solutions manual engineering electromagnetics 7e by william h hayt jr and john a download pdf: engineering electromagnetics by william h hayt john a buck. Engineering Electromagnetics - 7th Edition - William H. Hayt -. Solution Engineering Electromagnetics, William Hayt, eBook - norinkgibipen.cf Engineering . Engineering Electromagnetics - 7th Edition - William H. Hayt - Solution Manual. Arsh Khan. CHAPTER 1 Given the vectors M = −10ax + 4ay − 8az and N.
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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. Hayt Jr. Buck, The current edition is now useful for a full-fledged two-semester course on electromagnetics and is as well-suited for the communications engineer as for the electrical engineer. Adam H. William Hayt. Download with Google Download with Facebook or download with email. Hayt and Jr John A. The capacitance is thus.
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These dimensions are suitable for the actual sketch if symmetry is considered. As a check, compute the capacitance per meter both from your sketch and from the exact formula. The two axes are displaced by 2.
As a check on the accuracy, compute the capacitance per meter from the sketch and from the exact expression: The drawing is shown below. Use of the drawing produces: A solid conducting cylinder of 4-cm radius is centered within a rectangular conducting cylinder with a cm by cm cross-section.
In this case NQ is the number of squares around the full perimeter of the circular conductor, or four times the number of squares shown in the drawing. NV is the number of squares between the circle and the rectangle, or 5. The inner conductor of the transmission line shown in Fig. The axes are displaced as shown. Some improvement is possible, depending on how much time one wishes to spend. For the coaxial capacitor of Problem 6.
From Problem 6. A two-wire transmission line consists of two parallel perfectly-conducting cylinders, each hav- ing a radius of 0. A V battery is connected between the wires. No, since the charge density is not zero. It is known that both Ex and V are zero at the origin. Our two equations are: Both plates are held at ground potential. The problem did not provide information necessary to determine this. No for V1 V2. Only V2 is, since it is given as satisfying all the boundary conditions that V1 does.
The others, not satisfying the boundary conditions, are not the same as V1. Consider the parallel-plate capacitor of Problem 7. Both plates are at ground potential. The planes are parallel, and so we expect variation in potential in the direction normal to them. Using the two boundary condtions, our general potential function can be written: From Eq.
Repeat Problem 7.
Engineering Electromagnetics By William Hayt Ebook
Therefore, the solutions to parts a and b are unchanged from Problem 7. The two conducting planes illustrated in Fig. The medium surrounding the planes is air. For region 1, 0. Use 1 dV 2. Then The capacitance will be Qnet If the inner sphere is at V and the outer sphere at 0 V: What resistance is measured between the two perfect conductors?
Two coaxial conducting cones have their vertices at the origin and the z axis as their axis. Cone A has the point A 1, 0, 2 on its surface, while cone B has the point B 0, 3, 2 on its surface. This is equivalent to the continuity of the voltage derivatives: This situation is the same as that of Fig. The solution is found from Eq. Using thirteen terms,. The series converges rapidly enough so that terms after the sixth one produce no change in the third digit.
The four sides of a square trough are held at potentials of 0, 20, , and 60 V; the highest and lowest potentials are on opposite sides. Find the potential at the center of the trough: Here we can make good use of symmetry. The solution for a single potential on the right side, for example, with all other sides at 0V is given by Eq. Since we want V at the center of a square trough, it no longer matters on what boundary each of the given potentials is, and we can simply write: In Fig.
Solve for the potential at the center of the trough: The series solution will be of the form: From part a, the radial equation is: Functions of this form are called circular harmonics.
Referring to Chapter 6, Fig. Construct a grid, 0. Work to the nearest volt: The drawing is shown below, and we identify the requested voltage as 38 V. Use the iteration method to estimate the potentials at points x and y in the triangular trough of Fig. Work only to the nearest volt: The result is shown below. The mirror image of the values shown occur at the points on the other side of the line of symmetry dashed line.
Use iteration methods to estimate the potential at point x in the trough shown in Fig. The result is shown below, where we identify the voltage at x to be 40 V.
Note that the potentials in the gaps are 50 V. Using the grid indicated in Fig. The voltages at the grid points are shown below, where VA is found to be 19 V. Conductors having boundaries that are curved or skewed usually do not permit every grid point to coincide with the actual boundary. Figure 6. The other two distances are found by writing equations for the circles: The four distances and potentials are now substituted into the given equation: Using the method described in Problem 7.
Again, work to the nearest volt: The result is shown below, with values for the original grid points underlined: Use a computer to obtain values for a 0. Work to the nearest 0. Values along the vertical line of symmetry are included, and the original grid values are underlined.
The Biot-Savart method was used here for the sake of illustration. I will work this one from scratch, using the Biot-Savart law. It is also possible to work this problem somewhat more easily by using Eq. Each carries a current I in the az direction. Find H at the origin: Determine the side length b in terms of a , such that H at the origin is the same magnitude as that of the circular loop of part a.
Applying Eq. We need an expression for H in cartesian coordinates. A disk of radius a lies in the xy plane, with the z axis through its center. Find H at any point on the z axis. The Biot-Savart law reads: Calculate H at P 0, 0, 3: Since the limits are symmetric, the integral of the z component over y is zero. Find H in spherical coordinates a inside and b outside the sphere. Three uniform cylindrical current sheets are also present: The sketch below shows one of the slabs of thickness D oriented with the current coming out of the page.
For example, if the sketch below shows the upper slab in Fig. Thus H will be in the positive x direction above the slab midpoint, and will be in the negative x direction below the midpoint. We are now in a position to solve the problem. Find H at: This point lies within the lower slab above its midpoint. Referring to Fig. Since 0. Consider this situation as illustrated in Fig. There sec. The only way to enclose current is to set up the loop which we choose to be rectangular such that it is oriented with two parallel opposing segments lying in the z direction; one of these lies inside the cylinder, the other outside.
The loop is now cut by the current sheet, and if we assume a length of the loop in z of d, then the enclosed current will be given by Kd A. Thus H would not change with z. There would also be no change if the loop was simply moved along the z direction.
Suppose the rectangular loop was drawn such that the outside z-directed segment is moved further and further away from the cylinder. 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.
Engineering Electromagnetics 7th Edition William H. Hayt Solution Manual
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.Find H at: Buck Textbook Table of. Two 16 copper conductors 1.
Find s on both sections 1 and 2: The loop is now cut by the current sheet, and if we assume a length of the loop in z of d, then the enclosed current will be given by Kd A. To browse Academia.
Their centers are at the origin. Since E is at the perfectly-conducting surface, it will be normal to the surface, so we may write: