Engineering electromagnetics 7th edition pdf


 

Engineering Electromagnetics - 7th Edition - William H. Hayt - Solution Manual. Arsh Khan. CHAPTER 1 Given the vectors M = −10ax + 4ay − 8az and N. Engineering Electromagnetics. Sixth Edition. William H. Hayt, Jr.. John A. Buck. Textbook Table of Contents. The Textbook Table of Contents is your starting. Thank you very much for reading engineering electromagnetics 7th edition. As you 7th edition William Hayt John A Buck DRILL PROBLEMS SOLUTION PDF.

Author:DARRON PITMON
Language:English, Spanish, Indonesian
Country:Solomon Islands
Genre:Art
Pages:446
Published (Last):06.07.2016
ISBN:713-8-41721-751-7
Distribution:Free* [*Register to download]
Uploaded by: ROBIN

77348 downloads 178815 Views 36.50MB PDF Size Report


Engineering Electromagnetics 7th Edition Pdf

Engineering electromagnetics / William H. Hayt, Jr., John A. Buck. — 8th ed. p. cm. . as a faculty member, and then became coauthor for the sixth and seventh . Engineering Electromagnetics 7th Edition William H. Hayt Solution Manual. The BookReader requires JavaScript to be enabled. Please check that your browser. engineering electromagnetics 7th edition william h engineering electromagnetics 7th edition pdf pdf. Engineering Electromagnetics - 7th Edition - William H.

A significant emerging area of research activity involves multiphysics processes, and contributions in this area are particularly Read more Applied Mathematical Modelling focuses on research related to the mathematical modelling of engineering and environmental processes, manufacturing, and industrial systems. A significant emerging area of research activity involves multiphysics processes, and contributions in this area are particularly encouraged. This influential publication covers a wide spectrum of subjects including heat transfer, fluid mechanics, CFD, and transport phenomena; solid mechanics and mechanics of metals; electromagnets and MHD; reliability modelling and system optimization; finite volume, finite element, and boundary element procedures; modelling of inventory, industrial, manufacturing and logistics systems for viable decision making; civil engineering systems and structures; mineral and energy resources; relevant software engineering issues associated with CAD and CAE; and materials and metallurgical engineering. Applied Mathematical Modelling is primarily interested in papers developing increased insights into real-world problems through novel mathematical modelling, novel applications or a combination of these. Papers employing existing numerical techniques must demonstrate sufficient novelty in the solution of practical problems. Papers on fuzzy logic in decision-making or purely financial mathematics are normally not considered. Research on fractional differential equations, bifurcation, and numerical methods needs to include practical examples. Population dynamics must solve realistic scenarios. Papers in the area of logistics and business modelling should demonstrate meaningful managerial insight. Submissions with no real-world application will not be considered.

What force per unit area does each sheet exert on the other? First, we recognize from symmetry that only a z component of E will be present. The superposition integral for the z component of E will be: For the charged disk of Problem 2. The development is as follows: Find E at the origin if the following charge distributions are present in free space: An electric dipole discussed in detail in Sec. Using rectangular coordinates, determine expressions for the vector force on a point charge of magnitude q: Find the equation of the streamline passing through the point 1,3, An empty metal paint can is placed on a marble table, the lid is removed, and both parts are discharged honorably by touching them to ground.

An insulating nylon thread is glued to the center of the lid, and a penny, a nickel, and a dime are glued to the thread so that they are not touching each other. The assembly is lowered into the can so that the coins hang clear of all walls, and the lid is secured. The outside of the can is again touched momentarily to ground. 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. Find D and E everywhere. As a result, the calculation is nearly the same as before, with the only change being the limits on the total charge integral: 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? This will be the same computation as in part b, except the gaussian surface now lies at 20 mm.

Find D everywhere: 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: This is. A spherical surface of radius 3 mm is centered at P 4, 1, 5 in free space. Use the results of Sec. 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: 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: 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. 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. 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: Since the charge density is uniform and is spherically-symmetric, the angular coordinates do not matter. Again, the angles do not matter because of the spherical symmetry. We write in general: 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. The net potential function for the two charges would in general be: We use the superposition integral form: A point charge Q is located at the origin.

The result may be checked by conversion to spherical coordinates. The potential is expressed in spherical, rectangular, and cylindrical coordinates respec- tively as: Substitute P directly to obtain: This will be just The electron therefore moves approximately along a streamline. Where does it leave the plate and in what direction is it moving at the time? Considering the result of part a, we would expect the exit to occur along the bottom edge of the plate.

