 5 ( λ) ω r 1 + 3sin ( λ) sin α sin ( λ) cos 6 ( λ) Note we have used v = vcos( α); v = vsin( α) (α is the pitch angle) v + 1 v = v cos ( α ) + v sin ( α ) 1 v sin ( α) = v 1 1 sin ( α ) = v 1 1 B sin α B = v 1+ 3sin λ cos 6 ( λ) ( ) 1 sin α p 3.16
17 Now the angular drift (around the earth) which occurs in one bounce period is Φ = v D ds (ds along vector B) rcos( λ) v (because bounce period is T b = ds /v, so distance covered in a bounce period = T b = v D ds / v, and angular distance is as given. The angular drift, averaged over a bounce, is then ω D = Φ πt B Using expressions derived before, we can write this as ω D = 3v ω r I I 1 = 3E qb R E L I I 1 with kinetic energy E = mv /, and I = cos 4 ( λ ) 1 + sin ( λ) 1 sin ( α ) 1 + 3sin 3sin α ( ( λ) ) 1 / cos 6 ( λ) λ max 1 / 1+ 3sin ( λ) cos 6 ( λ) Numerically it is found that I ( α ) / I 1 ( α ) sin( α ), i.e. varying from.35 for a = particles to.5 for a = 9 particles. Then the bounce averaged drift period dλ T D = 1 ω D 5 /( LE) with T D in minutes and E in MeV. Typical values (5 kev, L = 6) is about an hour. The Ring Current around the earth The drifting bounced orbits just discussed are represented as below in a Mercator projection. Magnetic storms are accompanied by a decrease in the horizontal field intensity at the earth. A westward current around the earth would do this. This is expected from trapped particles. Several MA are carried. p 3.17
18 Northern reflection Equator West +  East Southern reflection We know the drift velocity v R + v B = m q B B v + v = 1 B v ω c B + v from which we derived the averaged angular frequency in the equatorial direction ω D 3v ω r 3E qb r i.e. the current density J = nqv D nqr ω D 3nE B r φ ˆ with n the number density and E the energy of the particles. Then the total current is given by i.e. Idl = J dv I 3E tot πb r where E tot is the total energy associated with the particles. Then the change in field at the center of this loop is B = µ I a 3µ E tot 4πB r 3 z p 3.18
19 The total perturbation must account for another contribution, namely the diamagnetic current due to the cyclotron motion. This is calculated by noticing that the diamagnetic contribution is equivalent to a ring of dipoles of radius r and total magnetic moment µ. Because r >> r c, the cyclotron radius, we can estimate the field at the center of the dipole ring as (from previous notes, B( r,λ) = µ M ( 1 / ) ) 4πr sin λ and B diamagnetic = µ 4πr 3 µ = µ E tot 4πB r 3 z µ = mnv B B = E tot B B = E tot B z Note the individual dipoles are aligned with the magnetic field direction The total perturbed field is then B 3µ E tot 4πB r z + µ E tot 3 4πB r z µ E tot 3 πb r z = E tot 3 M z where we have used B = µ M 4πr 3 Non Uniform E (finite Larmor radius) Time varying E (polarization drift) Time Varying B (magnetic moment) p 3.19
20 Adiabatic Invariants Appendix  Some Geometry If C is a space curve defined by the function r ( u), then dr / du is a vector in the direction of the tangent to C. If the scalar u is the arc length s measured from some field point C, then dr / ds is a unit tangent vector to C, and is called T. The rate at which T changes with respect to s is a measure of the curvature of C and is given by dt / ds. The direction of dt/ds at any point on C is normal to the curve at that point. If N is a unit vector in this normal direction, it is called the unit normal. Then dt / ds = κn, where κ is called the curvature, and ρ = 1/κ is called the radius of curvature. The position vector at any point is Therefore r = x( s)ˆ i + y( s) j ˆ + z( s) k ˆ T = dr ds = dx ds i ˆ + dy ds j ˆ + dz ds ˆ k dt ds = d x i ˆ + d y j ˆ + d z k ˆ ds ds ds p 3.
21 But dt ds =κn ; κ = dt ds ; ρ = 1 κ = d x ds + d y ds + d z ds 1 p 3.1
cos 6 λ m sin 2 λ m Mirror Point latitude Equatorial Pitch Angle Figure 5.1: Mirror point latitude as function of equatorial pitch angle.
