Electromagnetic Waves, Antennas, and RF Examples: From a 1D Wave to a Field
AntennasElectromagneticWaves2
2026-09-15 01:36:17
2026-09-15 01:36:17
Example
Electromagnetic Waves: From a 1D Wave to a Field
shorten title
electromagnetic waves
radio waves
scalar field
vector field
one-dimensional wave
spatial field
time-dependent field
electric field
plane wave
transverse wave
position vector
field snapshot
time history
exercises
worked solutions
GPS
RF
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\section*{Electromagnetic Waves, Antennas, and RF Examples: From a 1D Wave to a Field}
This companion article provides self-study exercises for EM01, \emph{From a 1D Wave to a Field}. The exercises remain deliberately within the conceptual scope of EM01: scalar and vector fields, spatial and temporal dependence, field components, field magnitude, snapshots, time histories, simple vector waves, and local measurements. Gradient, divergence, curl, dot products, cross products, and Maxwell's equations are reserved for later lessons.
All exercises are stated first. Complete worked solutions follow in Part II.
The central conceptual bridge is
\begin{equation}
\boxed{
u(t)
\longrightarrow
u(x,t)
\longrightarrow
\psi(x,y,z,t)
\longrightarrow
\mathbf E(x,y,z,t).}
\end{equation}
A scalar field assigns one scalar value to each point in its domain. A vector field assigns a vector to each point. A fixed-location sensor samples a field through time as
\begin{equation}
\boxed{\mathbf E(\mathbf r_0,t).}
\end{equation}
These ideas are the mathematical starting point for later radio-wave, antenna, and GPS analysis \cite{Griffiths2017,OpenStaxV2,FeynmanV2,MIT802}.
\section*{How to use this problem set}
Attempt all exercises in Part I before reading Part II. For every field expression, ask four questions in order:
\begin{enumerate}
\item What are the independent variables?
\item Does the field return a scalar or a vector?
\item Which directions describe the vector itself?
\item Which coordinates describe where the field changes?
\end{enumerate}
Keeping these questions separate avoids many early errors when reading electromagnetic wave notation.
\section*{Part I: Exercises}
\section*{Exercise 1: Scalar field or vector field?}
The figure below contrasts the two basic possibilities introduced in EM01.
\begin{center}
\includegraphics{EM01E_fig01_scalar_vs_vector_field.png}
\vspace{0.45em}
\textbf{Figure.}
A scalar field assigns one number to each point. A vector field assigns a vector with magnitude and direction to each point.
\end{center}
Classify each expression as scalar-valued or vector-valued:
\begin{align}
\psi(x,y)&=3x-y,\\
T(x,y,z,t)&=T_0+az,\\
\mathbf E(x,y)&=(2x)\hat{\mathbf x}+(y)\hat{\mathbf y},\\
p(x,t)&=p_0\cos(kx-\omega t),\\
\mathbf A(z,t)&=\hat{\mathbf y}A_0\sin(kz-\omega t).
\end{align}
For each vector field, identify the basis directions that appear explicitly.
\section*{Exercise 2: Evaluate a scalar field}
Let
\begin{equation}
\psi(x,y,z)=2x-y+z^2.
\end{equation}
Find the field value at
\begin{equation}
(x,y,z)=(3,4,-2).
\end{equation}
Is the result a scalar or a vector?
\section*{Exercise 3: Evaluate a vector field and its magnitude}
Let
\begin{equation}
\mathbf E(x,y,z)
=(2x)\hat{\mathbf x}
-(y)\hat{\mathbf y}
+(3z)\hat{\mathbf z}.
\end{equation}
At
\begin{equation}
(x,y,z)=(1,-2,2),
\end{equation}
find:
\begin{enumerate}
\item[(a)] $E_x$, $E_y$, and $E_z$;
\item[(b)] the vector $\mathbf E$;
\item[(c)] the magnitude $|\mathbf E|$.
\end{enumerate}
\section*{Exercise 4: A one-dimensional wave is already a field}
Consider
\begin{equation}
u(x,t)=A\cos(kx-\omega t).
\end{equation}
Explain why this is a scalar field even though it depends on two independent variables.
Then evaluate $u/A$ when
\begin{equation}
kx-\omega t=\frac{2\pi}{3}.
\end{equation}
\section*{Exercise 5: Snapshot versus time history}
The two panels below show two different views of the same wave field.
\begin{center}
\includegraphics{EM01E_fig02_snapshot_vs_time_history.png}
\vspace{0.45em}
\textbf{Figure.}
A spatial snapshot holds time fixed; a time history holds position fixed.
