電磁學題解ch

1、4-28Solutions for Chapter 4 Problems1.Current Continuity and Relaxation TimeP4.1: How long does it take for charge density to drop to 1% of its initial value in polystyrene?Polystyrene has er = 2.56 and s = 10-17 S/m.Solving for t we gett = 10.4x106 sec = 120 daysP4.2: At a particular point in a sla

2、b of silver, a charge density of 109 C/m3 is introduced. Plot rv versus time for a duration of 10 relaxation times.For silver we have er = 1 and s = 6.2x107 S/mThe relaxation time is: The plot is obtained with the following MATLAB routine.% MLP0402% Plot rhov vs time% for a charge placed in silverrh

3、ovo=1e9; %initial charge density, C/m3er=1; %relative permittivity of silversig=6.2e7; %conductivity, S/meo=8.854e-12;tau=er*eo/sig;t=tau/10:tau/10:10*tau;rho=rhovo*exp(-t./tau);subplot(2,1,1)plot(t,rho)xlabel(t (s)ylabel(rhov (C/m3)grid onsubplot(2,1,2)loglog(t,rho)xlabel(t (s)ylabel(rhov (C/m3)Fig

4、. P4.02grid onP4.3: A current density is given by J = re-.01tar A/m2. Find the charge density after 10 seconds if it has an initial value of zero.At t = 0, rv = 0 = 200 + C, C = -200.So we have P4.4: At t = 0 seconds, 60.0 mC is evenly distributed throughout a 2.00 cm diameter pure silicon sphere. (

5、a) Find the initial charge density. (b) How long does it take to drop to 10% of its initial value? (c) What will be the final surface charge density?(a) Volume (b) For pure silicon, (c) 2.Wave FundamentalsP4.5: A propagating electric field is given by(a) Determine the attenuation constant, the wave

6、frequency, the wavelength, the propagation velocity and the phase shift. (b) How far must the wave travel before its amplitude is reduced to 1.0 V/m?(a) (b) P4.6: A 10.0 MHz magnetic field travels in a fluid for which the propagation velocity is 1.0x108 m/sec. Initially, we have H(0,0)=2.0 ax A/m. T

7、he amplitude drops to 1.0 A/m after the wave travels 5.0 meters in the y direction. Find the general expression for this wave.The general expression for the wave is: Finally,P4.7: MATLAB: Modify the simple wave program in MATLAB 4.1 to include attenuation. Generate a plot for the case where the ampl

8、itude is 4 V/m, the attenuation constant is 0.001 Np/m, and the frequency is 1 MHz. Take your snapshot in time at 0 seconds, and let your phase shift be 0.% M-File: MLP0407% This is a modification of ML0401 that% includes attenuation. It plots a wave (in vac)% versus position (in z direction) for a

9、fixed time.% Wentworth, 1/19/03% Variables:% atten attenuation constant% Eo wave amplitude (V/m)% f frequency (Hz)% omega angular frequency (rad/sec)% t time snapshot (sec)% phi phase constant (degrees)% phir phase constant (radians)% c speed of light in vacuum (m/s)% lambda wavelength (m)% B phase

10、constant (1/m)% E electric field intensity% z positionclc %clears the command windowclear %clears variables% Initialize Variablesatten=0.001;Eo=4;f=1e6;t=0;phi=0;phir=phi*pi/180;c=2.998e8;lambda=c/f;B=2*pi/lambda;omega=2*pi*f;% Perform Calculationz=0:4*lambda/100:4*lambda;E=Eo*exp(-atten.*z).*cos(om

11、ega*t-B*z+phir);% Generate the Plotplot(z,E)axis(tight) %sets axes min & max data valuesgridxlabel(z(m)ylabel(E(V/m)Fig. P4.7P4.8: MATLAB: Modify the traveling wave program in MATLAB 4.2 to include attenuation. Use the parameters from problem P4.7, except for the fixed time of course.% M-File: MLP04

12、08% This program illustrates a traveling wave including% attenuation. It modifies ML0402.% Wentworth, 1/19/03% Variables:% atten attenuation (Np/m)% Eo wave amplitude (V/m)% f frequency (Hz)% omega angular frequency (rad/sec)% t time snapshot (sec)% phi phase constant (degrees)% phir phase constant

