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Transcript
Electricity and Magnetism AP Physics 2012-13 San Dieguito Academy Equations we will be using for the following sections 2 Electric Forces Charles Coulomb suspected that electric forces were similar to gravitational forces. He set about to find out if this was so with the use of a torsional balance. He placed a small charge on a pith ball. This pith ball was placed on a horizontal bar. The bar was suspended by a thread. Next, he inserted another slightly charged pith ball near by. Measuring the deflection of the horizontal bar he was able to determine that indeed. where k= 9 x 109 N m2 /C2 The unit of charge “q” is measured in Coulombs. On an elementary level, this is a large number of protons or electrons. One Coulomb = 6.24 x 1018 protons or the same number of electrons. Electric forces compared to gravitational forces are very strong. Let us take for example one Mole of Hydrogen atoms and individually separate all of the electrons from the protons. (Remember that a hydrogen atom consists of only one proton and one electron. We will place the protons in one bucket and the electrons in another. Sample Problem: These buckets will be placed one meter apart. Lets find the electrical force acting between the two buckets. 3 Lets do a few more example force problems. Imagine you have two coulombs of positive charge separated by 300 meters from another 2 coulombs of positive charge. What is the force on each group of charge? Electric forces are summative. Find the net force on the 1 Coulomb charge in each of the examples below. 4 Gravity and Potential Energy In the past when we have dealt with gravitational potential energy we looked at small objects being lifted up a small distance from the surface of the Earth. The PE gained was just equal to the work done to lift the object. The Work done was positive because the force applied and the change in displacement were in the same direction. We could use the equation for work W = F d What happens when the distances are much greater? The force of gravity is no longer constant but is governed by Newtons Universal law of gravitation Fg = G m1 m2 / r2. We now must integrate to find the total work done using W = ∫ Fapplied dx. Notice that the Force we must apply is a restraining force against gravity. This force is in the opposite direction to the change in displacement (dx). The angle between the two is 180 degrees . Before = After Definition of work Fg and Dot product Cos 180 = -1 Constants move out 5 r with exponents Integrating Evaluating 6 A Suppose a 10 kg ball of krypton falls to Earth from infinitely far away. How fast is it going upon impact? The Earth has a mass of 6.0 x 1024 kg and a mean radius of 6.38 x 106 meters. Let us use Conservation of Energy. Before = After 0 = ( KE ) + ( 0 = ( ) ) + ( ) vimpact = ____________ It makes sense that PEg must go negative if the KE of the krypton is to go positive. B. Now let us imagine placing this krypton in a very low (read at the surface) circular orbit. Find the speed of the ball using Fnet and centripetal acceleration Fnet Raw Eq ( = m ) = ( Substitutions ( ) = ( Answer ac ) ( ) ( ) ) v= ( ) What is the net energy for this low flying satellite? Total Energy = ( PEg ) + ( K E ) Raw equation Substitutions Answer = ( ) + ( = ( ) ) ( Energy net = ( ) ) 7 Electrical Potential Energy We can also do them using energy. However we have to look potential energy in a slightly different manner. In the past it was just equal to the work put in to move an object. This was Fg Let us look at charges first. Imagine that you have two protons that are infinitely far apart. They have no potential energy. The charge on the right is pushed closer and closer to the one on the left. The force that is applied must be equal but opposite that due to the electrical repulsion Fapplied = - Felect Notice that this charge moves against the repulsive force due to the other. As we do this we are doing work (W = force x distance) The work we do goes into potential energy as is shown below. Important: the force is working with the change in displacement. The Angle between the two is zero degrees.. 8 Here is the math. Notice: Unlike gravity, if both charges have the same sign, the PE will be positive. 9 Lets ask some questions. 1. What does it mean when you have a negative answer for PE? It just means that you have less energy in the system than in your original state. This energy was transferred into some other form, for example KE or heat. 2. When we bring two positive charges together they gain potential energy. Suppose one charge is much greater than the other one. Does it make any difference which charge is moved? No! All that counts is how far apart they were to begin with (PE original) and how far apart they are in the end. 3. What if we bring four charges A, B, C and D together, how do we calculate the net PE? You must find all the possible charge combinations. In this case that includes PEab PEac PEad PEbc PEbd PEcd 10 Lets do a few sample problems. Find the Potential energy in each case below. 11 4. When we bring charges together, where is the energy stored? This is a tough one. The easiest answer is that it is stored in the combined electric field that they create. So, if energy can be stored in fields, we should probably study what a field looks like. This is very hard to do, as fields such as gravitational, electrical and magnetic ones are invisible. Almost two hundred years ago, one man by the name of Michael Faraday tried to this when he came up with what he called lines of force. His assumption was that all positive charges have invisible streamers emerging from the positive charge. They went straight out in all directions. Where they were close together, the field was stronger. And, the field had a direction that matched these lines of force. He then went on the state that negative charges were the same, except that the lines of force would aim toward the negative charges. From this he formulated the following rules: 1. Lines of electrical force have a beginning (positive charge) and an end (negative charge) 2. Lines of force can bend but cannot cross each other. 3. Lines of force have a direction at any location that reflects the sum forces acting on a test charge it that point by all other charges. 12 Let us see what results if two charges are placed near each other. A positive and a negative charge Two positive charges What Faraday showed was that while electric fields are very common, it is not common that one encounters an electric field that is constant in direction and intensity. His theory did argue for a way to produce areas that have absolutely no electric field. This is in what is called a “Faraday cage.” Though he could not prove it mathematically (Faraday had only three years of formal education) he argued that if one had a conducting sphere, charges would spread out in such a manner as to be as far apart as possible. That meant the charges would be evenly distributed on the outside surface of the sphere. 