# 6.11 Electromagnetic Potentials

Our discussion of Faraday’s and Ampère’s laws revealed two aspects of the link between time-varying electric and magnetic fields. We now examine the implications of this interconnection on the electric scalar potential V and the vector magnetic potential A.

In the static case, Faraday’s law reduces to

∇×E=0, (static case)(6.62)

which states that the electrostatic field E is conservative. According to the rules of vector calculus, if a vector field E is conservative, it can be expressed as the gradient of a scalar. Hence, in Chapter 4, we defined E as

E=−∇V. (electrostatics)(6.63)

In the dynamic case, Faraday’s law is

∇×E=−∂B∂t.(6.64)

In view of the relation B=∇×A, Eq. (6.64) can be expressed as

∇×E=−∂∂t(∇×A),(6.65)

which can be rewritten as

∇×E+∂A∂t=0. (dynamic case)(6.66)

Let us for the moment define

E′=E+∂A∂t.(6.67)

Using this definition, Eq. (6.66) becomes

∇×E′=0.(6.68)

Following the same logic that led to Eq. (6.63) from Eq. (6.62), we define

E′=−∇V.(6.69)

Upon substituting Eq. (6.67) for E′ in Eq. (6.69) and then solving for E, we have

Equation (6.70) reduces to Eq. (6.63) in the static case.

When the scalar potential V and the vector potential A are known, E can be obtained from Eq. (6.70), and B can be obtained from

Next we examine the relations between the potentials, V and A, and their sources: the charge and current distributions ρ v and J in the time-varying case.

# 6.11.1 Retarded Potentials

Consider the situation depicted in Fig. 6-16. A charge distribution ρ v exists over a volume υ′ embedded in a perfect dielectric with permittivity ε. Were this a static charge distribution, then from Eq. (4.58a) the electric potential V(R) at an observation point in space specified by the position vector R would be

V(R)=1 4 π ε∫υ′ρ v(R i)R′d υ′,(6.72)

where R i denotes the position vector of an elemental volume Δ υ′ containing charge density ρ v(R i), and R′=|R−R i| is the distance between Δ υ′ and the observation point. If the charge distribution is time-varying, we may be tempted to rewrite Eq. (6.72) for the dynamic case as

V(R,t)=1 4 π ε∫υ′ρ v(R i,t)R′d υ′,(6.73)

but such a form does not account for “reaction time.” If V 1 is the potential due to a certain distribution ρ v 1 and ρ v 1 were to suddenly change to ρ v 2, it will take a finite amount of time before V 1 a distance R′ away changes to V 2. In other words, V(R,t) cannot change instantaneously. The delay time is equal to t′=R′/u p, where u p is the velocity of propagation in the medium between the charge distribution and the observation point. Thus, V(R,t) at time t corresponds to ρ v at an earlier time, that is, (t−t′). Hence, Eq. (6.73) should be rewritten as

and V(R,t) is appropriately called the retarded scalar potential. If the propagation medium is vacuum, u p is equal to the velocity of light c.

# Figure 6-16 Electric potential V(R) due to a charge distribution ρ v over a volume υ′.

Figure 6 16 Full Alternative Text

Similarly, the retarded vector potential A(R,t) is related to the distribution of current density J by

This expression is obtained by extending the expression for the magnetostatic vector potential A(R) given by Eq. (5.65) to the time-varying case.

# 6.11.2 Time-Harmonic Potentials

The expressions given by Eqs. (6.74) and (6.75) for the retarded scalar and vector potentials are valid under both static and dynamic conditions and for any type of time dependence of the source functions ρ v and J. Because V and A depend linearly on ρ v and J, and as E and B depend linearly on V and A, the relationships interconnecting all of these quantities obey the rules of linear systems. When analyzing linear systems, we can take advantage of sinusoidal-time functions to determine the system’s response to a source with arbitrary time dependence. As was noted in Section 1-7, if the time dependence is described by a (nonsinusoidal) periodic time function, it can always be expanded into a Fourier series of sinusoidal components, and if the time function is nonperiodic, it can be represented by a Fourier integral. In either case, if the response of the linear system is known for all steady-state sinusoidal excitations, the principle of superposition can be used to determine its response to an excitation with arbitrary time dependence. Thus, the sinusoidal response of the system constitutes a fundamental building block that can be used to determine the response due to a source described by an arbitrary function of time. The term time-harmonic is often used in this context as a synonym for “steady-state sinusoidal time-dependent.”

In this subsection, we derive expressions for the scalar and vector potentials due to time-harmonic sources. Suppose that ρ v(R i,t) is a sinusoidal-time function with angular frequency ω given by

ρ v(R i,t)=ρ v(R i)cos(ω t+ϕ).(6.76)

Phasor analysis, which was first introduced in Section 1-7 and then used extensively in Chapter 2 to study wave propagation on transmission lines, is a useful tool for analyzing time-harmonic scenarios. A time harmonic charge distribution ρ v(R i,t) is related to its phasor ρ˜v(R i) as

ρ v(R i,t)=R e ρ˜v(R i)e j ω t,(6.77)

Comparison of Eqs. (6.76) and (6.77) shows that in the present case ρ˜v(R i)=ρ v(R i)e j ϕ.