13、(radians)% c speed of light in vacuum (m/s)% lambda wavelength (m)% B phase constant (1/m)% E electric field intensity% z positionclc %clears the command windowclear %clears variables% Initialize Variablesatten=0.001Eo=4;f=1e6;t=1;phi=0;phir=phi*pi/180;c=2.998e8;lambda=c/f;B=2*pi/lambda;omega=2*pi*f

14、;% Perform Calculationz=0:4*lambda/100:4*lambda;E=Eo.*exp(-atten.*z).*cos(omega*t-B*z+phir);% Generate a Reference Frameplot(z,E)axis(0 4*lambda -2*Eo 2*Eo)gridxlabel(z(m)ylabel(E(V/m)pause% Make the Moviet=0:1/(40*f):1/f;for n=1:40; E=Eo.*exp(-atten.*z).*cos(omega*t(n)-B*z+phir); plot(z,E) axis(0 4

15、*lambda -2*Eo 2*Eo) grid title(General Wave Equation); xlabel(z(m); ylabel(E(V/m); M(:,1)=getframe;endFig. P4.8 (at end of animation)3. Faradays Law and Transformer EMFP4.9: The magnetic flux density increases at the rate of 10 (Wb/m2)/sec in the z direction. A 10 cm x 10 cm square conducting loop,

16、centered at the origin in the x-y plane, has 10 ohms of distributed resistance. Determine the direction (with a sketch) and magnitude of the induced current in the conducting loop.Fig. P4.9I=I=10mA clockwise (when viewedFrom +z)P4.10: A bar magnet is dropped through a conductive ring. Indicate in a

17、sketch the direction of the induced current when the falling magnet is just above the plane of the ring and when it is just below the plane of the ring, as shown in Figure 4.22.Refer to Figure P4.10.When the north pole first goes through the loop, flux is increasing and the current induced to oppose

18、 this change in flux is as shown.Fig. P4.10When the south pole is exiting the loop, flux is decreasing and the current induced acts to oppose this change in flux.P4.11: Considering Figure 4.7, suppose the area of a single loop of the pair is 100 cm2, and the magnetic flux density is constant over th

19、e area of the loops but changes with time as where Bo = 4.0 mWb/m2 and c = 0.30 Np/sec. Determine VR at 1, 10, and 100 seconds.at t = 1 sec, VR = 17.8 mVat t = 10 sec, VR = 1.20 mVat t = 100 sec, VR = 2.25x10-18 VP4.12: Sometimes a transformer is used as an impedance converter, where impedance is gi

20、ven by v/i. Find an expression for the impedance Z1 seen by the primary side of the transformer in Figure 4.11 that has a load impedance Z2 terminating the secondary.We have and P4.13: A 1.0 mm diameter copper wire is shaped into a square loop of side 4.0 cm. It is placed in a plane normal to a magn

21、etic field increasing with time as B = 1.0 t Wb/m2 az, where t is in seconds. (a) Find the magnitude of the induced current and indicate its direction in a sketch. (b) Calculate the magnetic flux density at the center of the loop resulting from the induced current, and compare this with the original

22、 magnetic flux density that generated the induced current at t = 1.0 sec.We find the distributed resistance of the loop and work the problem assuming this resistance is lumped in one spot as shown in the figure.(a) The induced current is Vemf divided by the distributed resistance of the wire loop.Fi

23、g. P4.13 (note that this answer has no time dependence)(b) The field at the center of the loop from a single arm of the loop is found from Eqn. (3.7):So P4.14: The mean length around a nickel core of a transformer like the one shown in Figure 4.11 is 16 cm, and its cross sectional area is 1 cm2. The

24、re are 30 turns on the primary side and 45 on the secondary side. If the current on the primary side is 1.0 sin20px106t mA, (a) calculate the amplitude of the magnetic flux in the core in the absence of the output windings. (b) With the output windings in place, calculate i2.(a) (b) P4.15: A triangu

25、lar wire loop has its vertices at the points (2, 0, 0), (0, 3, 0) and (0, 0, 4), with dimensions in meters. A time-varying magnetic field is given by B = 4t ay Wb/m2 (t in seconds). If the wire has a total distributed resistance of 2 W, calculate the induced current and indicate its direction in a c

