Homework 7 Due 18 November at 6:00 pm

Homework 7 Due 18 November at 6:00 pm 1. Maxwell s Equations Quasi-statics o a An air core, N turn, cylinrical solenoi of length an raius a, carries a current I Io cos t. a. Using Ampere s Law, etermine the magnetic flux ensity B an magnetic fiel intensity H for this solenoi, assuming that the solenoi is very long an there is no fringing. Note that, while you may know the answer to this question by inspection, show all steps. That is, set up an evaluate the line integral, fin the current enclose by the line, etc. b. Evaluate the time erivative of the magnetic flux ensity B t. B c. From the point form of Faraay s Law, we know that E t or 1E E r E E re E B r 1 r where we have r r r r t use the fact that the magnetic flux ensity B B ()to r rop the x an y irecte terms from the right han sie. We can use this information to also rop 1 E the same terms from the left han sie. re r r B. So far, we r t have shown that the electric fiel E can have, at most a or r irecte component. In fact, E E () r only because the electric fiel in this case has to be in the same irection as the currents that prouce the solenoi B fiel. Why this must be is a bit subtle, but maybe best seen by looking at what woul happen if the core material was iron instea of air. If that was the case, then we woul observe ey currents being create in the iron that mimic the current in the solenoi wining in an attempt to reuce the applie fiel as much as possible. Fiels an Waves I Homework 7 Fall 009 K. A. Connor November 009 1

Recall that currents inuce in the copper pipe or copper plate by the strong permanent magnet (emo one in class) we in a irection to prouce a fiel in the opposite irection. To qualitative show this for the solenoi, the en view of the solenoi has been reprouce below with some example ey currents shown as ashe lines. Also shown is a generic circular current inucing eies in a flat plate. Now, since there is only one component for E E () r, Faraay s Law simplifies to 1 r re r B t. Solve this equation for E E () r. Note that your solution must be a function of raius r an cannot be a constant like B since E must have a curl.. The solutions to parts a an c are the quasi-static fiel. However, to etermine the accuracy of this solution, we nee to use the E just foun to fin a correction to B. This correction to the original fiel must also be a function of r. To start this process, etermine the electric flux ensity D from E an then fin its time erivative D. This is the isplacement current ensity. t e. Use Ampere s Law, following the same steps as in part a, to fin the new contribution to the magnetic fiel intensity H. Note that, since the isplacement current ensity is istribute throughout the region 0 r a, evaluating the right han sie of Ampere s Law is a bit more complicate than it was for part a. Once you have the new H, also fin the new B. f. At this point, we can represent our fiel solutions as B B0 B an E E 1 where the subscripts inicate the orer of the corrections. For quasi-statics, we B shoul have that 1. Evaluate this ratio an etermine the range of B0 frequencies for which the quasi-static solution (i.e. B B an 0 E E ) is 1 reasonably accurate. Hint: There is an extensive iscussion of quasi-statics for two other geometries in the class notes, Unit 9, pages 17-4. Fiels an Waves I Homework 7 Fall 009 K. A. Connor November 009

. Plane Waves an Transmission Lines R1 VAMPL = 5 FREQ = 3GH V1 50 V T1 LOSSY V R 50 0 A transmission line is riven by a 5V, 3GHZ voltage source. Both the source an loa resistances are 50Ohms. The length of the line is 10cm. Now that we have aresse fining capacitance an inuctance, we shoul begin such a problem by first escribing the configuration of the transmission line. Assume that we have a microstrip line for which there are many, many tools available to etermine the line characteristics. The parameters of the line are as follows: r 4., ielectric thickness = 0.79375mm, the trace with is 1.5mm, the trace thickness is 0.035mm. A view of the line is shown below. Note that this shows a loss tangent, but we will assume lossless conitions. 0 a. Using this information an one of the impeance calculators foun on the Resources webpage, etermine the characteristic impeance Z o, the elay time t, the capacitance an inuctance per unit length for this line. You shoul fin that the line is matche to the source an loa. b. Use your answers to part a to set up the PSpice simulation shown above. Do timeomain analysis an show roughly 3-5 perios of the source an loa voltages. c. Write the phasor expressions for the voltage an current on the line an verify that your expression is consistent with your PSpice solution. Now, will use the information have on voltage an current to fin the electric an magnetic fiels that exist in the region between the plates. For this analysis, we will assume that the plates are infinite so that there is no fringing an the fiel structure is that of an ieal parallel plate. This approach will give us a goo answer for the mile of the fiel region, but probably not near the eges of the plates. We will first begin by fining the electric fiel. We will not fin the magnetic fiel irectly. Rather, we will fin it from the electric fiel. Fiels an Waves I Homework 7 Fall 009 K. A. Connor November 009 3