Two point charges, 1 nC at 0, 0, 0. Use q. Under what conditions does the answer agree with Eq. We perform a line integral of Eq. Using Eq. Find the total stored energy by applying a Eq. Four 0. The latter is just the potential at the square center arising from the original four charges, times the new charge value, or 4. Find the total current passing through each of these surfaces: Assume that a uniform electron beam of circular cross-section with radius of 0.

The requirement is that we have constant current throughout the beam path. This is found using the equation of continuity: Final answer: Assuming that there is no transformation of mass to energy or vice-versa, it is possible to write a continuity equation for mass. If we assume that the cube is an incremental volume element, determine an approximate value for the time rate of change of density at its center.

We may write the continuity equation for mass as follows, also invoking the divergence theorem: In all three cases mentioned in part a, the conductance is one-half the original value if all dimensions are reduced by one-half. This is easily shown using the given formula for conductance. Using data tabulated in Appendix C, calculate the required diameter for a 2-m long nichrome wire that will dissipate an average power of W when V rms at 60 Hz is applied to it: Thus 3.

Show that the ratio of the current densities in the two materials is independent of a and b. The total current passing radially outward through the medium between the cylinders is 3 A dc. Find, in terms of the given parameters: A hollow cylindrical tube with a rectangular cross-section has external dimensions of 0.

Converting all measurements to meters, the tube resistance over a 1 m length will be: We know that E will be z-directed, but the conductivity varies with x.

We therefore expect no z variation in E, and also note that the line integral of E between the bottom and top plates must always give V0.

Find the equation of the conductor surface: Since E is at the perfectly-conducting surface, it will be normal to the surface, so we may write: There are two reasons for this: Setting the given potential function equal to 0 and 60 and simplifying results in: For each line charge, this will be: The potential at any point is now: We just set the potential exression of part a equal to V to obtain: At a certain temperature, the electron and hole mobilities in intrinsic germanium are given as 0.

If the electron and hole concentrations are both 2. With the electron and hole charge magnitude of 1. Electron and hole concentrations increase with temperature. A semiconductor sample has a rectangular cross-section 1. The material has electron and hole densities of 1. Atomic hydrogen contains 5. The origin lies in region 1. Since the magnitude is negative, the normal component points into region 1 from the surface. So in region 1: With the dielectric gone, re-calculate E, D, Q, and the energy stored in the capacitor.

In the absence of friction in removing the dielectric, the charge and energy have returned to the battery that gave it. Capacitors tend to be more expensive as their capacitance and maximum voltage, Vmax , increase. Which of these dielectrics will give the largest CVmax product for equal plate areas: Trying this with the given materials yields the winner, which is barium titanate.

With the battery left connected, the plates are moved apart to a distance of 10d. Determine by what factor each of the following quantities changes: Remains the same, since the battery is left connected. Therefore, E has decreased by a factor of 0.

As Q is reduced by 0. This is also consistent with D having been reduced by 0. D will, as usual, be x-directed, originating at the top plate and terminating on the bottom plate. The key here is that D will be constant over the distance between plates. This can be understood by considering the x-varying dielectric as constructed of many thin layers, each having constant permittivity. The permittivity changes from layer to layer to approximate the given function of x. The approximation becomes exact as the layer thicknesses approach zero.

Repeat Problem 6. The ordering of parameters is changed over that in Problem 6. Remains the same, since with the battery disconnected, the charge has nowhere to go. Therefore, V0 has increased by a factor of Calculate the capacitance per square meter of surface area if: A parallel-plate capacitor is made using two circular plates of radius a, with the bottom plate on the xy plane, centered at the origin. Potential V0 is on the top plate; the bottom plate is grounded.

Two coaxial conducting cylinders of radius 2 cm and 4 cm have a length of 1m. For a spherical capacitor, we know that: Show that the capacitance per unit length of a cylinder of radius a is zero: With reference to Fig. Two 16 copper conductors 1. A 2 cm diameter conductor is suspended in air with its axis 5 cm from a conducting plane. Let the potential of the cylinder be V and that of the plane be 0 V. Find the surface charge density on the: The cylinder will image across the plane, producing an equivalent two-cylinder problem, with the second one at location 5 cm below the plane.

This is a quick one if we have already solved 6. These dimensions are suitable for the drawing. The sketch is shown below. The capacitance is thus. 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. 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.

Engineering Electromagnetics 7th Edition William H. Hayt Solution Manual

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.

You might also like: THE C ANSWER BOOK 2ND EDITION

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: We write in general: 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. The net potential function for the two charges would in general be: We use the superposition integral form: A point charge Q is located at the origin. The result may be checked by conversion to spherical coordinates. The potential is expressed in spherical, rectangular, and cylindrical coordinates respec- tively as: Substitute P directly to obtain: This will be just The electron therefore moves approximately along a streamline.

