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Book
Schaum's Outline of Electromagnetics, 5th Edition

by Mahmood Nahvi, Joseph A. Edminister

Tough test questions? Missed lectures? Not enough time?

Fortunately, there's Schaum's.

More than 40 million students have trusted Schaum's to help them succeed in the classroom and on exams. Schaum's is the key to faster learning and higher grades in every subject. Each Outline presents all the essential course information in an easy-to-follow, topic-by-topic format. You also get hundreds of examples, solved problems, and practice exercises to test your skills.

This Schaum's Outline gives you:

• Hundreds of supplementary problems to reinforce knowledge

• Concise explanations of all electromagnetic concepts

• Information on current density, capacitance, magnetic fields, inductance, electromagnetic waves, transmission lines, and antennas

• New section on transmission line parameters

• New section illustrating the use of admittance plane and Smith Chart

• New section on impedance transformation

• New chapter on sky waves, attenuation and delay effects in troposphere, line-of-sight propagation, and other relevant topics

• Support for all major textbooks for courses in electromagnetics

PLUS: Access to revised Schaums.com website with 24 problem-solving videos, and more.

Schaum's reinforces the main concepts required in your course and offers hundreds of practice questions to help you succeed. Use Schaum's to shorten your study time and get your best test scores!

Book Chapter
17. Antennas
Source – Schaum's Outline of Electromagnetics, 5th Edition

17. Antennas

Maxwell's equations as examined in Chapter 14 predict propagating plane waves in an unbounded source-free region. In this chapter the propagating waves produced by current sources or antennas are examined; in general, these waves have spherical wavefronts and direction-dependent amplitudes. Because free-space conditions are exclusively assumed throughout the chapter, the notation for the permittivity, permeability, propagation speed, and characteristic impedance of the medium can omit the subscript 0; likewise the wave number (phase shift constant) of the radiation will be written β = ωμϵ = ω/u.

Book Chapter
8. Capacitance and Dielectric Materials
Source – Schaum's Outline of Electromagnetics, 5th Edition

8. Capacitance and Dielectric Materials

Dielectric materials become polarized in an electric field, with the result that the electric flux density D is greater than it would be under free-space conditions with the same field intensity. A simplified but satisfactory theory of polarization can be obtained by treating an atom of the dielectric as two superimposed positive and negative charge regions, as shown in Fig. 8-1(a). Upon application of an E field, the positive charge region moves in the direction of the applied field and the negative charge region moves in the opposite direction. This displacement can be represented by an electric dipole moment, p = Qd, as shown in Fig. 8-1(c).

Figure 8-1  
08x01

For most materials, the charge regions will return to their original superimposed positions when the applied field is removed. As with a spring obeying Hooke's law, the work done in the distortion is recoverable when the system is permitted to go back to its original state. Energy storage takes place in this distortion in the same manner as with the spring.

A region Δv of a polarized dielectric will contain N dipole moments p. Polarization P is defined as the dipole moment per unit volume:

P=limΔ v → 0 NpΔv (C / m2)

This suggests a smooth and continuous distribution of electric dipole moments throughout the volume, which, of course, is not the case. In the macroscopic view, however, polarization P can account for the increase in the electric flux density, the equation being

D=ϵ0E+P

This equation permits E and P to have different directions, as they do in certain crystalline dielectrics. In an isotropic, linear material, E and P are parallel at each point, which is expressed by

P=χeϵ0E(isotropic material)

where the electric susceptibility χe is a dimensionless constant. Then,

D=ϵ0(1+χe)E=ϵ0ϵrE(isotropic material)

where ϵr ≡ 1+χe is also a pure number. Since D=ϵE (Section 4.4),

ϵr=ϵϵ0

whence ϵr is called the relative permittivity. (Compare Section 1.6.)

EXAMPLE 1.

Find the magnitudes of D and P for a dielectric material in which E = 0.15 MV/m and χe = 4.25.

Since ϵr=χe+1=5.25,

D=ϵ0 ϵr E=10−936π(5.25) (0.15×106)=6.96 μC / m2P=χeϵ0E=10−936π(4.25) (0.15×106)=5.64 μC / m2

Book Chapter
7. Electric Current
Source – Schaum's Outline of Electromagnetics, 5th Edition

7. Electric Current

Electric current is the rate of transport of electric charge past a specified point or across a specified surface. The symbol I is generally used for constant currents and i for time-variable currents. The unit of current is the ampere (1 A = 1 C/s; in the SI system, the ampere is the basic unit and the coulomb is the derived unit).

Ohm's law relates current to voltage and resistance. For simple dc circuits, I = V/R. However, when charges are suspended in a liquid or a gas, or where both positive and negative charge carriers are present with different characteristics, the simple form of Ohm's law is insufficient. Consequently, the current density J (A/m2) receives more attention in electromagnetics than does current I.

Book Chapter
3. Electric Field
Source – Schaum's Outline of Electromagnetics, 5th Edition

3. Electric Field

The concepts of electric force and field intensity were introduced in Chapter 1. This chapter elaborates further on those concepts and formulates them using vector notations, a necessary framework in electromagnetics. In doing so, it expands upon, and refers to, some examples and problems from Chapter 1.

