Engineering electromagnetics by hayt pdf

For region 1, 0. Then The capacitance will be Qnet The region between the spheres is filled with a perfect dielectric. Concentric conducting spheres have radii of 1 and 5 cm. The potential of the inner sphere is 2V and that of the outer is -2V. Find: a V r : We use the general expression derived in Problem 7.

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. Integrate again to find: Let the volume charge density in Fig. Use a development similar to that of Sec. The solution is found from Eq. Using thirteen terms, and. The series converges rapidly enough so that terms after the sixth one produce no change in the third digit. Thus, quoting three significant figures, 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.

Functions of this form are called circular harmonics. The Biot- Savart method was used here for the sake of illustration. A current filament of 3ax A lies along the x axis. Each carries a current I in the az direction. The parallel filamentary conductors shown in Fig. For the finite-length current element on the z axis, as shown in Fig. Since the limits are symmetric, the integral of the z component over y is zero.

Let a filamentary current of 5 mA be directed from infinity to the origin on the positive z axis and then back out to infinity on the positive x axis. The problem asks you to find H at various positions. Before continuing, we need to know how to find H for this type of current configuration. The sketch below shows one of the slabs of thickness D oriented with the current coming out of the page. The problem statement implies that both slabs are of infinite length and width.

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. Reverse the current, and the fields, of course, reverse direction.

We are now in a position to solve the problem. This point lies within the lower slab above its midpoint. Thus the field will be oriented in the negative x direction. Referring to Fig. Since 0. 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. If we assume an infinite cylinder length, there will be no z dependence in the field, since as we lengthen the loop in the z direction, the path length over which the integral is taken increases, but then so does the enclosed current — by the same factor. Thus H would not change with z. There would also be no change if the loop was simply moved along the z direction.

Again, using the Biot-Savart law, we note that radial field components will be produced by individual current elements, but such components will cancel from two elements that lie at symmetric distances in z on either side of the observation point. 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. We therefore must conclude that the field outside is zero.

Between the cylinders, we are outside the inner one, so its field will not contribute. Inner and outer currents have the same magnitude. We can now proceed with what is requested: a PA 1. We obtain 2. A current filament on the z axis carries a current of 7 mA in the az direction, and current sheets of 0. We require that the total enclosed current be zero, and so the net current in the proposed cylinder at 4 cm must be negative the right hand side of the first equation in part b.

Symmetry does help significantly in this problem. As a consequence of this, we find that the net current in region 1, I1 see the diagram on the next page , is equal and opposite to the net current in region 4, I4. Also, I2 is equal and opposite to I3. H from all sources should completely cancel along the two vertical paths, as well as along the two horizontal paths.

Assuming the height of the path is. Therefore, H will be in the opposite direction from that of the right vertical path, which is the positive x direction. Answer: No. Reason: the limit of the area shrinking to zero must be taken before the results will be equal. The value of H at each point is given. Each curl component is found by integrating H over a square path that is normal to the component in question. Along each segment, the field is assumed constant, and so the integral is evaluated by summing the products of the field and segment length 4 mm over the four segments.

If so, what is its value? Their centers are at the origin. This problem was discovered to be flawed — I will proceed with it and show how. The reader is invited to explore this further. 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. A solid nonmagnetic conductor of circular cross-section has a radius of 2mm.

All surfaces must carry equal currents. Use an expansion in cartesian coordinates to show that the curl of the gradient of any scalar field G is identically equal to zero. Thus, using the result of Section 8. The solenoid shown in Fig. 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. Make use of Eq. Solve these equations perhaps with the help of an example given in Section 7. 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.

The wire carries a current of 6 mA, flowing in the az direction from B to C. A filamentary current of 15 A flows along the entire z axis in the az direction.

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.

We wish to find the force acting to split the outer cylinder, which means we need to evaluate the net force in one cartesian direction on one half of the cylinder. Since the outer cylinder is a two-dimensional current sheet, its field exists only just outside the cylinder, and so no force exists.

