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Unit 13 - Physics 225 Relativity and Math Applications...

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Unformatted text preview: Physics 225 Relativity and Math Applications Spring 2010 Unit 13 Maxwell’s Equations N.C.R. Makins University of Illinois at Urbana-Champaign ©2010 Physics 225 13.2 13.2 Physics 225 13.3 Unit 13: Maxwell’s Equations Exercise 13.1: 3D Integration Warmup Last week we learned how to setup and calculate 3D integrals quantitatively. We’ll need those skills today and next week, so let’s start with a warmup problem. To help you out, the full-blown steps for 3D integration are on the back tear-off pages again, and here’s a short summary: A1. A2. A3. B1. Pick your coordinate system ri (i = 1,2,3) Pick your IPs = integration parameters uj (j = 1,…,n) Parametrize : provide constraint functions ri(uj) and bounds on all uj Express the differential d in terms of duj (Remember, d is the answer to this question: “If I move a little bit du in all of my integration paramters u , how much length/area/volume do I sweep out?”) B2. Write the form of the full n-dimensional integral including the bounds on uj B3. Applying your constraint functions, work on the integrand F ⊗ d until it is written in terms of only the IPs uj, their differentials duj, and constants C. Integrate! j j Also, we’ve collected a number of Tips & Tricks to address common complications: #1. When chosing your coordinate system ri, if one choice is best for the integrand F and a different choice is best for the integration region , choose the latter. ! #2. If the generic vector of position r appears in your integrand, remember that its form is quite different in Cartesian, spherical, and cylindrical coordinates. #3. If your integral gives a vector result, you must split it into 3 separate integrals, one for each component. (i.e., vectors add by components.) ˆ ˆˆˆ #4. If the curvilinear unit vectors r , s , ! , or ! appear in your integrand, you cannot pull them out of the integral as constants because they’re not constant → replace them with their Cartesian equivalents. And here’s the next one, for our warmup problem on the next page: Tips & Tricks #5: Cylindrical coordinates are wonderful for objects whose axis of symmetry lies along the z axis, and spherical coordinates are great for spherical objects centered at the origin. These coordinate systems are dreadful, however, if the cylinder or sphere over which you’re integrating is not centered on the origin, or if the cylinder’s axis is not in the z direction. What to do!? If you encounter such a problem, the solution is simple: just shift and/or rotate the coordinate system so that the object is centered/oriented in the best way for spherical/cylindrical coordinates! 13.3 Physics 225 13.4 (a) A solid sphere of radius R has a total mass M that is distributed uniformly throughout its volume. M 3M (Its volume mass density is thus a constant: ! = .) The sphere’s center is not located at the = V 4" R 3 origin, but on the x axis at the point (+2R,0,0). Calculate the moment of inertia I of the sphere. The ! ˆ2 ˆ formula for the moment of inertia of any object around an axis a is I a = " dm r ! a , where dm ˆ represents the differential of mass = all the little pieces of mass in the object. (For the case of a sphere, ˆˆ you’ll get the same answer for any axis, of course, as it’s spherically symmetric, so just use a = z .) All good? Then on to the next piece of trickery, which brings us closer to one of our goals for today: electromagnetic field calculations. Tips & Tricks #6: If your integral gives you a field as a result — i.e. a result that is a function, not a constant — you must be careful to distinguish between field point and source point coordinates. Let’s !! ! say you are trying to calculate an electric field E (r ) . The argument r on which your result will depend is called the field-point; it will be expressed as some triplet of coordinates ri, such as (x,y,z). Meanwhile, your integral will sweep over different coordinates, likely the coordinates of all the sources of the electric field (e.g. charges). We must be careful to distinguish these source-point coordinates, so we flag them with some different notation. Primes are often used for the source-point coordinates: e.g. (x′, y′, z′). I personally find it more helpful to denote them with some subscript indicating what they represent physically. If we are integrating over charges q, for example, you could label your source-point coordinates (xq, yq, zq). Whatever notation you choose, the critical point is this: The source-point coordinates ( xq , yq , zq ) vary over the integral, while the field-point coordinates ( x, y, z ) do not and can be treated as constants. 