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Engineering Electromagnetics 8th Edition William H. Hayt Original Engineering Electromagnetics - 6th Edition [William H. Hayt,Buck Engineering Electromagnetics 6e Pdf. Version, [version]. Download, Stock, [quota]. Total Files, 1. File Size, Engineering Electromagnetics: William H. Hayt Professor Emeritus Buck Author. Solutions of engineering electromagnetics 6th edition william h hayt Solutions of engineering electromagnetics 6th edition william h.
University of Errata, by Chris Mack, chris lithoguru. While teaching out of this Engineering Electromagnetics ; Recommended text. Optional text. Joseph A.
Edminister, Shaum's Outlines on Hayt, John A. Buck, ISBN: Engineering Electromagnetics. Engineering Electromagnetics, by W. Applied Electromagnetism, by L. Shen and J. Homework No. EE ; To understand various engineering applications of electromagnetics Hayt W. Required: by William H. We would expect Hz outside to decrease as the Biot-Savart law would imply but the same amount of current is always enclosed no matter how far away the outer segment is. With our rectangular path set up as in part a, we have no path integral contributions from the two radial segments, and no contribution from the outside z-directed segment.
Find H everywhere: The construction is similar to that of the toroid of round cross section as done on p.
Inner and outer currents have the same magnitude. We can now proceed with what is requested: We obtain 2. A balanced coaxial cable contains three coaxial conductors of negligible resistance. Assume a solid inner conductor of radius a, an intermediate conductor of inner radius bi , outer radius bo , and an outer conductor having inner and outer radii ci and co , respectively. The intermediate conductor carries current I in the positive az direction and is at potential V0.
These expressions will be the current in each conductor divided by the appropriate cross-sectional area. The results are: I az Inner conductor: Calculate H at: A solid conductor of circular cross-section with a radius of 5 mm has a conductivity that varies with radius.
The value of H at each point is given. We use the approximation: Each curl component is found by integrating H over a square path that is normal to the component in question. The x component of the curl is thus: To do this, we use the result of Problem 8. First, construct the rectangle with one side along the z axis, and with the opposite side lying at any radius outside the cylinder. This leaves only the path segment that coindides with the axis, and that lying parallel to the axis, but outside.
The integral is: Let the surface have the ar direction: Their centers are at the origin. We integrate counter-clockwise around the strip boundary using the right-hand convention , where the path normal is positive ax. When evaluating the curl of G using the formula in spherical coordinates, only one of the six terms survives: Integrals over x, to complete the loop, do not exist since there is no x component of H.
The path direction is chosen to be clockwise looking down on the xy plane. A long straight non-magnetic conductor of 0. Here we use H outside the conductor and write: A solid nonmagnetic conductor of circular cross-section has a radius of 2mm.
The current density will be: This is part a over again, except we change the upper limit of the radial integration: This is entirely outside the current distribution, so we need B there: All surfaces must carry equal currents.
We proceed as follows: Thus, using the result of Section 8. We can then write: The simplest form in this case is that involving the inverse hyperbolic sine. Since we have parallel current sheets carrying equal and opposite currents, we use Eq. Compute the vector magnetic potential within the outer conductor for the coaxial line whose vector magnetic potential is shown in Fig. By expanding Eq. Use Eq. The initial velocity in x is constant, and so no force is applied in that direction.
We integrate once: Integrating a second time yields the z coordinate: Have 1 1 K. Make use of Eq. Solve these equations perhaps with the help of an example given in Section 7. We begin by visualizing the problem. The positions are then found by integrating vx and vy over time: A circular orbit can be established if the magnetic force on the particle is balanced by the centripital force associated with the circular path.
In either case, the centripital force must counteract the magnetic force. A rectangular loop of wire in free space joins points A 1, 0, 1 to B 3, 0, 1 to C 3, 0, 4 to D 1, 0, 4 to A.
This will be the vector sum of the forces on the four sides. Note that by symmetry, the forces on sides AB and CD will be equal and opposite, and so will cancel. Find the total force on the rectangular loop shown in Fig. Uniform current sheets are located in free space as follows: Find the magnitude of the total force acting to split the outer cylinder apart along its length: A planar transmission line consists of two conducting planes of width b separated d m in air, carrying equal and opposite currents of I A.
