Electromagnetic Waves, Antennas, and RF Examples: From a 1D Wave to a Field

This companion article provides self-study exercises for EM01, 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

(1)

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

(2)

These ideas are the mathematical starting point for later radio-wave, antenna, and GPS analysis [1, 2, 3, 4].

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:

1. What are the independent variables?
2. Does the field return a scalar or a vector?
3. Which directions describe the vector itself?
4. Which coordinates describe where the field changes?

Keeping these questions separate avoids many early errors when reading electromagnetic wave notation.

Part I: Exercises

Exercise 1: Scalar field or vector field?

The figure below contrasts the two basic possibilities introduced in EM01.

Classify each expression as scalar-valued or vector-valued:

ψ(x,y)	= 3x − y,	(3)
T(x,y,z,t)	= T0 + az,	(4)
E(x,y)	= (2x)x + (y)y,	(5)
p(x,t)	= p0 cos(kx − ωt),	(6)
A(z,t)	= yA0 sin(kz − ωt).	(7)

For each vector field, identify the basis directions that appear explicitly.

Exercise 2: Evaluate a scalar field

Let

(8)

Find the field value at

(9)

Is the result a scalar or a vector?

Exercise 3: Evaluate a vector field and its magnitude

Let

(10)

At

(11)

find:

1. E_x, E_y, and E_z;
2. the vector E;
3. the magnitude |E|.

Exercise 4: A one-dimensional wave is already a field

Consider

(12)

Explain why this is a scalar field even though it depends on two independent variables.

Then evaluate u∕A when

(13)

Exercise 5: Snapshot versus time history

The two panels below show two different views of the same wave field.

For

(14)

find:

1. the spatial snapshot at t = 0;
2. the time history at x = 0;
3. the time history at a general fixed location x = x₀.

Explain in words what is held fixed in each case.

Exercise 6: Position vector and compact field notation

A point in space has coordinates

(15)

1. Write its position vector r.
2. Compute |r|.
3. Explain what the notation E(r,t) means in terms of x, y, and z.

Exercise 7: Uniform or nonuniform?

Classify each field as spatially uniform or spatially nonuniform at a fixed time:

E1(r,t)	= E0 cos(ωt)x,	(16)
E2(z,t)	= E0 cos(kz − ωt)x,	(17)
ψ3(x,y)	= 5,	(18)
ψ4(x,y)	= x2 + y2.	(19)

Which expressions can still vary with time even when they are spatially uniform?

Exercise 8: Field direction versus propagation direction

Consider

(20)

Identify:

1. the axis along which the field vector points;
2. the direction in which the wave pattern propagates;
3. the coordinate along which the phase varies;
4. whether the field and propagation directions are parallel or perpendicular.

Exercise 9: Evaluate a vector wave at selected phases

Let

(21)

Find E and |E| when the phase is

1. 0;
2. π∕2;
3. π;
4. 3π∕2.

Explain the physical meaning of a negative x component.

Exercise 10: Read vector components from notation

Consider

(22)

At

(23)

find:

1. all three components;
2. the complete vector;
3. the vector magnitude.

Exercise 11: A sensor samples a field locally

The figure below represents a spatially varying vector field and a sensor fixed at one location.

Suppose

(24)

A sensor is fixed at x = x₀.

1. Write the time history measured at the sensor.
2. What variable still changes in that measured expression?
3. Does this one sensor, by itself, provide the entire spatial field at one instant? Explain.

Exercise 12: Contours are not trajectories

A scalar field is

(25)

1. Find the field value at (3, 4).
2. Write the equation of the contour on which ψ = 25.
3. Describe the geometric shape of that contour.
4. Explain why the contour does not represent a material object or a field vector moving along that curve.

Exercise 13: Same field, different questions

Let

(26)

Find:

1. the spatial snapshot at t = 0;
2. the time history at (x,y) = (2, 3);
3. the field value at (x,y,t) = (2, 3,π∕(2ω));
4. whether ψ is scalar-valued or vector-valued.

Exercise 14: Synthesis - read a three-dimensional vector field

Consider the illustrative field

(27)

At an event where

(28)

find:

1. E_x, E_y, and E_z;
2. the complete vector E;
3. the magnitude |E|;
4. which spatial coordinate appears in the phase;
5. whether the field vector points along the same direction as the coordinate of variation at this event.

Repeat parts (a)–(c) when

(29)

Explain what this example teaches about separating vector direction from spatial dependence.

Part II: Complete Worked Solutions

Solution 1: Scalar field or vector field?

The classification depends on what each function returns, not on how many independent variables it has.

1. (30)

   returns one number, so it is a scalar field.

2. (31)

   returns one temperature value, so it is a scalar field.

3. (32)

   contains basis directions and returns a vector. It is a vector field with explicit x and y components.

4. (33)

   returns one pressure value, so it is a scalar field.

5. (34)

   is vector-valued. Its explicit basis direction is y.

Thus the number of coordinates in the argument does not decide whether a field is scalar or vector-valued.