Where does it leave the plate and in what direction is it moving at the time? Considering the result of part a, we would expect the exit to occur along the bottom edge of the plate.

Two point charges, 1 nC at 0, 0, 0. Use q. Under what conditions does the answer agree with Eq. We perform a line integral of Eq. Using Eq. Find the total stored energy by applying a Eq. Four 0. The latter is just the potential at the square center arising from the original four charges, times the new charge value, or 4.

Find the total current passing through each of these surfaces: Assume that a uniform electron beam of circular cross-section with radius of 0. The requirement is that we have constant current throughout the beam path. This is found using the equation of continuity: Final answer: Assuming that there is no transformation of mass to energy or vice-versa, it is possible to write a continuity equation for mass.

If we assume that the cube is an incremental volume element, determine an approximate value for the time rate of change of density at its center. We may write the continuity equation for mass as follows, also invoking the divergence theorem: In all three cases mentioned in part a, the conductance is one-half the original value if all dimensions are reduced by one-half. This is easily shown using the given formula for conductance.

Using data tabulated in Appendix C, calculate the required diameter for a 2-m long nichrome wire that will dissipate an average power of W when V rms at 60 Hz is applied to it: Thus 3. Show that the ratio of the current densities in the two materials is independent of a and b. The total current passing radially outward through the medium between the cylinders is 3 A dc. Find, in terms of the given parameters: A hollow cylindrical tube with a rectangular cross-section has external dimensions of 0.

Converting all measurements to meters, the tube resistance over a 1 m length will be: We know that E will be z-directed, but the conductivity varies with x. We therefore expect no z variation in E, and also note that the line integral of E between the bottom and top plates must always give V0.

Find the equation of the conductor surface: Since E is at the perfectly-conducting surface, it will be normal to the surface, so we may write: There are two reasons for this: Setting the given potential function equal to 0 and 60 and simplifying results in: For each line charge, this will be: The potential at any point is now: We just set the potential exression of part a equal to V to obtain: At a certain temperature, the electron and hole mobilities in intrinsic germanium are given as 0.

If the electron and hole concentrations are both 2.

With the electron and hole charge magnitude of 1. Electron and hole concentrations increase with temperature. A semiconductor sample has a rectangular cross-section 1. The material has electron and hole densities of 1. Atomic hydrogen contains 5. The origin lies in region 1. Since the magnitude is negative, the normal component points into region 1 from the surface. So in region 1: With the dielectric gone, re-calculate E, D, Q, and the energy stored in the capacitor.

In the absence of friction in removing the dielectric, the charge and energy have returned to the battery that gave it. Capacitors tend to be more expensive as their capacitance and maximum voltage, Vmax , increase. Which of these dielectrics will give the largest CVmax product for equal plate areas: Trying this with the given materials yields the winner, which is barium titanate. With the battery left connected, the plates are moved apart to a distance of 10d.

Determine by what factor each of the following quantities changes: Remains the same, since the battery is left connected. Therefore, E has decreased by a factor of 0. As Q is reduced by 0. This is also consistent with D having been reduced by 0.

D will, as usual, be x-directed, originating at the top plate and terminating on the bottom plate. The key here is that D will be constant over the distance between plates. This can be understood by considering the x-varying dielectric as constructed of many thin layers, each having constant permittivity. The permittivity changes from layer to layer to approximate the given function of x. The approximation becomes exact as the layer thicknesses approach zero.

Repeat Problem 6.

The ordering of parameters is changed over that in Problem 6. Remains the same, since with the battery disconnected, the charge has nowhere to go. Therefore, V0 has increased by a factor of Calculate the capacitance per square meter of surface area if: A parallel-plate capacitor is made using two circular plates of radius a, with the bottom plate on the xy plane, centered at the origin.

Potential V0 is on the top plate; the bottom plate is grounded. Two coaxial conducting cylinders of radius 2 cm and 4 cm have a length of 1m. For a spherical capacitor, we know that: Show that the capacitance per unit length of a cylinder of radius a is zero: With reference to Fig. Two 16 copper conductors 1. A 2 cm diameter conductor is suspended in air with its axis 5 cm from a conducting plane.

Let the potential of the cylinder be V and that of the plane be 0 V. Find the surface charge density on the: The cylinder will image across the plane, producing an equivalent two-cylinder problem, with the second one at location 5 cm below the plane. This is a quick one if we have already solved 6. These dimensions are suitable for the drawing. The sketch is shown below. The capacitance is thus. 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.

Applied Mathematical Modelling - Journal - Elsevier

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. 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.

Applied Mathematical Modelling

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.

Engineering Electromagnetics 7th Edition William H. Hayt Solution Manual

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:

Related Posts:


Copyright © 2019 nvrehs.info. All rights reserved.
DMCA |Contact Us