# Electromagnetics in FE Electrical Exam

Electromagnetics forms the backbone of modern electrical engineering, and its mastery is essential for success in the Fundamentals of Engineering (FE) Electrical exam. According to the NCEES® FE Electrical exam course guidelines, electromagnetics is a crucial topic that aspiring engineers must understand.

In this study guide, we will delve into all the ins and outs of electromagnetics in FE electrical exam, explore its key concepts, and provide valuable insights to help you excel in this fundamental area of electrical engineering. Let’s discuss electromagnetic electromagnetism accounts necessary to ace the FE Electrical exam.

## Introduction to Electromagnetics

Electromagnetism is a fundamental branch of physics that deals with the interplay between electric and magnetic fields and the effects they produce. It encompasses a wide range of phenomena, including the behavior of charged particles, the generation of magnetic fields, and the propagation of electromagnetic waves.

The principles of electromagnetism find applications in numerous modern-day technologies, some of which include:

**Electrical Power Generation and Distribution**– Electromagnetic generators, such as alternators and dynamos, convert mechanical energy into electrical energy. This electricity is then transmitted and distributed across power grids, enabling the functioning of electrical appliances.**Electromagnetic Waves and Communication**– Electromagnetic waves, including radio waves, microwaves, infrared, visible light, ultraviolet, X-rays, and gamma rays, are used for wireless communication, radio and television broadcasting, satellite communication, and medical imaging.**Electric Motors and Transformers**– Electric motors utilize the interaction between electric and magnetic fields to convert electrical energy into mechanical energy, enabling their use in various applications, from household appliances to industrial machinery. Based on electromagnetic induction, transformers efficiently transfer electrical energy between different voltage levels.**Magnetic Resonance Imaging (MRI)**– MRI machines use strong magnetic fields and radio waves to generate detailed images of the internal structures of the human body, aiding in medical diagnosis.

## Electrostatics

Electrostatics is a branch of physics that studies stationary electric charges and the forces they exert on each other. It focuses on the behavior of electric fields, the concept of charge, and the principles governing static electricity.

### Basics of Electrostatics

Let’s discuss the fundamental concepts of electrostatics that you must know to prepare for electromagnetics in the FE Electrical exam.

#### Electrostatics vs. Electricity

**Electrostatics (Static Charge)**– Deals with stationary charges and the forces between them.**Electricity (Moving Charge)**– Focuses on the flow of charges, such as in electric currents.

#### Charge

A charge is a fundamental property of matter that can be positive or negative.

- A
**positive charge**represents an excess of protons or a deficiency of electrons. - A
**negative charge**represents an excess of electrons or a deficiency of protons.

Charges are quantized, meaning they exist in discrete elementary charge (e) units, where e ≈ 1.6 x 10^-19 Coulombs (C).

Mathematically,

**q = ne **(Where n is the number of protons or electrons).

#### Nature of Charges

**Like Charges**– Similar charges repel each other (e.g., positive-positive or negative-negative).**Unlike Charges**– Opposite charges attract each other (e.g., positive-negative).

### Coulomb’s Law

Coulomb’s Law describes the electrostatic force between two point charges. It states that the force is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.

**For charges in air/vacuum:**

F = (k * q1 * q2) / r^2

**For charges in a dielectric medium:**

F = (1 / (4πε)) * (q1 * q2) / r^2

Where,

- F: Electrostatic force between the charges.
- k: Coulomb’s constant ≈ 8.99 x 10^9 Nm^2/C^2.
- q1, q2: Magnitudes of the charges.
- r: Distance between the charges.
- ε: Permittivity of the medium. ε ≈ 8.85 x 10^-12 C^2/(Nm^2) for vacuum or air.

Consider two charges (positive and negative), each with a magnitude of 2.0 μC (micro coulombs), placed 40 cm apart. A negative charge of -4.0 μC is placed at their center. Let’s calculate the electric force on the negative charge.