13 He then argued that there should be no lines of force inside the sphere as they would be repelled by the charges on the opposite side of the sphere. He was correct but it takes more than just intuition to demonstrate that something must be true. As time went by, scientists came to what Faraday was describing using lines of force as what we call a “field”. In the case of charges, the electric field and in the case of masses the gravitational field. Mathematical equations for gravitational and electric fields Imagine that you, the victim, are standing on the surface of the Earth, the acting mass. There is a force between the two of you that is given by Newton’s Universal Law of Gravitation. Suddenly you disappear leaving the earth all by itself, the Earth would still have gravity. There just would not be any force. We call this the 14 Earth’s gravitational field. Its strength is measured by the following equation: Lets look at the gravitational field strength on the surface of the Earth. Remember that m = 6 x 1024 kg, r = 6.38 x 106 meters. When we plug into the field equation we get (a familiar number) It is not an accident that “g” the gravitational field strength has the same units and magnitude as “g” the acceleration due to gravity. Field can be viewed as the potential to exert a force. The strength of the force will be the field strength multiplied by your mass. If Fgravitational = “g” m and Then “g” m = Reducing to “g” = a. 15 Fgravitational = m a ma Electric Fields As with mass and gravitational fields, the electric charge has an associated field. We seldom us the electrical force equation with the constant “k” rather we use the form below. Below are two charges. One is +2 Coul and the other is -2 Coul. Find the field strength and direction at each location. A ___________ at _____ B ___________ at _____ C ___________ at _____ D ___________ at _____ E ___________ at _____ Most fields are too complex to easily calculate. This is especially true of electric fields. There are several reasons why this is so. 1. You are often dealing with many, many individual charges. Just one coulomb of electrons is 6,000,000,000,000,000,000 electrons for which one has to account. 2. Electrons are in constant motion. It is like herding cats to keep them stationary. 3. Electrons flow over conductors so that they have the least possible potential energy in any given situation. This means that they are seldom evenly distributed except in the most simple situations (spheres, sheets, and very long straight wires.) 16 As a result, we will restrict ourselves to studying cases where distributions are simple to work with. In spite of this you will find it impressive how much can be done. Doing a simple situation the hard way using calculus (Before Gauss) Lets look a situation where there is a very, very long wire that has a positive charge distributed along it. This charge is distributed uniformly with an average charge per unit length of . We want to find the net electric field at some point P that is a distance D away from the wire. To approach this by using calculus we will break the wire into really, really small pieces of length dx. The contribution of just one of these is shown below. where dq = dx. Notice that dE can be broken up into x and y components. 17 When you add all of the contributions due to each segment dx you get A very interesting development occurs. The first integral will reduce to zero since each dx to the right of point P has a corresponding dx to the left side of P that cancels it out. The result is the following: The integral is still pretty messy because we have variables x, R, and . Cos = D / R So R = D / Cos therefore R = D sec Tan = X / D X = D Tan Differentiate X and d x = D d (tan ) d x = D sec2 d We can now substitute into the equation and we get 18 Giving Most physics books at this level do not use K = 9 x 109 Rather they use k = 1/(4 eo) where eo is the permittivity of free space and is equal to 8.85 x 10-12 they do this for reasons that will soon become apparent. This therefore gives us E at any distance D away from a long line of charge as Now, you should get two things from this exercise. The first is that it appears that the electric field goes down as 1 over the radius as you move away from a line of charge. Secondly, if this is one of the easiest situations for which calculus can be used to find an electric field. We are going to be in for trouble. There must be a better way! And there is It is called Gauss’s Law. Fredrick Gauss was one of the most gifted mathematicians of all time. And, although his mathematical prowess was phenomenal his law is extremely simple. 19 What Gauss did was to imagine that there was something called Flux. Flux was the amount of field that is intercepted by a surface. Looking at fields from the point of view of lines of force, the field intensity would be represented by the density of the lines. The first thing he realized was that if you had a single charge in space and were to enclose it by a spherical surface all the lines that emanated from the charge would be intercepted by the sphere’s surface. Place a bigger sphere and even though the intensity of the field would be less at its surface, the increase in intercepting surface would compensate so that the net recorded flux would be the same as seen by the smaller sphere. 20 Fredrick went one step further. He imagined that a drunk person was trying to enclose the charge within a sphere, but this person’s sphere was squished into a peanut shape as shown below. What he observed was that even where a line emerged and then was later re-intercepted by the peanut shape, it would eventually have to reemerge once again. Subtracting ingoing flux from outgoing flux and you are left with the same net flux that a spherical shape saw. The only factor that actually changed the net flux through a closed surface was the total net charge inside it. More net charge inside, more net flux . This is very powerful relationship. Lets look back for a second at the nature of Flux. The net flux through any surface is determined by three things. A. B. C. the size of the surface that is intercepting the field. the intensity of the field being intercepted. The angle between the field and the intercepting surface. 21 This third characteristic is very important. Look at the three surfaces below. The surface on the left is parallel to the field and intercepts none of it. The middle one is at an angle to the field and intercepts some of the field lines. The last one cuts straight across the field. It gets a maximum amount of the field. For ease of measurement, a surface’s orientation is given by drawing a normal to the surface. This means that for maximum flux we want an angle of zero degrees between surface and field. Gauss therefore defined electric flux by the following equation = E dot A = E A cos () Where is the flux E the electric field intensity A the surface area is the angle between the two Armed with this, Gauss postulated that for a closed surface and flux is equal to the net charge inside divided by a special constant eo which we have seen before. 22 Gauss’s law is extremely easy to use if you recognize one fact, and that is that shape matters. 1. pick a shape where E is constant at the surface. 2. pick a shape maximizes field interception (cos = 0) or 3. pick a shape that minimizes field interception (cos = 90) Lets look at four simple situations. 