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# Gen Physics II Noncalc PHYS 2020

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PHYS2020 General Physics II Course Lecture Notes Section V Dr Donald G Luttermoser East Tennessee State University Edition 31 Abstract These Class notes are designed for use of the instructor and students of the course PHYS 2020 General Physics taught by Dr Donald Lutternioser at East Tennessee State University These notes make reference to the College Physics Enhanced 7th Editizm 2006 textbook by Serway Faughn and Vuille V Magnetism A Magnets 1 Ancient people discovered a rock that when placed close to a sim ilar rock the rocks would move either closer or farther apart from each other gt these rocks were called loadstones or magnets 2 Like poles repel each other and unlike poles attract each other a Magnets have to be asymmetrical for this to work gt preferably the shape of a bar b One pole is called north N the other south 3 Like the electric eld magnets have a magnetic eld called the eld a The B eld points away from the north pole hence is analogous to a electric charge b The B eld points towards from the south pole hence is analogous to a electric charge 4 Unlike the electric eld there are no monopoles in magnetism gt magnets are always dipoles 2 poles V1 V 2 PHYS 2020 General Physics II Efield lz A 1 W monopole 1 pole dipolefield B Magnetic Field of the Earth 2 Wes 1 The Earth has a magnetic eld a Currently the magnetic N pole corresponds to the S ge ographic pole which is why your compass s N points to geographic N remember like poles repel opposites at tract b The magnetic S pole corresponds to the N geographic pole c The magnetic axis is not aligned with the spin axis 2 Devices used to measure the direction of the magnetic eld are called compasses The designation N and S for either pole on a compass was assigned based on which end pointed to the Earth s north pole gt N seeks Earth s North pole 3 The Earth s magnetic eld is generated by electric currents in the liquid outer core of the planet which is composed of mostly nickel and iron The electric currents arise from the Earth s rota tion The details of the geomagnetic eld source is still not well understood Donald G Luttermoser E TSU 4 The magnetic eld of the Earth ips polarity 16 reverses di rection in random intervals over time a b c d It has ipped 171 times over the past 76 million years The average period between ips is about 104 to 105 years with a peak average period of 200000 years The last ip occurred 780000 years ago We know this from the volcanic rocks near the Mid Atlantic Ridge As the sea oor spreads as mantle material ows upward at that location the rock solidi es when it hits the water The metallic compounds in the lava freeze the direction of the magnetic eld as the liquid lava turns to solid basalt rock Geomagnetic eld measurements over the past few hun dred years show that the B eld strength of the Earth is decreasing i Today the Earth s magnetic eld is 10 weaker than it was when Carl Gauss started keeping tabs on it in 1845 ii We may be on our way to another polarity ip within the next few thousand years C Magnetic Fields 1 When moving through a magnetic eld a charged particle expe riences a force a b The force has a maximum value when the charge moves J to the B eld lines The force becomes zero when the charge moves parallel to the eld V4 Q Q PHYS 2020 General Physics II This force is given by the equation F qu sin 6 F BE W4 qv sin 6 39 VQ i F FB E magnetic force ii q E electric charge of the particle moving in the B eld iii 1 E velocity of the particle iv B E strength of the magnetic eld v 6 E angle between 17 and The 81 units for magnetic eld is tesla i The magnetic eld is also measured in weber Wb per square meter 1T1me2 where a weber is the unit of magnetic ux see Vl of the notes ii Units of B i Wb i N N Bir4747 47 T l m2 Cms Am iii Note that the cgs units for magnetic eld strength is still Widely used especially in astronomy and geology gt the gauss 1Tu o Donald G Luttermoser E TSU V 5 iv Laboratory magnets can reach as high as 25000 G 25 T The Earth s magnetic eld is about 05 G 2 Though we wrote Eqs V l and V Z as scalar equations in reality they are 3 dimensz39onal vector equations gt 13 is always in a direction that is J to the plane de ned by the U and g a vectors In upper level physics F 117 X B a vector cross product a F 1 plane defined q by the B amp v 39gt e vectors V a The maximum force on a charged particle traveling through a B eld is Fm qu V B This will only occur when the 13 g and U are ail J to each other b The righthand rule is used to see which direction F points with respect to vectors g and U i For a positive charge treat 17 as the reference vec tor and curl the ngers of your right hand towards g keeping your thumb pointing out from your hand The direction your thumb is pointing is the direc tion of 13 V6 PHYS 2020 General Physics II ii For a negative charge do as above but swap 17 with g so that g is now the reference vector Example V71 Problem 195 Page 651 from the Serway Faughn amp Vuille textbook At the Equator near the Earth s sur face the magnetic field is approximately 500 uT northward and the electric field is about 100 NC downward in fair weather Find the gravitational electric and magnetic forces on an electron with an instantaneous velocity of 600 X 106 ms directed to the east in this environment As an additional question what is the total force on the electron and what is the final acceleration on the electron Solution Let s de ne 2 as the unit vector in the upward direction 02 in the East direction and g in the North direction The mass of an electron is me 911 X 10 31 kg and the charge is e l60 X 10 19 C The various forces are then equal to the following For the gravitational force g7 points downward opposite to the 2 direction therefore 13 mg m g 2 911 x 1031 kg 980 msg 2 893 x 1030 N 2 or in the downward direction For the electric force E points downward opposite the 2 di rection therefore 12 qE e E 2 160 x 10719 o 100 NC 2 160 x 1017 N 2 or in the upward direction For the magnetic force 17 points eastward in the positive 0 direction and g points northward in the positive 7 direction Donald G Luttermoser E TSU V 7 Using the right hand rule for a negative charge we choose g as the reference vector and curl it into 17 Doing this we curl our ngers from North to East and our right thumb points in the downward direction hence negative 2 Since we have used the right hand rule to determine our direction we will just use the absolute value of our quantities in Eq V l Since E J 17 6 90quot so a FB qvBsin6 2 160 x 1019 C600 x 106 ms500 x 106 T sin9002 7 480 x 1017 N 2 or in the downward direction We can now calculate the total force on this electron as a Rat FQF FB 893 x 1030 N 2 160 x 1017 N 2 480 x 1017 N 2 320 x 1017 N 2 or the electron rnoves downward with an acceleration of 130 320 x 1017 N 2 a me 911 x 1m31 kg l 351 x 1013 ms22 358 x 10129 23 or 358 trillion surface gravitiesl Note that the gravitational force is negligible with respect to the electric and the magnetic forces in the above calculation of the total force V8 PHYS 2020 General Physics II D Magnetic Force on a CurrentCarrying Conductor 1 A current carrying Wire also experiences a force if it passes through a magnetic eld Inside the Wire charges are moving along the length of the Wire 677 X X X X X X X an The total force Ftot acting on the Wire will be equal to the sum of all of the forces on the individual charges Fmd owing in the Wire Plot NEnd 7 where N is the number of charges in the Wire Since on average all of the charges are moving in the same direction the total force will be the same direction as the average forces on the individual particles a The number of charges in the Wire a unitless number is determined with N My v 5 where n is the particle density of charges m 3 A is the cross sectional area of the Wire m2 and E is the length of the Wire Donald G Luttermoser E TSU b c d e The forces on the individual particles is just determined from Eq V l Find m B 81110 V45 where W is the drift velocity