26、arefully drawn sketch.Fig. P4.15Vemf = -(4)(4)=-16V4.Faradays Law and Motional EMFP4.16: Referring to Figure 4.23, suppose a conductive bar of length h = 2.0 cm moves with velocity u = -1.0 m/s ar towards an infinite length line of current I = 4.0 A. Find an expression for the voltage from one end o

27、f the bar to the other when r reaches 10 cm and indicate which end is positive.In Figure P4.16, an imaginary circuit has been chosen. For the chosen circulation direction, we have the sign for Vemf as shown. Then, Therefore, the bottom of the bar is positive.Fig. P4.16P4.17: Suppose we have a conduc

28、tive bar moving along a pair of conductive rails as in Figure 4.12, only now the magnetic flux density is B = 4.0ax + 3.0az Wb/m2. If R = 10. W, w = 20. cm, and uy = 3.0 m/s, calculate the current induced and indicate its direction. (clockwise when viewed from the +z axis)P4.18: The radius r of a pe

29、rfectly conducting metal loop in free space, situated in the x-y plane, increases at the rate of (pr)-1 m/sec. A break in the loop has a small 2.0 ohm resistor across it. Meanwhile, there exists a magnetic field B = 1.0 az T. Determine the current induced in the loop, and show in a sketch the direct

30、ion of flow.Here weve assumed dS = -dSaz to get iind and Vemf as shown. Our approach will be to find f, then Vemf = -df/dt.Fig. P4.18P4.19: Rederive Vemf for the rectangular loop of Figure 4.16 if the magnetic field is now B = Boaz.Fig. P4.19aWe see in Figure P4.19a that for the line sections.Fig. P

31、4.19bFor the section we have:, so for the section, the contributions to Vemf cancel. This will also be the case for the section, and therefore Vemf = 0; no current is induced.P4.20: In Figure 4.16, replace the rectangular loop with a circular one of radius a and rederive Vemf.P4.21: A conductive rod

32、, of length 6.0 cm, has one end fixed on a grounded origin and is free to rotate in the x-y plane. It rotates at 60 revolutions per second in a magnetic field B = 100. mT az. Find the voltage at the end of the bar.Fig. P4.21We can confirm the sign by observing that a positive charge placed in the mi

33、ddle of the bar would move to the ungrounded end by the Lorentz force equation.P4.22: Consider the rotating conductor shown in Figure 4.24. The center of the 2a diameter bar is fixed at the origin, and can rotate in x-y plane with B = Boaz. The outer ends of the bar make conductive contact with a ri

34、ng to make one end of the electrical contact to R; the other contact is made to the center of the bar. Given Bo = 100. mWb/m2, a = 6.0 cm, and R = 50. W, determine I if the bar rotates at 1.0 revolution per second.Figure P4.22 indicates one of the paths for the circulation integral.Fig. P4.22P4.23:

35、A Faraday Disk Generator is similar to the rotating conductor of P4.22, only now the rotating element is a disk instead of a bar. Derive an expression of the Vemf produced by a Faraday Disk Generator, and using the parameters given in problem 4.22, find I.Worked exactly as P4.22.P4.24: Consider a sl

36、iding rail problem where the conductive rails expand as they progress in the y direction as shown in Figure 4.25. If w = 10. cm and the distance between the rails increases at the rate of 1.0 cm in the x direction per 1.0 cm in the y direction, and uy = 2.0 m/sec, find the Vemf across a 100. W resis

37、tor at the instant when y = 10. cm if the field is Bo = 100. mT.First we modify the figure so that the top rail is horizontal and all the spreading occurs via the bottom rail. As before, our approach will be to find f and then df /dt. We have:Now, notice that x and y are not independent and are in f

38、act related: x=y+wSo we haveAlternate Method:Fig. P4.245.Displacement CurrentP4.25: Suppose a vector field is given asVerify that the divergence of the curl of this vector field is equal to zero.P4.26: Suppose a vector field is given byVerify that the divergence of the curl of this vector field is e

39、qual to zero.P4.27: A pair of 60 cm2 area plates are separated by a 2.0 mm thick layer of ideal dielectric characterized by er = 9.0. If a voltage v(t) = 1.0 sin (2px103t) V is placed across the plates, determine the displacement current.P4.28: Plot the loss tangent of seawater (s = 4 S/m and er = 8