As we have,

q1 = +2.0 μC = 2.0 x 10^-6 C

q2 = -2.0 μC = 2.0 x 10^-6 C

q3 = -4.0 μC = -4.0 x 10^-6 C

r = 40 cm = 0.4 m

**Using Coulomb’s Law:**

*Force on q3 by q1:*

*Since q3 lies at the center of both charges, we will take r= r/2 = 0.2 m*

F1 = (k * q1 * q3) / r^2

F1 = (8.99 x 10^9) * [(2.0 x 10^-6) * (4.0 x 10^-6)] / (0.2)^2

F1 = 1.798 N

*Force on q3 by q2:*

*Since q3 lies at the center of both charges, we will take r= r/2 = 0.2 m*

F2 = (k * q2 * q3) / r^2

F2 = (8.99 x 10^9) * [(2.0 x 10^-6) * (4.0 x 10^-6)] / (0.2)^2

F2 = 1.798 N

**Resultant Force:**

*Since q1 pulls the q3 (like charges attract each other) and q2 pushes the q3 (Unlike charges repel each other). Both forces are directed towards q1, so:*

F = F1 + F2

F = 3.596 N

Therefore, the electric force on the charge q3 is 3.596 N.

### Electric Field and Electric Field Intensity:

#### Electric Field

An electric field is a region where an electric charge experiences a force.

It is the force experienced per unit positive charge placed at a given point in space.

E = F/q

#### Electric Field Intensity

Electric Field Intensity (E) at a point is the force experienced by a test charge (q0) placed at that point.

The formula for electric field intensity due to a point charge (Q) is given by:

E = (k * Q) / r^2

**For positive charges:**

- E points radially outward from the charge (away from it).

**For negative charges:**

- E points radially inward toward the charge (toward it).

### Gauss’s Law and Electric Flux

#### Electric Flux

Electric flux (ΦE) represents the number of electric field lines passing through a closed surface. It depends on the strength of the electric field and the surface area and is defined as a dot product between the electric field and the area from which the electric field or electric line of forces are passing through.

Mathematically,

Φ = E. ΔA = EΔA cosθ

### Gauss’s Law

Gauss’s Law states that the total electric flux through a closed surface is proportional to the total charge enclosed by that surface.

Mathematically, Gauss’s Law is expressed as:

ΦE = (Qenclosed) / ε0

Where:

- ΦE: Electric flux.
- Qenclosed: Total charge enclosed by the closed surface.
- ε0: Permittivity of free space ≈ 8.85 x 10^-12 C^2/(Nm^2).

### Impact of Air and Dielectric Medium (Capacitor Use-Case)

#### Capacitor

A capacitor consists of two conductive plates separated by a distance (d). The presence of air or a dielectric material between the plates affects its behavior.

#### Capacitance with Air

The capacitance (C) of a capacitor with air as the dielectric is given by:

C = (ε0 * A) / d

Where:

- A: Area of the plates.
- d: Separation between the plates.
- B. Capacitance with Dielectric Medium:

The capacitance (C) of a capacitor with a dielectric medium of thickness (d) is given by:

C = (ε * ε0 * A) / d

Where:

- ε: Permittivity of the dielectric material.

Consider a capacitor with air between the plates. The area of each plate is 100 cm^2, and the separation between them is 2 mm. Let’s calculate the capacitance.

We have,

A = 100 cm^2 = 0.01 m^2

d = 2 mm = 0.002 m

**Using the formula:**

C = (ε0 * A) / d

C = (8.85 x 10^-12 C^2/(Nm^2)) * 0.01 m^2 / 0.002 m

C ≈ 4.43 x 10^-9 F (Farads)

Therefore, the capacitance of the capacitor with air as the dielectric is approximately 4.43 nanofarads.

## Magnetostatics

Magnetostatics is a branch of physics that studies magnetic fields and the interactions between magnetic fields and electric currents. It deals with stationary or slowly varying magnetic fields.

### Basics of Magnetostatics

Let’s discuss the fundamental concepts of magnetostatics that you must know to prepare for electromagnetics in FE Electrical exam.

#### Magnetic Force

The magnetic force is experienced by a moving charged particle or a current-carrying conductor in a magnetic field. It acts perpendicular to the velocity or current direction and the magnetic field direction.

#### Poles

**North Pole**– The north pole of a magnet is the end that points toward the Earth’s north pole when the magnet is freely suspended.**South Pole**– The south pole of a magnet is the end that points toward the Earth’s south pole when the magnet is freely suspended.

#### Nature of Poles

**Like Poles**– Similar magnetic poles repel each other (e.g., north-north or south-south).**Unlike Poles**– Opposite magnetic poles attract each other (e.g., north-south).

