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):

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

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 $I_{\text{enc}}$:

• 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):

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:

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:

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:

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 $\eta = 98.10\%$.