A sie view of the line: x. Assuming that the phasor voltage on the line creates an electric fiel in the insulator region between the plates, fin the electric fiel E. Note the coorinate system shown above an that this fiel is a uniform plane wave. e. From E fin D an then etermine the surface charge ensity S on each conuctor using the bounary conition on the normal component of D. f. Determine the intrinsic impeance of the ielectric meium. g. From your answers for E an, etermine the magnetic fiel intensity H h. Show that your answer for H is consistent with your expression for the current on the line (from part c) by evaluating the bounary conition for the tangential component of H. i. Using your expressions for E an H, etermine the average power ensity (Poynting vector) an then fin the total power flowing in the line (see below). j. Show that the total power flowing in the line is equal to the power elivere to the loa resistor from your transmission line analysis. Note: Because the configuration aresse here is not an ieal parallel plate structure, we woul have to solve for the fiels using a numerical metho to answer this question irectly. However, we can aress an equivalent ieal structure to see that the voltages an fiels are properly connecte. To answer parts i an j, use the expressions in table.1 of Ulaby for inuctance an capacitance per unit length for a parallel plate structure. Ajust the values for an w (keeping the value of the same) to obtain the ientical values for Z o, l, an c for the transmission line. Then answer the questions. Fiels an Waves I Homework 7 Fall 009 K. A. Connor November 009 4

Solution: Problem 1 h o a a. The integration path for Ampere s Law is shown above. The fiel outsie the solenoi is assume to be ero because the solenoi is infinitely long. The NI magnetic fiel is only in the -irection so that: H l H h I h enclose or NI o NI H an B oh b. The time erivative is B on I on Io sin t t t 1 re B c. The electric fiel is foun from r r. Multiply through by r an t integrate re B oni r r sin t an re r t r o NI sint or r o NI finally E sin t. Note that this fiel is not a constant, but increases linearly with raius. Look at the iscussions mentione from the class notes to see similar behavior.. The isplacement current ensity is D E r o oni o cos t t t e. We now have the first orer term E E 1 an woul like to fin the secon orer correction to the original magnetic fiel B B B 0 ue to the isplacement current ensity. This current is ifferent from the original current in the solenoi winings because it is istribute throughout the core region 0 r a. Thus, the integral will be a little more involve, although it is funamentally the same. First we raw a slightly ifferent integration path. Fiels an Waves I Homework 7 Fall 009 K. A. Connor November 009 5

h r a o The integration path now goes from the -axis to the raius r. H l H h H r h I h r NI o o () 0 () enclose cos tr or r o oni r o o NI H () r cost so we have that B () r cos t f. The ratio B r c oo 1 gives us where c is the spee of B0 a light. We also note that for the quasi-static solution to work, we see that the c imensions must be much smaller than a wavelength. a Fiels an Waves I Homework 7 Fall 009 K. A. Connor November 009 6

Problem a. Microstripline Parameters from http://www.mantaro.com/resources/impeance_calculator.htm#microstrip_impe ance Note that the impeance of the line is roughly 50 Ohms, so everything is matche. Fiels an Waves I Homework 7 Fall 009 K. A. Connor November 009 7

b. PSpice Simulation The transmission line parameters can be foun in the iagram above. The lossless assumption is use so that resistance an conuctance are both ero. The line length is 0.1m. 3.0V.0V 1.0V -0.0V -1.0V -.0V -3.0V 0s 0.1ns 0.ns 0.3ns 0.4ns 0.5ns 0.6ns 0.7ns 0.8ns 0.9ns 1.0ns V(R1:) V(T1:B+) Time 3.0V The output (re) shows the elay from the line. This can be eliminate from the plot by showing the same amount of time but later. The first plot shows us that the corresponing input an output peaks are as inicate by the arrows below..0v 1.0V -0.0V -1.0V -.0V -3.0V 10.0ns 10.1ns 10.ns 10.3ns 10.4ns 10.5ns 10.6ns 10.7ns 10.8ns 10.9ns 11.0ns V(R1:) V(T1:B+) Time Fiels an Waves I Homework 7 Fall 009 K. A. Connor November 009 8

j c. The phasor form of the voltage an current on the line. V() V e 5. e V I Z e j 5. e j. e j () 05. Z 50 o 50 o 110x10 9 6 1 lc ( 3x10 ). 7x10 10. 7 j Note that both V in an V out for the line in PSpice have amplitues equal to.5v which checks with this result.. Since the plate separation is 0.79375mm, the electric fiel in the center of the 5. structure is roughly equal to E e j e j x () 315. j j e. The electric flux ensity in the ielectric is D () 315. e 13. e so j that S D x () 13. o e with the positive charge on the top plate an negative on the bottom. o 377 f. 184 4. 315. j j g. Hy () e 0. 0135e h. The bounary conition is for the tangential component of the magnetic fiel j JS () 0. 0135e To check against the total current, we nee the equivalent parallel plate structure, evelope in the following. i. For a parallel plate transmission line, the parameters are given in Table -1 of w Ulaby, where c o, l l o an Zo. We can fin the w c w equivalent parallel plate structure to the microstripline we have been analying, by fixing at its present value an then solving for w an. The results are w=3.7mm an 6. o. Both parameters change in reasonable irections from the values we use above. Charges cover more area than just the original with of 1.5mm an there is fiel in regions with no ielectric so that the average value of shoul be smaller. These changes also prouce a new value for o 377 33. Using these numbers, we can multiply the current 6. ensity in part h by the new value of w to obtain j j. I () wj() (.). 37 0 0135e 0 05e as above. For the average power S E ensity P 1 315. ave. 0. The area between the plates is (3.7)(.8)=.96 ( 33) so that the power is P = 0.063 Watts. 1 (.) 5 j. From the original transmission line, the power is P 0. 063 which 50 agrees with the previous result. Fiels an Waves I Homework 7 Fall 009 K. A. Connor November 009 9 x o