13.3 Magnetic Fields, Induction, Transformers, and Motor Efficiency

13.3 Magnetic Fields, Induction, Transformers, and Motor Efficiency

Core FE Exam Principle: Time-varying magnetic fields induce electric fields (Faraday's Law), forming the operational basis for electrical inductors, generators, and transformers. Ideal transformers transform AC voltage linearly while transforming impedance quadratically by (a^2 = (N_1/N_2)^2).

Magnetic Fields, Biot-Savart Law, and Ampere's Law

Magnetic fields are created by moving electric charges or current-carrying conductors. The magnetic field is characterized by:

• Magnetic Flux Density ((\mathbf{B})): Vector field measured in Teslas (T) or Weber/(\text{m}^2) ((\text{Wb/m}^2)).
• Magnetic Field Strength ((\mathbf{H})): Measured in Amperes per meter (A/m).
• Constitutive Relation: $\mathbf{B} = \mu \mathbf{H} = \mu_0 \mu_r \mathbf{H}$ where (\mu_0 = 4\pi \times 10^{-7} \ \text{H/m}) is the magnetic permeability of free space and (\mu_r) is relative permeability.

Biot-Savart Law

The Biot-Savart Law computes the differential magnetic flux density (d\mathbf{B}) generated by a current element (I d\mathbf{l}) at distance (r):

$d\mathbf{B} = \frac{\mu_0 I}{4\pi} \frac{d\mathbf{l} \times \hat{\mathbf{r}}}{r^2}$

For a long, straight, infinitely extended conductor carrying current (I), the magnetic flux density at radial distance (r) simplifies to:

$B = \frac{\mu_0 I}{2\pi r}$

Ampere's Circuital Law

Ampere's Law states that the line integral of magnetic field density (\mathbf{B}) around any closed path equals (\mu_0) times the total enclosed current (I_{\text{enc}}):

$\oint \mathbf{B} \cdot d\mathbf{l} = \mu_0 I_{\text{enc}}$

• Long Air-Core Solenoid: For a solenoid with (N) turns over length (L) (turn density (n = N/L)): $B = \mu_0 n I = \mu_0 \left(\frac{N}{L}\right) I$
• Toroid: For a toroidal coil of mean radius (r): $B = \frac{\mu_0 N I}{2\pi r}$

Magnetic Forces on Charges and Conductors

• Lorentz Force on Moving Charge: (\mathbf{F} = q \left( \mathbf{E} + \mathbf{v} \times \mathbf{B} \right))
• Force on Current-Carrying Wire: (\mathbf{F} = I \left( \mathbf{L} \times \mathbf{B} \right) \implies F = I L B \sin\theta)

Electromagnetic Induction, Faraday's Law, and Lenz's Law

Magnetic Flux ((\Phi))

Magnetic flux (\Phi) measures the total magnetic field passing normally through a surface area (A):

$\Phi = \iint \mathbf{B} \cdot d\mathbf{A} = B A \cos\theta \quad [\text{Webers, Wb} = \text{T}\cdot\text{m}^2]$

Faraday's Law of Induction

Faraday's Law states that any change in magnetic flux linkage through a conducting loop induces an electromotive force (EMF, (e)) proportional to the time rate of flux change:

$e = -N \frac{d\Phi}{dt} \quad [\text{Volts, V}]$

where (N) is the number of turns in the coil. The negative sign represents Lenz's Law.

Lenz's Law: The direction of an induced voltage or current always creates a magnetic field that opposes the original change in magnetic flux that produced it.

Motional EMF

When a straight conductor of length (L) moves at velocity (v) perpendicular to a uniform magnetic field (B), an EMF is induced across its terminals:

$e = B L v \sin\theta \quad [\text{Volts, V}]$

Ideal Transformers and Impedance Transformation

An ideal transformer consists of two high-permeability coupled windings around a common magnetic core, assuming zero winding resistance, zero core losses ((P_{\text{core}} = 0)), and zero magnetic flux leakage.

                    Ideal Transformer Schematic
                   N1 : N2  (Turns Ratio a = N1/N2)
             +-------+  ||  +-------+
             |       )| || |(       |
     +  I1   | Primary) || |(Secondary I2   +  
    V1  ---> | Coil  )| || |( Coil  <---   V2 Load Z_L
     -       |       )| || |(       |       -  
             +-------+  ||  +-------+

Primary-to-Secondary Transformation Equations

Let (N_1) be primary turns, (N_2) secondary turns, and (a = \frac{N_1}{N_2}) the turns ratio.

1. Voltage Transformation Ratio: $\frac{V_1}{V_2} = \frac{N_1}{N_2} = a \implies V_1 = a V_2$
2. Current Transformation Ratio: $\frac{I_1}{I_2} = \frac{N_2}{N_1} = \frac{1}{a} \implies I_1 = \frac{I_2}{a}$
3. Apparent Power Conservation: $S_1 = V_1 I_1 = (a V_2) \left(\frac{I_2}{a}\right) = V_2 I_2 = S_2$

Impedance Transformation (Reflected Load Impedance)

When a load impedance (\mathbf{Z}_L = \mathbf{Z}_2) is connected across the secondary terminals, the equivalent impedance (\mathbf{Z}_1') reflected to (seen from) the primary side is:

$\mathbf{Z}_1' = \frac{\mathbf{V}_1}{\mathbf{I}_1} = \frac{a \mathbf{V}_2}{\left(\frac{\mathbf{I}_2}{a}\right)} = a^2 \left(\frac{\mathbf{V}_2}{\mathbf{I}_2}\right) = a^2 \mathbf{Z}_2 = \left(\frac{N_1}{N_2}\right)^2 \mathbf{Z}_L$

Exam Tip: Impedance transforms with the square of the turns ratio (a^2). Stepping up voltage ((a < 1)) decreases reflected impedance, while stepping down voltage ((a > 1)) increases reflected impedance.