13.4 Physics 225 13.5 Time to practice! You learned in Physics 212 that the electric field of a point charge q located at the origin is kq/r2, where k = 1/4πε0. We also know that this electric field points radially outwards from the charge q. Now let’s consider the more general case of a point charge located not at the origin but at ! some source point rq . In complete vector notation, the electric field of such a point charge is !! !! r ! rq E (r ) = kq ! ! ! 3 . This formula is called Coulomb’s Law. See how that classic 1/r2 dependence is r ! rq still there despite the cube in the denominator? We needed to use the trick of “vector/|vector|3” to encode the direction information. (b) Does this vector formula make sense? Take the specific case of a point charge q located at the ! ! source point rq = ( xq , yq , zq ) = (0, 0, a ) on the z-axis. What is the electric field E at the field point ! r = ( x, y, z ) = (b, 0, 0) on the x-axis? Your answer must be in vector form, so provide all three Cartesian ! components of the field E = ( Ex , Ey , Ez ) . And please, Draw! A! Sketch! OK, now on to integration! To calculate the electric field of an extended charge distribution, we simply integrate the electric field components from all the little pieces of charge in the distribution. We’ll use the symbol dq to refer to all these little pieces of charge. Here is Coulomb’s Law in its most general form: !! E (r ) = !! r # rq 1 dq ! ! 4!" 0 $ | r # rq |3 Alternatively, we can use a different strategy. The electric potential of a point charge at the origin (relative to infinity) is kq/r. We can integrate electric potential instead, then take its gradient to find the electric field. Usually this is much easier, as it involves a scalar integral not a vector integral! ! V (r ) = dq 1 !! $ | r # rq | 4!" 0 ! ! E = !"V Naturally, we’re going to try both. Excellent practice! 13.5 Physics 225 13.6 Consider a very large plate of radius R lying in the xy plane and carrying a uniform surface charge density σ. We will be treating this as an infinitely large plate by taking the limit R → ∞, so it doesn’t matter if you treat it as a circular plate of radius R or a square plate of sides 2R. Your task: calculate the !! electric field E (r ) produced by this plate. First, let’s try the potential route → set up and calculate dq 1 ! the electric potential V (r ) = !! $ | r # rq | . But first … 4!" 0 (c) We must make a key observation to make our life very much simpler! This is going to be an infinite plate, so some symmetry considerations can help us … In the space below, convince yourself and !! ! your colleagues that the field E (r ) and potential V (r ) (or anything else!) caused by this plate can only depend on the coordinate z, not x or y. (d) Since V can only depend on z, we can pick any x and y values we choose for our field-point → clearly we will choose x = 0 and y = 0, i.e. we will pick our field point to be on the z-axis without loss of generality.1 Now you’re good to go! Using Cartesian coordinates (as shown in the figure), set up the dq 1 ! integral V (r ) = !! $ | r # rq | for field points on the z-axis and for a square plate of charge density σ 4!" 0 with corners at (±R,±R,0). Don’t calculate it, just set it up. In case it’s unclear, dq here is just σ dA → we are sweeping over a surface, hence over little elements of area dA each with a little bit of charge σ dA. 1 This is equivalent to Tips & Tricks #5: shift your coordinate system to put the z-axis through your field point. 13.6 Physics 225 13.7 (e) To calculate your integral, it’s much better to use cylindrical coordinates. This also gives us some experience switching coordinate systems. Rewrite your integral for V in cylindrical coordinates, i.e., using source-point coordinates sq and φq instead of xq and yq. (The field point coordinate remains simply z.) Remember, the area element dA you need will be of a different form in cylindrical coordinates. This time, you can do the integral → go for it! ! ! (f) Now take the gradient to obtain the electric field: E = !"V . Take the limit R → ∞ … your result should match the familiar electric field of an infinite plate from your 212 formula sheet! (g) Check your answer’s direction → From the same symmetry principles you used on the previous page, the electric field of an infinite plate must point in the z direction, it cannot have any x or y components. Is it completely clear that this is so? If not ask your instructor! 