### Magnetic Force (F = qVB sinθ)

Consider a charged particle with charge q moving with velocity v in a magnetic field B. The formula gives the magnetic force (F) experienced by the charged particle:

F = qVB sinθ

Where:

- F: Magnetic force
- q: Charge of the particle
- V: Velocity of the particle
- B: Magnetic field
- θ: Angle between the velocity vector and the magnetic field vector

#### Step-by-Step Derivation

Start with the definition of magnetic force (F) acting on a charged particle moving in a magnetic field:

F = q(v x B)

Rewrite the cross product (v x B) in magnitude and direction form:

v x B = vB sinθ

Substitute the expression for the cross product in the original equation:

F = q(vB sinθ)

Thus, the magnetic force on a charged particle moving in a magnetic field is given by F = qVB sinθ.

Consider a proton (q = 1.6 x 10^-19 C) is moving with a velocity of 2 x 10^5 m/s in a magnetic field of 0.5 T. If the angle between the velocity vector and the magnetic field vector is 60°, let’s calculate the magnetic force experienced by the proton.

We have,

q = 1.6 x 10^-19 C

v = 2 x 10^5 m/s

B = 0.5 T

θ = 60°

**Using the formula:**

F = qVB sinθ

F = (1.6 x 10^-19 C)(2 x 10^5 m/s)(0.5 T)(sin 60°)

**Simplifying:**

F ≈ 1.6 x 10^-19 x 2 x 10^5 x 0.5 x 0.866

F ≈ 1.39 x 10^-12 N

Therefore, the magnetic force experienced by the proton is approximately 1.39 x 10^-12 Newtons.

### Magnetic Force on a Conductor (F = BIL sinθ)

Consider a current-carrying conductor of length L placed in a magnetic field B. The formula gives the magnetic force (F) experienced by the conductor:

F = BIL sinθ

Where:

- F: Magnetic force
- B: Magnetic field
- I: Current flowing through the conductor
- L: Length of the conductor
- θ: Angle between the current direction and the magnetic field direction

#### Derivation

Start with the expression for the force (F) experienced by an infinitesimal segment (dl) of a current-carrying conductor in a magnetic field (B):

dF = Idl x B

Rewrite the cross product (Idl x B) in magnitude and direction form:

Idl x B = IdlB sinθ

Integrate over the entire length (L) of the conductor to find the total force (F):

F = ∫(IdlB sinθ)

The angle (θ) between the current direction and the magnetic field direction is constant for a straight conductor. Thus, it can be taken out of the integral:

F = IBL sinθ ∫dl

The integral of dl over the length of the conductor gives the length itself (L):

F = BIL sinθ

Hence, the magnetic force on a current-carrying conductor in a magnetic field is given by F = BIL sinθ.

Consider a 2-meter-long wire carrying a current of 5 A is placed in a magnetic field of 0.3 T. If the angle between the current direction and the magnetic field direction is 30°, calculate the magnetic force experienced by the wire.

We have,

I = 5 A

L = 2 m

B = 0.3 T

θ = 30°

**Using the formula:**

F = BIL sinθ

F = (0.3 T)(5 A)(2 m)(sin 30°)

**Simplifying:**

F = 0.3 x 5 x 2 x 0.5

F = 1.5 N

Therefore, the magnetic force experienced by the wire is 1.5 Newtons.

### Torque in a Current-Carrying Coil in Magnetic Field (T = BIAN cosα)

Consider a coil with N turns, area A, and current I placed in a magnetic field B. The formula gives the torque (T) experienced by the coil:

T = BIAN cosα

Where:

- T: Torque
- B: Magnetic field
- I: Current flowing through the coil
- A: Area of the coil
- N: Number of turns in the coil
- α: Angle between the normal to the coil and the magnetic field direction

#### Derivation

**Step 1 – Define the magnetic moment (μ) of the loop.**

The magnetic moment (μ) of a current-carrying loop is defined as the product of the current (I) and the area (A) of the loop. Mathematically:

μ = IA

**Step 2 – Determine the torque on one side of the loop.**

Consider one side of the loop with length ‘a’ perpendicular to the magnetic field (i.e., α = 0). The magnetic force (F) on this side due to the magnetic field (B) is given by:

F = BIL

where:

L = Length of the side of the loop (in meters, m)