Next, we express the retarded charge density ρ v(R i,t−R′/u p) in phasor form by replacing t with (t−R′/u p) in Eq. (6.77):

ρ v(R i,t−R′/u p)=R e ρ˜v(R i)e j ω(t−R′/u p)=R e ρ˜v(R i)e−j ω R′/u p e j ω t=R e ρ˜v(R i)e−j k R′e j ω t,(6.78)

where

k=ω u p(6.79)

is called the wavenumber or phase constant of the propagation medium. (In general, the phase constant is denoted by the symbol “β”, but for lossless dielectric media, it is commonly denoted by the symbol “k” and called the wavenumber.) Similarly, we define the phasor V˜(R) of the time function V(R,t) according to

V(R,t)=R e V˜(R)e j ω t.(6.80)

Using Eqs. (6.78) and (6.80) in Eq. (6.74) gives

R e V˜(R)e j ω t=R e 1 4 π ε∫υ′ρ˜v(R i)e−j k R′R′e j ω t d υ′.(6.81)

By equating the quantities inside the square brackets on both sides of Eq. (6.81) and canceling the common e j ω t factor, we obtain the phasor-domain expression

For any given charge distribution, Eq. (6.82) can be used to compute V˜(R). Then the resultant expression can be used in Eq. (6.80) to find V(R,t). Similarly, the expression for A(R,t) given by Eq. (6.75) can be transformed into

A(R,t)=R e A˜(R)e j ω t(6.83)

with

where J˜(R i) is the phasor function corresponding to J(R i,t).

The magnetic field phasor H˜ corresponding to A˜ is given by

H˜=1 μ∇×A˜.(6.85)

Recall that differentiation in the time domain is equivalent to multiplication by j ω in the phasor domain, and in a nonconducting medium (J=0), so Ampère’s law given by Eq. (6.41) becomes

Hence, given a time-harmonic, current-density distribution with phasor J˜, Eqs. (6.84) to (6.86) can be used successively to determine both E˜ and H˜. The phasor vectors E˜ and H˜ also are related by the phasor form of Faraday’s law:

# Example 6-8: Relating E to H

In a nonconducting medium with ε=16 ε 0 and μ=μ 0, the electric field intensity of an electromagnetic wave is

E(z,t)=x ˆ 10 sin(10 10 t−k z) (V/m).(6.88)

Determine the associated magnetic field intensity H and find the value of k.

Solution: We begin by finding the phasor E˜(z) of E(z,t). Since E(z,t) is given as a sine function and phasors are defined in this book with reference to the cosine function, we rewrite Eq. (6.88) as

E(z,t)=x ˆ 10 cos(10 10 t−k z−π/2) (V/m)=R e E˜(z)e j ω t,(6.89)

with ω=10 10 (rad/s) and

E˜(z)=x ˆ 10 e−j k z e−j π/2=−x ˆ j 10 e−j k z.(6.90)

To find both H˜(z) and k, we will perform a “circle”: We will use the given expression for E˜(z) in Faraday’s law to find H˜(z); then we will use H˜(z) in Ampère’s law to find E˜(z), which we will then compare with the original expression for E˜(z); and the comparison will yield the value of k. Application of Eq. (6.87) gives

H˜(z) =−1 j ω μ∇×E˜=−1 j ω μ x ˆ y ˆ z ˆ∂/∂x∂/∂y∂/∂z−j 10 e−j k z 0 0=−1 j ω μ y ˆ∂∂z(−j 10 e−j k z)=−y ˆ j 10 k ω μ e−j k z.(6.91)

So far, we have used Eq. (6.90) for E˜(z) to find H˜(z), but k remains unknown. To find k, we use H˜(z) in Eq. (6.86) to find E˜(z):

E˜(z)=1 j ω ε∇×H˜=1 j ω ε−x ˆ∂∂z−j 10 k ω μ e−j k z=−x ˆ j 10 k 2 ω 2 μ ε e−j k z.(6.92)

Equating Eqs. (6.90) and (6.92) leads to

k 2=ω 2 μ ε,

or

k=ω μ ε=4 ω μ 0 ε 0=4 ω c=4×10 10 3×10 8=133(rad/m).(6.93)

With k known, the instantaneous magnetic field intensity is then given by

H(z,t)=R e H˜(z)e j ω t=R e−y ˆ j 10 k ω μ e−j k z e j ω t=y ˆ 0.11 sin(10 10 t−133 z) (A/m).(6.94)

We note that k has the same expression as the phase constant of a lossless transmission line [Eq. (2.49)].