If this cylinder possessed a finite thickness, then we would need to include its self-force, since there would be an interior field and a volume current density that would spatially overlap.

Two infinitely-long parallel filaments each carry 50 A in the az direction. Find the force exerted on the: a filament by the current strip: We first need to find the field from the current strip at the filament location. A current of 6A flows from M 2, 0, 5 to N 5, 0, 5 in a straight solid conductor in free space. An infinite current filament lies along the z axis and carries 50A in the az direction.

The rectangular loop of Prob. Find the vector torque on the loop, referred to an origin: a at 0,0,0 : The fields from both current sheets, at the loop location, will be negative x-directed. This filament carries a current of 3 A in the ax direction. An infinite filament on the z axis carries 5 A in the az direction. Assume that an electron is describing a circular orbit of radius a about a positively-charged nucleus. The hydrogen atom described in Problem 16 is now subjected to a magnetic field having the same direction as that of the atom.

What are these decreases for the hydrogen atom in parts per million for an external magnetic flux density of 0. We first write down all forces on the electron, in which we equate its coulomb force toward the nucleus to the sum of the centrifugal force and the force associated with the applied B field.

With the field applied in the same direction as that of the atom, this would yield a Lorentz force that is radially outward — in the same direction as the centrifugal force. Finally, 1m e2 a 2 B 2. Calculate the vector torque on the square loop shown in Fig. So we must use the given origin. 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: a the magnetic flux density is 0.

Let its center lie on the z axis and let a dc current I flow in the az direction in the center conductor. Find a H everywhere: This result will depend on the current and not the materials, and is: I 1. Point P 2, 3, 1 lies on the planar boundary boundary separating region 1 from region 2. The core shown in Fig. A coil of turns carrying 12 mA is placed around the central leg.

We now have mmf The flux in the center leg is now In Problem 9. Using this value of B and the magnetization curve for silicon steel,. 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. The reluctance of each gap is now 0. 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. I will leave the answer at that, considering the lack of fine resolution in Fig. This field differs from H2 only by the negative x component, which is a non-issue since the component is squared when finding the energy density.

A toroidal core has a square cross section, 2. The currents return on a spherical conducting surface of 0. Find the inductance of the cone-sphere configuration described in Problem 9. The inductance is that offered at the origin between the vertices of the cone: 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. A rectangular coil is composed of turns of a filamentary conductor. Find the mutual inductance in free space between this coil and an infinite straight filament on the z axis if the four corners of the coil are located at a 0,1,0 , 0,3,0 , 0,3,1 , and 0,1,1 : In this case the coil lies in the yz plane.

Find the mutual inductance of this conductor system in free space: a the solenoid of Fig. We first find the magnetic field inside the conductor, then calculate the energy stored there. It may be assumed that the magnetic field produced by I t is negligible. The location of the sliding bar in Fig. The rails in Fig. Develop a function of time which expresses the ohmic power being delivered to the loop: First, since the field does not vary with y, the loop motion in the y direction does not produce any time-varying flux, and so this motion is immaterial.

Then D 1. 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.

Neglect fields outside the dielectric. The equation is thus not valid with these fields. A 10 GHz radar signal may be represented as a uniform plane wave in a sufficiently small region. Then the power factor is P. Note that in Problem Perfectly-conducting cylinders with radii of 8 mm and 20 mm are coaxial. The external and internal regions are non-conducting. The inner and outer dimensions of a copper coaxial transmission line are 2 and 7 mm, respectively.

The dielectric is lossless and the operating frequency is MHz. 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.

Most microwave ovens operate at 2. A good conductor is planar in form and carries a uniform plane wave that has a wavelength of 0.