Solution 2: Evaluate a scalar field

Substitute

(35)

into

(36)

Then

ψ(3, 4,−2)	= 2(3) − 4 + (−2)2	(37)
	= 6 − 4 + 4	(38)
	= 6.	(39)

Therefore

(40)

The result is a scalar because the field itself is scalar-valued.

Solution 3: Evaluate a vector field and its magnitude

The components are

(41)

At (1,−2, 2),

Ex	= 2,	(42)
Ey	= −(−2) = 2,	(43)
Ez	= 3(2) = 6.	(44)

Hence

(45)

The magnitude is

|E|	=	(46)
	=	(47)
	≈ 6.63.	(48)

Therefore

(49)

Solution 4: A one-dimensional wave is already a field

The function

(50)

assigns one scalar value u to every pair (x,t). It is therefore a scalar field on one spatial dimension plus time.

At phase

(51)

we have

	= cos	(52)
	= −.	(53)

Thus

(54)

The fact that u depends on two variables does not make it vector-valued.

Solution 5: Snapshot versus time history

Starting from

(55)

set t = 0 for a spatial snapshot:

(56)

This compares many positions at one instant.

Set x = 0 for a time history at the origin:

u(0,t)	= A cos(−ωt)	(57)
	= A cos(ωt),	(58)

because cosine is even. Thus

(59)

At a general fixed location x = x₀,

(60)

The snapshot holds time fixed. The time history holds position fixed.

Solution 6: Position vector and compact field notation

The position vector is

(61)

Its magnitude is

|r|	= m	(62)
	= m	(63)
	≈ 4.58 m.	(64)

Therefore

(65)

The notation

(66)

is compact notation for a vector field whose spatial dependence can be written as

(67)

The symbol r packages the three spatial coordinates into one position vector.

Solution 7: Uniform or nonuniform?

For

(68)

there is no spatial coordinate in the value, so the field is spatially uniform. It can still vary with time.

For

(69)

the value changes with z, so it is spatially nonuniform.

For

(70)

the value is the same everywhere, so the field is spatially uniform.

For

(71)

the value changes from point to point, so the field is spatially nonuniform.

Thus

(72)

while

(73)

Only E₁ among the spatially uniform examples explicitly varies with time.

Solution 8: Field direction versus propagation direction

The field is

(74)

The basis vector 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

(75)

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

(76)

(77)

These directions are perpendicular in this illustrative transverse wave.

Solution 9: Evaluate a vector wave at selected phases

The field is

(78)

where

(79)

At 𝜃 = 0,

(80)

so

(81)

At 𝜃 = π∕2,

(82)

At 𝜃 = π,

(83)

At 𝜃 = 3π∕2,

(84)

A negative x component means the vector points in the −x direction. It does not mean the vector magnitude is negative.

Solution 10: Read vector components from notation

The components are

(85)

At

(86)

we get

Fx	= 2 + 0.5 = 2.5,	(87)
Fy	= 2(−1) = −2,	(88)
Fz	= −3(0.5) = −1.5.	(89)

Thus

(90)

The magnitude is

|F|	=	(91)
	=	(92)
	≈ 3.54.	(93)

Therefore

(94)

Solution 11: A sensor samples a field locally

The full field is

(95)

At the fixed sensor position x = x₀,

(96)

The position x₀ 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.

Solution 12: Contours are not trajectories

The field is

(97)

At (3, 4),

ψ(3, 4)	= 32 + 42	(98)
	= 9 + 16	(99)
	= 25.	(100)

Thus

(101)

The contour ψ = 25 satisfies

(102)

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.

Solution 13: Same field, different questions

The scalar field is

(103)

At t = 0,

(104)

so the spatial snapshot is

(105)

At the fixed point (2, 3),

(106)

At

(107)

we have

(108)

so

(109)

The field is scalar-valued because it returns one number for each (x,y,t).

Solution 14: Synthesis - read a three-dimensional vector field

The field is

(110)

with

(111)

At 𝜃 = 0,

(112)

Therefore

(113)

and

(114)

Its magnitude is

(115)

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

(116)

we have

(117)

Thus

(118)

so

(119)

with

(120)

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.

Common mistakes

• 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.
• Mistake: confusing a negative vector component with a negative magnitude. Magnitude is nonnegative; the sign belongs to a component direction.
• Mistake: confusing the basis direction of a vector with the coordinate along which the field varies.
• Mistake: treating a spatial snapshot and a time history as the same graph. One holds time fixed; the other holds position fixed.
• Mistake: interpreting contour lines as trajectories. They only connect equal scalar-field values.
• Mistake: assuming one fixed sensor measures the entire spatial field at one instant.

What EM01E reinforces

The exercises reinforce the sequence

(121)

A scalar field returns one scalar value at each event. A vector field returns a vector with components such as

(122)

Its magnitude is

(123)

A field’s vector direction and its direction of spatial variation are separate ideas. A fixed sensor samples the field locally as

(124)

These concepts prepare the way for EM02, where the vector mathematics will be developed more systematically before Maxwell’s equations are introduced.

References