**Step 3 – Calculate the torque on one side of the loop.**

The torque (τ) on this side of the loop is given by:

τ = F * a = BIL * a

**Step 4 – Find the torque on the entire loop.**

Since the loop has two sides of length ‘a’ and two sides of length ‘b’, the total torque (T_total) on the entire loop is:

T_total = 2τ_a + 2τ_b

= 2 * BIL * a + 2 * BIL * b

= 2BIL(a + b)

**Step 5 – Substitute A = ab (area of the loop) to get the final expression.**

The area (A) of the loop is given by the product of the lengths of its sides (a and b):

A = ab

Now, substitute this value into the total torque expression:

T_total = 2BIL(a + b)

= 2BI(ab + b²)

= 2BI(a + b²/a)

**Step 6 – Take the limit as b approaches zero.**

In the limit, as b approaches zero, the loop becomes infinitesimally tiny and approaches the shape of a planar coil. At this limit, we define the number of turns per unit length as N = 1/a (number of turns per meter). Thus, b²/a approaches zero.

T_total = 2BI(a + b²/a)

= 2BI(a + 0)

= 2BIA

= 2μ

**Step 7 – Rewrite the expression in terms of the angle (α) between the magnetic field and the normal to the plane of the loop.**

Since μ = IA, we have:

T_total = 2μ

Recall that the magnetic moment (μ) can be written as:

μ = IA

Now, rewrite the expression for the total torque in terms of α:

T_total = 2IA

Using the definition of the dot product of two vectors:

A • B = |A| |B| cos α

where A • B is the dot product of vectors A and B, |A| and |B| are their magnitudes, and α is the angle between them.

We can write B as:

B = |B| (cos α) n̂

where n̂ is the unit vector in the direction of the magnetic field B.

Substitute B in terms of |B|, (cos α), and n̂:

T_total = 2I |B| (cos α) n̂ A

Since n̂ only shows a unit vector in the direction of B and has magnitude “1”, in scaler form:

T = 2BIA cos α

We will assume only one arm of torque, so divide on both sides by 2 to consider torque on one side of the coil.

T = BIA cos α

For N number of turns in a coil.

**T = BIAN cos α**

This is the derived expression for the magnetic torque on a current-carrying loop in a magnetic field.

Consider a coil with 100 turns, an area of 0.02 m^2, and a current of 2 A is placed in a magnetic field of 0.5 T. If the angle between the normal to the coil and the magnetic field direction is 45°, let’s calculate the torque experienced by the coil.

We have,

N = 100

I = 2 A

A = 0.02 m^2

B = 0.5 T

α = 45°

**Using the formula:**

T = BIAN cosα

T = (0.5 T)(100)(2 A)(0.02 m^2)(cos 45°)

**Simplifying:**

T = 0.5 x 100 x 2 x 0.02 x 0.707

T ≈ 1.414 Nm

Therefore, the torque experienced by the coil is approximately 1.414 Newton meters.

## Electromagnetism

Magnetism is intimately connected to electricity through the relationship between electric currents and magnetic fields. When an electric current flows through a conductor, it produces a magnetic field around it. This phenomenon is known as electromagnetism and forms the basis of how electricity produces magnetism.

Let’s look at some crucial aspects of electromagnetics in the FE Electrical exam.

### Conventional Current

In the context of conventional current flow (positive to negative), the magnetic field around a straight conductor can be determined using the right-hand rule. If you grasp the conductor with your right hand, such that your thumb points in the direction of the current, the fingers will curl in the direction of the magnetic field lines.

The strength of the magnetic field produced by a straight conductor is directly proportional to the current flowing through it. As the current increases, the magnetic field strength increases. The magnetic field lines form concentric circles around the conductor, with the conductor’s center acting as the axis.

### Induced Current

Induced current refers to creating a current in a conductor due to changes in the magnetic field through the conductor. Faraday’s Law of Electromagnetic Induction governs this phenomenon and is fundamental to understanding the relationship between electricity and magnetism.

According to Faraday’s Law, a changing magnetic field through a wire loop induces an electromotive force (EMF) and, subsequently, an induced current in the loop. The induced current flows in a direction that opposes the change in the magnetic field, as described by Lenz’s Law.