1. Field at a location r near a point charge (show that the integral E dA really gives Q/eo) We will pick a spherical surface centered over the charge with radius R. Why? Because E is always coming out of the surface at zero degrees (cos = 1) and because the field intensity will be the same at all points (E is a constant) 23 2. Find the Field at point r near a charged conducting sphere of radius R Charge density in coul/m2 To show that outside of it the field is the same as if all the charge were at a point in the center. We will pick a spherical surface with radius r>R that is concentric with the conducting sphere. Why? Because of symmetry, E is always coming out of the surface at zero degrees (cos = 1) and because the field intensity will be the same at all points (E is a constant) 24 3. Use Gauss’ law to show that there is no electric field inside of a conducting sphere (Faraday Cage) We will pick a spherical surface with radius r < R and centered over the charged sphere. Why? Because of symmetry, E is always coming out of the surface at zero degrees (cos = 1) and because the field intensity will be the same at all points (E is a constant) 25 Find the field at point P above a line of charge We will pick a Cylindrical surface parallel to the line of charge with flat circular ends Why? There are three distinct surfaces. The ends have no net field going through them ( = 90 and Cos = 0) and can be cast aside. That leaves the cylinder. Because of symmetry E is always coming out of the cylinder at zero degrees (cos = 1) and the field intensity will be the same at all points (E is a constant) 26 Field at a point P above a sheet of charge (density in Coul/m2) We will pick a Cylindrical surface cutting directly across the sheet of charge with ends P above and below the surface. Why? There are three distinct surfaces. The cylinder has no net field going through it ( = 90 and Cos = 0) and can be cast aside. That leaves the circular ends. Because of symmetry E is always coming out of the ends at zero degrees (cos = 1) and the field intensity will be the same at all points on each end (E is a constant) Notice that the Electric field is constant above and below a large sheet of charge. This will be very important soon 27 5. Imagine two enormous sheets of charge. One is positively charged, the other negatively charged. Using Gauss’s Law show that the field between the plates is /e o Show that there is not field either above or below the two plates. 28 5. One last situation which we shall encounter often is the insulating sphere (radius r) of uniform charge density throughout. The sphere has a net charge Q on it. What is the electric field strength at some point at a radius R inside the sphere? We will draw a sphere of radius R inside the larger sphere Why? Because all charge outside this sphere we can neglect. And, because the field due to the charge inside the sphere will cause a field that is outward directed ( = 0 and E constant) We must determine the fraction of the total charge that is inside the Gaussian surface. Qtotal x volume of small sphere divided by the volume of the large sphere 29 Gravity and Gauss’s Law Several students have asked if there is a way to determine the gravitational field strength “g” inside the Earth. We can, but we must modify Gauss’s law for gravity. First we must find out what the gravitational equivalent for eo is for mass. Remember that for electricity, we changed from k = 9E9 into 1/(4 eo ) This gave us an eo which was equal to _____________________ We will do the same thing for G =6.67E-11 by converting it to 1/(4 xo ) This gives us xo which is equal to ______________________ Now it is just a matter of substituting into the general equation. 30 Force – Field – Potential – Potential Energy Definite Indefinite Up to this point we have dealt fairly extensively with forces, fields, and Potential Energy. The way they are related to each other is shown below. Definite Indefinite In cases where the masses are point masses these relationships become: Definite Indefinite In cases where we are dealing with point charges these become 31 In cases where we are dealing with line charges we use Gauss’ Law to determine the field strength “E” and then move back to force and PE. Definite Indefinite In cases where we are dealing with sheets of charge we use Gauss’ Law to determine the field strength and then move back to force and then to PE Definite Indefinite Notice that there is a fourth unassigned ellipse in the diagram. It is about to become very important as we move into circuits and then into magnetism. Up until now we have dealt with charges at rest. It has been easy to determine their total charge or their relative charge density. When charge starts to flow, keeping track of every electron is very hard to do. It becomes well nigh impossible to determine exactly how much Potential Energy is in a system of moving particles. 32 Fortunately, for most things we do it is not really necessary to know the system’s total energy. It will suffice just to know relative energy densities at various locations. We call this electric potential or Voltage. One Volt is what you have when charges are placed close enough together to have an average energy density of 1 joule per coulomb of charge. The Relationships between Potential – Potential Energy – Field – and Force are given below. We will be using it on the following pages. F F = q E E Field Force V = - E dr U = F dr U U = q V V Potential P.E. and F F = m g g Field Force U = F dr Pot = U U = m Pot Pot Potential P.E. 33 g dr Now lets look at some specific examples where we apply what we have learned F F = m g g Field Force U = F dr Pot = U . U = m Pot g dr Pot Potential P.E. Imagine that you have a 10 kg mass that is held stationary in a constant gravitational field of strength g = 9.8 N/meter which is directed downward. What is the force that is needed to hold the mass there? Remember that the relationship between force and field is F = _____ ______ Substituting we get F = _____ ______ Or F = ________ directed _____ Now we decide to move the mass upwards a distance of 5 meters. How much Potential Energy is invested in this move? The relationship between Force and PE (Ug) is Ug = ∫ _____ _____ Substituting we get Ug = ____ ____ ∫ ______ Or Ug = ________ 34 Gravitational Potential is by definition the amount of Work (Joules) per unit of mass (1 kg) that is required to move an object from point A to point B. If from the preceding information, we want to find the potential needed to move a 1 kg mass upwards in the gravitational field given we use one the following general equations: F F = m g g Field Force U = F dr Pot = U U = m Pot Pot Potential P.E. PE to Potential g dr Or Field to Potential Ug = ________ x Potential Potential = ∫______ _______ ______ = ________ x Potential Potential = _______ ∫ _______ Potential = _____________ Potential = ____________ In most cases we will find that it is easier to find potential by using field strength Potential = ∫______ _______ Where the field is constant we can reduce the above equation to just Potential = ______ _______ 35 The Case of the Constant Electric Field (what we find in a standard Capacitor). Imagine that you have a +5 Coulomb charge that is held stationary in a constant Electrical field of strength E = 10 N/Coulomb which is directed downward. Question one: If this is the field created by two parallel plates. One above the picture and one below, what would be the sign of the