13.7 Physics 225 13.8 (h) And now a tour-de-force: ready for the vector integral? You bet! Set up and integrate the electric !! !! r # rq 1 field of the large plate using Coulomb’s Law: E (r ) = dq ! ! . Use all the symmetry 4!" 0 $ | r # rq |3 simplifications we’ve discussed: restrict yourself to field-points on the z-axis, and also, realize that you only have to calculate the z-component of this field Ez(0,0,z). About that last point: it is very instructive to set up the full integral for all three components … but before you actually do any of the integrals, stare at them carefully → you should see at once that only the z-component integral will survive, the others will integrate to zero. You can use either Cartesian or cylindrical coordinates for this task. I suggest cylindrical, if only for practice with this somewhat unfamiliar system. If you use Cartesian, you will need the following integrals (which you could obtain yourself using integration-by-parts and trig substitution): "x% 1 dx x dx ( ( x 2 + a 2 ) = a tan!1 $ a ' ! ( x 2 + a 2 )3/2 = 2 2 2 #& a a +x 13.8 Physics 225 13.9 Interlude: Maxwell’s Equations At last it is time to encounter the full might-and-beauty of Maxwell’s Equations, the four equations that define all of electromagnetic theory. Drum roll please … here they are! !! # !"E = $0 ! !! $B !"E =# $t !! !"B = 0 ! !! ! $E ! " B = µ0 J + µ0# 0 $t This electromagnetic theory, with its quantum-mechanical extension QED = Quantum ElectroDynamics, is arguably the most precisely-tested theory in the world. Experiment and theory have been pushed to more than 7 significant digits, and they still agree. The reason that the four Maxwell equations are “all you need” comes from a mathematical principle known as the Helmoltz Theorem: The behavior of any vector field is defined entirely by its divergence and curl. ! In other words: if you know the curl of a vector field E and you know its divergence as well, then you know everything about it.2 Today, we’ll consider a special case of E&M called statics → problems in steady-state where nothing is changing with time. In this special case, the two time-derivatives in Maxwell’s equations disappear, which simplifies things considerably: !! # !"E = $ !!0 !"E = 0 !! !"B = 0 !! ! ! " B = µ0 J We will learn what that unfamiliar “J” symbol means in the last of Maxwell’s equations in a bit. But first, let’s write down the complete solution of these static equations! The complete solution is provided by two multi-dimensional integrals. You’ve just worked with Coulomb’s Law; the other one is the Biot-Savart Law for the magnetic field. Here they are: !! E (r ) = !! 1 ! r $ rs #(rs ) ! ! 3 dVs | r $ rs | 4!" 0 % !! !! µ0 ! ! r # rs J (rs ) " ! ! 3 dVs B( r ) = 4! $ | r # rs | 2 To be perfectly precise, there is one additional piece of information needed, which is boundary conditions on the field. Don’t worry, we’ll address this soon. 13.9 Physics 225 13.10 Exercise 13.2: The Divergence Equations Everything you’ve learned in Physics 212 so far about the behavior of electric and magnetic fields is encoded in Maxwell’s Equations. With a little practice, you can “read off” Gauss’ Law, Ampere’s Law, Faraday’s Law, and more from Maxwell’s 4 equations. To accomplish this, all you need is the collection of vector-calculus theorems we studied several week’s ago: the Gauss-Green-Stokes Theorems (GGS). (a) The GGS theorems all relate the integral of “some derivative” of a field over a region R to the integral of the field itself over the region’s boundary ∂R. We studied three incarnations of GGS: ! Volume ?! dV = " ! " Volume ! ?! dA ! Surface ! ?! dA = " ! "Surface ! ?! dl ! b ! ! ! ?! dl = ?!(b ) " ?!(a ) ! ! a Can you!fill in the question marks? The elements you need are E (some vector field) !! ! " E (its divergence) !! ! " E (its curl) V (some scalar field) ! !V (its gradient) … where do they go? Two things to note: ! ! The differential elements dl and dA are vectors, so you can never just multiply them with another vector → you must always specify either a dot-product or a cross-product when multiplying two vectors. ! The GGS theorems are mathematical theorems. Though we’re using the symbols E and V for our vector and scalar fields, they’re not meant to represent electric field and electric potential – or any other physical observable – in this context. So off you go: write down the three GGS theorems! Memorization is not needed here, just think physically. You’ve done many exercises exploring the physical significance of grad, div, and curl. Use that physical insight to figure out the form of our three GGS theorems → they’re pretty obvious really! 