Unit 4: Electromagnetic Induction

1. Electric Flux, Magnetic Flux, and Electromotive Force (EMF)

Electric Flux (Φ_E)

Electric flux measures the total number of electric field lines passing normally through a given surface area in an electric field.

Definition: The surface integral of the electric field vector E over a closed or open surface A is defined as the electric flux:
Φ_E = ∫ E · dA = ∫ E cos θ dA

The SI unit of electric flux is Volt-meter (V·m) or Newton-meter² per Coulomb (N·m²/C).

Magnetic Flux (Φ_B)

Magnetic flux measures the total number of magnetic field lines passing perpendicular through a surface.

Definition: The surface integral of the magnetic field vector B over a surface area A is defined as the magnetic flux:
Φ_B = ∫ B · dA = B A cos θ

where θ is the angle between the magnetic field vector B and the area vector A (which is normal to the surface).

• Maximum Flux: Occurs when the surface is perpendicular to magnetic field lines (θ = 0°, cos 0° = 1), so Φ_B = B A.
• Zero Flux: Occurs when the surface is parallel to magnetic field lines (θ = 90°, cos 90° = 0), so Φ_B = 0.
• SI Unit: Weber (Wb) or Tesla-meter² (T·m²). 1 Wb = 1 T·m².

Electro-motive Force (EMF)

Electromotive Force (e or ε) is the work done per unit charge by a non-electrostatic source in moving a charge around a closed conductive loop.

Mathematical Definition: e = ∲ f_nc · dl = -dΦ_B / dt

Despite being called a 'force', EMF is not a mechanical force; it is an electric potential difference measured in Volts (V), where 1 V = 1 Joule per Coulomb (J/C).

2. Faraday's Law of Induction (Integral and Differential Forms)

Fundamental Statement

Faraday's Law of Electromagnetic Induction states that whenever there is a change in the magnetic flux linked with a closed circuit, an electromotive force (EMF) is induced in the circuit. The magnitude of the induced EMF is directly proportional to the time rate of change of magnetic flux through the circuit.

1. Integral Form of Faraday's Law

The induced EMF around a closed path boundary C enclosing a surface S is given by:

e = -dΦ_B / dt = -d/dt [ ∫_S B · dA ]

Since the induced EMF is also defined as the line integral of the induced electric field E along the closed contour C:

∲_C E · dl = -d/dt ∫_S B · dA

If the contour C is stationary with respect to time, the time derivative can be brought inside the integral as a partial derivative:

∲_C E · dl = -∫_S (∂B / ∂t) · dA

2. Differential Form of Faraday's Law

To obtain the point or differential form, apply Stokes' Theorem to convert the line integral of the electric field into a surface integral:

∲_C E · dl = ∫_S (∇ × E) · dA

Substituting this relation into the integral form gives:

∫_S (∇ × E) · dA = -∫_S (∂B / ∂t) · dA

Rearranging the integrals over the same surface S:

∫_S [ (∇ × E) + (∂B / ∂t) ] · dA = 0

Since this equality holds for any arbitrary surface S, the integrand itself must vanish identically everywhere:

Differential Form: ∇ × E = -∂B / ∂t

Exam Observation: In electrostatics, ∇ × E = 0 (conservative electric field). In time-varying electromagnetic fields, ∇ × E ≠ 0, showing that induced electric fields are non-conservative and form closed loops.

3. Lenz's Law and Conservation of Energy

Lenz's Law Statement

Lenz's Law: The direction of the induced current (or induced EMF) is always such that it produces a magnetic field that opposes the change in magnetic flux that produced it.

This law accounts for the negative sign in Faraday's equation: e = -dΦ_B / dt.

Proof of Conservation of Energy

Lenz's law is a direct consequence of the Principle of Conservation of Energy:

1. Consider a bar magnet moving towards a stationary conducting coil with its North pole facing the loop.
2. As the magnet approaches, the magnetic flux through the loop increases.
3. According to Lenz's law, an induced current flows in a counter-clockwise direction (when viewed from the magnet side), creating an induced North pole facing the approaching magnet.
4. Like poles repel; hence, a repulsive mechanical force acts on the approaching magnet.
5. To keep moving the magnet towards the coil, external mechanical work must be performed against this repulsive force.
6. This mechanical work is converted directly into electrical energy, which dissipates as Joule heat (I²R) in the coil circuit.

Important Note: If the induced current were in the opposite direction (attracting the magnet), the magnet would accelerate towards the coil without external work, generating electrical energy continuously out of nothing, which violates the Law of Conservation of Energy.