charge will we find on the top plate? _______________ (draw it in the diagram) What sign of charge will we find on the bottom plate? _______________ (draw it in the diagram) Question two: What is the force that is needed to hold the charge stationary? The general relationship between force and field is: Question three: We move the charge upwards a distance of 5 meters, How much PE (Ue) does this require? The relationship between force and Ue is: F = ______ ______ Ue = ∫ ______ ______ Susbstituting we get Substituting we get F = ______ _______ Ue = _____ ______ ∫ ______ F = __________ directed _____ Ue = _____________ 36 Question four: Using the relationship between field (E) and potential (V) we can find the potential (Voltage) difference between the charge’s original location and its new location 5 meters higher. The general equation is _______ = - ∫ _______ _______ Because E is constant this reduces to a general equation of _______ = - _______ _______ Substituting values gives Or _______ = - _______ _______ Potential Difference = _______________ Volts. The Electron Volt Imagine an electron (mass = _______________ and charge of _______________) is held stationary in a uniform electric field of 0.5 Newtons / Coulomb. Question one: How far and in what direction would we have to move the electron until it had gained a potential of one volt? The relationship between Field and Potential is ________ = _________ _________ Because E is constant, this integral reduces to ________ = _________ _________ which gives a required displacement of d = __________ meters _______ (direction) 37 After moving this distance the electron has gained PE (Ue) which is given by the relationship ________ = _______ ________ Substituting The Energy gained is ________ = _______ ________ ______________ Joules is called an electron Volt (eV) This doesn’t seem to be a lot of energy. However, after moving the electron this distance we let the electron go. It gains kinetic energy as it looses Potential Energy. Using conservation of energy we can find the speed of the electron after it has lost one electron volt. Before = After _________ = _________ ______ Joules = ___ _____ _____ Substituting values ______ Joules = ___ _____ _____ This gives a final velocity of ____________ directed _________ 38 Television and Electron Tubes The old fashioned television set works by firing electrons from the back end of a vacuum tube to the screen at the front. The screen at the front end is coated with a phosphorous that glows when it is hit hard enough by these electrons. It takes a lot of energy for electrons to do this. Most televisions accelerate electrons across voltages of about 15,000 Volts. Lets look at what is happening. Screen Electron gun 15,000 Volts 1. The gun consists of two sheets. One with a hole in the center of it. Draw the sign of the charges that collect on each sheet. 2. Draw the field that is created by these sheets of charge. 3. The center of the back sheet is heated to a temperature where electrons begin to boil off. They are attracted by the other sheet How many electron volts do they have when they leave the back sheet? (trick question) 39 4. How fast are they moving as they impact on the front plate? a. General equation b. Substitutions c. Answer. 5. The electric field (E) created by the two sheets is uniform (constant) if the distance between the two sheets is two centimeters, what is the strength of the field that is created? a. General equation b. Substitutions c. Answer. 40 Aiming Electrons in a TV Tube The way a television paints a picture is by Accurately aiming the electron stream that is emerging from the electron gun at the back end of the TV tube. Early on it was thought that the best way was to use the same type of mechanism as is found in the electron gun itself. If one wanted to move the beam upwards it would be passed between horizontal plates as shown. 6. What sign charge should be placed on the top plate to do this? ________ 7. The electron is in this vertical electric field for a horizontal distance of 4 cm. By the time it has traveled this horizontal distance we want the electron to be deflected vertically by a distance of 1 cm. Let us assume that the electrons have a speed of ______________ m/sec as found in problem 4 and a mass of 9.11E-31 kg. 41 8. How much force must the electrons experience during the 2 cm under which it is being pulled upwards? 9. How strong of an Electric field do we need to make this deflection? 10. Assuming the plates are 4 cm apart, what voltage is needed to create this electric field 42 43 Magnetism AP Physics 2013 San Dieguito Academy 44 Magnetic phenomena have been known in the west for at least two-thousand years. In ancient Greece naturally occurring magnets were found to attract small amounts of iron from out of sand. The area from which many of these stones came was called Magnesia. The stones eventually were given the name magnetite. The first recorded magnetic compass comes from China in 271 CE The use of compasses was one of the great reasons for the age of exploration 1300 – 1600 in Europe. This is especially true as regards sea travel where the compass in combination with sextants and the first accurate watches allowed sailors to determine their locations accurately. Here is what they knew about magnets. 1. Magnets have two poles north and south. 2. A magnet’s north pole attracts another magnet’s south pole (opposites attract). 3. A magnets north pole repels another magnets north pole (likes repel). 4. The Earth itself is a magnet. 5. Iron rods if heated to red hot temperatures and then allowed to cool become magnetized. It was also known that iron filings arranged themselves forming curved patterns at the poles of magnets and curving outward and backward connecting these poles. During the 1800’s Michael Faraday developed the concept of lines of force based on these patterns.. These are the rules Faraday used for magnets. 1. 2. 3. 4. 5. Magnetic field lines emerge from the North Pole of a magnet. They circle around and enter the South Pole of a magnet. Where the lines are closer the field is stronger The direction of the lines represents the direction of the field Magnetic field lines continue even inside the magnet Below is a diagram of the Earth’s magnetic field. Along the field lines on the right side of the diagram are rectangles representing compass magnets. 45 Faraday developed his lines of force for charges as well as for magnets there were several major differences between the two. 1. 2. Lines of force start on positive charges and end on negative charges. With magnets, lines of force have no beginning and no ending. On this and the next page are several charge and magnetic situations, Draw in the lines of force. 46 47 In 1800 C.E. Oersted discovered, accidentally, that a current flowing in a wire created its own magnetic field. This field went did not go in the direction of the wire. Rather it seemed to be at all points perpendicular to the wire Faraday developed his lines of force to try to explain this. 48 Below is the rule for the creation of magnetic fields by flowing currents. 