The outer conductor thickness is 0. Use information from Secs. The coax is air-filled. 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 direction. The electric field is given by the appropriate form of Eq. We find the two components of Hs separately, using the two components of Es. Specifically, the x component of Es is associated with a y component of Hs , and the y component of Es is associated with a negative x component of Hs.

Similarly, a positive z component for E requires a negative y component for H. Therefore, 10! With the dielectric constant greater for x-polarized waves, the x component will lag the y component in time at the output. Suppose that the length of the medium of Problem Given the general elliptically-polarized wave as per Eq. What percentage of the incident power density is transmitted into the copper?

We need to find the reflection coefficient. A uniform plane wave in region 1 is normally-incident on the planar boundary separating regions 1 and 2.

There are two possible answers. First, using Eq. The field in region 2 is then constructed by using the resulting amplitude, along with the attenuation and phase constants that are appropriate for region 2.

Also, the intrinsic impedance Pi Try measuring that. A MHz uniform plane wave in normally-incident from air onto a material whose intrinsic impedance is unknown. Measurements yield a standing wave ratio of 3 and the appearance of an electric field minimum at 0.

A 50MHz uniform plane wave is normally incident from air onto the surface of a calm ocean. Within the limits of our good conductor approximation loss tangent greater than about ten , the reflected power fraction, using the formula derived in part a, is found to decrease with increasing frequency.

The transmitted power fraction thus increases. Calculate the fractions of the incident power that are reflected and trans- mitted.

The total electric field in the plane of the interface must rotate in the same direction as the incident field, in order to continually satisfy the boundary condition of tangential electric field continuity across the interface. Therefore, the reflected wave will have to be left circularly polarized in order to make this happen. The transmitted field will be right circularly polarized as the incident field for the same reasons.

A left-circularly-polarized plane wave is normally-incident onto the surface of a perfect conductor. Determine the standing wave ratio in front of the plate.

Repeat Problem A uniform plane wave is normally incident from the left, as shown. Thus, at 2. In this case we use 2. MathCad was used in both cases. The slabs are to be positioned parallel to one another, and the combination lies in the path of a uniform plane wave, normally-incident. The slabs are to be arranged such that the air spaces between them are either zero, one-quarter wavelength, or one-half wavelength in thickness.

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In this experiment, you will find out which materials let electricity flow through them conductors and which ones prevent electricity from flowing through them insulators. A solid conducting cylinder of 4-cm radius is centered within a rectangular conducting cylinder with a cm by cm cross-section. 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.

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. 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. Then The capacitance will be Qnet 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. 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.

In Fig. 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. 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: 0. The four distances and potentials are now substituted into the given equation:.

Using the method described in Problem 7. 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. 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. 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. 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.

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. This point lies within the lower slab above its midpoint.

Referring to Fig. Since 0. 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. 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. Inner and outer currents have the same magnitude.

We can now proceed with what is requested: a PA 1. 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.

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.

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. This leaves only the path segment that coindides with the axis, and that lying parallel to the axis, but outside. Their centers are at the origin. 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. A solid nonmagnetic conductor of circular cross-section has a radius of 2mm.

All surfaces must carry equal currents. Thus, using the result of Section 8. The simplest form in this case is that involving the inverse hyperbolic sine. 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.

Make use of Eq. Solve these equations perhaps with the help of an example given in Section 7. 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.

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. 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. The rectangular loop of Prob. 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. So we must use the given origin. Then M 0. At radii between the currents the path integral will enclose only the inner current so, 3. Find a H everywhere: This result will depend on the current and not the materials, and is: I 1.

The core shown in Fig. A coil of turns carrying 12 mA is placed around the central leg. We now have mmf 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. The reluctance of each gap is now 0.

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. A toroidal core has a square cross section, 2. The currents return on a spherical conducting surface of 0.

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. The rings are coplanar and concentric. We use the result of Problem 8. Now for the right hand side. The location of the sliding bar in Fig. The rails in Fig. Then D 1. Now B 2. 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. Thus 1.

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? 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. 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.