of the charges in the wire given by Eq Ill 6 and 17d points in the same direction as If the B eld is J to the length of the wire met quBsin90 nA quBXnAE but I nquA see Eq Ill 5 so met 71qu AXE 5 01quot FM B I z v 7 If the B eld is in some arbitrary direction with respect to the length of the wire hence current then 1331am V8 where 6 is the angle between the g vector and the E vector For this force equation the direction of 13 with respect to g and E is found from the right hand rule following what was written for a point charge however here we use IE in place of U Since I is automatically de ned as the direction that positive charge ows we take with I as the reference vector and curl it towards g Your right thumb then indicates the direction of 13 V9 PHYS 2020 General Physics II This is how speakers work a current is passed through a wire wrapped around a magnet and the wire is connected to a di aphragm usually paper in cheaper speakers surrounded by a cone to direct the sound waves a As the current is turned on and off in the wire the wire continuously feels a force cycle between an on and off7 state gt causes the wire to move back and forth b Since the coil is connected to the diaphragm the diaphragm moves as well causing sound waves to form c The sound propagates in the direction that the cone is pointing see Figure 1913 in your textbook Example V72 A wire carries a steady current of 240 A A straight section of the wire is 0750 m long and lies along the x axis within a uniform magnetic field of magnitude 160 T in the posi tive z direction If the current is in the 06 direction what is the magnitude of the force on the section of the wire Solution Here i 32 and E L92 with B 160 T and L 0750 m Curling E into E has our right thumb pointing in the y direction ie g The magnitude of the force is found with Eq V 8 with 6 900 the angle between and F BIL sin6 160 T240 A0750 m sin 900 288 N Using this magnitude with the above derived direction gives us a magnetic force vector of F 288Ng Donald G Luttermoser E TSU E Torque on a Current Loop Galvanorneters and Electric Motors 1 Consider a rectangular loop carrying current I in the presence of an external B eld in the plane of the loop 3 A k t a B I 1 into page I gt v gt F B 2 r 39 x b 1 out of page At I 34 gt E L I t a B I a I a Forces on the a lengths are zero in the diagram above since g E FE length b Forces on the b lengths are 131 F2 BIZ v9 2 Viewing this loop on its side rotated about the x axis gives 2 h F1 47 a2 Hl current gt 0 Q current out of page d KJ d into page 1 2 h gt F2 gt PHYS 2020 General Physics II a From this viewpoint let s rotate the loop about the origin 0 and about the y axis above The magnitude hence a positive number of the maxi mum torque that will occur on the loop will take place b when I is farthest from the z axis 2 6 12 as drawn above a a Tm F1d1 F2d2 F1 F2 BM BM BIab c Since the area of the loop is A ab we see that 3 If we let the loop rotate about the y axis then gt Ah 2 F gt 1 B I Iout I a2 39 E d1 3 sin e I e L I r x 61 6 32 a Iquot l39 d2 2 sm 9 gt F2 V ll 739 BIAsin 4 If we have more than 1 loop in the B eld gt ie a loop with V 12 N turns we get 739 NBIA sin6 Donald G Luttermoser E TSU Example V73 Problem 1922 Page 653 from the Ser way amp Faughn textbook A current of 170 mA is maintained in a single circular loop with a circumference of 200 m A magnetic field of 0800 T is directed parallel to the plane of the loop What is the magnitude of the torque exerted by the magnetic field on the loop Solution The magnitude of torque for a single loop in a magnetic eld is given by Eq V 11 or Eq V 12 by setting N 1 Here 6 is the angle between the eld and the normal line of the plane de ned by the loop The B eld is parallel to the loop hence plane so 6 900 since the normal 1 line of the loop plane is at a right angle to the B eld We now need to determine the cross sectional area of the loop The circumference of the loop is given as C 27W 200 m From this we can calculate the area of the loop as follows C 200 m 100 i 7 m rgi 27T 7T 1 2 1 A7Tr27rlt mgt mg 7T 7T We can now calculate the magnitude of the torque from Eq V 11 739 B A sin6 100 0800 T170 x 103 A lt7 m2 sin 900 7T 7 433 x 103 Nm V 14 PHYS 2020 General Physics II 5 Galvanometers use torque due to an internal magnetic eld to deduce current in an electric circuit a When current passes through the coil of a galvanometer situated in a B eld produced by two permanent magnets inside the galvanometer the magnetic torque causes the coil to twist The angle through which the coil rotates olt current through it b A needle e indicator is connected to the coil which points at a calibrated scale gt the needle s motion indi cates the amount of current 6 Likewise passing a current through a loop embedded in a mag netic eld can impart a torque on the loop which causes the loop to rotate gt an electric motor a The current cannot be allowed to continue in the same direction since once the loop rotates through the 6 0 con guration the force goes to zero at that point b Once the loop rotates past this direction due to its in ertia the torque acts to slow the motion ie the force reverses direction causing the rotation to stop c A continuous rotation can be maintained if the direction of the current changes every half rotation This can be accomplished automatically if using an alternating cur rent AC what your wall outlets deliver 1 A direct current DC can be used if the current can be made to change direction in the motor itself gt this is accomplished mechanically in such DC electric motors with split ring contacts called commutators and brushes as shown in Figure 1917 in the textbook Donald G Luttermoser E TSU F Motion of a Charged Particle in a BField 1 Assume a charged particle charge moves J to a B eld a Since the force on the particle follows the right hand rule 13 J g and 13 J U gt the particle is forced to move in a circular path b The magnitude of the force always points towards the center of the circular path a c If U is not perpendicular to B the particle moves in a helical path along the eld lines gt VW lt l 1 39 v W q gt F V lt lt 2 Since the particle is in circular motion there must be a cen tripetal force e the magnetic force on the particle 2 F qu v13 7 a Here F6 mug7 is the centripetal force b The radius of curvature of the path is then m1 7 7 V 14 T qB PHYS 2020 General Physics II Example V74 Problem 1928 Page 654 from the Serway amp Faughn textbook A cosmic ray proton in interstellar space has an energy of 100 MeV and executes a Circular orbit having a radius equal to that of Mercury39s orbit around the Sun 580 X 1010 m What is the magnetic field in that region of space Solution Since the path is circular the particle moves J to the B eld hence the magnetic force supplies the centripetal acceleration so FCFB 2 m qu r B qr From General Physics I we can write the linear momentum of p m1 iZm and the kinetic energy of this proton is 100 MeV Convert ing this to Joules gives this proton as 160 x 1019 J KE1 16v 00X0elt 18V gt 160 x 1012 J Noting that the mass of a proton is m 167 X 10 27 kg and the charge of a proton is q e 160 X 10 19 C we now have all we need to determine the B eld from the equation above B i 727 2mKE Q71 Q71 Q71 2167 x 1027 kg160 x 1012 J 160 x 1019 C580 x 1010 m 788 x 1012 T Donald G Luttermoser ETSU V 17 G Ampere s Law 1 If a wire is grasped in the right hand with the thumb pointing in the direction of the current the ngers will curl in the direction of g gt right hand rule for Wires a Hans Oersted gured this out in 1819 b The B eld curves around a straight wire carrying a cur rent B field c Shortly after Oested s discovery scientists determined the following relation between the strength of the B eld and the current in a wire MOI B V 15 27W i 7 E distance L from wire ii uo E permeability of free space uo 47T X 10 7 TmA 2 Ampere s Circuit Law ZBH Ag do VlG PHYS 2020 General Physics II which is the algebraic form of the more general calculus form given by f duol where the symbol means line integral and d means an in nitesimally small segment of the closed path we are integrating or