1. Thumb __________________________________________________________ 2. Fingers __________________________________________________________ I call it the hitch-hiker’s rule for creating a field. Pretend you are hitching a ride in the direction of the current (use your right hand) or in the direction of the moving electrons (left hand). The direction of the field is given by the curl of your fingers. . Interesting Properties of Magnetism Force a charge to move and the charge creates a magnetic field around itself. Using what we have seen on the previous page, we can produce a field that looks very much like that of a bar magnet. To do this we coil wire into what is called a “solenoid” In the picture below draw the field lines. 49 Move a charge through an existing field and it experiences a force on itself (and an equal but opposite force on the field is moving through) The strange thing about this force is that the force is perpendicular to the direction of the charge’s motion and perpendicular to the magnetic field. To find the direction of the force we use the following rule: 1. Arm __________________________________________________________ 1. Fingers ________________________________________________________ 1. Thumb _________________________________________________________ We call this the young fascist rule for charges moving in a B field. Remember use the right arm for positive charges and the left arm for negative charges. Describe the path of the moving charge in each of the following. 50 Given a positive charge moving through a magnetic field in each of the cases below. Find the direction of the missing component Remember that there is only a force on a charge when it has a component of its velocity that is perpendicular to the B field. In the picture to the right electrons are shot through a B field from the left and are bent downward. What direction is the magnetic field? ______________________________ Hand rules will give you directions but they do not give you magnitudes. The strength of the force is given below. A moving charge a moving current Fmag = q v Fmag = I l x B x B q is charge measured in Coulombs B is magnetic field measured in Tesla field measured in meters. v is velocity measured in m/sec L is the length of wire that is in the B 51 Motors In 1820 the Danish physics professor Hans Christian Oersted made the remarkable discovery that an electric current in a wire will deflect a magnetic compass. Soon afterwards it was demonstrated that the converse was also true: a strong magnetic field would also exert a force on a wire if the current flowing through the wire was moving perpendicular to the magnetic field. If a current were to flow through a wire placed between the two poles of a powerful magnet as shown below, the wire would be pushed upward. If either the current or the magnetic poles were to be reversed the force on the wire would then be downward. North South Coil current in wire moving toward the viewer Oersted This relationship between electric currents and magnetic fields has been used to great advantage in what we call electric motors. One simple motor consists of just an electromagnet, a disk, and an iron rod. The electromagnet attracts the iron rod, pulling the disk in a counter clockwise direction until the iron is at its closest position to the electromagnet. At this point the current in the electromagnet is turned off and the disk continues turning, pulling the rod upwards, until a complete rotation has been made. Once again the switch is closed the electromagnet turns on and the disk is pulled around by the magnetic force. There are two basic problems with this motor. The first is that you must manually switch the current on and off once every revolution and the second is that 52 the motor only pulls during half of a turn and then coasts the second half. A remedy to the switching problem came in the form of a device called a commutator. A commutator is a strip of conducting metal which is placed around portions of the motor axle. A copper brush rubs over the axle and alternately passes current to the axle when touching copper and stops passing current when not touching the copper. An improvement on this motor is the one shown below. It consists of a rotating armature with an electromagnet wound on it. This armature is placed between the poles of a magnet (called the field magnet) and current is passed through it so that the arm nearest the north pole of the field magnet is also north and the arm nearest the south pole of the field magnet is also south. Repulsion pushes the armature around this time in a clockwise motion. Notice that in figure B not only is this motor working on repulsion between like poles but that it is also acting on attraction between unlike poles. This motor needs two brushes and two metal contacts on its commutator. This allows current to flow into and out of the motor at all times. When the armature makes a 180 degree turn the brushes will change from one metal contact to another effectively reversing the flow of current and reversing the poles on the armature. This motor has the added advantage that it is under power during the full cycle of each revolution. 53 Another similar motor is shown below. Notice that this motor has no bulky arms to move. Rather its armature windings are around the shaft of the armature. It actually is an improved version of the wire launcher shown in the very first picture of this handout. The starter motor of your car is wound this way. Without the added inertia of the arms of the previous motor, this version can accelerate to full speed very rapidly. In the picture above, which direction is the force on the wire that is closest to the south pole of the magnet? _______________________ Which direction is the force on the wire closest to the North pole of the magnet? _______________________ Which direction is the force on the wire on the back side of the motor (that goes between the previous two wires? _______________________ 54 Lets Do Some Calculations The rail gun. It has been the ambition of many to accelerate objects to high speeds using magnetism. One method to do this is the rail gun. Basically it consists of two tracks that are electrified to a large potential difference. The tracks rest upon very large magnets as shown below. If a conducting bar is placed upon the tracks a current will flow. Because the current is flowing at an angle perpendicular to the B field it experiences a force. For simplicity sake let us assume that B = 0.5 Tesla, resistance of the system = 1, the distance between the rails is 1.5 meters, and the mass of the bar is 5 kg. a. In what direction will the bar want to travel? b. What is the net force on the bar? __________________ Equation: Substitutions Answer __________________ c. If the track is 200 meters long, how fast will the bar be moving when it reaches the end? 55 2. It has been considered using a rail gun to launch the space shuttle. If the shuttle has a mass of 6000 kg, How long would the tracks of this gun have to be? Assume the required launch velocity is 8000 m/sec. 3. During a solar storm electrons stream from the sun covering the distance between the Sun and the Earth in about 24 hours. They hit the outer layers of the Earth’s magnetic field where the magnetic field is about 1 x 10-6 Tesla. At this point answer the following: In which direction would they turn (right, left, up, down) ___________________ a. What would be the radius of curvature of the electrons? Equation: Substitution