Chapter 5 The Inner Magnetosphere 5.1 Trapped Particles The motion of trapped particles in the inner magnetosphere is a combination of gyro motion, bounce motion, and gradient and curvature drifts. In
More informationLectures on basic plasma physics: Hamiltonian mechanics of charged particle motion
Lectures on basic plasma physics: Hamiltonian mechanics of charged particle motion Department of applied physics, Aalto University March 8, 2016 Hamiltonian versus Newtonian mechanics Newtonian mechanics:
More informationCHARGED PARTICLE MOTION IN CONSTANT AND UNIFORM ELECTROMAGNETIC FIELDS
CHARGED PARTICLE MOTION IN CONSTANT AND UNIFORM ELECTROMAGNETIC FIELDS In this and in the following two chapters we investigate the motion of charged particles in the presence of electric and magnetic
More informationPROBLEM SET. Heliophysics Summer School. July, 2013
PROBLEM SET Heliophysics Summer School July, 2013 Problem Set for Shocks and Particle Acceleration There is probably only time to attempt one or two of these questions. In the tutorial session discussion
More informationPenning Traps. Contents. Plasma Physics Penning Traps AJW August 16, Introduction. Clasical picture. Radiation Damping.
Penning Traps Contents Introduction Clasical picture Radiation Damping Number density B and E fields used to increase time that an electron remains within a discharge: Penning, 936. Can now trap a particle
More informationqb 3 B ( B) r 3 e r = 3 B r e r B = B/ r e r = 3 B ER 3 E r 4
Magnetospheric Physics  Homework solution, /8/14 18 Gradient and curvature drift (a) A single proton has a parallel and perpendicular energy of 1 kev. Compute (B B) /B 3 and determine the instantaneous
More informationCandidacy Exam Department of Physics February 6, 2010 Part I
Candidacy Exam Department of Physics February 6, 2010 Part I Instructions: ˆ The following problems are intended to probe your understanding of basic physical principles. When answering each question,
More informationModels of ocean circulation are all based on the equations of motion.
Equations of motion Models of ocean circulation are all based on the equations of motion. Only in simple cases the equations of motion can be solved analytically, usually they must be solved numerically.
More informationSolid State Physics FREE ELECTRON MODEL. Lecture 17. A.H. Harker. Physics and Astronomy UCL
Solid State Physics FREE ELECTRON MODEL Lecture 17 A.H. Harker Physics and Astronomy UCL Magnetic Effects 6.7 Plasma Oscillations The picture of a free electron gas and a positive charge background offers
More informationRotational & RigidBody Mechanics. Lectures 3+4
Rotational & RigidBody Mechanics Lectures 3+4 Rotational Motion So far: point objects moving through a trajectory. Next: moving actual dimensional objects and rotating them. 2 Circular Motion  Definitions
More informationExam 2 Solutions. Note that there are several variations of some problems, indicated by choices in parentheses.
Exam 2 Solutions Note that there are several variations of some problems, indicated by choices in parentheses. Problem 1 Part of a long, straight insulated wire carrying current i is bent into a circular
More informationKinematics (2)  Motion in Three Dimensions
Kinematics (2)  Motion in Three Dimensions 1. Introduction Kinematics is a branch of mechanics which describes the motion of objects without consideration of the circumstances leading to the motion. 2.
More informationSpace Physics. An Introduction to Plasmas and Particles in the Heliosphere and Magnetospheres. MayBritt Kallenrode. Springer
MayBritt Kallenrode Space Physics An Introduction to Plasmas and Particles in the Heliosphere and Magnetospheres With 170 Figures, 9 Tables, Numerous Exercises and Problems Springer Contents 1. Introduction
More informationEvery magnet has a north pole and south pole.
Magnets  Intro The lodestone is a naturally occurring mineral called magnetite. It was found to attract certain pieces of metal. o one knew why. ome early Greek philosophers thought the lodestone had
More informationDynamics of Relativistic Particles and EM Fields
October 7, 2008 1 1 J.D.Jackson, Classical Electrodynamics, 3rd Edition, Chapter 12 Lagrangian Hamiltonian for a Relativistic Charged Particle The equations of motion [ d p dt = e E + u ] c B de dt = e
More informationUniform Circular Motion:Circular motion is said to the uniform if the speed of the particle (along the circular path) remains constant.