summing over a b C 01 ZBH A6 means that we take the sum over all products B H A6 around a closed path EU is the B eld component that is tangential to the B eld circular path A6 is a path segment along the B eld This law Eq V l6 is usually just referred to as Ampere s law It is one of the basic equations of electromagnetism gt B 1 x A Side View gt B m Faceon View Donald G Luttermoser E TSU 3 We can now use Ampere s circuit law to derive Eq V l5 Am pere s general law a Starting with Eq V l6 we see that since B is perfectly circular around a wire BH is uniform along AZ b A6 is just individual segments of the complete circle de ned by B gt summing up all of these segments around the entire circle is just the circumference of the circle 2 A6 C 27W so BH 27W no or I Uo B B ll 27W which is Eq V45 4 We can now bring 2 wires close together and calculate the effect of one of the wire s B eld on the other A gt Ii 1 a 39 a d B2 l F1 outof page 39 I I r gt 12 39 l rd a b c PHYS 2020 General Physics II For Wire 2 located a distance d from Wire 1 012 27rd B2 The force on Wire 1 from B2 is given by Eq V 7 o 0 F1 B2118 2 I18 a 128 7rd 27rd where s is the length of a given segment in the line We can rewrite this as the force per unit length 5 101112 3 27rd 39 V 17 The force between 2 parallel wires carrying a current is used to de ne the SI units of current gt ampere a If 2 long wires 1 meter apart carry the same current and the forcelength on each Wire is 20 X 10 7 Nm then the current is de ned to be 10 A b If a conductor carries a steady current of 10 A then through any cross section of wire 10 C of charge passes by every second Example V75 Problem 1942 Page 655 from the Ser way amp Faughn textbook A long straight wire lies on a horizontal table and carries a current of 120 uA In a vacuum a proton moves parallel to the wire opposite the current with a constant velocity of 230 X 104 ms at a constant distance d above the wire Determine the value of d You may ignore the magnetic field due to the Earth Solution Since the proton moves at constant velocity the net force acting on it is zero Since the current carrying Wire produces a B eld it Donald G Luttermoser ETSU V 21 will supply a force on the proton Also this force will be pointing upward 2 direction as dictated from the right hand rule since the proton is moving opposite to the current and g points J to the wire hence U and parallel to the table top As such there must be some counterbalancing force pointing downward that also is acting on the proton to keep the net force zero Since we are ignoring the GE s B eld the only other force available is the gravitational force acting on the proton from the 69 which points downward just as we need Hence we have FBFg qumg B go The B eld is being generated by the current I in the wire as such we also can use Eq V l5 to determine B with 7 d for our case here No B 27rd Solving these two simultaneous equations for d gives mg i No go i 27rd qvLOI qv d uo 27ng 27ng 160 x 1049 C230 x 104 rns 27T167 x 1027 kg980 msg 4w x 107 TmA120 x 106 A V22 PHYS 2020 General Physics II H Solenoids 1 A solenoid is a device that produces a dipole magnetic eld from electricity owing through a coil of wire see Figure 1932 in the textbook Solenoids also are called electromagnets The strength of the B eld inside the solenoid where the eld is relatively parallel is given by where n N E is the number of turns in the coil per unit length a We can derive the solenoid equation Eq V l8 through Ampere s law as follows Consider the B eld of an in nitely long solenoid The B eld inside this solenoid is uniform and parallel to the axis of the solenoid b The B eld outside the in nite solenoid is zero since the eld lines can never leave an in nitely long cylinder As such we can set B 0 on the outside c Taking a rectangular path in our closed loop through one side of the solenoid as shown in Figure 1935 in the text book the sides of the rectangle that are J 639 90quot to the B eld produce no terms in the sum of Eq V 16 since B AF BA cos BAZcosQOO 0 d Since B 0 outside the solenoid B A6 0 on that side of the rectangle independent of the fact that 6 00 for that part of the loop e As such the only side that has a non zero value is the one interior to the solenoid If we set the length of that side Donald G Luttermoser E TSU f to L we can rewrite the sum in Eq V 16 as 2 BH AK BL Using this in Eq V l6 we can write BL uoNI where here we have to introduce N the number of turns in length L since the current I in Eq V 16 is the total of all currents in a given closed loop g Since n NL we can now write N B Hoff don which is Eq V l8 Example V76 Problem 1948 Page 656 from the Ser way amp Faughn textbook It is desired to construct a solenoid that has a resistance of 500 9 at 200C and that produces a magnetic field of at its center of 400 X 10 2 T when it carries a current of 400 A The solenoid is to be constructed from copper wire having a diameter of 0500 mm If the radius of the solenoid is to be 100 cm determine a the number of turns of the wire needed and b the length the solenoid should have Solution a Using Eq Ill 10 with p 170 X 10 8 Qrn for copper from Table 171 from the textbook the required length of wire to be used is L L E Rlt7TD24gt 2 p PHYS 2020 General Physics II 500 2 7r 0500 x 103 m2 4 170 X 10 8 Q In 577 m The total number of turns on the solenoid 16 the number of times this length of Wire will go around a 100 cm radius cylinder L L 577 m Nii 919 C 27W 20100 X 10 2 m Solution b is From B iron the number of turns per unit length on the solenoid is B 110 i 400 x 102 T 4w x 107 TmA400 A 796 X 103 turnsm Thus the required length for the solenoid is LS a 919 turns 115 115 n 796x103turnsm 0 m I Magnetic Domains As we now have seen electric currents traveling in curved paths produce a magnetic eld a It is electric currents in the hot ionized liquid Ni Fe core of the Earth that produces its dipole magnetic eld b Since electrons are in orbit around the nucleus of an atom indiVidual atoms also produce their own internal magnetic eld Donald G Luttermoser E TSU 2 The orientation of the B elds generated by individual atoms is generally random gt the magnetic effect produced by the electrons orbiting the nucleus is either zero or very small in most materials Besides orbiting the nucleus electrons also have an intrinsic spin which also produces a magnetic eld a Typically the orientations of these spin B elds are ran dom too producing no net B eld to the macroscopic world most substances are not magnets b However certain strongly magnetic materials such as iron Fe cobalt Co and nickel Ni have electron spin mag netic elds that do not completely cancel out gt these materials are called ferromagnetic ferro from the Latin name for iron c In ferromagnetic materials strong coupling occurs be tween neighboring atoms forming large groups of atoms called domains in which the spins are aligned d In unmagnetized substances the domains are randomly oriented But when an external eld is applied many of the the domains line up with the external eld gt this causes the material to become magnetized Materials that can retain a macroscopic magnetic eld after they have been pumped by an external magnetic eld are called hard magnetic materials gt permanent magnets Those materials that lose their magnetic eld once the external magnet is removed are called 50ft magnetic materials Fe and Ni are examples The unalignment of the magnetic domains occur PHYS 2020 General Physics II due to thermal motions of the atoms in the domains PHYS2020 General Physics II Course Lecture Notes Section IX Dr Donald G Luttermoser East Tennessee State University Edition 31 Abstract These Class notes are designed for use of the instructor and students of the course PHYS 2020 General Physics taught by Dr Donald Lutternioser at East Tennessee State University These notes make reference to the College Physics Enhanced 7th Editizm 2006 textbook by Serway Faughn and Vuille IX Electromagnetic Radiation Photons A Properties of Electromagnetic Radiation 1 Maxwell showed in 1865 that electric phenomena were related to magnetic phenomena gt Maxwell s