40、1) versus log of frequency from 1 Hz to 1 GHz. At what frequency is the magnitude of the displacement current density equal to the magnitude of the conduction current density?(Plot this in Figure P4.28)Solving for f when tand = 1: f = 890MHz% MLP0428clearclcfor i=1:9 for j=1:10 m=(i-1)*10+j; f(m)=j*

41、10(i-1); tand(m)=889e6/f(m); endendloglog(f,tand)xlabel(frequency (Hz)ylabel(loss tangent)grid onFig. P4.28P4.29: A 1.0 m long coaxial cable of inner conductor diameter 2.0 mm and outer conductor diameter 6.0 mm is filled with an ideal dielectric with er = 10.2. A voltage v(t) = 10.cos(6px106 t) mV

42、is placed on the inner conductor and the outer conductor is grounded. Neglecting fringing fields at the ends of the coax, find the displacement current between the inner and outer conductors.so Now evaluate id with the given parameters:7.Lossless TEM WavesP4.30: Suppose in free space that E(z,t) = 5

43、.0e-2zt ax V/m. Is the wave lossless? Find H(z,t).Since the wave has an attenuation term (e-2zt) it is clearly not lossless.This integral is solved by parts where we let We arrive at:P4.31: An electric field propagating in a lossless non-magnetic media is characterized byFind the wave amplitude, fre

44、quency, propagation velocity, wavelength, and the relative permittivity of the media. (b) Find H(y,t).Inserting the given values we find: P4.32: A magnetic field propagating in free space is given byFind f, b, l, and E(z,t).so inserting the given values we find: P4.33: Find the instantaneous express

45、ion for E for the magnetic field of problem P4.6. We can write this wave as , and can reinsert the values at the end.First we have Assuming a nonmagnetic medium, we have or er = 9. Now we evaluate the curl of H:So Solving for E:Now we can evaluate the variables:SoWe can use a trigonometric identity

46、to simplify this answer. Given thatwe can let where b=cosy, c=siny, and tany=c/b. Now, since b/c = 251/56, then y=12.6 and A=257. ThusThis result is easier to obtain using the phasor approach of the next section.P4.34: Given, at some point distant from a point source at the origin in free space,find

47、 frequency, phase constant and H(r,t).Assume we can ignore the first term in the far-field for large r, then we haveInserting the given values we findP4.35: In a lossless, non-magnetic media, the magnetic field at some point distant from a source at the origin is given byfind the relative permittivi

48、ty of the media, the frequency and phase constant of the wave, and E(r,t).In the far-field, we will ignore the first term (has 1/r) leaving us withNow, Plugging in the given values we arrive at:.P4.36: Suppose, in a non-magnetic medium of relative permittivity 3, thatDetermine b and H(y,t).Inserting

49、 the given values we arrive at:8.Time-Harmonic Fields and PhasorsP4.37: Show that We first show that :Expanding the left side we have:Now expanding the right side we have:So since we have :thenP4.38: Derive the differential phasor form of (a) Gausss Law, and (b) Amperes Circuit Law.P4.39: Find H(y,t

50、) in problem P4.31(b) using phasors.Plugging in the values from P4.31, we findP4.40: Find E(z,t) in problem P4.32 using phasors.Plugging in the values from P4.32, we findP4.41: In free space,Find H(z,t).The phasor version of E is: Inserting the appropriate values we then find:P4.42: Find H(y,t) in p

51、roblem P4.36 using phasors.In phasor form the field of P4.36 is: Now apply Inserting the appropriate values we then find:P4.43: MATLAB: In MATLAB 4.4, a polar plot of the phasor corresponded to a location on a sine wave for a particular time, at a fixed position in space. You can also make a polar p

52、hasor plot for a snapshot in time, where you change position. Modify MATLAB 4.4 to provide an animation of the phasor versus sine wave as you change the position.% M-File: MLP0443% This program modifies ML0404, generating a% position plot animation synchronized with% a polar plot.% Wentworth, 1/19/03% Variables:% Eo field amplitude (V/m)% E electric field intensity (V/m)% E E-field for movie% f frequency (Hz)% theta angle% theta1 angle for movie% T period