Circular Motion: Uniform Circular Motion:Circular motion is said to the uniform if the speed of the particle (along the circular path) remains constant. Angular Displacement: Scalar form:?s = r?θ Vector
More informationYell if you have any questions
Class 36: Outline Hour 1: Concept Review / Overview PRS Questions Possible Exam Questions Hour : Sample Exam Yell if you have any questions P361 Before Starting All of your grades should now be posted
More informationPhys 7221 Homework # 8
Phys 71 Homework # 8 Gabriela González November 15, 6 Derivation 56: Torque free symmetric top In a torque free, symmetric top, with I x = I y = I, the angular velocity vector ω in body coordinates with
More informationCBE 6333, R. Levicky 1. Orthogonal Curvilinear Coordinates
CBE 6333, R. Levicky 1 Orthogonal Curvilinear Coordinates Introduction. Rectangular Cartesian coordinates are convenient when solving problems in which the geometry of a problem is well described by the
More informationCHAPTER 11 RADIATION 4/13/2017. Outlines. 1. Electric Dipole radiation. 2. Magnetic Dipole Radiation. 3. Point Charge. 4. Synchrotron Radiation
CHAPTER 11 RADIATION Outlines 1. Electric Dipole radiation 2. Magnetic Dipole Radiation 3. Point Charge Lee Chow Department of Physics University of Central Florida Orlando, FL 32816 4. Synchrotron Radiation
More informationSession 6: Analytical Approximations for Low Thrust Maneuvers
Session 6: Analytical Approximations for Low Thrust Maneuvers As mentioned in the previous lecture, solving nonkeplerian problems in general requires the use of perturbation methods and many are only
More informationGauss s Law & Potential
Gauss s Law & Potential Lecture 7: Electromagnetic Theory Professor D. K. Ghosh, Physics Department, I.I.T., Bombay Flux of an Electric Field : In this lecture we introduce Gauss s law which happens to
More informationChapter 6. Quantum Theory of the Hydrogen Atom
Chapter 6 Quantum Theory of the Hydrogen Atom 1 6.1 Schrodinger s Equation for the Hydrogen Atom Symmetry suggests spherical polar coordinates Fig. 6.1 (a) Spherical polar coordinates. (b) A line of constant
More informationMOTION IN TWO OR THREE DIMENSIONS
MOTION IN TWO OR THREE DIMENSIONS 3 Sections Covered 3.1 : Position & velocity vectors 3.2 : The acceleration vector 3.3 : Projectile motion 3.4 : Motion in a circle 3.5 : Relative velocity 3.1 Position
More informationExam 2 Solutions. ε 3. ε 1. Problem 1
Exam 2 Solutions Problem 1 In the circuit shown, R1=100 Ω, R2=25 Ω, and the ideal batteries have EMFs of ε1 = 6.0 V, ε2 = 3.0 V, and ε3 = 1.5 V. What is the magnitude of the current flowing through resistor
More informationMagnetic Fields. or I in the filed. ! F = q! E. ! F = q! v! B. q! v. Charge q as source. Current I as source. Gauss s Law. Ampere s Law.