equations or laws a Electric elds originate on positive charges and terminate on negative charges The electric eld due to a point charge can be determined by applying Coulomb s law WELB Ea i This is similar to Gauss s law given in Eq 1 7 ii p 1 V is the electric charge density charge per unit volume iii The del symbol is from vector calculus and is de ned by 34743 8xx Byy 822 7 in Cartesian coordinates where 8 is the partial derivative symbol from calculus iv The 67 operation is called the divergence in higher mathematics it is the dot product of the del operator and the vector in question here the electric eld vector b Magnetic eld lines always form closed loops gt they do not begin or end anywhere gt there are no magnetic monopoles Mathematically this is given by Maxwell s 1X1 2 c d PHYS 2020 General Physics II equation for the divergence of the B eld 9 g 0 the Zero simply means that there are no magnetic monopoles A varying B eld induces an emf and hence electric eld gt Faraday s law re expressed in Maxwell s form The vx operation is called the curl in higher mathe matics Whereas the divergence produces a scalar the curl produces a vector Magnetic elds are generated by moving charges or cur rents gt Ampere s law re expressed in Maxwell s form x uoj where J olt BqBt is the current density Maxwell s last 2 laws allow electromagnetic waves e radiation to selfpropagate at a velocity of a b 1 c 299792 x 108 ms IX1 m c is called the speed of light since visible light is a form of electromagnetic radiation An oscillating electric charge produces an E eld that varies in time which produces a B eld that varies in time which produces a new E eld that varies in time Donald G Luttermoser ETSU IX 3 and so on E eld wavecrest y x 4 dilf39ection H 0 wave B eld A propagatlon c The ratio of the maximum magnitude E eld to the max imum magnitude of the B eld of an E M wave is Emax Bum c IX 2 d We will see in the Optics portion of this course that the above value of c is the value of the speed of light in a vacuum Light slows when it enters a medium ie air though not by much glass etc e Even though the solution to Maxwell s equations clearly showed that light or any E M radiation can self propagate it took another 50 years before Einstein demonstrated this with the Special Theory of Relativity in 1905 Prior to this light was assumed to propagate on a medium in space called the Ether f In the late 1800s Michelson and Morley tried to measure the Earth s motion through the Ether and was unable to detect it either the Ether moved with the Earth or the Ether didn t exist Einstein chose the latter and used this fact as one of the underlying principles to relativity IX 4 PHYS 2020 General Physics II g In the Special Theory of Relativity Einstein also showed that this vacuum speed of light represents the fastest speed in the Universe i Energy 69 E M radiation gravitational always travels at this speed in a vacuum no slower no faster ii Matter must always travel slower than c it can never go this speed nor any faster iii This results from the following reduced mass equa tion derived in the relativistic momentum equation from relativity mo where m0 is the mass an object has when at rest m and v is the velocity of the mass iv As can be seen in this equation if 1 ltlt 6 then m m mO the mass remains relatively constant at the rest mass value v However as v gt c U262 gt 11 vgc2 gt 0 and nally moMl 1902 gt 00 The mass gets in n ity bigl Since an in nite amount of energy would be required to push even the smallest mass to the speed of light in a vacuum it is impossible for even the smallest mass to be accelerated to the speed of light Matter can be accelerated to speeds very close to c but not right at cl Donald G Luttermoser E TSU vi This is not just a technical barrier such as the sound barrier it is impossible for anything with mass even an electron to be accelerated to U 6 vii Also note that if v gt c then we get a negative number in the square root gt an imaginary num ber which is a meaningless quantity for velocity in our physical Universe 3 EM waves also behave like particles a Sometimes joking called a waviclel b C 01 As such Planck called a packet of electromagnetic radia tion a photon in the early 1900s and this term has been used ever since to describe E M radiation The photoelectric effect demonstrates this particle picture nicely If a photons hits a certain type of metal it can knock an electron off of an atom in the metal and produce an electric current Einstein won his Nobel Prize for guring this out He never won a Nobel Prize for either the Special or General Theories of Relativity since at the time of their development they could not be fully tested due to the lack of technology At this point we need to distinguish between two types of radiation The word radiation simply means something owing from one point to another This something however can either be particles of energy or particles of matter i The ow of photons energy particles is called ra diation IX 6 PHYS 2020 General Physics II ii Also the motion of atomic and subatomic mat ter particles given off during radioactivity events is called radiation This was discovered by the Curies in the early 1900s The Curies discovered three types of particle radiation 0 Alpha particles have positive charge and were later determined to be helium He nuclei As such He nuclei are often referred to as alpha par ticles 0 Beta particles have negative charge and were later found to be the simple electron o Gamma particles have no charge and were later realized to be high energy photons These parti cles were renamed gamma rays and represent the highest energy form of E M radiation 4 E M waves are transverse waves since the electric and magnetic elds are J to the direction of propagation of the wave and to each other 5 E M waves carry both energy and momentum despite have no mass a The energy is proportional to frequency and inversely pro portional to wavelength E by IX3 i E is the energy Sl unit of J cgs unit of erg of the E M wave Donald G Luttermoser E TSU ii V Greek nu is the frequency unit of HZ ls of the EM wave Note that the textbook uses f for a photon s frequency just as we did in the case of sound However it is more common to use V for the frequency of an E M wave We will just use the variable f as the frequency of waves requiring a medium 66 sound water waves etc iii h Planck s constant 662620 X 10 34 Js 662620 x 1027 ergs iv There is another form of Planck s constant the angular frequency w 27w version h called h bar h27T 105459 x 1034 Js 105459 x 10 27 ergs where a photon s energy is given by E ha in terms of angular frequency v Also note that the amplitude of the wave is re lated to the intensity of the E M radiation which is related to the total number of photons The am plitude hence intensity has nothing to do with the energy of the wave vi The frequency of the wave is related to the wave length 66 the distance between wavecrests with y m6 As such we also can write the energy of a photon as E 7 IX5 PHYS 2020 General Physics II b The momentum of a photon E M wave is P X IX 6 01quot p i IX7 c Note that if we use Eq IX 3 in Eq IX 6 we get E 197 or Epc IX 8 6 Example 1X71 An electromagnetic wave in vacuum has an electric field amplitude of 220 Vm Calculate the amplitude of the corresponding magnetic field Solution We just need to use Eq IX 2 and solve for Bum BWE 733x10 7T 733nT c 300 X 108 ms Example 1X72 Compare the wavelength energy and momen tum of a radio wave at 108 Hz eg a TV signal to that of visible light at 500 nm the yellow part of the spectrum Solution First use Eq IX 4 to get the wavelength of the radio wave Vradio VB 108 HZ c 300 X 108 ms A 7 300 R W 108 s 1 m AV 500 nm 500 x 109 m 500 x 107 m AR 300m 6 7 6 1 Donald G Luttermoser ETSU IX 9 gt the radio wave is 6 million times longer than the Visible light wave Now use Eqs IX 3 and IX 5 to compare the energies ER hVR 663 x 1034 J s108 s4 663 x 1026 J be 663 x 1034 Js300 x 108 ms 19 E 7 398 10 J V Av 500 x 107 m X ER X 10726 J 7 167 1 EV 398gtlt 1019 J X 0 or EV 600 X 106ER gt the Visible light photon is 6 million times more