Take the current in the top plate in the positive z direction, and so the bottom plate current is directed along negative z. Find the force exerted on the: Compute the vector torque on the wire segment using: The rectangular loop of Prob.
Find the vector torque on the loop, referred to an origin: They will add together to give, in the loop plane: Assume that an electron is describing a circular orbit of radius a about a positively-charged nucleus. Calculate the vector torque on the square loop shown in Fig.
We observe two things here: So we must use the given origin. Find H in a material where: Then M 0.
At radii between the currents the path integral will enclose only the inner current so, 3. Find the magnitude of the magnetization in a material for which: Compute for: Find a H everywhere: This result will depend on the current and not the materials, and is: The normal component of H1 will now be: HT 2 tangential component of H2 at the boundary: The core shown in Fig. A coil of turns carrying 12 mA is placed around the central leg.
Find B in the: We now have mmf The air gap reluctance adds to the total reluctance already calculated, where 0. In Problem 9. Using this value of B and the magnetization curve for silicon. Using Fig.
A toroidal core has a circular cross section of 4 cm2 area. The mean radius of the toroid is 6 cm. There is a 4mm air gap at each of the two joints, and the core is wrapped by a turn coil carrying a dc current I1. I will use the reluctance method here.
The reluctance of each gap is now 0. We are not sure what to use for the permittivity of steel in this case, so we use the iterative approach. From Fig. Then, in the linear material, 1. This is still larger than the given value of.
The result of 0. A toroid is constructed of a magnetic material having a cross-sectional area of 2. There is also a short air gap 0. This is d 0. In this case we use the magnetization curve, Fig. A toroidal core has a square cross section, 2. The currents return on a spherical conducting surface of 0. We thus perform the line integral of H over a circle, centered on the z axis, and parallel to the xy plane: From Problem 9.
Second method: Use the energy computation of Problem 9. The core material has a relative permeability of A coaxial cable has conductor dimensions of 1 and 5 mm. Find the inductance per meter length: The interfaces between media all occur along radial lines, normal to the direction of B and H in the coax line.
B is therefore continuous and constant at constant radius around a circular loop centered on the z axis. In this case the coil lies in the yz plane. The rings are coplanar and concentric. We use the result of Problem 8. That solution is reproduced below: Determine this force: Now for the right hand side. The location of the sliding bar in Fig. Find the voltmeter reading at: Have 0. The rails in Fig. In this case, there will be a contribution to the current from the right loop, which is now closed.
Develop a function of time which expresses the ohmic power being delivered to the loop: This will be Id 0. Then D 1. Now B 2. We set the given expression for Jd equal to the result of part c to obtain: Find the total displacement current through the dielectric and compare it with the source current as determined from the capacitance Sec.
The parallel plate transmission line shown in Fig. Note that B as stated is constant with position, and so will have zero curl. We use the expression for input impedance Eq. The Hard Way: Thus 1.
Find L, C, R, and G for the line: A transmitter and receiver are connected using a cascaded pair of transmission lines. At the operating frequency, Line 1 has a measured loss of 0. The link is composed of 40m of Line 1, joined to 25m of Line 2. At the joint, a splice loss of 2 dB is measured. If the transmitted power is mW, what is the received power? The total loss in the link in dB is 40 0. Suppose a receiver is rated as having a sensitivity of -5 dBm — indicating the minimum power that it must receive in order to adequately interpret the transmitted data.
Consider a transmitter having an output of mW connected to this receiver through a length of transmission line whose loss is 0. What is the maximum length of line that can be used? From this result, we subtract the maximum dB loss to obtain the receiver sensitivity: For this impedance to equal 50 ohms, the imaginary parts must cancel. If so what are they?
At the input end of the line, a DC voltage source, V0 , is connected. Here, the line just acts as a pair of lossless leads to the impedance. In a circuit in which a sinusoidal voltage source drives its internal impedance in series with a load impedance, it is known that maximum power transfer to the load occurs when the source and load impedances form a complex conjugate pair.