Magnetic Fields Charge q as source Gauss s Law Electric field E F = q E Faraday s Law AmpereMaxwell Law Current I as source Magnetic field B Ampere s Law F = q v B Force on q in the field Force on q v
More informationLecture 2. Lecture 1. Forces on a rotating planet. We will describe the atmosphere and ocean in terms of their:
Lecture 2 Lecture 1 Forces on a rotating planet We will describe the atmosphere and ocean in terms of their: velocity u = (u,v,w) pressure P density ρ temperature T salinity S up For convenience, we will
More informationLecturer: Bengt E W Nilsson
9 3 19 Lecturer: Bengt E W Nilsson Last time: Relativistic physics in any dimension. Lightcone coordinates, lightcone stuff. Extra dimensions compact extra dimensions (here we talked about fundamental
More informationElectric Potential Lecture 5
Chapter 23 Electric Potential Lecture 5 Dr. Armen Kocharian Electrical Potential Energy When a test charge is placed in an electric field, it experiences a force F = q o E The force is conservative ds
More informationThe total luminosity of a disk with the viscous dissipation rate D(R) is
Chapter 10 Advanced Accretion Disks The total luminosity of a disk with the viscous dissipation rate D(R) is L disk = 2π D(R)RdR = 1 R 2 GM Ṁ. (10.1) R The disk luminosity is half of the total accretion
More informationM273Q Multivariable Calculus Spring 2017 Review Problems for Exam 3
M7Q Multivariable alculus Spring 7 Review Problems for Exam Exam covers material from Sections 5.5.4 and 6.6. and 7.. As you prepare, note well that the Fall 6 Exam posted online did not cover exactly
More informationPh1a: Solution to the Final Exam Alejandro Jenkins, Fall 2004
Ph1a: Solution to the Final Exam Alejandro Jenkins, Fall 2004 Problem 1 (10 points)  The Delivery A crate of mass M, which contains an expensive piece of scientific equipment, is being delivered to Caltech.
More informationChapter 28. The Magnetic Field
Chapter 28 The Magnetic Field Magnetic Field Force Exerted by a Magnetic Field Point Charge in a Magnetic Field Torques on Current Loops MFMcGrawPHY 2426 Ch28aMagnetic Field  Revised 10/03/2012 2 Magnetic
More informationElastic Scattering. R = m 1r 1 + m 2 r 2 m 1 + m 2. is the center of mass which is known to move with a constant velocity (see previous lectures):
Elastic Scattering In this section we will consider a problem of scattering of two particles obeying Newtonian mechanics. The problem of scattering can be viewed as a truncated version of dynamic problem
More informationMaxwell s equations and EM waves. From previous Lecture Time dependent fields and Faraday s Law
Maxwell s equations and EM waves This Lecture More on Motional EMF and Faraday s law Displacement currents Maxwell s equations EM Waves From previous Lecture Time dependent fields and Faraday s Law 1 Radar
More informationPhysics 8.02 Exam Two Equation Sheet Spring 2004
Physics 8.0 Exam Two Equation Sheet Spring 004 closed surface EdA Q inside da points from inside o to outside I dsrˆ db 4o r rˆ points from source to observer V moving from a to b E ds 0 V b V a b E ds
More informationDownloaded from
Question 4.1: A circular coil of wire consisting of 100 turns, each of radius 8.0 cm carries a current of 0.40 A. What is the magnitude of the magnetic field B at the centre of the coil? Number of turns
More informationPhysical Processes in Astrophysics
Physical Processes in Astrophysics Huirong Yan Uni Potsdam & Desy Email: hyan@mail.desy.de 1 Reference Books: Plasma Physics for Astrophysics, Russell M. Kulsrud (2005) The Physics of Astrophysics, Frank
More informationPhysics 214 Solution Set 4 Winter 2017
Physics 14 Solution Set 4 Winter 017 1. [Jackson, problem 1.3] A particle with mass m and charge e moves in a uniform, static, electric field E 0. (a) Solve for the velocity and position of the particle
More informationIntroduction to Vector Calculus (29) SOLVED EXAMPLES. (d) B. C A. (f) a unit vector perpendicular to both B. = ˆ 2k = = 8 = = 8