energetic than the radio photon Finally use Eq IX 6 to compare the momenta pp WP AV 53900 007m 1 7 167 10 pv hAV AR 300m x or pV 600 x 106m B The Electromagnetic Spectrum 1 The EM spectrum and spectra in general is de ned to be the intensity or flux of an E M wave as a function of wavelength frequency or energy a 1n spectroscopy very small units of length are often used i 1 micrometer pm or u 10 6 m ii 1 nanometer nm 10 9 m iii 1 Angstrom 10 10 m b Visible light ranges from 40 X 10 7 m to 70 X 10 7 m or 04 um 07 pm or 400 nm 700 nm or 4000 A 7000 A IX 10 PHYS 2020 General Physics II The continuous E M spectrum called the continuum is split into different regimes Below we list the E M spectrum from shortest to longest wavelengths with distinct wavelength dividing lines Often however these dividing lines are fuzzy such that one scientist might call a long IR photon a microwave whereas another would refer to it as a microwave a Gamma rays Highest energy highest frequency and shortest wavelengths A lt 001 nm These types of pho tons are given off by nuclei during high energy nuclear reactions and by extremely hot gas T gt 30 X 108 K over 300 million degrees Kelvinl b Xrays 001 nm lt A lt 10 nm These photons are given off by low energy nuclear reactions and very hot gas 30 X 105 K lt T lt 30 x 108 K c Ultraviolet UV 10 nm lt A lt 400 nm These photons are given off by electron transitions in atoms typically involving the ground state and by hot gas 7000 K lt T lt 30 x 105 d Visible visual 400 nm lt A lt 700 nm These photons are given off by electron transitions in atoms and warm gas 4000 K lt T lt 7000 e Infrared IR 700 nm lt A lt 1 mm These photons are given off by electron transitions in atoms between excited levels and molecules Cool gas 3 K lt T lt 4000 K can also thermally emit these photons f Microwaves 1 mm lt A lt 10 cm These photons are given off by electronic devices or from cold gas 003 K lt T lt 3 K Donald G Luttermoser E TSU g IX 11 Radio waves A gt 10 cm These photons have the lowest energy lowest frequency and longest wavelengths and can be created by electrical circuits and from extremely cold gas T lt 003 The longest radio waves A gt 1 km are sometimes just referred to as long waves 3 Some spectral regions are subdivided into smaller bands In the case of visible light we refer to those bands as colors since our eyes perceive them that way The wavelength delineation which are approximate for each color from shortest to longest wave lengths are a b C d e f 4 The visible region is the spectrum is the narrowest Violet 400 nm lt A lt 450 nm Blue 450 nm lt A lt 490 nm Green 490 nm lt A lt 520 nm Yellow 520 nm lt A lt 590 nm Orange 590 nm lt A lt 630 nm Red 630 nm lt A lt 700 nm It corre sponds to the wavelengths to which the human eye is sensitive a b The Earth s atmosphere is transparent assuming clouds are not blocking your view at visible wavelengths The human eye is most sensitive to the green yellow part of the visible spectrum between 500 570 nm which is also the wavelengths where the Sun emits the peak of its in tensity in the E M spectrum natural selection at work IX 12 PHYS 2020 General Physics II Example 1X73 The eye is most sensitive to light of wavelength 550 X 10 7 m which is in the gree yellow region of the visible elec tromagnetic spectrum What is the frequency of this light Solution We just need to use Eq IX 4 c 300 X 108 ms 14 C The Formation of Spectra and the Doppler Effect of Light 1 In the 1860s Kirchhoff realized that there are 3 types of spectra that objects emit which depend upon the state and orientation the object is in gt Kirchhoff s Laws a Law 1 A luminous opaque object solid liquid or gas emits light at all wavelengths M spectrum thus pro ducing a continuous spectrum a complete rainbow of colors without any spectral lines Continuous Spectrum Donald G Luttermoser ETSU IX 13 b Law 2 A rari ed luminous gas emits light whose spec trum shows an emissionline spectrum a series of bright spectral lines against a dark background I Emission Lines Alit c Law 3 If white light ie a continuum from a luminous A source passes through a lower density gas an absorption line spectrum appears a series a dark spectral lines among the colors of the continuous spectrum Continuum Absorption 7 Lines IX 14 PHYS 2020 General Physics II 1 Kirchhoff s laws can be summarized with the following cartoon Star i Continuum Source Kirchhoff s Radiation Laws 2 The Doppler effect for EM Radiation a The spectrum of an object will be blueshifted if it is approaching the observer b The spectrum of an object will be redshifted if it is receding from the observer c The wavelength shift in a spectral line is given by AA 7 UT A0 7 c 7 where AA A AO negative shift blueshift AO rest lab wavelength UT radial re line of sight velocity IX 9 of the object and c speed of light Example 1X74 We observe a hydrogen spectral line of Polaris the North Star with a wavelength of 656248 A which in the lab oratory is measured to be at 656285 A What is the radial velocity of Polaris Donald G Luttermoser E TSU IX 15 Solution We are given A 656248 A and A0 656285 A so AA 656248 A 656285 A 037 A AA 037 A 7 c A0 656285 A 5638 x 10 5 300 x 108 ms 169 x 104 ms 169 kms Polaris is moving towards us negative sign and the line was blueshifted at 169 kms 0 300 x 108 ms D Blackbody Radiation 1 Late in the 1800s and in the early part of the 20th century Boltz mann Planck and others investigated E M radiation that was given off by hot objects a In general matter can absorb some radiation e pho tons converted to thermal energy re ect some and trans mit some of the energy b The color of cool objects objects that don t emit their own visible light is dictated by the wavelengths of light they either re ect absorb or transmit i A blue sweater is blue because the material re ects blue light from either room lights or the Sun more effectively than the other colors of the rainbow ii Coal is black because it absorbs visible light and re ects very little IX 16 2 OD PHYS 2020 General Physics II iii Glass is transparent because visible light is al most completely transmitted through the glass with little absorption and re ection To make the physics a little less complicated these scientists invented the concept of an ideal or perfect radiator a A hypothetical body that completely absorbs every kind of E M radiation that falls on it b This absorption continues until an equilibrium tempera ture is reached c At that point all incoming radiation is immediately re radiated away as soon as it is absorbed note that this is Lot the same as being reflected 1 Such a perfect radiator or absorber is called a blackbody A blackbody radiator has an energy ux F de ned as the energy emitted per unit area each second and related to the in tensity that is radiated away which is proportional to the 4th power of temperature via a a 567 X 10 8 7Vm2K4 is the Stefan Boltzmann con stant IX10 b The units of energy flux is VVm2 in the 81 system and ergcm2s in the cgs system c The temperature in Eq IX 10 is in units of Kelvin and is the equilibrium or effective temperature 1 Eq IX 10 is the StefanBoltzmann law Donald G Luttermoser E TSU 4 The total brightness or luminosity L of a blackbody is just the ux integrated over all of the surface of the object For a spherical object the surface area is 477 R2 where R is the radius of the spherical blackbody so the luminosity is L 477 R2 F 47TUR2 T4 IX11 Note that if we treat stars as blackbodies we can eliminate the constants in the above equation by dividing both sides by solar values L 4770132 T4 LG i 47TUREDT L 7 ltRgt2ltTgt4 Lo 7 Re To In 1893 Wien discovered a simple relationship between T of a IX 12 blackbody and the wavelength where the maximum amount of light is emitted gt Wien s displacement law usually called Wien s law for short i 2897 x 103 m K A max T 7 IX13 or in the other wavelength units we have discussed 02897 cm K Am f 2897 x 106 nmK T 2897 x 107 A K T IX 17 IX 18 PHYS 2020 General Physics II 6 Blackbody radiators being in thermal equilibrium emit contin uous spectra that are called Planck curves Planck Curve Flux Example 1X75 A star has