\end{center}
For
\begin{equation}
u(x,t)=A\cos(kx-\omega t),
\end{equation}
find:
\begin{enumerate}
\item[(a)] the spatial snapshot at $t=0$;
\item[(b)] the time history at $x=0$;
\item[(c)] the time history at a general fixed location $x=x_0$.
\end{enumerate}
Explain in words what is held fixed in each case.
\section*{Exercise 6: Position vector and compact field notation}
A point in space has coordinates
\begin{equation}
x=2\,\text{m},
\qquad
y=-1\,\text{m},
\qquad
z=4\,\text{m}.
\end{equation}
\begin{enumerate}
\item[(a)] Write its position vector $\mathbf r$.
\item[(b)] Compute $|\mathbf r|$.
\item[(c)] Explain what the notation $\mathbf E(\mathbf r,t)$ means in terms of $x$, $y$, and $z$.
\end{enumerate}
\section*{Exercise 7: Uniform or nonuniform?}
Classify each field as spatially uniform or spatially nonuniform at a fixed time:
\begin{align}
\mathbf E_1(\mathbf r,t)&=E_0\cos(\omega t)\hat{\mathbf x},\\
\mathbf E_2(z,t)&=E_0\cos(kz-\omega t)\hat{\mathbf x},\\
\psi_3(x,y)&=5,\\
\psi_4(x,y)&=x^2+y^2.
\end{align}
Which expressions can still vary with time even when they are spatially uniform?
\section*{Exercise 8: Field direction versus propagation direction}
Consider
\begin{equation}
\mathbf E(z,t)
=\hat{\mathbf x}E_0\cos(kz-\omega t).
\end{equation}
\begin{center}
\includegraphics{EM01E_fig03_field_vs_propagation_direction.png}
\vspace{0.45em}
\textbf{Figure.}
The field vector points along the $x$ direction while the phase varies and propagates along $z$.
\end{center}
Identify:
\begin{enumerate}
\item[(a)] the axis along which the field vector points;
\item[(b)] the direction in which the wave pattern propagates;
\item[(c)] the coordinate along which the phase varies;
\item[(d)] whether the field and propagation directions are parallel or perpendicular.
\end{enumerate}
\section*{Exercise 9: Evaluate a vector wave at selected phases}
Let
\begin{equation}
\mathbf E(z,t)
=\hat{\mathbf x}(12\,\text{V/m})\cos(kz-\omega t).
\end{equation}
Find $\mathbf E$ and $|\mathbf E|$ when the phase is
\begin{enumerate}
\item[(a)] $0$;
\item[(b)] $\pi/2$;
\item[(c)] $\pi$;
\item[(d)] $3\pi/2$.
\end{enumerate}
Explain the physical meaning of a negative $x$ component.
\section*{Exercise 10: Read vector components from notation}
Consider
\begin{equation}
\mathbf F(x,y,t)
=(x+t)\hat{\mathbf x}
+(2y)\hat{\mathbf y}
-(3t)\hat{\mathbf z}.
\end{equation}
At
\begin{equation}
x=2,
\qquad
y=-1,
\qquad
t=0.5,
\end{equation}
find:
\begin{enumerate}
\item[(a)] all three components;
\item[(b)] the complete vector;
\item[(c)] the vector magnitude.
\end{enumerate}
\section*{Exercise 11: A sensor samples a field locally}
The figure below represents a spatially varying vector field and a sensor fixed at one location.
\begin{center}
\includegraphics{EM01E_fig04_sensor_sampling_field.png}
\vspace{0.45em}
\textbf{Figure.}
A sensor fixed at $\mathbf r_0$ records the local field through time as $\mathbf E(\mathbf r_0,t)$.
\end{center}
Suppose
\begin{equation}
\mathbf E(x,t)
=\hat{\mathbf y}E_0\cos(kx-\omega t).
\end{equation}
A sensor is fixed at $x=x_0$.
\begin{enumerate}
\item[(a)] Write the time history measured at the sensor.
\item[(b)] What variable still changes in that measured expression?
\item[(c)] Does this one sensor, by itself, provide the entire spatial field at one instant? Explain.
\end{enumerate}
\section*{Exercise 12: Contours are not trajectories}
A scalar field is
\begin{equation}
\psi(x,y)=x^2+y^2.
\end{equation}
\begin{enumerate}
\item[(a)] Find the field value at $(3,4)$.