Practical Transformer Characteristics and Efficiency

Real transformers deviate from ideal behavior due to physical loss mechanisms:

1. Winding Copper Losses ((P_{\text{copper}})): (I^2 R) heating losses in primary and secondary conductor windings: $P_{\text{copper}} = I_1^2 R_1 + I_2^2 R_2$
2. Core Iron Losses ((P_{\text{core}})):
   • Hysteresis Loss: Energy required to continually reorient magnetic domains in the AC iron core.
   • Eddy Current Loss: Circular induced currents in the conductive core (mitigated by thin laminated steel sheets).
3. Transformer Efficiency ((\eta)): $\eta = \frac{P_{\text{out}}}{P_{\text{in}}} \times 100\% = \frac{P_{\text{out}}}{P_{\text{out}} + P_{\text{core}} + P_{\text{copper}}} \times 100\%$ where (P_{\text{out}} = V_2 I_2 \cos\theta_L).

Comprehensive Worked Engineering Example

Problem Statement

A step-down distribution transformer supplies power to an industrial manufacturing load. The transformer details and load parameters are:

• Primary winding turns (N_1 = 2400), secondary turns (N_2 = 120).
• Primary supply voltage (V_1 = 4800 \text{ V}) (RMS, 60 Hz).
• Secondary load impedance (\mathbf{Z}_L = 4.0 + j 3.0 \ \Omega).
• Transformer internal losses: Constant core loss (P_{\text{core}} = 150 \text{ W}), primary winding resistance (R_1 = 2.50 \ \Omega), and secondary winding resistance (R_2 = 0.0060 \ \Omega).

Calculate:

1. The turns ratio (a), secondary voltage (V_2), secondary current magnitude (I_2), and primary current magnitude (I_1).
2. The equivalent load impedance (\mathbf{Z}_1') reflected to the primary side.
3. The active output power (P_{\text{out}}), total copper loss (P_{\text{copper}}), total losses (P_{\text{loss}}), and overall operating efficiency (\eta).

Step-by-Step Solution

Step 1: Compute Turns Ratio, Voltages, and Currents

• Turns ratio (a): $a = \frac{N_1}{N_2} = \frac{2400}{120} = 20.0$
• Secondary terminal voltage (V_2): $V_2 = \frac{V_1}{a} = \frac{4800 \text{ V}}{20.0} = 240.0 \text{ V}$
• Magnitude of secondary load impedance (|\mathbf{Z}_L|): $|\mathbf{Z}_L| = \sqrt{4.0^2 + 3.0^2} = \sqrt{16.0 + 9.0} = 5.00 \\ \Omega$
• Secondary load power factor angle: (\theta_L = \arctan(3.0 / 4.0) = 36.87^circ), (pf_L = \cos(36.87^circ) = 0.800 \text{ lagging}).
• Secondary current magnitude (I_2): $I_2 = \frac{V_2}{|\mathbf{Z}_L|} = \frac{240.0 \text{ V}}{5.00 \\ \Omega} = 48.00 \text{ A}$
• Primary current magnitude (I_1): $I_1 = \frac{I_2}{a} = \frac{48.00 \text{ A}}{20.0} = 2.400 \text{ A}$

Step 2: Compute Reflected Load Impedance on Primary Side

• Using the impedance scaling law (\mathbf{Z}_1' = a^2 \mathbf{Z}_L): $\mathbf{Z}_1' = (20.0)^2 \times (4.0 + j 3.0 \\ \Omega) = 400 \times (4.0 + j 3.0) = 1600 + j 1200 \\ \Omega$
• Reflected impedance magnitude: (|\mathbf{Z}_1'| = 400 \times 5.00 = 2000 \ \Omega).

Step 3: Compute Output Power, Losses, and Efficiency

• Active output power (P_{\text{out}}): $P_{\text{out}} = V_2 I_2 \cos\theta_L = 240.0 \text{ V} \times 48.00 \text{ A} \times 0.800 = 9216.0 \text{ W} = 9.216 \text{ kW}$
• Total winding copper loss (P_{\text{copper}}): $P_{\text{copper}} = I_1^2 R_1 + I_2^2 R_2 = (2.400 \text{ A})^2 \times 2.50 \\ \Omega + (48.00 \text{ A})^2 \times 0.0060 \\ \Omega$ $P_{\text{copper}} = (5.760 \times 2.50) + (2304.0 \times 0.0060) = 14.40 \text{ W} + 13.824 \text{ W} = 28.224 \text{ W}$
• Total losses (P_{\text{loss}}): $P_{\text{loss}} = P_{\text{core}} + P_{\text{copper}} = 150.0 \text{ W} + 28.224 \text{ W} = 178.224 \text{ W}$
• Operating Efficiency (\eta): $\eta = \frac{P_{\text{out}}}{P_{\text{out}} + P_{\text{loss}}} \times 100\% = \frac{9216.0 \text{ W}}{9216.0 \text{ W} + 178.224 \text{ W}} \times 100\%$ $\eta = \frac{9216.0}{9394.224} \times 100\% = 98.10\%$

Final Answer: (a = 20), (V_2 = 240 \text{ V}), (I_2 = 48.0 \text{ A}), (I_1 = 2.40 \text{ A}), reflected primary impedance (\mathbf{Z}1' = 1600 + j 1200 \ \Omega), output power (P{\text{out}} = 9.216 \text{ kW}), and efficiency (\eta = 98.10%).