Book Chapter
4. Electric Flux
Source – Schaum's Outline of Electromagnetics, 5th Edition

4. Electric Flux

With charge density defined as in Section 3.5, it is possible to obtain the net charge contained in a specified volume by integration. From

dQ=ρ dv(C)

it follows that

Q=∫vρ dv  (C)

In general, ρ will not be constant throughout the volume v.

EXAMPLE 1.

Find the charge in the volume defined by 1 ≤ r ≤ 2 m in spherical coordinates, if

ρ=5 cos2ϕr4 (C/m3)

By integration,

Q=∫0 2π∫0 π∫1  2(5 cos2 ϕr4) r2sin θ dr dθ dϕ=5π C

Book Chapter
14. Electromagnetic Waves
Source – Schaum's Outline of Electromagnetics, 5th Edition

14. Electromagnetic Waves

Some wave solutions to Maxwell's equations have already been encountered in the Solved Problems of Chapter 13. The present chapter will extend the treatment of electromagnetic waves. Since most regions of interest are free of charge, it will be assumed that charge density ρ = 0. Moreover, linear isotropic materials will be assumed, with D = ϵE, B = μH, and J = σE.

Book Chapter
6. Electrostatics: Work, Energy, and Potential
Source – Schaum's Outline of Electromagnetics, 5th Edition

6. Electrostatics: Work, Energy, and Potential

A charge Q experiences a force F in an electric field E. In order to maintain the charge in equilibrium, a force Fa must be applied in opposition (Fig. 6-1):

Figure 6-1  
06x01

Work is defined as a force acting over a distance. Therefore, a differential amount of work dW is done when the applied force Fa produces a differential displacement dl of the charge—that is, moves the charge through the distance dℓ = |dl|. Quantitatively,

dW=Fa⋅dl=−QE⋅dl

Note that when Q is positive and dl is in the direction of E, dW = −QE dℓ < 0,="" indicating="" that="">work was done by the electric field. [Analogously, the gravitational field of the earth performs work on a (positive) mass M as it is moved from a higher elevation to a lower one.] On the other hand, a positive dW indicates work done against the electric field (cf. lifting the mass M).

Component forms of the differential displacement vector are as follows:

dl=dxax+dyay+dzaz (Cartesian)dl=drar+r dϕaϕ+dzaz (cylindrical)dl=drar+r dθaθ+rsinθ dϕaϕ (spherical)

The corresponding expressions for dℓ were displayed in Section 2.6.

EXAMPLE 1.

An electrostatic field is given by E = (x/2 + 2y)ax + 2xay (V/m). Find the work done in moving a point charge Q = −20 μC(a) from the origin to (4, 0, 0) m, and (b) from (4, 0, 0) m to (4, 2, 0) m.

  1. The first path is along the x axis, so that dl = dxax.

    dW=− QE·d l=(20×10− 6) (x2+2y)dxW=(20×10− 6)∫0 4(x2+2y)dx=80 μJ

  2. The second path is in the ay direction, so that dl = dyay.

    W=(20×10− 6) ∫0 2​2x dy=320 μJ

Book Chapter
11. Forces and Torques in Magnetic Fields
Source – Schaum's Outline of Electromagnetics, 5th Edition

11. Forces and Torques in Magnetic Fields

A charged particle in motion in a magnetic field experiences a force at right angles to its velocity, with a magnitude proportional to the charge, the velocity, and the magnetic flux density. The complete expression is given by the cross product

F=QU×B

Therefore, the direction of a particle in motion can be changed by a magnetic field. The magnitude of the velocity, U, and consequently the kinetic energy, will remain the same. This is in contrast to an electric field, where the force F = QE does work on the particle and therefore changes its kinetic energy.

If the field B is uniform throughout a region and the particle has an initial velocity normal to the field, the path of the particle is a circle of a certain radius r. The force of the field is of magnitude F = |Q| UB and is directed toward the center of the circle. The centripetal acceleration is of magnitude ω2r = U2/r. Then, by Newton's second law,

|Q| UB=m U2r     or     r=mU|Q| B

Observe that r is a measure of the particle's linear momentum, mU.

EXAMPLE 1.

Find the force on a particle of mass 1.70 × 10−27 kg and charge 1.60 × 10−19 C if it enters a field B = 5 mT with an initial speed of 83.5 km/s.

Unless directions are known for B and U0, the particle's initial velocity, the force cannot be calculated. Assuming that U0 and B are perpendicular, as shown in Fig. 11-1,

F=|Q| UB=(1.60×10−19) (83.5×103) (5×10−3)=6.68×10−17 N
Figure 11-1  
11x01

EXAMPLE 2.

For the particle of Example 1, find the radius of the circular path and the time required for one revolution.

r=mU|Q| B=(1.70×10−27) (83.5×103)(1.60×10−19) (5×10−3)=0.177 mT=2πrU=13.3 μs

Book Chapter
5. Gradient, Divergence, Curl, and Laplacian
Source – Schaum's Outline of Electromagnetics, 5th Edition

5. Gradient, Divergence, Curl, and Laplacian

In electromagnetics we need indicators for how a field, whether a scalar or a vector, changes within a segment of space or integrates over that segment. In this chapter we present three operators for such purposes: gradient, divergence, and curl. The gradient provides a measure of how a scalar field changes. For vector fields we use the divergence and the curl. For convenience, we may start with the Cartesian coordinate system. (However, note that the above operators are definable and usable in all three coordinate systems.)

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