a temperature of 10000 K and a radius of 20 Re what is its energy flux and wavelength of maximum flux What is its luminosity with respect to the Sun Q Note that Re 696 X 108 m and T9 5800 K Solution For this problem we Will assume that stars are blackbody ra diators We then use the Stefan Boltzmann laW Eq lX lO to determine the energy ux of this star F 567 x 108 W 11172164 10 000 K4 567 x 108 W 11172164 104 K4 567 x 108 W 11172164 1016 K4 567 x 108 W at2 The wavelength of maximum brightness is given by Wien s laW Donald G Luttermoser E TSU IX 19 Eq lX 13 using nm for wavelength instead of m 2897 x 106 nm K A 29 UV 139 ht m 10 000 K 1g Finally the luminosity is given by Eq IX 12 i 5131 L5 7 Bo T5 7 201292 lt10000K 159 5800 K 7 400 884 3500 3500 LG 4 400 1724 E Spectroscopy The Study of Spectra 1 In 1814 Joseph von Fraunhofer discovered about 600 dark lines in the solar spectrum gt spectral lines The darkest he labeled from A in the red to H in the blue note that a K line was added later In 1859 Gustav Kirchhoff and Robert Bunsen discovered that each element contained a unique set of lines in their spectra gt spectral lines are the ngerprints of chemical elements and com pounds This is the basis behind spectral analysis A device that breaks white light into its component colors hence displaying spectral lines if they are present is called a spectro graph or spectroscope Bohr early in the 20th century demonstrated that absorption lines result from photons being absorbed by an atom causing an electron to jump from a lower to a higher energy state IX 20 PHYS 2020 General Physics II a Electrons can only orbit nuclei in allowed orbits or states This is the basis of the branch of modern physics called quantum mechanics b Bohr s model atom corresponds to the hydrogen atom which will be discussed in X of these notes Likewise an emission line results when an electron jumps from an upper to a lower level in an atom or molecule In both cases the energy separation between the two levels must be equal to the energy of the photon that is either absorbed or emitted following AE Eu Eg hm IX14 where E is the energy of the upper level Eg the energy of the lower level Dug is the frequency of the photon that is absorbed or emitted and hVug is the energy of this photon The interaction of E M radiation and matter will be discussed in more detail in the next section of these notes emission absorption PHYS2020 General Physics II Course Lecture Notes Section I Dr Donald G Luttermoser East Tennessee State University Edition 31 Abstract These Class notes are designed for use of the instructor and students of the course PHYS 2020 General Physics taught by Dr Donald Lutternioser at East Tennessee State University These notes make reference to the College Physics Enhanced 7th Editizm 2006 textbook by Serway Faughn and Vuille I Electric Forces 85 Electric Fields A Properties of Electric Charge 1 2 Benjamin Franklin was the rst to realize that there are two types of electric charge a Positive charge q b Negative charge q Like charges repel one another whereas unlike charges attract each other Electric charge is always conserved in any type of reaction or process In 1909 Robert Milliken discovered Via the Millikan OilDrop Experiment that if an object is charged its charge is always in multiples of the fundamental unit of charge e a Charge is said to be quantized gt q ie i26i36 etc b e 160219 X 10 19 C C E Coulomb in the 81 system 480325 X 10 10 esu esu E electrostatic unit in the cgs system c Elementary particles i Electron q e ii Proton q e iii Neutron q 0 l2 PHYS 2020 General Physics II B Insulators and Conductors 1 t Conductors are materials in which electric charges move freely 216 they have low internal r652395tan66 a Copper b Aluminum c Silver Insulators are materials in which electric charges do Lot freely move i they have high internal r652395tan66 a Glass b Rubber Semiconductors are materials that lie in between these other two gt if controlled amounts of foreign atoms are added to semi conductors their electrical properties can be changed by orders of magnitude a Silicon b Germanium The Earth can be considered to be an in nite reservoir of or for electrons 3 b It can accept g supply an unlimited number of electrons When a conductor is connected to the Earth 6 9 con ducting Wire or copper pipe it is said to be grounded gt lightning rods Donald G Luttermoser E TSU 5 An object can be charged in one of two ways a Conduction Charge exchange through contact i If one charged object comes in contact with as sec ond object charge can move from the charged ob ject to the uncharged object ii Rubbing two different materials together called frictional work can produce negative charge on one object and a positive charge on the other b Induction Charge exchange with no contact i Charge one object ii This charge produces an electric eld to form iii This electric eld can then induce charges to mi grate on a second object gt the second object po larizes see Figure 154 in the textbook C Coulomb s Law 1 Coulomb s law states that two electric charges experience a force between them such that a It is inversely proportional to the square of the separation 7 between the 2 particles along the line that joins them b It is proportional to the product of the magnitudes of the charges ql and qg on the two particles c It is attractive if the charges are of opposite sign and re pulsive if the charges have the same sign PHYS 2020 General Physics II 2 Mathematically the Coulomb force Fe is a k6 E Coulomb s constant 89875 x 109 N m2C2 SI units 1000 dyne cm2esu2 cgs units b The sign of the electric force will depend upon the orienta tion of the de ned coordinate system for a given problem whether a given force is repulsive 69 both q s are positive or both q s are negative or attractive 69 the charges are opposite in sign c f is the unit vector in the radial direction 3 Coulomb s law like Newton s law of gravity F Gmlmgrg is a A eld force law gt there is no physical contact be tween the particles b And an inversesquare law gt the strength of the elec tric force falls off as the inverse of the distance squared 4 If more than two charged particles exist in a system then the Coulomb force exerted on one particle is the summation of all of the Coulomb forces between that particle and the rest of the particles in the ensemble s N N N FEZE ZEQZF2 1 2 239 j k where E is the component Coulomb forces of all N particles in the x direction F is the component forces in the y direction and Fk is the component forces in the z direction Eq 1 2 is known as the principle of superposition Donald G Luttermoser E TSU Example 171 Problem 153 Page 525 from the Serway amp Faughn textbook An alpha particle charge 206 is sent at high speed toward a gold nucleus charge 796 What is the elec trical force acting on the alpha particle when it is 20 X 10 14 m from the gold nucleus Solution Use Coulomb s Law 16 Eq l l where 11 26 qg 796 7 20 x 1014 m and e 160 x 1019 C 26796 r lQ1l lQ2l Fe 7 kg l l 72 72 kg N n12 20 79160 x 1019 C2 9 83999 X 10 C2 gt i 20 x 1014 m 91 N repulsion The Coulomb force is a repulsion force here since both particles have charges of the same sign D The Electric Field 1 An electric charge emits an electric eld which always points away from a positive charge and points towards a negative l l 1 1 Efield Efield points away points to charge charge l 6 PHYS 2020 General Physics II a Assume we take a small ie negligible positive test charge go and place it near a larger positively charged object Q Q lo 9 b The direction of E ie the electric eld at a point is de ned to be the direction of the electric force that would be exerted on a small ie negligible with respect to the charge on the larger object qO ltlt Q positive charge placed at that point c Note that if qO starts to become comparable in size in terms of charge to Q the electric eld of Q will be altered by go 2 Hence the electric eld is de ned by the electric force exerted on a charged particle by another charged particle or object a E E 7 1 3 go a Note that from this equation E always points in the same direction that the force