Introduction to Vector Calculus (9) SOLVED EXAMPLES Q. If vector A i ˆ ˆj k, ˆ B i ˆ ˆj, C i ˆ 3j ˆ kˆ (a) A B (e) A B C (g) Solution: (b) A B (c) A. B C (d) B. C A then find (f) a unit vector perpendicular
More informationPhysics Jonathan Dowling. Final Exam Review
Physics 2102 Jonathan Dowling Physics 2102 Final Exam Review A few concepts: electric force, field and potential Electric force: What is the force on a charge produced by other charges? What is the force
More informationChapter 27 Magnetic Field and Magnetic Forces
Chapter 27 Magnetic Field and Magnetic Forces Lecture by Dr. Hebin Li Goals for Chapter 27 To study magnets and the forces they exert on each other To calculate the force that a magnetic field exerts on
More informationSolutions for the Practice Final  Math 23B, 2016
olutions for the Practice Final  Math B, 6 a. True. The area of a surface is given by the expression d, and since we have a parametrization φ x, y x, y, f x, y with φ, this expands as d T x T y da xy
More informationExam 2 Solutions. = /10 = / = /m 3, where the factor of
PHY049 Fall 007 Prof. Yasu Takano Prof. Paul Avery Oct. 17, 007 Exam Solutions 1. (WebAssign 6.6) A current of 1.5 A flows in a copper wire with radius 1.5 mm. If the current is uniform, what is the electron
More informationConservation of Angular Momentum
Physics 101 Section 3 March 3 rd : Ch. 10 Announcements: Monday s Review Posted (in Plummer s section (4) Today start Ch. 10. Next Quiz will be next week Test# (Ch. 79) will be at 6 PM, March 3, Lockett6
More informationMagnetic Materials. 1. Magnetization 2. Potential and field of a magnetized object
Magnetic Materials 1. Magnetization 2. Potential and field of a magnetized object 3. Hfield 4. Susceptibility and permeability 5. Boundary conditions 6. Magnetic field energy and magnetic pressure 1 Magnetic
More informationMap projections. Rüdiger Gens
Rüdiger Gens 2 Outline! Relevant terms! Why map projections?! Map projection categories " Projection surfaces " Features preserved from distortions! Map projection examples! Right choice Relevant terms!
More informationPLANAR KINETICS OF A RIGID BODY FORCE AND ACCELERATION
PLANAR KINETICS OF A RIGID BODY FORCE AND ACCELERATION I. Moment of Inertia: Since a body has a definite size and shape, an applied nonconcurrent force system may cause the body to both translate and rotate.
More informationThe dynamics of high and low pressure systems
The dynamics of high and low pressure systems Newton s second law for a parcel of air in an inertial coordinate system (a coordinate system in which the coordinate axes do not change direction and are
More informationYell if you have any questions
Class 36: Outline Hour 1: Concept Review / Overview PRS Questions Possible Exam Questions Hour : Sample Exam Yell if you have any questions P361 efore Starting All of your grades should now be posted
More informationr( θ) = cos2 θ ω rotation rate θ g geographic latitude   θ geocentric latitude   Reference Earth Model  WGS84 (Copyright 2002, David T.
1 Reference Earth Model  WGS84 (Copyright 22, David T. Sandwell) ω spheroid c θ θ g a parameter description formula value/unit GM e (WGS84) 3.9864418 x 1 14 m 3 s 2 M e mass of earth  5.98 x 1 24 kg
More informationSOLUTIONS TO THE FINAL EXAM. December 14, 2010, 9:00am12:00 (3 hours)
SOLUTIONS TO THE 18.02 FINAL EXAM BJORN POONEN December 14, 2010, 9:00am12:00 (3 hours) 1) For each of (a)(e) below: If the statement is true, write TRUE. If the statement is false, write FALSE. (Please
More informationDust devils, water spouts, tornados
Balanced flow Things we know Primitive equations are very comprehensive, but there may be a number of vast simplifications that may be relevant (e.g., geostrophic balance). Seems that there are things
More informationMoment of inertia. Contents. 1 Introduction and simple cases. January 15, Introduction. 1.2 Examples
Moment of inertia January 15, 016 A systematic account is given of the concept and the properties of the moment of inertia. Contents 1 Introduction and simple cases 1 1.1 Introduction.............. 1 1.
