|Language:||English, Indonesian, Arabic|
|ePub File Size:||30.43 MB|
|PDF File Size:||9.26 MB|
|Distribution:||Free* [*Registration needed]|
Electromagnetics Hayt 7th Edition Solution Manual [PDF] [EPUB] A First Ross 8th Edition Solutions Manual A First Course in Probability Ross. Engineering electromagnetics 7th edition - william h. hayt - solution manual. 1. CHAPTER 1 Given the vectors M = −10ax + 4ay − 8az and. Engineering Electromagnetics 7th Edition William H. Hayt Solutions Manual Fundamentals of Engineering Electromagnetics David K. Cheng www. Javier.
Please visit my Blog to find the book you are looking for and download it for free. Engineering Electromagnetics. Sign In. Update Cancel. You dismissed this ad. The feedback you provide will help us show you more relevant content in the future. 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.
Study smarter, not harder.
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 The device is carefully disassembled with insulating gloves and tools.
All coins were insulated during the entire procedure, so they will retain their original charges: Again, since the coins are insulated, they retain their original charges. First, from part b, the point charge will now lie inside. We integrate over the surface to find: We just integrate the charge density on that surface to find the flux that leaves it. Determine the total flux leaving: The total flux through the cylindrical surface and the two end caps are, in this order: Of the 6 surfaces to consider, only 2 will contribute to the net outward flux.
These fluxes 28 The net outward flux becomes: Find D and E everywhere. From the symmetry of the configuration, we surmise that the field will be everywhere z-directed, and will be uniform with x and y at fixed z. As a result, the calculation is nearly the same as before, with the only change being the limits on the total charge integral: To find the charge we integrate: The gaussian surface is a spherical shell of radius 1 mm.
The enclosed charge is the result of part a. Does this indicate a continuous charge distribution? If so, find the charge density variation with r. This will be the same computation as in part b, except the gaussian surface now lies at 20 mm. Find D everywhere: Since the charge varies only with radius, and is in the form of a cylinder, symmetry tells us that the flux density will be radially-directed and will be constant over a cylindrical surface of a fixed radius.
Noting that the charges are spherically-symmetric, we ascertain that D will be radially-directed and will vary only with radius. Volume charge density is located as follows: A cube is defined by 1 x,y, z 1. We note that there is no z component of D, so there will be no outward flux contributions from the top and bottom surfaces. This is Q. State whether the divergence of the following vector fields is positive, negative, or zero: A spherical surface of radius 3 mm is centered at P 4, 1, 5 in free space.
Use the results of Sec.
Study smarter, not harder.
Using the appropriate surface integrals, evaluate the net outward flux of F2 through the closed surface consisting of the hemisphere of part a and its circular base in the xy plane. Show that div D is zero everywhere except at the origin.
Using the formula for divergence in spherical coordinates see problem 3. Find div D everywhere: Physical reasoning then 35 This is in the region where the second field expression is valid. The total surface charge should be equal and opposite to the total volume charge.
Repeat Problem 3. Evaluate the surface integral side for the corresponding closed surface: Note that since the x component of D does not vary with x, the outward fluxes from the front and back surfaces will cancel each other.
The same is true for the left 36 This leaves only the top and bottom surfaces, where the fluxes are: We use the divergence theorem and first evaluate the surface integral side. A little thought is in order here: We could just as well position the two points at the same z location and the problem would not change. Halfway along this line is a point of symmetry in the field make a sketch to see this.
This means that when starting from either point, the initial force will be the same. This is also found by going through the same procedure as in part a, but with the direction roles of P and Q reversed.
Note that with the positive-x-directed field, the expended energy in moving the charge from 0 to x would be negative. As a look ahead, we can show by taking its curl that E is conservative. We therefore expect the same answer for all three paths. We obtain: Note that the field is conservative, so we would get the same result by integrating along a two-segment path over x and y as shown: Since the charge density is uniform and is spherically-symmetric, the angular coordinates do not matter.
The potential function for r 0. Again, the angles do not matter because of the spherical symmetry. We write in general: We need to find a potential function for the combined charges which is zero at M.
The total potential function will be the sum of the three. Combining the integration constants, we obtain: Three identical point charges of 4 pC each are located at the corners of an equilateral triangle 0. How much work must be done to move one charge to a point equidistant from the other two and on the line joining them?
We choose a path along which motion occurs in one coordinate direction at a time. If the potential at the origin is V, find V at P 4, 1, 3: The net potential function for the two charges would in general be: Assume a zero reference at infinity.
We use the superposition integral form:Find the potential at the center of the trough: Again, the angles do not matter because of the spherical symmetry.
The attachment point is found by transforming yL to yin1 , where the former point is located at 0.
Tangential velocity is angular velocity times the perpendicular distance from the rotation axis. With the length of the line at 2. The ordering of parameters is changed over that in Problem 6. An asymmetric slab waveguide is shown in Fig. The result is shown below, with values for the original grid points underlined:
- AMERICAN HEADWAY 3 TEACHERS BOOK SECOND EDITION
- NCERT ACCOUNTANCY BOOK CLASS 11 SOLUTIONS PDF
- HUMAN ANATOMY AND PHYSIOLOGY MARIEB 8TH EDITION EBOOK
- BASIC MECHANICAL ENGINEERING BOOK BY VENUGOPAL
- CONTROL SYSTEMS ENGINEERING NISE 6TH EDITION PDF
- 8 MINUTE MEDITATION EBOOK
- ECOLOGY SECOND EDITION CAIN PDF
- HORACE KEPHART CAMPING AND WOODCRAFT DOWNLOAD
- AUTOMOBILE GEARBOX PDF