13.10 Physics 225 13.11 Next let’s consider Maxwell’s divergence equations, neither of which involves any time-derivatives: !! # !"E = $0 and !! !"B = 0 (b) The equation for the divergence of the electric field is called Gauss’ Law, for good reason: apply !! the appropriate GGS theorem to ! " E = # / $ 0 to obtain the familiar Physics 212 integral form of Gauss’ Law. (Should you integrate this equation over a volume, an area, or a curve?) !! (c) Now integrate the other divergence equation, ! " B = 0 over an arbitrary volume, using GGS. What statement can you make about the magnetic flux through a particular type of surface? (d) Consider your answer to part (c), which comes directly from Maxwell’s equations. What does this teach us about: the geometric shape of magnetic field lines? the existence (or not) of “north/south” magnetic charges? 13.11 Physics 225 13.12 Exercise 13.3: Current Densities and Ampere’s Law So Maxwell’s divergence equations tell us about the point sources (“charges”) from which the electric and magnetic fields emanate. They tell us in particular that there are no such sources for the magnetic field … so where does the magnetic field come from? Well, we know from Physics 212 that the magnetic field is produced by currents. In Physics 212, the currents we have dealt with have pretty-much all been confined to thin wires. But to obtain a general theory of magnetism, we must consider current distributions → cases where the current is spread throughout a region of space, rather than simply confined to a simple one-dimensional path. Thus we !! must introduce the volume current density J (r ) , which is the current-version of the volume charge ! ! density !(r ) . Here is the definition of J , both in symbols and in words: !! ! The volume current density J (r ) !! dI J (r ) ! is the current per unit dA" area-perpendicular-to-flow. ! ! The units of J are thus amps-per-square-meter. You may be wondering why this J is current per unit area while ρ is charge per unit volume. Here’s the reason: If you have a region of space (say a cube) that contains some charge density ρ, you do a volume integral of ρ over the cube to obtain the total ! charge Q inside the cube → no problem. But if your cube contains a current density J , how do you obtain the total current I through the cube? Well, unless you specify a direction this question is ! meaningless! For example, if J points to the right overall, the total current out of the right-hand face of your cube would be positive, while the total current out of the top face would be zero. So we never calculate a “total current I” … we always calculate “the total current I through the surface S” → total current are always calculated in reference to a particular surface. Once you !! I = " J ! dA have chosen your surface, the formula is: Surface (a) To clarify all this, let’s work a simple example. Consider a long bar of metal with a square cross-section of side-length a. The central axis of this bar coincides with the z-axis of your coordinate system, and the bar carries a !! ˆ uniform current density J (r ) = J z where J is a constant. What is the total current I through the surface shown, which is a large flat square of side b > a that is parallel to the xy-plane? b a b J a 13.12 Physics 225 13.13 (b) That wasn’t too hard. Let’s try a more abstract example. Consider a region of space occupied by !! x 2 + y2 ˆ z , where J0 is a constant with the same units as J, and a is a this current density: J (r ) = J 0 a2 constant with units of length. Picture this current flow in your mind, or maybe draw a sketch. Your task !! is to calculate the total current I = " J ! dA which passes through a circle of radius R that is centered Surface on the z-axis and lies parallel to the xy-plane. Perfect practice in multidimensional integration! (c) Remember to check the units of your answer for I → do they make sense? 