When a magnetic field through a loop of wire changes, the magnetic flux (ΦB) linked with the loop also changes. The magnetic flux is the product of the magnetic field strength (B) and the area (A) perpendicular to the magnetic field.

- If the magnetic field increases, the induced current creates a magnetic field in the opposite direction to oppose the increase.
- If the magnetic field decreases, the induced current creates a magnetic field in the same direction to oppose the decrease.

This behavior of induced currents allows for various applications, such as in generators, transformers, and inductive sensors. It also explains the working principle behind electric motors, where the interaction between the induced magnetic field and a permanent magnetic field generates mechanical motion.

We will discuss Faraday’s law in detail in the following sections. Let’s discuss Ampere’s law and its application before moving toward induction and Faraday’s law.

### Ampere’s Law – Explained

To understand Ampere’s law in electromagnetics for the FE Electrical exam, we must understand the concept of the magnetic field and Biot Savart’s law.

Magnetic Field and Biot-Savart’s Law

**Magnetic Field (B)** – A magnetic field is a region in which magnetic forces are experienced. It is produced by moving charges or currents.

**Like electric flux, we also have magnetic flux as a dot product of magnetic field (B) and area (ΔA), which is given as Φm = B. ΔA = BΔA cosθ*

**Biot-Savart’s Law** – Biot-Savart’s Law gives the magnetic field at a point due to a small current element (dI) in a wire. It is given by:

dB = (μ0 / 4π) * (dI × r) / r^3

Where:

- dB: Infinitesimal magnetic field at a point
- dI: Infinitesimal current element in the wire
- r: Distance vector from the current element to the point
- μ0: Permeability of free space (≈ 4π × 10^-7 T·m/A)

#### Ampere’s Law

Ampere’s Law relates the magnetic field around a closed loop (C) to the total current enclosed by the loop. It is expressed as:

∮B · dl = μ0 * I_enclosed

Where:

∮: Integration around a closed loop

- B: Magnetic field
- dl: Infinitesimal vector element along the loop
- I_enclosed: Total current enclosed by the loop

##### Derivation

Consider a closed loop C and divide it into small segments dl.

Apply Biot-Savart’s Law to find the magnetic field contribution dB at each point on the loop due to the corresponding current element dI.

The magnetic field dB and dl are vectors, so their dot product (dB · dl) represents the dB component along dl.

Sum up all the contributions by integrating around the closed loop:

∮B · dl = ∮(dB · dl)

= ∮[(μ0 / 4π) * (dI × r) / r^3] · dl

= (μ0 / 4π) * ∮[dI × dl / r^3]

Apply the cross-product identity dI × dl = I * dl, where I is the total current enclosed by the loop:

∮B · dl = (μ0 / 4π) * ∮[I * dl / r^3]

The integral on the right-hand side is equal to the total current enclosed:

∮B · dl = (μ0 / 4π) * I_enclosed

Hence, Ampere’s Law states that the line integral of the magnetic field around a closed loop is equal to μ0 times the total current enclosed by the loop.

#### Application of Ampere’s Law – Solenoid and Toroid

##### Solenoid

A solenoid is a long, cylindrical coil of wire with many closely spaced turns. It generates a nearly uniform magnetic field inside.

**Magnetic Field Inside a Solenoid:**

By applying Ampere’s Law to a solenoid, we find that the magnetic field inside the solenoid is approximately:

B = μ0 * n * I

Where:

- B: Magnetic field inside the solenoid
- μ0: Permeability of free space
- n: Number of turns per unit length
- I: Current flowing through the solenoid

Consider a solenoid with 200 turns per meter, a length of 0.4 meters, and a current of 2 Amperes. Let’s calculate the magnetic field inside the solenoid.

We have,

n = 200 turns/m

L = 0.4 m

I = 2 A

**Step 1** – Determine the number of turns in the solenoid.

The number of turns (N) in the solenoid can be calculated by multiplying the turns per meter (n) by the length (L) of the solenoid.

N = n * L

N = 200 turns/m * 0.4 m

N = 80 turns

**Step 2** – Calculate the total current enclosed by the solenoid.

The total current (I_total) enclosed by the solenoid is given by the product of the current per turn (I) and the number of turns (N).

I_total = I * N

I_total = 2 A * 80 turns

I_total = 160 A-turns

**Step 3** – Use Ampere’s Law to find the magnetic field inside the solenoid.

Ampere’s Law states that the magnetic field (B) inside a solenoid is given by the product of the permeability of free space (μ0), the number of turns per unit length (n), and the total current enclosed (I_total).