4. Self-Inductance (L) and Mutual Inductance (M)

Self-Inductance (L)

Self-inductance is the property of an electric circuit by virtue of which it opposes any change in the current flowing through itself by inducing a counter-EMF.

The total magnetic flux linkage (NΦ_B) in a coil is directly proportional to the current (I) flowing through it:

NΦ_B = L I

Self-Inductance Coefficient: L = NΦ_B / I
Induced EMF: e = -L (dI / dt)

The SI unit of inductance is the Henry (H). One Henry is defined as the self-inductance of a circuit in which an induced EMF of 1 Volt is produced when current changes at the rate of 1 Ampere per second (1 H = 1 V·s/A).

Mutual Inductance (M)

Mutual inductance is the property of two neighboring coils by virtue of which a changing current in one coil (primary) induces an EMF in the secondary coil.

The flux linkage in coil 2 (secondary) due to current in coil 1 (primary) is:

N₂ Φ₂₁ = M I₁

Mutual Inductance Coefficient: M = (N₂ Φ₂₁) / I₁
Induced EMF in Secondary: e₂ = -M (dI₁ / dt)

By reciprocity principle: M₁₂ = M₂₁ = M.

5. Self-Inductances in Series and Parallel

Inductors in Series

Consider two inductors with self-inductances L₁ and L₂ connected in series.

Case 1: No Magnetic Linkage (M = 0)

Total EMF e = e₁ + e₂ = -L₁ (dI/dt) - L₂ (dI/dt) = -(L₁ + L₂) (dI/dt)

L_eq = L₁ + L₂

Case 2: With Mutual Coupling (M ≠ 0)

• Aiding Flux (Currents in same sense): L_eq = L₁ + L₂ + 2M
• Opposing Flux (Currents in opposite sense): L_eq = L₁ + L₂ - 2M

Inductors in Parallel

Consider two inductors L₁ and L₂ connected in parallel without mutual coupling (M = 0).

Total current I = I₁ + I₂. Differentiating with respect to time:

dI/dt = (dI₁/dt) + (dI₂/dt)

Since voltage across both inductors is equal (e = -L₁ dI₁/dt = -L₂ dI₂/dt = -L_eq dI/dt):

(e / L_eq) = (e / L₁) + (e / L₂)

1 / L_eq = (1 / L₁) + (1 / L₂) → L_eq = (L₁ L₂) / (L₁ + L₂)

6. Self-Inductance Calculations (Solenoid, Coaxial Cylinders, Toroid)

1. Long Solenoid

Consider a long solenoid of length l, cross-sectional area A, total turns N, and turn density n = N / l.

1. Magnetic field inside long solenoid: B = μ₀ n I = μ₀ (N / l) I
2. Flux through a single turn: Φ₁ = B A = (μ₀ N I A) / l
3. Total flux linkage for N turns: Φ_total = N Φ₁ = (μ₀ N² A I) / l
4. Using L = Φ_total / I:

L = (μ₀ N² A) / l = μ₀ n² A l

If filled with a magnetic core of relative permeability μ_r, then L = (μ₀ μ_r N² A) / l.

2. Coaxial Cylinders

Consider two coaxial thin cylindrical shells of length l with inner radius a and outer radius b carrying equal and opposite current I.

1. Magnetic field in the region between cylinders (a < r < b) using Ampere's Law: B=(μ₀ I) / (2π r)
2. Consider a strip of width dr and length l between the cylinders. The area element dA = l dr.
3. Flux through this strip: dΦ = B dA = [ (μ₀ I) / (2π r) ] l dr
4. Integrating from r = a to r = b:
   Φ = ∫_a^b [ (μ₀ I l) / (2π) ] (dr / r) = [ (μ₀ I l) / (2π) ] ln(b / a)
5. Since L = Φ / I:

L = [ (μ₀ l) / (2π) ] ln(b / a)

3. Toroid

Consider a toroid of N turns having inner radius r₁, outer radius r₂, and rectangular cross-section of height h.

1. Magnetic field at radius r inside core: B = (μ₀ N I) / (2π r)
2. Flux through an elemental strip of width dr and height h: dΦ = B (h dr) = [ (μ₀ N I h) / (2π) ] (dr / r)
3. Integrating from r₁ to r₂:
   Φ₁ = [ (μ₀ N I h) / (2π) ] ln(r₂ / r₁)
4. Total flux linkage Φ_total = N Φ₁ = [ (μ₀ N² I h) / (2π) ] ln(r₂ / r₁)
5. Dividing by I gives self-inductance:

Rectangular Cross-section: L = [ (μ₀ N² h) / (2π) ] ln(r₂ / r₁)
Circular Cross-section (mean radius r_avg, area A): L ≈ (μ₀ N² A) / (2π r_avg)

7. Mutual Inductance Between Two Coaxial Solenoids

Consider two thin, closely wound coaxial solenoids of length l. Inner solenoid S₁ has radius r₁, number of turns N₁. Outer solenoid S₂ has radius r₂ (r₂ > r₁), number of turns N₂.