More informationThe Virial Theorem, MHD Equilibria, and ForceFree Fields
The Virial Theorem, MHD Equilibria, and ForceFree Fields Nick Murphy HarvardSmithsonian Center for Astrophysics Astronomy 253: Plasma Astrophysics February 10 12, 2014 These lecture notes are largely
More informationSENIOR_ 2017_CLASS_12_PHYSICS_ RAPID REVISION_1_ DERIVATIONS IN FIRST FIVE LESSONS Page 1
INDIAN SCHOOL MUSCAT Department of Physics Class XII Rapid Revision 1 DERIVATIONS IN FIRST FIVE LESSONS 1) Field due to an infinite long straight charged wire Consider an uniformly charged wire of infinite
More informationExperiment V Motion of electrons in magnetic field and measurement of e/m
Experiment V Motion of electrons in magnetic field and measurement of e/m In Experiment IV you observed the quantization of charge on a microscopic bead and measured the charge on a single electron. In
More informationDynamics Rotating Tank
Institute for Atmospheric and Climate Science  IACETH Atmospheric Physics Lab Work Dynamics Rotating Tank Large scale flows on different latitudes of the rotating Earth Abstract The large scale atmospheric
More informationThe magnetic force acting between nucleons
The magnetic force acting between nucleons *Valerio Dallacasa and + Norman D. Cook * University of Verona, Verona, Italy + University of Southern California, Los Angeles, California, USA, and Kansai University,
More informationElectromagnetic Theory
Summary: Electromagnetic Theory Maxwell s equations EM Potentials Equations of motion of particles in electromagnetic fields Green s functions LienardWeichert potentials Spectral distribution of electromagnetic
More informationUniversity of Alabama Department of Physics and Astronomy. Problem Set 6
University of Alabama Department of Physics and Astronomy PH 16 LeClair Fall 011 Instructions: Problem Set 6 1. Answer all questions below. Show your work for full credit.. All problems are due 14 October
More informationTheory English (Official)
Q31 Large Hadron Collider (10 points) Please read the general instructions in the separate envelope before you start this problem. In this task, the physics of the particle accelerator LHC (Large Hadron
More information7a3 2. (c) πa 3 (d) πa 3 (e) πa3
1.(6pts) Find the integral x, y, z d S where H is the part of the upper hemisphere of H x 2 + y 2 + z 2 = a 2 above the plane z = a and the normal points up. ( 2 π ) Useful Facts: cos = 1 and ds = ±a sin
More informationTranslational vs Rotational. m x. Connection Δ = = = = = = Δ = = = = = = Δ =Δ = = = = = 2 / 1/2. Work
Translational vs Rotational / / 1/ Δ m x v dx dt a dv dt F ma p mv KE mv Work Fd / / 1/ θ ω θ α ω τ α ω ω τθ Δ I d dt d dt I L I KE I Work / θ ω α τ Δ Δ c t s r v r a v r a r Fr L pr Connection Translational
More informationPHY Fall HW6 Solutions
PHY249  Fall 216  HW6 Solutions Allen Majewski Department Of Physics, University of Florida 21 Museum Rd. Gainesville, FL 32611 October 11, 216 These are solutions to Halliday, Resnick, Walker Chapter
More informationAccelerator Physics. G. A. Krafft Jefferson Lab Old Dominion University Lecture 2
Accelerator Physics G. A. Krafft Jefferson Lab Old Dominion University Lecture 2 Fourvectors Fourvector transformation under z boost Lorentz Transformation v ' v v v v 0 0 3 ' 1 1 ' v v 2 2 v ' v v 3
More informationexample consider flow of water in a pipe. At each point in the pipe, the water molecule has a velocity
Module 1: A Crash Course in Vectors Lecture 1: Scalar and Vector Fields Objectives In this lecture you will learn the following Learn about the concept of field Know the difference between a scalar field
More informationEDP with strong anisotropy : transport, heat, waves equations
EDP with strong anisotropy : transport, heat, waves equations Mihaï BOSTAN University of AixMarseille, FRANCE mihai.bostan@univamu.fr Nachos team INRIA Sophia Antipolis, 3/07/2017 Main goals Effective
More informationDistance travelled time taken and if the particle is a distance s(t) along the xaxis, then its instantaneous speed is:
Chapter 1 Kinematics 1.1 Basic ideas r(t) is the position of a particle; r = r is the distance to the origin. If r = x i + y j + z k = (x, y, z), then r = r = x 2 + y 2 + z 2. v(t) is the velocity; v =
More informationOrbital Motion in Schwarzschild Geometry
Physics 4 Lecture 29 Orbital Motion in Schwarzschild Geometry Lecture 29 Physics 4 Classical Mechanics II November 9th, 2007 We have seen, through the study of the weak field solutions of Einstein s equation
More informationChapter 7: Circulation and Vorticity
Chapter 7: Circulation and Vorticity Circulation C = u ds Integration is performed in a counterclockwise direction C is positive for counterclockwise flow!!! Kelvin s Circulation Theorem The rate of change
More informationChapter 6: Vector Analysis
Chapter 6: Vector Analysis We use derivatives and various products of vectors in all areas of physics. For example, Newton s 2nd law is F = m d2 r. In electricity dt 2 and magnetism, we need surface and
More informationMomentum Circular Motion and Gravitation Rotational Motion Fluid Mechanics
Momentum Circular Motion and Gravitation Rotational Motion Fluid Mechanics Momentum Momentum Collisions between objects can be evaluated using the laws of conservation of energy and of momentum. Momentum
More informationThree particles, a, b, and c, enter a magnetic field as shown in the figure. What can you say about the charge on each particle?