B = μ0 * n * I_total

B = (4π × 10^-7 T·m/A) * 200 turns/m * 160 A-turns

Step 4: Calculate the magnetic field inside the solenoid.

B ≈ 4π × 10^-7 T·m/A * 200 turns/m * 160 A-turns

B ≈ 4π × 10^-7 T·m/A * 32000 A-turns/m

B ≈ 4π × 10^-7 T·m/A * 32000 A/m

B ≈ 4π × 10^-7 T

B ≈ 1.26 × 10^-6 T

Therefore, the magnetic field inside the solenoid is approximately 1.26 μT (microteslas).

##### Toroid

A toroid is a doughnut-shaped coil of wire. It generates a nearly uniform magnetic field inside.

**Magnetic Field Inside a Toroid:**

By applying Ampere’s Law to a toroid, we find that the magnetic field inside the toroid is approximately:

B = μ0 * n * I

Where:

- B: Magnetic field inside the toroid
- μ0: Permeability of free space
- n: Number of turns per unit length
- I: Current flowing through the toroid

### Faraday’s Law of Electromagnetic Induction and Lenz’s Law

#### Faraday’s Law

Faraday’s Law states that a changing magnetic field through a wire loop induces an electromotive force (EMF) and, consequently an induced current. It is expressed as:

EMF = -dΦB / dt

Where:

- EMF: Electromotive force induced in the loop
- dΦB / dt: Rate of change of magnetic flux through the loop

#### Lenz’s Law

Lenz’s Law states that the direction of the induced current in a closed loop is such that it opposes the change in magnetic flux that produced it.

### Induction

Induction is the process of generating an electromotive force (EMF) and an induced current in a circuit due to changes in the magnetic field through the circuit. It is based on Faraday’s Law and is an essential electrical transformer and generator principle.

#### Self-Induction

Self-induction occurs when a changing current in a coil induces an EMF in the same coil. The magnitude of the induced EMF is given by:

EMF = -L * dI / dt

Where:

- EMF: Electromotive force induced in the coil
- L: Self-inductance of the coil
- dI / dt: Rate of change of current in the coil

#### Mutual Induction

Mutual induction occurs when a changing current in one coil induces an EMF in an adjacent coil. The magnitude of the induced EMF is given by:

EMF = -M * dI1 / dt

Where:

- EMF: Electromotive force induced in the adjacent coil
- M: Mutual inductance between the two coils
- dI1 / dt: Rate of change of current in the first coil

### Motional EMF

A Motional EMF occurs when a conductor moves through a magnetic field, resulting in an induced EMF. The formula for the motional EMF is:

EMF = B * L * v

Where:

- EMF: Electromotive force induced in the conductor
- B: Magnetic field strength
- L: Length of the conductor perpendicular to the field
- v: Velocity of the conductor

## Electromagnetic Wave

The Electromagnetic wave is another vital topic to understand electromagnetics in FE Electrical exam. These waves are a form of energy that can propagate through space without requiring a medium for transmission.

They consist of oscillating electric and magnetic fields that are mutually perpendicular and perpendicular to the direction of wave propagation. This unique characteristic enables electromagnetic waves to travel through a vacuum, such as outer space.

The propagation of electromagnetic waves without a medium can be understood by Maxwell’s equations (Discussed in the following section), which describe the fundamental principles of electromagnetism. These equations were developed by James Clerk Maxwell in the 19th century and provide a mathematical framework for understanding the behavior of electric and magnetic fields.

### Light as an Electromagnetic Wave (Maxwell’s Theory)

James Clerk Maxwell’s theory of electromagnetism unified the previously separate theories of electricity and magnetism, and it predicted the existence of electromagnetic waves, including light. Maxwell’s equations mathematically describe the generation, propagation, and interaction of electromagnetic waves.

Maxwell’s equations reveal that oscillating electric charges can create time-varying electric fields, generating time-varying magnetic fields. The changing magnetic fields then induce further changes in the electric fields, resulting in a self-sustaining wave of coupled electric and magnetic fields propagating through space. This wave is known as an electromagnetic wave.