Answer b. How many eV does the average electron have? Equation: Substitutions Answer 4. Not only are electrons shot out in solar storms but so too are protons. Explain how things would be different and calculate. In which direction would they turn? 56 __________________ What is the radius of curvature for the protons? __________________ How much energy in eV’s would each proton have? __________________ 5. You are given the diagram of the typical DC motor shown below. There are 80 windings around the armature. The armature has a radius of 2 cm. And a length of 6 cm. The magnetic field has a strength of 0.2 Tesla. Find the following: Current in the coils is 1 Amp a. In each picture draw the direction of the magnetic field. b. Determine the magnitude and direction of the force on the armature. Equation: Substitutions Answer c. Determine the magnitude of the torque on the armature. Equation: Substitutions 57 Answer d. If the armature is cylindrical and has a mass of 300 grams find its angular acceleration. Equation: Substitutions Answer 6. A cyclotron is essentially a very powerful electromagnet over which electrons (and other charged particles) are passed. They orbit in a circular pattern with a radius based upon their speed. If we want electrons to move at 99 percent of the speed of light (c = 3 x 108 m/sec) with an orbital radius of 2 meters, how strong must the magnetic field be? 58 High Energy Particle Physics (neglecting relativity for the moment) 1. Lets look at a situation where we have particles of unknown mass, charge and speed. We want to find out everything we can about them. The first thing we can do is determine if they are charged by passing them through either a magnetic or electric field as shown below. Explain how you can determine the charge in each case. Magnet If the charge is neutral then A B If the charge is positive then A If the charge is negative then Capacitor B A B 2. Now that we know the net charge of the particles, (imagine that they have a positive charge) the next question is how fast are they moving if we do not know their mass or exact quantity of charge. This is where a device called the velocity selector comes in. It is simply the combining of the magnet and the capacitor in the picture above. Let us assume that the magnetic field is fixed with a strength of B and the Electric field is variable. Show how by varying the E Field you can select only particles of one specific speed? 59 We can go a step further. The E field in the capacitor is related to the d V across the capacitor by the following equation: Suppose that the distance between the plates is “d” the equation for the velocity of the emerging particles is v= Notice that all positively charged particles of the same speed will be selected regardless of their mass or magnitude of their charges 60 The mass spectrometer 1. Imagine a bunch of particles moving of different masses but having the same velocity how can we determine their individual masses. To do this we use what is called a mass spectrometer. It is a very simple device. Suppose we have the positively charged particles from the problem above with a known velocity. They are shot into a known magnetic field that is perpendicular to them. There is no electric field present. Less massive particles will land at point “a” but more massive particles will land further out. For example it point “b” What direction is the magnetic field directed? _____________________ The force acting upon the particles by the magnetic field is given by the following general equation Since the path of the particles is circular we can use F = m a to find the mass of the particles in terms of known quantities (Spectrometer magnetic field strength Bs, particle velocity v, path diameter “D” and charge q. Show your work below. 61 Maxwell’s Equations First Law Second Law Third Law Fourth Law 62 Developing Maxwell’s Equations Maxwell’s equations represent probably the high point of classical physics. They tie together electricity and magnetism. They predict the speed of light and the nature of most electromagnetic phenomena: reflection, refraction, diffraction and interference. They are what allowed us to develop radio and Television, computer hard drives and motors. They explain alternating current, why an electric transformer can raise or lower voltage, and how electric generators work. Interestingly, it only took four equations to do all of this. Maxwell’s first law we know pretty well already. It is called Gauss’ Law. Remember that all this really says is that you can discover the quantity of the net charge inside of a closed surface by measuring the electric field intercepted at the surface. Maxwell’s second law is the magnetic equivalent for Gauss’ Law It is based on the assumption that there are no magnetic monopoles. That for every north pole there must be a south pole. The result is that the sum of all magnetic fields intercepted by a closed surface must therefore be zero. This doesn’t sound like an extremely useful statement but it does come in helpful from time to time. 63 Maxwell’s third law is also called Ampere’s Law. We will start by learning the short version of it. Basically Ampere’s Law state that any current will create a magnetic field that is in a plane that is perpendicular to the flow of the charge. Don’t let this equation intimidate you. Imagine that in the picture below we are looking at a cross-section of a wire that is carrying a current (I) towards us (out of the page) This current creates a magnetic field (B) that is oriented in a counterclockwise direction. You can use the hitch hikers rule to show this. Now, the field will get weaker as we get farther from the wire. This is because the energy in the field (the field line) must be spread over a greater distance (circumference = 2 r) Ampere’s law takes advantage of this fact. The left side of the equation is what is called a line integral. Draw a circle around the wire with some radius r. Add up all of the small components of the magnetic field B in small increments dl until you make a complete circle. It does not matter what radius you pick, the answer will be the same. This answer is directly proportional to the amount of current flowing in the wire. The proportionality constant is called . In a vacuum it is called o. or the magnetic permittivity of free space. o.= 4 E 10-7 Tesla Meters/Ampere. dl B B B Ampere proved that it wasn’t necessary to draw a perfect circle around the wire. Any shape will do. Just add up the components of magnetic field that are in the direction of shape you have drawn and you will get the total current inside the shape 64 you have drawn. Sample problem 1. You have a wire that has a current of 10 Amps flowing through it, What is the magnitude of the magnetic field it creates at a distance of 1 cm from the wire’s center? (ans. approx. 