\item[(b)] Write the equation of the contour on which $\psi=25$.
\item[(c)] Describe the geometric shape of that contour.
\item[(d)] Explain why the contour does not represent a material object or a field vector moving along that curve.
\end{enumerate}
\section*{Exercise 13: Same field, different questions}
Let
\begin{equation}
\psi(x,y,t)=xy\cos(\omega t).
\end{equation}
Find:
\begin{enumerate}
\item[(a)] the spatial snapshot at $t=0$;
\item[(b)] the time history at $(x,y)=(2,3)$;
\item[(c)] the field value at $(x,y,t)=(2,3,\pi/(2\omega))$;
\item[(d)] whether $\psi$ is scalar-valued or vector-valued.
\end{enumerate}
\section*{Exercise 14: Synthesis - read a three-dimensional vector field}
Consider the illustrative field
\begin{equation}
\mathbf E(x,y,z,t)
=\hat{\mathbf x}(4\,\text{V/m})\cos(kz-\omega t)
+\hat{\mathbf y}(3\,\text{V/m})\sin(kz-\omega t).
\end{equation}
At an event where
\begin{equation}
kz-\omega t=0,
\end{equation}
find:
\begin{enumerate}
\item[(a)] $E_x$, $E_y$, and $E_z$;
\item[(b)] the complete vector $\mathbf E$;
\item[(c)] the magnitude $|\mathbf E|$;
\item[(d)] which spatial coordinate appears in the phase;
\item[(e)] whether the field vector points along the same direction as the coordinate of variation at this event.
\end{enumerate}
Repeat parts (a)--(c) when
\begin{equation}
kz-\omega t=\frac{\pi}{2}.
\end{equation}
Explain what this example teaches about separating vector direction from spatial dependence.
\section*{Part II: Complete Worked Solutions}
\section*{Solution 1: Scalar field or vector field?}
The classification depends on what each function returns, not on how many independent variables it has.
\begin{enumerate}
\item[(a)]
\begin{equation}
\psi(x,y)=3x-y
\end{equation}
returns one number, so it is a scalar field.
\item[(b)]
\begin{equation}
T(x,y,z,t)=T_0+az
\end{equation}
returns one temperature value, so it is a scalar field.
\item[(c)]
\begin{equation}
\mathbf E(x,y)=(2x)\hat{\mathbf x}+y\hat{\mathbf y}
\end{equation}
contains basis directions and returns a vector. It is a vector field with explicit $x$ and $y$ components.
\item[(d)]
\begin{equation}
p(x,t)=p_0\cos(kx-\omega t)
\end{equation}
returns one pressure value, so it is a scalar field.
\item[(e)]
\begin{equation}
\mathbf A(z,t)=\hat{\mathbf y}A_0\sin(kz-\omega t)
\end{equation}
is vector-valued. Its explicit basis direction is $\hat{\mathbf y}$.
\end{enumerate}
Thus the number of coordinates in the argument does not decide whether a field is scalar or vector-valued.
\section*{Solution 2: Evaluate a scalar field}
Substitute
\begin{equation}
x=3,
\qquad
y=4,
\qquad
z=-2
\end{equation}
into
\begin{equation}
\psi=2x-y+z^2.
\end{equation}
Then
\begin{align}
\psi(3,4,-2)
&=2(3)-4+(-2)^2\\
&=6-4+4\\
&=6.
\end{align}
Therefore
\begin{equation}
\boxed{\psi(3,4,-2)=6.}
\end{equation}
The result is a scalar because the field itself is scalar-valued.
\section*{Solution 3: Evaluate a vector field and its magnitude}
The components are
\begin{equation}
E_x=2x,
\qquad
E_y=-y,
\qquad
E_z=3z.
\end{equation}
At $(1,-2,2)$,
\begin{align}
E_x&=2,\\
E_y&=-(-2)=2,\\
E_z&=3(2)=6.
\end{align}
Hence
\begin{equation}
\boxed{
\mathbf E
=2\hat{\mathbf x}
+2\hat{\mathbf y}
+6\hat{\mathbf z}.}
\end{equation}
The magnitude is
\begin{align}
|\mathbf E|
&=\sqrt{2^2+2^2+6^2}\\
&=\sqrt{44}\\
&\approx6.63.
\end{align}
Therefore
\begin{equation}
\boxed{|\mathbf E|\approx6.63.}
\end{equation}
\section*{Solution 4: A one-dimensional wave is already a field}
The function
\begin{equation}
u(x,t)
\end{equation}
assigns one scalar value $u$ to every pair $(x,t)$. It is therefore a scalar field on one spatial dimension plus time.