13.13 Physics 225 13.14 We’re now ready to recover another 212 result from Maxwell’s equations. The relevant one is: ! !! ! $E ! " B = µ0 J + µ0# 0 $t Since we are restricting our attention today to static, or steady-state problems where nothing is changing with time, we lose the last term from the equation above, which simplifies things considerably. (d) In the steady-state case, apply the appropriate!GGS theorem to the differential equation above to ! obtain an integral relation for the magnetic field B in terms of the current density J . (Should you integrate the differential equation over a volume, an area, or a curve?) Do you recognize the equation you have obtained? Be sure to express it in familiar Physics 212 form to make yourself very happy. Exercise 13.4: Solutions for High Symmetry via GGS So: you have applied GGS to Maxwell’s equations to recover both Gauss’ Law and Ampere’s Law in integral form. These GGS-type laws permit a very elegant solution technique for finding electric and magnetic fields in a special case: when the charge and current distributions have a high degree of symmetry. These solutions don’t work all the time, but when they do, they are much easier than performing brute-force Coulomb or Biot-Savart integrals! You saw this high-symmetry technique in 212; we’ll apply it now in a more general setting, and with more care. !! !! ! (a) Maxwell’s magneto-static equations ! " B = 0 and ! " B = µ0 J tell us a lot about the possible shapes of the magnetic field. Consider a thick wire of circular cross! ˆ section carrying a uniform current density J = J z , where J is a constant. The radius of the wire is a. Using only Maxwell’s magneto-static equations plus the symmetries of the given current distribution, construct enough arguments to convince a circuitcourt judge of the following: (see footnote3 for a hint …) !! ˆ The magnetic field B(r ) must point in the ! direction at all points in space. The magnitude of the magnetic field can only depend on the cylindrical radialcoordinate s. 3 a J Circuit-court judges are frequently convinced by arguments that exclude other possibilities … 13.14 Physics 225 13.15 !! ˆ (b) The judge is now convinced that B(r ) = B( s )!! in this particular problem. Use Maxwell’s !! ! magneto-static equation ! " B = µ0 J , plus your expertise with the GGS theorem, to prove that the µ Ja 2 µ Js magnitude of the magnetic field is B = 0 outside the wire and B = 0 inside the wire. 4 2s 2 a J !! !! ! Bravo! You are such an expert with the magneto-static equations ! " B = 0 and ! " B = µ0 J that you can find their solution for sufficiently symmetric current densities. (c) So now consider this problem: An infinitely-long solenoid of radius a and n turns per-unit-length carries a time-dependent current I (t ) = I 0 ei!t . (Remember from last week: what’s meant by this expression is that the physical current is the real part of this complex expression.) Use your magneto-static skills to determine the magnetic field created by this current, both inside and outside the solenoid. 5 Hint: Start by determining the B field outside the solenoid … z a I (t ) = I 0 ei!t 4 5 Hint: In case you didn’t recognize it, this is just an Ampere’s Law problem. Yet another Ampere’s Law problem … and one for which the answer is on your P212 formula sheet. 13.15 Physics 225 13.16 That was just a Physics 212 problem … but the magnetic field you wrote down is changing with time, and that means that it will create an induced electric field! How shall we calculate it? Since there’s no net charge-density in this problem (ρ = 0), the equations for the induced E-field are: ! !! !! $B !"E =# !"E = 0 $t If you compare these equations to the magnetostatic ones you’ve been working with, !! !! ! ! " B = µ0 J , !"B = 0 you see immediately that Induced E fields are produced by –∂B/∂t in exactly the same way that static B fields are produced by steady current-densities µ 0J. (d) Given the equations written above for the induced E-field, and using the B-field that you found in part (c), determine the direction and functional-dependence of the electric field induced by the changing magnetic field of the solenoid. z a !! (e) Finally, apply the equation for ! " E plus your expertise with the GGS theorem !! to calculate the induced electric field E (r , t ) both outside and inside the solenoid. I (t ) = I 0 ei!t Physical Intuition Checkpoint: Induced fields are pretty magical when you first encounter them: they seem to appear out of nowhere. What do they look like? How do they behave? Well you just calculated !! one → let’s build up some intuition about these mysterious fields. Visualize your induced E (r , t ) field as well as the current that caused it … does anything strike you about the field’s direction? phase? …? 