2 x 10-4 Tesla 65 Sample problem 2. It used to be that to measure the current in a wire one needed to cut the wire and then run the current through the current measuring device. New amp meters use a Amperes law, a collar which can wrap around the wire and a thing called a magnetometer in this collar with which to measure the magnetic field created by the current in the wire. Collar Magnetometer Wire B Hinge If the magnetometer on your amp meter reads (5.9 x 10-6 Tesla and it is 0.5 cm from the center of the wire, what current is running through the wire? 66 Sample Problem 3: The Solenoid The solenoid is a compact way of creating a pretty strong magnetic field if you do not have a massive current. A solenoid is just a coil of wire. It may be one layer thick or many layers. Let us assume that L is the length of the loop. And that N is the number of turns in the loop. You can use Ampere’s law to get a very good approximation of the field strength inside the solenoid. What we will do is to draw a rectangular loop. 1. Notice that in Ampere’s law, the integral B dl is a dot product. Because of this since the magnetic fields are horizontal, crossing sides A and C at 90 degrees, there is no contribution here. 2. We have placed side B at a great distance from the solenoid and can assume that even though the field is parallel to this side, it is so weak that there is almost no contribution here. 3. this leaves us with side D as the only side that is significant. Now let us substitute into Ampere’s Law I is increased by the number of turns ____ _____ _____ = ______________ B is a constant. Remove it from the integral ____ _____ _____ = _____ ______ The integral of dl is just L ____ _____ _____ = _____ ___________ Or B = ____________________ Notice that N/L is just the number of turns 67 per unit length (n) and the equation can be reduced to B = ____________________ Let us see what happens when you have a solenoid that is 10 cm long, with 1000 turns and you decide to run 1 Ampere of current through it. It turns out that if you were to place an iron core inside the solenoid the magnetic field is increased by a factor (k) of 200. In the last problem what would the new magnetic field be? How would taking into account the relative permeability (k) change the equation for the magnetic field strength of a solenoid B= 68 Sample problem 4: The magnetic field created by a toroid. If you want to make an even more powerful magnet you can bend a solenoid around on itself so that the magnetic field never leaves iron core inside the solenoid. This is what is called a toroidal solenoid. To calculate the field inside of the toroid we will draw a circular loop that is inside of the toroid itself. Explain why the loop should be circular Use Ampere’s law to determine the field (B) 69 Maxwell’s Fourth Law – Magnetic Induction Maxwell’s fourth law is often known as Faraday’s Law. It asserts that any change in a magnetic fields near a conductor will cause an induced voltage in the conductor and a resultant flow of charge through that conductor. One can look at this several ways. The simplest form of Faraday’s law is d Fm V dt Remember that is the amount of magnetic field intercepted a surface. = B A cos . The simplest example of how this works is a wire that is being pulled across a magnetic field as we did with the rail gun back on page 90. There a 1.5 meter long metal bar was sliding across a magnetic field B = 0.5 Tesla. If the bar is moving at a speed of 10 m/sec answer the following: Velocity A. Which rail will become positively charged?(think hand rule) B. What will be the potential difference between the two rails? (Faraday’s Law) 70 Notice that this voltage is directly opposed to the voltage that we used to get the bar moving in the first place. We call this back Emf. It is what limits the maximum speed that a rail gun can obtain. Let us see how this works. In the rail gun problem we placed a 12 Volts battery across the rails. In this problem we assumed a net system resistance of 0.01 Ohm which limited the initial current to _________ Amps. Since Faraday’s law predicts a counter voltage working against the battery it lowers the net Emf of the system. At maximum speed, the induced counter Emf will 12 Volts. 5000 V Velocity At what speed mill the bar max out? NASA makes a mistake. Several years ago the guys at NASA proposed a unique method for generating electrical power on board the international space station. (see picture below) The idea employed a direct application of Faraday’s law. What they were going to do was place a long wire 320 meters that would drop from the station towards the Earth. As the station and wire orbited hundreds of miles above the equator. Lets see what happens. A. Using hand rules, determine whether positive charge will congregate at the station or the end of the tether. B. Use Faraday’s law to determine the potential difference between the ship and the end of the tether. 71 C. It turns out that there are a couple of major problems with the plan. One of them is electrical in nature. Hint: think circuits. D. There is also a major mechanical problem. What is it? The Electric Generator. In the last problem we used Faraday’s law with a changing flux created as a bar swept out an area across a magnetic field. Let us consider a situation that on the surface of it sounds somewhat improbable. Imagine that you have a large magnet whose south pole is facing you. B = 1.5 Tesla. In front of the magnet is a circular conducting rubber band as shown below. Now lets imagine that the rubber band contracts rapidly. Its radius changes from 10 cm to 5 cm in one half of a second. A. In what direction will the current created by this shrinking go? B. What average voltage will this create? C. Now the rubber band suddenly expands again to its original size in the same amount of time. What direction will the current be going as it expands? 72 The above problem does sound improbable. However it is at the heart of how the generators that produce most of the electricity in the world today. Let me explain. Let us imagine a loop of wire as seen below. This loop is free to spin in a circular motion, around and around. If the loop is in front of a large magnet, from the magnet’s point of view the area of intercepted magnetic field first decreases, disappears and then grows again. 1 2 A B AB 1 2 South South 3 4 5 BA 3 4 South South BA 6 Side View 7 BA 7 South A 9 South South A A A A AB 8 South A A 9 AB 5 Top View 6 South 8 A A 1. In pictures 1,2,3 which way is the current flowing From A to B or B to A? __________ 2. In pictures 3,4,5, which way is the current flowing From A to B or B to A? __________ 3. In pictures 5,6,7, which way is the current flowing From A to B or B to A? __________ 4. In pictures 7,8,9, which way is the current flowing From A to B or B to A? __________ 5. If the loop is making 60 complete revolutions per second and has a radius of 10 cm and is in a B field of 1 Tesla, What is the average potential produced during pictures 1,2,3? 