At phase
\begin{equation}
kx-\omega t=\frac{2\pi}{3},
\end{equation}
we have
\begin{align}
\frac{u}{A}
&=\cos\left(\frac{2\pi}{3}\right)\\
&=-\frac12.
\end{align}
Thus
\begin{equation}
\boxed{u=-\frac{A}{2}.}
\end{equation}
The fact that $u$ depends on two variables does not make it vector-valued.
\section*{Solution 5: Snapshot versus time history}
Starting from
\begin{equation}
u(x,t)=A\cos(kx-\omega t),
\end{equation}
set $t=0$ for a spatial snapshot:
\begin{equation}
\boxed{u(x,0)=A\cos(kx).}
\end{equation}
This compares many positions at one instant.
Set $x=0$ for a time history at the origin:
\begin{align}
u(0,t)
&=A\cos(-\omega t)\\
&=A\cos(\omega t),
\end{align}
because cosine is even. Thus
\begin{equation}
\boxed{u(0,t)=A\cos(\omega t).}
\end{equation}
At a general fixed location $x=x_0$,
\begin{equation}
\boxed{u(x_0,t)=A\cos(kx_0-\omega t).}
\end{equation}
The snapshot holds time fixed. The time history holds position fixed.
\section*{Solution 6: Position vector and compact field notation}
The position vector is
\begin{equation}
\boxed{
\mathbf r
=2\hat{\mathbf x}
-1\hat{\mathbf y}
+4\hat{\mathbf z}\,\text{m}.}
\end{equation}
Its magnitude is
\begin{align}
|\mathbf r|
&=\sqrt{2^2+(-1)^2+4^2}\,\text{m}\\
&=\sqrt{21}\,\text{m}\\
&\approx4.58\,\text{m}.
\end{align}
Therefore
\begin{equation}
\boxed{|\mathbf r|\approx4.58\,\text{m}.}
\end{equation}
The notation
\begin{equation}
\mathbf E(\mathbf r,t)
\end{equation}
is compact notation for a vector field whose spatial dependence can be written as
\begin{equation}
\mathbf E(x,y,z,t).
\end{equation}
The symbol $\mathbf r$ packages the three spatial coordinates into one position vector.
\section*{Solution 7: Uniform or nonuniform?}
For
\begin{equation}
\mathbf E_1(\mathbf r,t)=E_0\cos(\omega t)\hat{\mathbf x},
\end{equation}
there is no spatial coordinate in the value, so the field is spatially uniform. It can still vary with time.
For
\begin{equation}
\mathbf E_2(z,t)=E_0\cos(kz-\omega t)\hat{\mathbf x},
\end{equation}
the value changes with $z$, so it is spatially nonuniform.
For
\begin{equation}
\psi_3(x,y)=5,
\end{equation}
the value is the same everywhere, so the field is spatially uniform.
For
\begin{equation}
\psi_4(x,y)=x^2+y^2,
\end{equation}
the value changes from point to point, so the field is spatially nonuniform.
Thus
\begin{equation}
\boxed{\mathbf E_1\text{ and }\psi_3\text{ are spatially uniform,}}
\end{equation}
while
\begin{equation}
\boxed{\mathbf E_2\text{ and }\psi_4\text{ are spatially nonuniform.}}
\end{equation}
Only $\mathbf E_1$ among the spatially uniform examples explicitly varies with time.
\section*{Solution 8: Field direction versus propagation direction}
The field is
\begin{equation}
\mathbf E(z,t)=\hat{\mathbf x}E_0\cos(kz-\omega t).
\end{equation}
The basis vector $\hat{\mathbf x}$ tells us that the field vector lies along the $x$ axis. Depending on the sign of the cosine, it points toward $+x$ or $-x$.
The phase
\begin{equation}
kz-\omega t
\end{equation}
has the right-moving form, so the pattern propagates toward increasing $z$.
The spatial coordinate appearing in the phase is $z$, so the phase varies along the $z$ direction.
Therefore
\begin{equation}
\boxed{\text{field direction: }\pm\hat{\mathbf x},}
\end{equation}
\begin{equation}
\boxed{\text{propagation direction: }+\hat{\mathbf z}.}
\end{equation}
These directions are perpendicular in this illustrative transverse wave.
\section*{Solution 9: Evaluate a vector wave at selected phases}
The field is
\begin{equation}
\mathbf E=\hat{\mathbf x}(12\,\text{V/m})\cos\theta,
\end{equation}
where
\begin{equation}
\theta=kz-\omega t.