exerted on a positively charged particle qO by a charged object that gives rise to E b E is measured in NC in SI units and dyneesu in the cgs system Donald G Luttermoser E TSU c The electric eld E is analogous to surface gravity e acceleration due to gravity in Newton s Theory of Grav itation since we can write Newton s 2nd law as 3 lu i g 7 where g is measured in Nkg rns2 and is nothing more than the gravitational force per unit mass e the strength of the gravitational eld Likewise the electric eld is the electrical force per unit charge Let the object that produces an E eld be given a charge labeled with q Then a test charge qO a distance r from q experiences a force from q described by Coulomb s law e Eq l l Fe keilql lgqol 7 7 Plugging this into Eq l 3 gives a second equation that describes the E eld a lql E kg 7572 7 a If q is positive e q gt 0 then E is positive and points radially away from the charge b If q is negative e q lt 0 then E is negative and points radially inward towards the charge An electron with a speed of 300 X 106 ms moves into a uniform electric field of 1000 NC The field is parallel Example 172 to the electron s motion How far does the electron travel before it is brought to rest Solution When the electron with mass me 911 X 10 31 kg enters the PHYS 2020 General Physics II electric eld it experiences a retarding force given by Eq 1 3 F eE negative since the E eld slows the electron Since this force slows the electron it produces a deceleration Using Newton s 2nd law we can write F eE 160 x 10 19 C1000 NC a We me 911x10 31 kg 176 x 1014 ms2 Using one of the 1 D equations of motion from General Physics I we can solve for the distance that the electron travels before coming to a stop Ax We have 1 0 6 comes to rest 1 300 X 106 ms and a has been calculated above giving 1 1 2an U2 U2 A o 06 2a 0 300 X 106 ms2 20 176 x 1014 ms2 When summing E elds from multiple charges it is best to draw a vector diagram rst at a point where the E eld is to be deter mined Draw the E eld vector of each charge in the direction as dictated by the rules above Then vectorially add the individual E elds together letting the diagram determine the sign of the E eld typically to the right and upwards Then add using the absolute value of the charge E it 15 Donald G Luttermoser E TSU where either or is selected based upon the direction of the E led vector see Example 1 3 below Example 173 Positive charges are situated at three corners of a rectangle as shown in the figure below Find the electric field at the fourth corner E2 E3 N r1 0 600 m q1 600 nC E1 Pill r3 lr20200m a l o 39o q2200nC q3500nC Solution From the geometry of the rectangle as shown in the gure above we have and r r r3 0600 m2 0200 m2 0400 m2 1 0200 m V2 t 1 i as an m 0600 m The components of the individual E eld vectors are gt 184 E1 E1I92 E192 E2 E2y E2fQ E3 E3102E3yg E3 cosq i Egsinq The resultant E eld at the vacant corner is ERE1E2E3 PHYS 2020 General Physics II with the resultant component vectors of 3 E1 E1x 0 E31 E1 E3cos 2391 2 cos b kelqil kequl 7 1 732 N 1112 600 x 109 o 500 x 109 o 899 1 97 184 lt X 0 o2 gt l 0600 m2 0400 m2 COS 256 No 3 E9 ZEjy 0E2yE3yE2E3sin j1 kel l kelgSI Sings 7 2 7 3 i 8 99 X 109N111 300 x 109 o 500 x 109 o 811118 40 7 o2 0200 m2 0400 m2 710 No Thus ER E352 Ey2 755 NC With E1 lt 0 and Ey gt 0 the resultant E eld lies in the second quadrant with an angle of E 710 N C n 1 7y 1 0 0 itan tan 256 NCgt 70 However this is with respect to the x axis 6 with respect to the 06 axis is then 6 1800 639 180O 70O 110 Hence we can express the answer in one of two ways ER 256 NCat 710 NCM 0 ER 755 NC at 1100 counterclockwise from the x axis Donald G Luttermoser E TSU E Electric Field Lines 1 Drawing electric eld lines a a The electric eld vector E is tangent to the electric eld lines at each point b The number of lines per unit area through a surface J to the lines is proportional to the strength of the electric eld in a given region i E is large when the eld lines are close together ii E is small when the eld lines are far apart 2 For a system of charged particles the following rules apply a The lines must begin on positive charges or at in nity and must terminate on negative charges or in the case of excess charge at in nity b The number of lines drawn leaving a positive charge or approaching a negative charge is proportional to the mag nitude of the charge 69 if a charge of q has n lines per unit volume leaving the charge a charge of 2q will have 2n lines per unit volume approaching the charge see EX l 4 c No 2 eld lines can cross each other Donald G Luttermoser E TSU Example 174 Problem 1528 Page 527 from the Serway amp Faughn textbook Figure P1528 shows the electric field lines for two point Charges separated by a small distance a Determine the ratio qlqg b What are the signs of ql and qg Solution a amp b The magnitude of qg is 3 times the magnitude of ql since 3 times as many lines emerge from qg as enter 11 Hence q2 3q1 Since the eld lines are emerging from qg it must have a positive charge and since they terminate on ql it must have a negative charge Thus 1 a q 3 and F Conductors in Electrostatic Equilibrium 1 A good conductor contains electrons that are not bound to any one atom gt free to move about the material 2 Electrostatic Equilibrium No net motion of charge occurs Within a conductor The following is true for such a conductor a b C The electric eld is zero everywhere inside the conductor Any excess charge on an isolated conductor resides en tirely on its surface The electric eld just outside a charged conductor is J to the conductor s surface l 14 PHYS 2020 General Physics II 1 On an irregular shaped conductor charge accumulates at its sharpest pomts charge accumulates conductor G The Electric Flux and Gauss s Law 1 The electric ux is a measure of the number of E eld lines that crosses a given area a An E eld whose lines penetrate a cross sectional or sur face area A J to A has an electric flux DE given by crgtE EA 16 b However if the E eld lines lie at an angle 6 with re spect to the normal line of area A see Figure 1525 in the textbook the electric flux is given by the more general formula a a DEEAEAcos6 1 7 where the is called the dot product operation and A has magnitude A the total cross sectional or surface area and direction given by the normal line of the area The angle 6 is the angle between that normal line and the E eld direction c When the area is constructed such that a closed surface is formed we shall adopt the convention that the flux lines Donald G Luttermoser E TSU 2 One of the most important law s in all of electromagnetism is passing into the interior of the volume are negative and those passing L1H of the interior of the volume are positive Gauss s law a b C d e If we use a sphere for our enclosing volume the sphere 2 If we place charge q at has a surface area of A 47W the center of this sphere then the eld lines will always be J to the surface of the sphere since they point radially outward Using Eq 1 4 in Eq 1 6 we can write the electric ux as crgtE EA 166 Am 4mg 7 independent of the distance from the charge In electromagnetism there is a constant related to the Coulomb constant called the permittivity of free space 60 This constant is given by c2 885 x 1012 W 1 8 1 60 47Tke Using this constant in our flux equation above we can write the electric flux as DE 47Tqu g 60 Using calculus we could show that this simple result is true for any closed surface even non symmetrical ones that surrounds any charge q gt such a surface is referred to as a gaussian surface This bit of mathematics was rst worked out by Gauss and is therefore called Gauss s law PHYS 2020 General Physics II i In words this law states The electric ux through any closed surface is equal to the net charge Q inside the surface divided by the permit tivity of free space 60 ii Mathematically it is given by Qinside Eo 13E l g Example 175 Problem 1539 Page 527 from the Ser way amp Faughn textbook An electric field of intensity 350 kNC is applied along the x axis Calculate the electric flux through a rectangular plane 0350 m wide and 0700 m long if a