Light is a specific type of electromagnetic wave that falls within a certain range of wavelengths in the electromagnetic spectrum. The oscillations of electric and magnetic fields in light waves occur at extremely high frequencies, giving rise to the phenomenon of visible light.

Therefore, light is considered an electromagnetic wave because it exhibits the same fundamental characteristics as other electromagnetic waves, including the ability to propagate without a medium and the capability to exhibit properties like reflection, refraction, interference, and diffraction.

### Maxwell’s Equations and their Applications

Maxwell’s equations form a set of four fundamental equations that describe the behavior of electric and magnetic fields. These equations are

**Gauss’s Law for Electric Fields:**

∇ · E = ρ / ε₀

This equation relates the electric field (E) to the electric charge density (ρ) and the electric constant (ε₀). It states that the electric field divergence is proportional to the charge density.

**Gauss’s Law for Magnetic Fields:**

∇ · B = 0

This equation states that the magnetic field (B) has no sources or sinks. The magnetic field divergence is always zero, indicating no magnetic monopoles.

**Ampere’s Law with Maxwell’s Addition:**

∇ × B = μ₀J + μ₀ε₀ ∂E / ∂t

This equation combines Ampere’s Law with the addition of Maxwell’s displacement current term. It relates the magnetic field (B) to the current density (J), the magnetic constant (μ₀), and the displacement current (ε₀ ∂E / ∂t), which accounts for the time-varying electric field contribution.

Maxwell’s equations help in:

- Understanding the behavior of electromagnetic waves, including light.
- Analysis of antennas and the propagation of radio waves.
- Modeling and design of electrical circuits and devices.
- Studying electromagnetic interactions and phenomena, such as reflection, refraction, and diffraction.

### Impact of Medium on Propagation of Electromagnetic Waves

The propagation of electromagnetic waves is influenced by the properties of the medium through which they travel. Let’s explore the impact of different mediums on electromagnetic wave propagation:

#### Vacuum (Empty Space)

In a vacuum, such as outer space, electromagnetic waves propagate without any attenuation or distortion. They maintain their speed, wavelength, and frequency as determined by the electromagnetic spectrum.

#### Air and Gases

Air and other gases have a relatively low density and do not significantly affect the propagation of electromagnetic waves. The presence of air or gas may cause minimal scattering or absorption, but it generally has a negligible impact on the overall propagation.

#### Transparent Materials (Lighter Mediums)

Transparent materials, such as glass or water, have a refractive index greater than 1. This index determines the speed of light in the medium, which is generally slower than in a vacuum. When electromagnetic waves pass from a vacuum into a lighter medium, they undergo refraction, bending toward the normal to surface. This bending is due to the change in the wave’s velocity.

#### Dense Materials (Denser Mediums)

Dense materials, such as diamond or lead, have a high refractive index. When electromagnetic waves pass from a vacuum into a denser medium, they undergo refraction, but this time they bend away from the normal to the surface. This bending is due to the decrease in the wave’s velocity.

#### Dielectric Materials

Dielectric materials, like plastics or ceramics, are non-conductive and have an electric permittivity greater than a vacuum. When electromagnetic waves encounter dielectric materials, they experience a decrease in velocity and a change in wavelength.

This change leads to effects like wave slowing down, polarization, and partial reflection and transmission at the interface between different dielectric mediums.

#### Conductive Materials (Metals)

Conductive materials, such as metals, have a high electrical conductivity. When electromagnetic waves encounter a conductive medium, the electric field induces free charges to move within the material, resulting in the absorption of the wave’s energy. As a result, electromagnetic waves are rapidly attenuated and do not propagate far into the material.

#### Insulating Materials

Insulating materials, like rubber or plastics, have low electrical conductivity. Electromagnetic waves passing through insulating materials experience minimal absorption or attenuation. However, some insulators may exhibit slight dispersion, where the speed of light varies with wavelength, leading to a slight broadening of the wave’s frequency spectrum.

**It’s important to note that the behavior of electromagnetic waves in different mediums is described by Maxwell’s equations, which consider the electrical and magnetic properties of the medium. These equations provide a mathematical framework for understanding wave propagation and the interactions between electromagnetic waves and different materials.*

## Waves and Antennas

Understanding Antenna operations and wave spectrums is another vital domain of electromagnetics in the FE Electrical exam. Let’s start with the different types of wave spectrums that you must know.