1 Three particles, a, b, and c, enter a magnetic field as shown in the figure. What can you say about the charge on each particle? 6 Determine the magnitude and direction of the force on an electron traveling
More informationA) Yes B) No C) Impossible to tell from the information given.
Does escape speed depend on launch angle? That is, if a projectile is given an initial speed v o, is it more likely to escape an airless, nonrotating planet, if fired straight up than if fired at an angle?
More information9.3 Worked Examples Circular Motion
9.3 Worked Examples Circular Motion Example 9.1 Geosynchronous Orbit A geostationary satellite goes around the earth once every 3 hours 56 minutes and 4 seconds, (a sidereal day, shorter than the noontonoon
More informationAAPT UNITED STATES PHYSICS TEAM AIP 2007
007 Semifinal Exam Solutions 1 AAPT UNITED STATES PHYSICS TEAM AIP 007 Solutions to Problems Part A Question 1 a. There is a high degree of symmetry present. Points b, c, and e are at the same potential;
More informationSummary: Curvilinear Coordinates
Physics 2460 Electricity and Magnetism I, Fall 2007, Lecture 10 1 Summary: Curvilinear Coordinates 1. Summary of Integral Theorems 2. Generalized Coordinates 3. Cartesian Coordinates: Surfaces of Constant
More informationRelativity II. Home Work Solutions
Chapter 2 Relativity II. Home Work Solutions 2.1 Problem 2.4 (In the text book) A charged particle moves along a straight line in a uniform electric field E with a speed v. If the motion and the electric
More informationEQUATIONS OF MOTION: NORMAL AND TANGENTIAL COORDINATES (Section 13.5)
EQUATIONS OF MOTION: NORMAL AND TANGENTIAL COORDINATES (Section 13.5) Today s Objectives: Students will be able to apply the equation of motion using normal and tangential coordinates. APPLICATIONS Race
More informationElectric and Magnetic Forces in Lagrangian and Hamiltonian Formalism
Electric and Magnetic Forces in Lagrangian and Hamiltonian Formalism Benjamin Hornberger 1/26/1 Phy 55, Classical Electrodynamics, Prof. Goldhaber Lecture notes from Oct. 26, 21 Lecture held by Prof. Weisberger
More informationEinstein s Equations. July 1, 2008
July 1, 2008 Newtonian Gravity I Poisson equation 2 U( x) = 4πGρ( x) U( x) = G d 3 x ρ( x) x x For a spherically symmetric mass distribution of radius R U(r) = 1 r U(r) = 1 r R 0 r 0 r 2 ρ(r )dr for r
More informationElectricity & Magnetism Study Questions for the Spring 2018 Department Exam December 4, 2017
Electricity & Magnetism Study Questions for the Spring 2018 Department Exam December 4, 2017 1. a. Find the capacitance of a spherical capacitor with inner radius l i and outer radius l 0 filled with dielectric
More informationChapter 5. Shallow Water Equations. 5.1 Derivation of shallow water equations
Chapter 5 Shallow Water Equations So far we have concentrated on the dynamics of smallscale disturbances in the atmosphere and ocean with relatively simple background flows. In these analyses we have
More informationMATH2000 Flux integrals and Gauss divergence theorem (solutions)
DEPARTMENT O MATHEMATIC MATH lux integrals and Gauss divergence theorem (solutions ( The hemisphere can be represented as We have by direct calculation in terms of spherical coordinates. = {(r, θ, φ r,
More informationApplied Nuclear Physics (Fall 2006) Lecture 19 (11/22/06) Gamma Interactions: Compton Scattering
.101 Applied Nuclear Physics (Fall 006) Lecture 19 (11//06) Gamma Interactions: Compton Scattering References: R. D. Evans, Atomic Nucleus (McGrawHill New York, 1955), Chaps 3 5.. W. E. Meyerhof, Elements
More informationSuggested solutions, FYS 500 Classical Mechanics and Field Theory 2015 fall