Derivation Steps:

1. Pass current I₁ through inner solenoid S₁.
2. The uniform magnetic field generated inside S₁ is:
   B₁ = μ₀ (N₁ / l) I₁
3. Since r₁ < r₂, the magnetic field B₁ exists only inside S₁ (area A₁ = π r₁²) and is zero outside S₁.
4. Flux through a single turn of outer solenoid S₂ due to B₁ is:
   Φ_21(single) = B₁ A₁ = [ μ₀ (N₁ / l) I₁ ] (π r₁²)
5. Total magnetic flux linkage with all N₂ turns of solenoid S₂:
   Φ₂₁ = N₂ Φ_21(single) = (μ₀ N₁ N₂ π r₁² I₁) / l
6. By definition, mutual inductance M₂₁ = Φ₂₁ / I₁:

M = (μ₀ N₁ N₂ A₁) / l

where A₁ = π r₁² is the cross-sectional area of the inner solenoid.

8. Coefficient of Coupling and Its Derivation

Definition

The coefficient of coupling (k) is a measure of the degree of magnetic linkage between two coils. It represents the fraction of magnetic flux produced by one coil that passes through another coil.

k = M / √(L₁ L₂)      (0 ≤ k ≤ 1)

• Tight Coupling (k = 1): 100% flux produced by coil 1 links with coil 2.
• Loose Coupling (0 < k < 1): Partial flux linkage.
• Zero Coupling (k = 0): No flux linkage (coils perpendicular or isolated).

Derivation

Consider two solenoids S₁ and S₂ of length l and cross-sectional area A wound over each other.

Self-inductance of coil 1: L₁ = (μ₀ N₁² A) / l

Self-inductance of coil 2: L₂ = (μ₀ N₂² A) / l

Multiplying L₁ and L₂:

L₁ L₂ = [ (μ₀ N₁² A) / l ] × [ (μ₀ N₂² A) / l ] = [ (μ₀ N₁ N₂ A) / l ]²

Taking the square root on both sides:

√(L₁ L₂) = (μ₀ N₁ N₂ A) / l

Notice that the right-hand side is equal to the ideal mutual inductance M_max when all flux links both coils:

M_max = √(L₁ L₂)

If only a fraction k of magnetic flux links coil 2, then actual mutual inductance is M = k M_max = k √(L₁ L₂).

Derived Formula: k = M / √(L₁ L₂)

9. Energy Stored in a Magnetic Field

Derivation of Stored Energy in an Inductor

When an increasing current flows through an inductor, an opposing back EMF e = -L (dI/dt) is produced. To maintain current flow, external energy must be supplied against this back EMF.

1. Work done in small time dt: dW = P dt = e I dt = (L dI / dt) I dt = L I dI
2. Total work done to establish current from 0 to steady value I:
   U = ∫ dW = ∫₀^I L I dI = (1/2) L I²

Energy Stored in Inductor: U = (1/2) L I²

Magnetic Energy Density (u_B)

Energy density is defined as the magnetic energy stored per unit volume inside the magnetic field region.

For a long solenoid of length l and area A:

1. Inductance L = (μ₀ N² A) / l
2. Magnetic field B = (μ₀ N I) / l → I = (B l) / (μ₀ N)
3. Substitute L and I into energy equation:
   U = (1/2) [ (μ₀ N² A) / l ] [ (B l) / (μ₀ N) ]² = (1/2) [ (μ₀ N² A) / l ] [ (B² l²) / (μ₀² N²) ] = (B² A l) / (2 μ₀)
4. Since volume V = A l:
   u_B = U / V = [ (B² A l) / (2 μ₀) ] / (A l)

Magnetic Energy Density: u_B = B² / (2 μ₀)

10. Transformer Principles and Energy Losses

Operating Principle

A transformer is a static electrical device that transfers electrical energy between two or more circuits through mutual induction, changing AC voltage levels while preserving frequency.

Transformation Ratio Equation:
V_s / V_p = N_s / N_p = I_p / I_s = K

• Step-up Transformer: N_s > N_p → V_s > V_p and I_s < I_p
• Step-down Transformer: N_s < N_p → V_s < V_p and I_s > I_p

Detailed Losses in Transformers