the plane is parallel to the yz plane b the plane is parallel to the my plane and c the plane contains the y axis and its normal makes an angle of 4000 with the x axis Solution a The ux through an area is given by Eq 1 7 where 6 is the angle between the direction of the eld E and the normal line of area A The magnitude of the area of the plane is A 0350 m 0700 m 0245 mg When the plane is parallel to the yz plane the normal line of the area lies parallel to the x axis so 6 0 and the ux is cIgtE EAcos639 350 x 103 NC0245 m2 cos 0quot 858 N mQC Solution b When the plane is parallel to the x axis as it must be when it is parallel to the my plane the normal line of the area is at a right Donald G Luttermoser E TSU angle to the x axis and hence E eld so 6 900 and cigtE EAcos639 350 x 103 NC0245 n12 cos 90quot Solution c Since the E eld is along the x axis and the normal of the area make an angle of 4000 with respect to the x axis 6 400 so crgtE EA cos6 350 x 103 NC0245 n12 cos 400 657 N mZo PHYS2020 General Physics II Course Lecture Notes Section X Dr Donald G Luttermoser East Tennessee State University Edition 31 Abstract These Class notes are designed for use of the instructor and students of the course PHYS 2020 General Physics taught by Dr Donald Lutternioser at East Tennessee State University These notes make reference to the College Physics Enhanced 7th Editizm 2006 textbook by Serway Faughn and Vuille X Interaction of Photons with Matter A The Classical Point of View 1 A system is a collection of particles that interact among them selves via internal forces and that may interact with the world outside via external elds a To a classical physicist a particle is an indivisible mass point possessing a variety of physical properties that can be measured i Intrinsic Properties These don t depend on the particle s location don t evolve with time and aren t in uenced by its physical environment eg rest mass and charge ii Extrinsic Properties These evolve with time in response to the forces on the particle eg posi tion and momentum b These measurable quantities are called observables c Listing values of the observables of a particle at any time gt specify its state A trajectory is an equivalent way to specify a particle s state d The state of the system is just the collection of the states of the particles comprising it 2 According to classical physics all properties intrinsic and ex trinsic of a particle could be known to in nite precision gt for instance we could measure the precise value of both position and momentum of a particle at the same time X1 X2 PHYS 2020 General Physics II Classical physics predicts the outcome of a measurement by cal culating the trajectory ie the values of its position and mo mentum for all times after some initial arbitrary time to of a particle tgt 2 to E trajectory X l where the linear momentum is by de nition 17 E m 17 X 2 with m the mass of the particle a Trajectories are state descriptors of Newtonian physics b To study the evolution of the state represented by the trajectory in Eq 1 1 we use Newton s Second Law APE Ar X3 ma where PE is the potential energy of the particle c To obtain the trajectory for t gt to one only need to know PE and the initial conditions gt the values of r and 17 at the initial time to 1 Notice that classical physics tacitly assumes that we can measure the initial conditions without altering the motion of the particle gt the scheme of classical physics is based on precise speci cation of the position and momentum of the particle From the discussion above it can be seen that classical physics describes a Determinate Universe gt knowing the initial con ditions of the constituents of any system however complicated we can use Newton s Laws to predict the future Donald G Luttermoser E TSU X 3 5 If the Universe is determinate then for every e ect there is a cause gt the principle of causality B The Quantum Point of View 1 The concept of a particle doesn t exist in the quantum world so called particles behave both as a particle and a wave gt waveparticle duality a The properties of quantum particles are not in general well de ned until they are measured b Unlike the classical state the quantum state is a conglom eration of several possible outcomes of measurements of physical properties c Quantum physics can tell you only the probability that you will obtain one or another property 1 An observer cannot observe a microscopic system without altering some of its properties gt the interaction is un avoidable The effect of the observer cannot be reduced to zero in principle or in practice 2 This is Lot just a matter of experimental uncertainties nature itself will not allow position and momentum to be resolved to in nite precision gt Heisenberg Uncertainty Principle HUP AxApIgt1 h in where h 662620 X 10 27 erg sec 6626 X 10 34 J sec is Planck s Constant a A06 is the minimum uncertainty in the measurement of the position in the x direction at time to X4 0 C d e f g h PHYS 2020 General Physics II Apr is the minimum uncertainty in the measurement of the momentum in the x direction at time to Similar constraints apply to the pairs of uncertainties Ay Apy and AZ Apz Position and momentum are fundamentally incompatible observables gt the Universe is inherently uncertain We can also write the HUP in terms of energy as AEAtZg X 5 This principle arises from geometry through a theorem known as the Schwarz inequality of triangles The details of this relationship is too di icult to cover in this course It is covered in our senior level Quantum Physics course The HUP strikes at the very heart of classical physics the trajectory gt if we cannot know the position and momentum of a particle at to we cannot specify the initial conditions of the particle and hence cannot calculate the trajectory Once we throw out trajectories we can no longer use New ton s Laws new physics must he invented Since Newtonian ie mechanics and Maxwellian ie thermo dynamics physics describe the macroscopic world so well physi cists developing quantum mechanics demanded that when ap plied to macroscopic systems the new physics must reduce to the old physics gt this Correspondence Principle was coined by Neils Bohr Donald G Luttermoser E TSU 4 Due to quantum mechanics probabilistic nature only statisti cal information about aggregates of identical systems can be ob tained Quantum mechanics can tell us nothing about the behav ior of individual systems Moreover the statistical information provided by quantum theory is limited to the results of measure ments gt thou shall not make any statements that can never be veri ed In the realm of the very small various quantities ie energy orbital angular momentum spin angular momentum are quan tized gt values of these parameters are not continuous but instead come in jumps or steps a When we are in the realm of electrons interacting with photons ie distances less than 10 9 m the laws of quantum mechanics describe the physics b When we are in the realm of the nucleus ie distances less than 10 14 m the laws of nuclear physics is de scribed with quantum chromodynamics Note that in this section of the notes we often will be using units of energy to describe masses through Einstein s equation of E meg 1 MeV 106 eV 178 X 10 30 kg Note that the mass of the electron is 911 X 10 31 kg 0511 MeV C Particles and Forces 1 What are the natural forces a Classical physics describes forces ie gravity and E M as elds This de nition rst arose with Faraday for the E M force X5

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Recurring Subscriptions: If you have canceled your recurring subscription on the day of renewal and have not downloaded any documents, you may request a refund by submitting an email to support@studysoup.com

Satisfaction Guarantee: If you’re not satisfied with your subscription, you can contact us for further help. Contact must be made within 3 business days of your subscription purchase and your refund request will be subject for review.

Please Note: Refunds can never be provided more than 30 days after the initial purchase date regardless of your activity on the site.