UNIVERSITETET I STAVANGER Institutt for matematikk og naturvitenskap Suggested solutions, FYS 500 Classical Mecanics and Field Teory 015 fall Set 1 for 16/17. November 015 Problem 68: Te Lagrangian for
More informationChapter 7. Geophysical Fluid Dynamics of Coastal Region
1 Lecture Notes on Fluid Dynamics (1.63J/2.21J) by Chiang C. Mei, 2006 Chapter 7. Geophysical Fluid Dynamics of Coastal egion [ef]: Joan Brown and six others (Open Univeristy course team on oceanography):
More informationThe Structure of the Magnetosphere
The Structure of the Magnetosphere The earth s magnetic field would resemble a simple magnetic dipole, much like a big bar magnet, except that the solar wind distorts its shape. As illustrated below, the
More informationChapter 9 Notes. x cm =
Chapter 9 Notes Chapter 8 begins the discussion of rigid bodies, a system of particles with fixed relative positions. Previously we have dealt with translation of a particle: if a rigid body does not rotate
More informationTridib s Physics Tutorials. NCERTXII / Unit 4 Moving charge and magnetic field
MAGNETIC FIELD DUE TO A CURRENT ELEMENT The relation between current and the magnetic field, produced by it is magnetic effect of currents. The magnetic fields that we know are due to currents or moving
More information1/25/2010. Circulation and vorticity are the two primary
Lecture 4: Circulation and Vorticity Measurement of Rotation Circulation Bjerknes Circulation Theorem Vorticity Potential Vorticity Conservation of Potential Vorticity Circulation and vorticity are the
More informationFoundation Course Unit. Fields and Waves. SOLUTIONS. Examiners: Dr. A. G. Polnarev Prof. F. Di Lodovico
Foundation Course Unit SEF006 Fields and Waves. SOLUTIONS Examiners: Dr. A. G. Polnarev Prof. F. Di Lodovico c Queen Mary University of London, 2016 Page 2 SEF006 (2016) SECTION A SOLUTIONS TO SECTION
More informationFLUX OF VECTOR FIELD INTRODUCTION
Chapter 3 GAUSS LAW ntroduction Flux of vector field Solid angle Gauss s Law Symmetry Spherical symmetry Cylindrical symmetry Plane symmetry Superposition of symmetric geometries Motion of point charges
More information### dv where ρ is density, R is distance from rotation axis, and dv is
Comments This is one of my favorite problem assignments in our Solid Earth Geophysics class, typically taken by junior and senior concentrators and by firstyear graduate students. I encourage students
More informationVU Mobile Powered by S NO Group All Rights Reserved S NO Group 2012
PHY101 Physics Final Term Solved MCQs (Latest) 1 1. A total charge of 6.3 10 8 C is distributed uniformly throughout a 2.7cm radius sphere. The volume charge density is: A. 3.7 10 7 C/m3 B. 6.9 10 6 C/m3
More informationTransmission line demo to illustrate why voltage along transmission lines is high
Transmission line demo to illustrate why voltage along transmission lines is high Connect to step down transformer 120V to 12V to lightbulb 12 V 6.5 A Lights up brightly Connect it to long fat wires Lights
More informationChapter 30 Solutions
Chapter 30 Solutions 30.1 B µ 0I R µ 0q(v/π R) R 1.5 T *30. We use the BiotSavart law. For bits of wire along the straightline sections, ds is at 0 or 180 to ~, so ds ~ 0. Thus, only the curved section
More informationThe Central Force Problem: Hydrogen Atom
The Central Force Problem: Hydrogen Atom B. Ramachandran Separation of Variables The Schrödinger equation for an atomic system with Z protons in the nucleus and one electron outside is h µ Ze ψ = Eψ, r
More informationLecture No 1 Introduction to Diffusion equations The heat equat
Lecture No 1 Introduction to Diffusion equations The heat equation Columbia University IAS summer program June, 2009 Outline of the lectures We will discuss some basic models of diffusion equations and
More information