Engineering electromagnetics 6th edition pdf

This section explains the PIN diode in detail. A light-sensitive diode that conducts current in one direction only. PIN photodiodes have a fast response time.

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A new co-author from Georgia Tech has come aboard for the sixth edition to help update the book. Designed for introductory courses in electromagnetics or electromagnetic field theory at the junior-level, and offered in departments of electrical engineering, the text is a widely respected, updated version that stresses fundamentals and problem solving and discusses the material in an understandable, readable way.

As in the previous editions, the book retains the scope and emphasis that have made the book very successful while updating all the problems. The Finite-Difference Time-domain FDTD method allows you to compute electromagnetic interaction for complex problem geometries with ease. The simplicity of the approach coupled with its far-reaching usefulness, create the powerful, popular method presented in The Finite Difference Time Domain Method for Electromagnetics.

This volume offers timeless applications and formulations you can use to treat virtually any material type and geometry. The Finite Difference Time Domain Method for Electromagnetics explores the mathematical foundations of FDTD, including stability, outer radiation boundary conditions, and different coordinate systems.

Elements of Electromagnetics. 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.

Specify an arrangement of slabs and air spaces such that a the wave is totally transmitted through the stack: In this case, we look for a combination of half- wave sections. Let the inter-slab distances be d1 , d2 , and d3 from left to right.

Two possibilities are i. Thus every thickness is one-quarter wavelength. The reflection coefficient for waves incident on the front slab thus gets close to unity, and approaches 1 as the number of slabs approaches infinity. The 50MHz plane wave of Problem Therefore, for s polarization,.

The fraction transmitted is then 0. Since the wave is circularly-polarized, the s-polarized component represents one-half the total incident wave power, and so the fraction of the total power that is reflected is.

The transmitted wave, while having all the incident p-polarized power, will have a reduced s-component, and so this wave will be right-elliptically polarized. A dielectric waveguide is shown in Fig. All subsequent reflections from the upper an lower boundaries will be total as well, and so the light is confined to the guide.

A Brewster prism is designed to pass p-polarized light without any reflective loss. The prism of Fig. In the Brewster prism of Fig. The light is incident from air, and the returning beam also in air may be displaced sideways from the incident beam. More than one design is possible here. Using the result of Example For this to work, the Brewster angle must be greater than or equal to the critical angle.

Using Eq. Over a certain frequency range, the refractive index of a certain material varies approximately linearly. The pulse will broaden and will acquire a frequency sweep chirp that is precisely linear with time. Additionally, a pulse of a given bandwidth will broaden by the same amount, regardless of what carrier frequency is used.