### Wave Spectrums

Wave spectrums refer to the classification of electromagnetic waves based on their wavelengths and frequencies. Here are the main types of wave spectrums:

#### Radio Waves

- Longest wavelengths and the lowest frequencies in the spectrum.
- Used in radio broadcasting, television transmission, and wireless communication.

#### Microwaves

- Have Shorter wavelengths than radio waves but longer than infrared waves.
- Used in microwave ovens, satellite communication, and radar systems.

#### Infrared Waves

- Have Longer wavelengths than visible light but shorter than microwaves.
- Used in remote controls, thermal imaging, and communication in short-range devices like IR sensors.

#### Visible Light

- The range of wavelengths that the human eye can detect.
- Used in illumination, display technologies (LED, LCD), and optical communication.

#### Ultraviolet (UV) Waves

- Shorter wavelengths than visible light but longer than X-rays.
- Applied in sterilization, fluorescent lighting, and UV photography.

#### X-Rays

- Shorter wavelengths and higher frequencies than UV waves.
- Used in medical imaging (X-ray radiography), airport security scanners, and industrial testing.

#### Gamma Rays

- Shortest wavelengths and highest frequencies in the spectrum.
- Applied in cancer treatment (radiotherapy), nuclear medicine, and sterilization processes.

### Antennas, Their Types, and Characteristics

Antennas are electric equipment used to catch electric signals and waves of different frequencies. They can vary based on their design, purpose, and frequency range. Here are some common types and characteristics:

Antenna Type | Description | Characteristics |

Dipole Antenna | Simplest form, consists of two conductive elements | Omnidirectional radiation pattern, commonly used in radio and TV broadcasting |

Parabolic Dish | Large dish-shaped reflector with a feed at its focus point | Highly directional, high gain, commonly used in satellite communication and long-range WiFi networks |

Patch Antenna | Flat antenna with a conductive patch on a dielectric substrate | Compact size, low-profile, used in mobile devices, WiFi routers, and GPS systems |

Helical Antenna | Spiral-shaped antenna with a wire wound around a cylinder | Circular polarization, used in satellite communication, wireless communication, and RFID systems |

### Receiver and Transmitter

In modern-day aviation and warfare, electric equipment used for communication rely on receivers and transmitters to exchange information. Here is an overview of how they work and their applications:

#### Receiver

A receiver is an electric equipment used in communication systems to capture and decode electromagnetic signals from the environment. It performs the following functions:

- Antenna: Captures incoming signals from the air or other sources.
- Amplification: Boosts the weak signals received from the antenna.
- Demodulation: Extracts the original information from the signals (voice, data, or video).
- Decoding/Decryption: Converts the modulated information into its original form for interpretation.
- Audio/Visual Output: Presents the decoded information for human consumption.

In aviation, receivers are used in aircraft for various purposes, including radio communication, navigation systems (GPS), weather radar, and collision avoidance systems.

They are also employed in communication systems for military aircraft, ground vehicles, ships, and soldiers on the battlefield. They allow secure communication, situational awareness, intelligence gathering, and coordination among military units.

#### Transmitter

A transmitter is an electric equipment used in communication systems to generate and send out electromagnetic signals. It performs the following functions:

**Signal Generation**– Converts the original information into modulated signals suitable for transmission.**Modulation**– Superimposes the information onto a carrier wave, adjusting properties like amplitude, frequency, or phase.**Amplification**– Increases the power of the modulated signals to ensure adequate transmission.**Transmission**– Sends the modulated signals through an antenna into the surrounding medium.- In aviation, transmitters are crucial for radio communication between aircraft and air traffic control, as well as for broadcasting navigation beacons. In warfare, transmitters are employed for military communications, radar systems, electronic warfare (jamming or deception), and various intelligence-gathering operations.

### Conclusion

ELectromagnetics plays a pivotal role in electrical engineering. The key topics discussed in this study guide are crucial to understanding electromagnetics in FE Electrical exam as per the NCEES® FE Electrical exam guidelines.

By understanding the principles of electromagnetism, you gain a deep comprehension of how electric and magnetic fields interact, how electromagnetic waves propagate, and how devices such as antennas and transformers function.

This knowledge empowers you to tackle complex electrical engineering problems confidently and lays the foundation for further specialization in power systems, telecommunications, and electronics.

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