The condition of maximum power transfer will be met if the input impedance to the line is the conjugate of the internal impedance. What average power is delivered to each load resistor? First, we need the input impedance. The parallel resistors give a net load impedance of 20 ohms. For the transmission line represented in Fig.
A ohm transmission line is 0. The line is operating in air with a wavelength of 0. Determine the average power absorbed by each resistor in Fig. The next step is to determine the input impedance of the 2. The power dissipated by the ohm resistor is now 1 V 2 1 Find s on both sections 1 and 2: A lossless transmission line is 50 cm in length and operating at a frequency of MHz.
Determine s on the transmission line of Fig. Note that the dielectric is air: With the length of the line at 2. To achieve this, the imaginary part of the total impedance of part c must be reduced to zero so we need an inductor. Continuing, for this value of L, calculate the average power: The power delivered to the load will be the same as the power delivered to the input impedance.
Using normalized impedances, Eq. A line drawn from the origin through this point intersects the outer chart boundary at the position 0. With a wavelength of 1. On the WTL scale, we add 0. A straight line is now drawn from the origin though the 0. A compass is then used to measure the distance between the origin and zin. This is close to the value of the VSWR, as we found earlier.
Problem This is close to the computed inverse of yL , which is 1. Now, the position of zL is read on the outer edge of the chart as 0. The point is now transformed through the line length distance of 1.
Drawing a line between this mark on the WTG scale and the chart center, and scribing the compass arc length on this line, yields the normalized input impedance. On the WTG scale, we read the zL location as 0. The distance is then 0. Using a compass, we set its radius at the distance between the origin and zL. What is s on the remainder of the line? This will be just s for the line as it was before. This would return us to the original point, requiring a complete circle around the chart one- half wavelength distance.
The distance from the resistor will therefore be: With the aid of the Smith chart, plot a curve of Zin vs. Then, using a compass, draw a circle beginning at zL and progressing clockwise to the positive real axis. The intersections of the lines and the circle give a total of 11 zin values.
The table below summarizes the results. A fairly good comparison is obtained. We mark this on the positive real axis of the chart see next page.
The load position is now 0. A line is drawn from the origin through this point on the chart, as shown.
We then scribe this same distance along the line drawn through the. A line is drawn from the origin through this location on the chart. Drawing a line from the chart center through this point yields its location at 0. Alternately, use the s scale at the bottom of the chart, setting the compass point at the center, and scribing the distance on the scale to the left.
This distance is found by transforming the load impedance clockwise around the chart until the negative real axis is reached. This distance in wavelengths is just the load position on the WTL scale, since the starting point for this scale is the negative real axis.
So the distance is 0. Transforming the load through this distance toward the generator involves revolution once around the chart 0. A line is drawn between this point and the chart center. This is plotted on the Smith chart below. We then set on the compass the distance between yL and the origin. The same distance is then scribed along the positive real axis, and the value of s is read as 2.
First we draw a line from the origin through zL and note its intersection with the WTG scale on the chart outer boundary. We note a reading on that scale of about 0. To this we add 0. A line drawn from the 0. This is at the zero position on the WTL scale. The load is at the approximate 0. The wavelength on a certain lossless line is 10cm. We begin by marking zin on the chart see below , and setting the compass at its distance from the origin.
First, use a straight edge to draw a line from the origin through zin , and through the outer scale. We read the input location as slightly more than 0. The line length of 12cm corresponds to 1.
Thus, to transform to the load, we go counter-clockwise twice around the chart, plus 0. A line is drawn to the origin from that position, and the compass with its previous setting is scribed through the line. A standing wave ratio of 2. Probe measurements locate a voltage minimum on the line whose location is marked by a small scratch on the line.
When the load is replaced by a short circuit, the minima are 25 cm apart, and one minimum is located at a point 7 cm toward the source from the scratch. Find ZL: This is a possible location for the scratch, which would otherwise occur at multiples of a half-wavelength farther away from that point, toward the generator.