13.16 Physics 225 13.17 Generic Procedure for Multi-Dimensional Integration " F (r )!!! d ! n i n ! Example problem: Calculate the total charge on a plate of charge density ! = Kxz 2 / y , where the plate lies in the xz-plane with corners at (0,0,0) and (a,0,b). A. Parametrize the integration region n Here “parametrize” means “express in mathematical form”. Here’s how: A1. Pick your coordinate system ri (where i = 1,2,3) Make a choice suitable to the integration region . Your options are ri = (x,y,z) Cartesian, (r,θ,φ) spherical, and (s,φ,z) cylindrical. Ex: Pick Cartesian as it matches the plate best → ri = (x,y,z) A2. Pick your IPs = integration parameters uj (where j = 1,…,n) To sweep the integral over a region of dimensionality n, you will need to integrate over n variables. These are your parameters of integration. They can be anything, so we’ll generically call them uj. However, in a large class of problems, these uj will be a subset of your coordinates. In other words, each uj will simply be one of your coordinates ri → very simple! We will concentrate on this class of problems today. Ex: We can sweep over the plate with the x and z coordinates, so we pick x and z as our IPs. A3. Parametrize : provide constraint functions ri(uj) and bounds on all uj To describe a region mathematically, you need both of these elements. First, constraint formulas: each coordinate ri of your chosen coordinate system must be expressed entirely as a function of your chosen integration parameters uj. These functions are called “constraints” because there are generally more coordinates ri (3) than there are IPs (1 for a path, 2 for an area, 3 for a volume). Second, you need to indicate the bounds on your IPs. The example will clarify: Ex: The constraints and bounds defining the plate are as follows: constraints on ri bounds on uj: x = param #1 (varies) x: 0 → a y = 1/3 z = param #2 (varies) z: 0 → b B. Write the integral in terms of your IPs uj, their differentials duj, and constants Before you can actually perform your integral, you must make sure that everything that varies during the integration is expressed entirely in terms of your chosen integration parameters uj. Otherwise, you have an ill-defined integral that you cannot do. The steps are on the flip side … 13.17 Physics 225 13.18 B1. Express the differential d in terms of duj In this step, you have to figure out the differential d of length/area/volume for your path/surface/volume. It will involve the product of the differentials duj of each of your integration parameters. How to figure it out? Use your sketch to answer the following question: “If I move a little bit duj in each of my integration parameters uj, how much length/area/volume do I sweep out?” The answer to that question is the ! differential dl or dA or dV that you seek. If ! your problem requires a vector differential ( dl or dA ) use the sketch again to figure out how to ˆ write it using the unit vectors ri of your coordinate system. Every region is different, there are no cooked formulas → use your sketch and your knowledge of geometry for this key step! Ex: We are sweeping over our plate with the IPs x and z. If we move a tiny bit dx and a tiny bit dz along the plate (at the same time), we sweep out a tiny rectangular area dx dz. So that’s our differential!of area: dA = dx dz. If the problem required the vector differential dA ,!we know that it must point ˆ perpendicular to the surface, so we would write dA = dx dz y . B2. Write the form of the full integral including the bounds on uj The integral will have n integral signs, where n is the dimensionality of . Each integral will sweep over one of your IP differentials duj within the bounds you found before. Ex: The total charge Q on the plate is " plate a b 0 0 ! dA = " dx " dz ! Kxz 2 / y . B3. Applying your constraint functions, write the final integrand F ⊗ d in terms of ONLY the IPs uj, their differentials duj, and constants You apply all of your constraint equations at this step, to restrict the integrand to the region . Also, beware of any dot- or cross-products appearing as ⊗: calculate them here. Finally, by “constants”, what we mean here are “things that don’t vary over the integral” … as noted above, anything that does vary must be expressed entirely in terms of our IPs. Ex: The integrand is σ dA where the surface charge density is ! = Kxz 2 / y . Restricting this to the plate, we apply the constraint y = 1/3 to obtain the final a integrand 3Kxz dx dz . Our final integral for Q is thus: 2 b ! dx ! dz ! 3Kxz 0 2 0 Check: is everything that varies over the integral properly written in terms of our IPs x and z and nothing else but constants? Yes it is, so we’re good. C. Integrate! ’Nuff said. Just remember to sanity-check your final answer against the original problem! If you’re supposed to get a vector, you must have a vector at the end … if you’re supposed to get a constant number, you can’t have any variables left at the end … etc. 13.18 ...
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