\end{equation}
At $\theta=0$,
\begin{equation}
\cos0=1,
\end{equation}
so
\begin{equation}
\boxed{\mathbf E=+12\hat{\mathbf x}\,\text{V/m},\qquad |\mathbf E|=12\,\text{V/m}.}
\end{equation}
At $\theta=\pi/2$,
\begin{equation}
\boxed{\mathbf E=\mathbf 0,\qquad |\mathbf E|=0.}
\end{equation}
At $\theta=\pi$,
\begin{equation}
\boxed{\mathbf E=-12\hat{\mathbf x}\,\text{V/m},\qquad |\mathbf E|=12\,\text{V/m}.}
\end{equation}
At $\theta=3\pi/2$,
\begin{equation}
\boxed{\mathbf E=\mathbf 0,\qquad |\mathbf E|=0.}
\end{equation}
A negative $x$ component means the vector points in the $-x$ direction. It does not mean the vector magnitude is negative.
\section*{Solution 10: Read vector components from notation}
The components are
\begin{equation}
F_x=x+t,
\qquad
F_y=2y,
\qquad
F_z=-3t.
\end{equation}
At
\begin{equation}
x=2,
\qquad
y=-1,
\qquad
t=0.5,
\end{equation}
we get
\begin{align}
F_x&=2+0.5=2.5,\\
F_y&=2(-1)=-2,\\
F_z&=-3(0.5)=-1.5.
\end{align}
Thus
\begin{equation}
\boxed{
\mathbf F
=2.5\hat{\mathbf x}
-2\hat{\mathbf y}
-1.5\hat{\mathbf z}.}
\end{equation}
The magnitude is
\begin{align}
|\mathbf F|
&=\sqrt{(2.5)^2+(-2)^2+(-1.5)^2}\\
&=\sqrt{12.5}\\
&\approx3.54.
\end{align}
Therefore
\begin{equation}
\boxed{|\mathbf F|\approx3.54.}
\end{equation}
\section*{Solution 11: A sensor samples a field locally}
The full field is
\begin{equation}
\mathbf E(x,t)=\hat{\mathbf y}E_0\cos(kx-\omega t).
\end{equation}
At the fixed sensor position $x=x_0$,
\begin{equation}
\boxed{
\mathbf E(x_0,t)
=\hat{\mathbf y}E_0\cos(kx_0-\omega t).}
\end{equation}
The position $x_0$ is now a constant. Time $t$ remains the changing independent variable.
A single sensor therefore records a time history at one location. It does not, by itself, give the complete spatial field at one instant because values at other positions are not simultaneously measured.
\section*{Solution 12: Contours are not trajectories}
The field is
\begin{equation}
\psi(x,y)=x^2+y^2.
\end{equation}
At $(3,4)$,
\begin{align}
\psi(3,4)
&=3^2+4^2\\
&=9+16\\
&=25.
\end{align}
Thus
\begin{equation}
\boxed{\psi(3,4)=25.}
\end{equation}
The contour $\psi=25$ satisfies
\begin{equation}
\boxed{x^2+y^2=25.}
\end{equation}
This is a circle of radius $5$ centered at the origin.
The contour simply identifies all points where the scalar field has the same value. It is not, by itself, the path of a particle, a material object, or a vector arrow moving through the field.
\section*{Solution 13: Same field, different questions}
The scalar field is
\begin{equation}
\psi(x,y,t)=xy\cos(\omega t).
\end{equation}
At $t=0$,
\begin{equation}
\cos0=1,
\end{equation}
so the spatial snapshot is
\begin{equation}
\boxed{\psi(x,y,0)=xy.}
\end{equation}
At the fixed point $(2,3)$,
\begin{equation}
\boxed{\psi(2,3,t)=6\cos(\omega t).}
\end{equation}
At
\begin{equation}
t=\frac{\pi}{2\omega},
\end{equation}
we have
\begin{equation}
\omega t=\frac{\pi}{2},
\end{equation}
so
\begin{equation}
\boxed{\psi=6\cos\left(\frac{\pi}{2}\right)=0.}
\end{equation}
The field is scalar-valued because it returns one number for each $(x,y,t)$.
\section*{Solution 14: Synthesis - read a three-dimensional vector field}
The field is
\begin{equation}
\mathbf E
=\hat{\mathbf x}(4\,\text{V/m})\cos\theta
+\hat{\mathbf y}(3\,\text{V/m})\sin\theta,
\end{equation}
with
\begin{equation}
\theta=kz-\omega t.