As a check, I will do the problem analytically. At the point X, indicated by the arrow in Fig. With the short circuit removed, a voltage minimum is found 5cm to the left of X, and a voltage maximum is located that is 3 times voltage of the minimum. Use the Smith chart to determine: Again, no Smith chart is needed, since s is the ratio of the maximum to the minimum voltage amplitudes.
Now we need the chart. This point is then transformed, using the compass, to the negative real axis, which corresponds to the location of a voltage minimum. On the chart, we now move this distance from the Vmin location toward the load, using the WTL scale. A line is drawn from the origin through the 0. We begin with the general phasor voltage in the line: We use the same equation for V z , which in this case reads: With a short circuit replacing the load, a minimum is found at a point on the line marked by a small spot of puce paint.
The 1m distance is therefore 3. Therefore, with the actual load installed, the Vmin position as stated would be 3. This being the case, the normalized load impedance will lie on the positive real axis of the Smith chart, and will be equal to the standing wave ratio.
The Smith chart construction is shown on the next page. This point is to be transformed to a location at which the real part of the normalized admittance is unity.
The stub is connected at either of these two points. The stub input admittance must cancel the imaginary part of the line admittance at that point. This point is marked on the outer circle and occurs at 0.
The length of the stub is found by computing the distance between its input, found above, and the short-circuit position stub load end , marked as Psc. The length of the main line between its load and the stub attachment point is found on the chart by measuring the distance between yL and yin2 , in moving clockwise toward generator.
In this case, everything is the same, except for the load- end position of the stub, which now occurs at the Poc point on the chart. This occurs at 0. The attachment point is found by transforming yL to yin1 , where the former point is located at 0. The lossless line shown in Fig. For the line to be matched, it is required that the sum of the normalized input admittances of the shorted stub and the main line at the point where the stub is connected be unity. So the input susceptances of the two lines must cancel.
This line is one-quarter wavelength long, so the normalized load impedance is equal to the normalized input admittance. The Smith chart construction is shown below. To cancel the input normalized susceptance of We therefore write 2. The two-wire lines shown in Fig. In this case, we have a series combination of the loaded line section and the shorted stub, so we use impedances and the Smith chart as an impedance diagram.
The requirement for matching is that the total normalized impedance at the junction consisting of the sum of the input impedances to the stub and main loaded section is unity. In the transmission line of Fig. First, the load voltage is found by adding voltages along the right side of the voltage diagram at the indicated times. The load voltage as a function of time is found by accumulating voltage values as they are read moving up along the right hand boundary of the chart.
The pulse amplitudes are calculated as follows: In the charged line of Fig. This problem accompanies Example Plots of the voltage and current at the resistor are then found by accumulating values from the left sides of the two charts, producing the plots as shown.
Plot the load resistor voltage as a function of time: With the left half of the line charged to V0 , closing the switch initiates at the switch location two voltage waves: The results are summarized as follows: A simple frozen wave generator is shown in Fig.
Determine and plot the load voltage as a function of time: Closing the switches sets up a total of four voltage waves as shown in the diagram below. With positive z travel, and with Es along positive x, Hs will lie along positive y. We use the fact that each to component of Es , there will be an orthogonal Hs component, oriented such that the cross product of Es with Hs gives the propagation direction. First, we note that if E at a given instant points in the negative x direction, while the wave propagates in the forward y direction, then H at that same position and time must point in the positive z direction.
We use the general formula, Eq. Using With Es in the positive y direction at a given time and propagating in the positive x direction, we would have a positive z component of Hs , at the same time.
Find the distance a uniform plane wave can propagate through the material before: In a non-magnetic material, we would have: Find both the power factor and Q in terms of the loss tangent: First, the impedance will be: I will demonstrate: We apply our equation to the result of part a: Note that in Problem Given, a MHz uniform plane wave in a medium known to be a good dielectric. Also, the specified distance in part f should be 10m, not 1km.
We use the good dielectric approximations, Eqs. Perfectly-conducting cylinders with radii of 8 mm and 20 mm are coaxial. From part a, we have 4. This will be 4.