73 Inducing currents by changing magnetic field strength Faraday’s law states induced voltages are given by the equation V = -d /dt where the flux = B dot A. Up to this point, we have induced currents in wires by changing the A (area) in the flux equation not the magnetic field strength B. However, you can just as effectively change flux by keeping the area constant and either increasing or decreasing the magnetic field strength that is intercepted. One way of doing this is to drop a magnet through a loop. Side View S S N N S N Top View (ring moving toward you) B B B B B B B B B B B B B B B In the first picture (left) the north end of the magnet is moving downward toward the ring. From the magnet’s point of view, the ring is moving upwards. The ring moves upwards through a magnetic field that is pointing outwards. Draw in the direction of the current that is generated. While the magnet is inside of the loop (center) there is no net field moving across the loop and there is no net current in the loop. As the magnet leaves the loop (right), magnetic field lines start crossing the loop again. Only, this time they are pointed inwards. Draw in the direction of the current that is generated. Lenz’s Law (the electromagnetic equivalent of Newton’s third law (for every action there is an equal but opposite reaction) states that an induced current will create its own magnetic field. This field will be opposed to the magnetic fields that created it. Lets go back to the first picture of the dropping magnet. Draw in the 74 magnetic fields created by the current in the loop. They have the same shape as created by a solid magnet if it were placed at the center of the loop. Draw in this magnet. Notice that the magnet has the north pole aiming upwards. What impact do you think this induced magnet will have on the falling of the real magnet? Electric Transformers A transformer is a device that can raise or lower the Voltage of electricity in a system. It is a simple instrument, just an iron bagel with wires wrapped around it. The principles upon which it works are just as subtle as the device is simple. Principle #1: Magnetic fields are current driven. Wrap a wire around a metal bagel and the current in the wire creates a magnetic field in the bagel. (if the current flows as is shown, draw the direction of the magnetic field inside of the bagel. Principle #2: You can increase the apparent quantity of the current that is flowing by increasing the number of loops. If one loop creates 1 B then two loops creates 2 B. Principle #3: For every action there is an equal but opposite reaction (Lenz’s Law) If there is a conducting hoop somewhere around the bagel, as the current in the original loop increases the increasing induced magnetic field will induce a current in the hoop that will create a magnetic field in opposition to this. Draw in the direction of the induced current in this hoop. Principle # 4 Energy is conserved. Let us look at a situation where there are two loops in the original circuit (left side of bagel) They will create twice the magnetic field as created by one loop. If there is only one loop on the right side of the bagel then it must generate twice the current as was in the original circuit in order to react cancel out the inducing magnetic field. The hoop on the right side of the bagel is a circuit in itself. This circuit is called the Secondary circuit. The circuit on the left side of the bagel is called the Primary circuit. According to the laws of conservation of energy, energy out (secondary) must be equal to the energy in (primary). The equation for this is Pin = Pout Iin Vin = Iout Vout 75 Remember that if there are twice as many turns in the primary as in the secondary circuit then there must be twice as much current in the secondary circuit as in the primary circuit. The power equation reduces to Vin = 2 Vout What we have is a voltage reducing transformer. If we want to increase voltage then we have more turns in the secondary than in the primary. The relationship between turns and voltage is given by Tsec / Tpri = Vsec / Vpri Sample problem. Lets say you want to make a neon sign transformer that raises 120 Volts to 1200 Volts. A. If the primary(A) has 20 turns, how many turns must the secondary (B) have? B. If the secondary draws 60 milliamps how much will the primary draw? 76 Inductance (L) Inductance ( L )is a measure of a coil’s ability (N = number of turns) to create magnetic flux as a given current (I)is flowing through it N m = L I Remember Faraday’s law? V = - d (Nm) / d t Well if we combine these two the result is V = - L dI /dt Or V L dI dt where inductance is measured in Henries and a Henry is a Volt Second per Amp Calculating Inductance is very difficult. Only for a few simple situations can inductance be relatively easily determined from a theoretical basis. (This is OK because inductance is easily determined experimentally) For a tightly packed air core solenoid inductance is L = o n2 l A Where o is the magnetic permeability of free space n is the number of turns per unit length (turns/meter) l is the length of the coil (meters) A is the end view area of the core 77 What is the importance of inductance? Energy Inductance has important effects in several ways. First it gives a direct way to determine the energy stored in a coil’s magnetic field. UB = 1/2 L I2 See how this compares to energy stored in an electric field in a capacitor UE = 1/2 C V2 This is sometimes given as UE = 1/2 q2 / C Or the energy of mechanical motion UKE = 1/2 m v2 Or the energy stored in a spring Usp = 1/2 k x 2 Simple Harmonic Motion Springs when stretched and set free to move will oscillate in SHM translating energy between KE and PEspring. It turns out that a conductor-inductor system will oscillate in SHM as well. These circuits are often referred to as LC circuits. The energy in these systems alternates between a capacitor and an inductor. B E E Remember how for a spring = (k/m). Well for an LC circuit = (1/LC). 78 It is important here to actually derive this equation. We start with the Energy equation U The general equation Remember how And = UE + UB UE =1/2 C V2 Ub =1/2 C V2 U Substituting we get Energy is time dependent = 1/2 q2/C dU/dt = differentiate and we get dU/dt = Remember L and C are constants + 1/2 L I2 d (1/2 q2/C q /C dq/dt 1/2 L I2 )/dt + + L I dI/dt however U is constant with time 0 = q /C dq /dt + L I dI /dt But wait! di/dt is the derivative of dq/dt 0 = q /C dq/dt + L I d2q/dt2 0 = And I = dq/dt q /C dq/dt + L dq/dt d2q/dt2 0 = This all reduces to q /C + L q’’ q’’ = - (1/ LC) q x’’ = - 2 bx Which is familiar For SHM In this case = ______________ Frequency = ___________________ 79 B E E Before we leave the topic of Simple Harmonic Motion and LC circuits lets look at one more aspect of how Faraday’s law plays a part here In the diagram at the top of the page we see in the figure at the left energy is stored in a capacitor and that the pressure (Voltage) across the capacitor is trying to push current through the system in a clockwise motion through the system. The net resistance of the wire in the coil is negligible. What one might expect is an extremely rapid discharge of the capacitor. This does not happen. Faraday’s law below explains why. d Fm V dt As the current starts to ramp up a back EMF is being created against it. The result is that current will slowly increases. Remember that current in the coil places energy in the magnetic field that is produced. This is what you see in the center picture at the top of the page. At this point there is no reason for a maintained current. So, how does the capacitor get re-charged as shown in the diagram at the right at the top of the page? Well, Faraday’s law comes to play again. d Fm V dt As the current starts to fall, the decreasing (-) magnetic field creates a positive voltage across the coil. This positive voltage starts to force current in a clockwise direction in the circuit. The current slowly declines for two reasons 1. ____________________________________________ 1. ____________________________________________ 80 81