\end{equation}
At $\theta=0$,
\begin{equation}
\cos0=1,
\qquad
\sin0=0.
\end{equation}
Therefore
\begin{equation}
E_x=4\,\text{V/m},
\qquad
E_y=0,
\qquad
E_z=0,
\end{equation}
and
\begin{equation}
\boxed{\mathbf E=4\hat{\mathbf x}\,\text{V/m}.}
\end{equation}
Its magnitude is
\begin{equation}
\boxed{|\mathbf E|=4\,\text{V/m}.}
\end{equation}
The only spatial coordinate appearing in the phase is $z$, so the field varies spatially along $z$ in this expression. At this event, the vector itself points along $x$, not along $z$.
At
\begin{equation}
\theta=\frac{\pi}{2},
\end{equation}
we have
\begin{equation}
\cos\left(\frac{\pi}{2}\right)=0,
\qquad
\sin\left(\frac{\pi}{2}\right)=1.
\end{equation}
Thus
\begin{equation}
E_x=0,
\qquad
E_y=3\,\text{V/m},
\qquad
E_z=0,
\end{equation}
so
\begin{equation}
\boxed{\mathbf E=3\hat{\mathbf y}\,\text{V/m},}
\end{equation}
with
\begin{equation}
\boxed{|\mathbf E|=3\,\text{V/m}.}
\end{equation}
The field direction can change even though the spatial dependence still enters through the coordinate $z$. This is exactly why vector direction and direction of spatial variation must be read separately from the notation.
\section*{Common mistakes}
\begin{itemize}
\item \textbf{Mistake:} deciding that a field is vector-valued merely because it depends on several variables. Scalar fields can depend on many spatial coordinates and time.
\item \textbf{Mistake:} confusing a negative vector component with a negative magnitude. Magnitude is nonnegative; the sign belongs to a component direction.
\item \textbf{Mistake:} confusing the basis direction of a vector with the coordinate along which the field varies.
\item \textbf{Mistake:} treating a spatial snapshot and a time history as the same graph. One holds time fixed; the other holds position fixed.
\item \textbf{Mistake:} interpreting contour lines as trajectories. They only connect equal scalar-field values.
\item \textbf{Mistake:} assuming one fixed sensor measures the entire spatial field at one instant.
\end{itemize}
\section*{What EM01E reinforces}
The exercises reinforce the sequence
\begin{equation}
\boxed{
u(x,t)
\longrightarrow
\psi(x,y,z,t)
\longrightarrow
\mathbf E(x,y,z,t).}
\end{equation}
A scalar field returns one scalar value at each event. A vector field returns a vector with components such as
\begin{equation}
\boxed{
\mathbf E
=E_x\hat{\mathbf x}
+E_y\hat{\mathbf y}
+E_z\hat{\mathbf z}.}
\end{equation}
Its magnitude is
\begin{equation}
\boxed{|\mathbf E|=\sqrt{E_x^2+E_y^2+E_z^2}.}
\end{equation}
A field's vector direction and its direction of spatial variation are separate ideas. A fixed sensor samples the field locally as
\begin{equation}
\boxed{\mathbf E(\mathbf r_0,t).}
\end{equation}
These concepts prepare the way for EM02, where the vector mathematics will be developed more systematically before Maxwell's equations are introduced.
\begin{thebibliography}{9}
\bibitem{Griffiths2017}
David J. Griffiths,
\emph{Introduction to Electrodynamics},
4th ed., Cambridge University Press, 2017.
\bibitem{OpenStaxV2}
Samuel J. Ling, Jeff Sanny, and William Moebs,
\emph{University Physics, Volume 2},
OpenStax, 2016,
chapters on electric fields, magnetic fields, and electromagnetic waves.
\bibitem{FeynmanV2}
Richard P. Feynman, Robert B. Leighton, and Matthew Sands,
\emph{The Feynman Lectures on Physics, Volume II},
Addison-Wesley, 1964,
chapters introducing electromagnetic fields.
\bibitem{MIT802}
Massachusetts Institute of Technology,
\emph{8.02 Physics II: Electricity and Magnetism},
MIT OpenCourseWare,
materials on electric and magnetic fields.
\bibitem{Crawford1968}
Frank S. Crawford, Jr.,
\emph{Waves},
Berkeley Physics Course, Volume 3,
McGraw-Hill, 1968.
\end{thebibliography}