The external and internal regions are non-conducting. We use J 1. Use The inner and outer dimensions of a copper coaxial transmission line are 2 and 7 mm, respec- tively.
The dielectric is lossless and the operating frequency is MHz. Calculate the resistance per meter length of the: A hollow tubular conductor is constructed from a type of brass having a conductivity of 1. The inner and outer radii are 9 mm and 10 mm respectively.
Calculate the resistance per meter length at a frequency of a dc: In this case the current density is uniform over the entire tube cross-section. We write: Therefore we can approximate the resistance using the formula: Most microwave ovens operate at 2.
Since the. A good conductor is planar in form and carries a uniform plane wave that has a wavelength of 0. Assuming the conductor is non-magnetic, determine the frequency and the conductivity: The outer conductor thickness is 0. Use information from Secs. The result is squared, terms collected, and the square root taken. Consider a left-circularly polarized wave in free space that propagates in the forward z direc- tion.
Since the wave propagates in the positive y direction and has equal x and z amplitudes, we identify the polarization as left circular. With the dielectric constant greater for x-polarized waves, the x component will lag the y component in time at the out- put.
Suppose that the length of the medium of Problem Describe the polarization of the output wave in this case: Given the general elliptically-polarized wave as per Eq. What percentage of the incident power density is transmitted into the copper?
We nevertheless proceed: Next we apply Eq. A uniform plane wave in region 1 is normally-incident on the planar boundary separating regions 1 and 2. There are two possible answers. Its intrinsic impedance is therefore approximated well by Eq. Solve for z to obtain ln 8. This is found using Eq. This is given by Eq. Pi Try measuring that. A MHz uniform plane wave in normally-incident from air onto a material whose intrinsic impedance is unknown. Determine the impedance of the unknown material: A 50MHz uniform plane wave is normally incident from air onto the surface of a calm ocean.
The transmitted power fraction thus increases. A left-circularly-polarized plane wave is normally-incident onto the surface of a perfect con- ductor. This is a standing wave exhibiting circular polarization in time. Determine the standing wave ratio in front of the plate. Repeat Problem We can now evaluate the phase shift for the three given cases. The slabs are to be positioned parallel to one another, and the combination lies in the path of a uniform plane wave, normally-incident. The slabs are to be arranged such that the air spaces between them are either zero, one-quarter wavelength, or one-half wavelength in thickness.
Specify an arrangement of slabs and air spaces such that a the wave is totally transmitted through the stack: In this case, we look for a combination of half-wave sections. Let the inter-slab distances be d1 , d2 , and d3 from left to right. Two possibilities are i. Thus every thickness is one-quarter wavelength. The impedances transform as follows: The 50MHz plane wave of Problem Next we need the angle of refraction, which means that we need to know the refractive index of seawater at 50MHz.
Therefore, for s polarization,. The fraction transmitted is then 0. The transmitted wave, while having all the incident p-polarized power, will have a reduced s-component, and so this wave will be right-elliptically polarized. A dielectric waveguide is shown in Fig. Find n0 in terms of n1 and n2: The prism of Fig. In the Brewster prism of Fig. The light is incident from air, and the returning beam also in air may be displaced sideways from the incident beam.
More than one design is possible here. Using the result of Example For this to work, the Brewster angle must be greater than or equal to the critical angle. Assume conductivity does not vary with frequency: In a good conductor: Over a small wavelength range, the refractive index of a certain material varies approximately.
All frequency components arrive simultaneously. Describe the pulse at the output of the second channel and give a physical explanation for what happened.
In fact, we may write in general: The pulse, if originally transform-limited at input, will emerge, again transform-limited, at its original width. Two aluminum-clad steel conductors are used to construct a two-wire transmission line.
The radius of the steel wire is 0. The dielectric is air, and the center-to-center wire separation is 4 in. Furthermore, the skin depth is considerably less than the aluminum layer thickness, so the bulk of the current resides in the aluminum, and we may neglect the steel.
Each conductor of a two-wire transmission line has a radius of 0. Pertinent dimensions for the transmission line shown in Fig.