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The Physics of Motors and Generators

Motors and generators are complementary devices. A motor converts electrical energy into mechanical energy (rotation), while a generator converts mechanical energy into electrical energy. Both rely on the interaction between magnetic fields and electric currents.

DC Motors: Force and Torque

When a current-carrying coil is placed in a magnetic field, the sides of the coil experience a Lorentz force (F = BIL). This creates a torque that causes the coil to rotate:

τ = N × B × I × A × sin θ

Where:

  • Ï„ (Torque) is measured in Newton-meters (N·m).
  • N is the number of turns in the coil.
  • B is the magnetic field strength (Tesla).
  • I is the current (Amperes).
  • A is the area of the coil (m²).
  • θ is the angle between the coil's normal and the field lines.

AC Generators: Electromagnetic Induction

An AC generator rotates a coil in a magnetic field to induce an alternating voltage. According to Faraday's Law, the induced EMF is the rate of change of magnetic flux linkage:

ε(t) = NBAω sin(ωt)

Where ω is the angular velocity (ω = 2π × frequency). The Peak EMF (ε₀) occurs when sin(ωt) = 1, so ε₀ = NBAω.

Power and Mechanical Relationships

For a rotating motor or generator, the relationship between torque (τ), power (P), and speed (ω) is:

P = τ × ω

If you know the speed in RPM (revolutions per minute), you must convert it to rad/s: ω = (RPM × 2π) / 60.

💡 Pro Tip: Back-EMF
As a motor spins, it also acts as a generator! This "Back-EMF" opposes the supply voltage, limiting the current. This is why a motor draws the most current when it first starts (zero RPM = zero Back-EMF).

Deep Dive: Worked Examples

✅ Example 1: DC Motor Torque

A rectangular coil (10 cm x 5 cm) with 200 turns is in a 0.5 T field. If a current of 3.0 A flows through it, calculate the maximum torque produced.

Given: N = 200, B = 0.5 T, I = 3.0 A, A = 0.1 × 0.05 = 0.005 m²
Formula: Ï„ = NBIA
Solution:
τ = 200 × 0.5 × 3.0 × 0.005
τ = 100 × 3.0 × 0.005 = 1.5 N·m

✅ Example 2: Generator Peak Voltage

An AC generator consists of a 500-turn coil of area 0.02 m² rotating at 3000 RPM in a 0.15 T magnetic field. Calculate the peak voltage output.

Given: N = 500, A = 0.02 m², RPM = 3000, B = 0.15 T
ω: ω = (3000 × 2π) / 60 = 50 × 2π = 314.16 rad/s
Formula: ε₀ = NBAω
Solution:
ε₀ = 500 × 0.15 × 0.02 × 314.16
ε₀ = 1.5 × 314.16 = 471.2 V

✅ Example 3: Motor Power Output

A small DC motor produces 0.25 N·m of torque while spinning at 1200 RPM. Calculate its mechanical power output in Watts.

Given: τ = 0.25 N·m, RPM = 1200
ω: ω = (1200 × 2π) / 60 = 20 × 2π = 125.66 rad/s
Formula: P = τ × ω
Solution:
P = 0.25 × 125.66 = 31.42 W

✅ Example 4: Motor Back-EMF and Current

A motor with internal resistance of 2.0 Ω is connected to a 12V DC source. When spinning at rated speed, it generates a back-EMF of 10.5V. Calculate the current when starting and at rated speed.

Starting: I = V / R = 12 / 2.0 = 6.0 A
Running: I = (V - εback) / R = (12 - 10.5) / 2.0 = 1.5 / 2.0 = 0.75 A
(Note the massive difference in current!)

✅ Example 5: Generator Efficiency

A generator produces 1.5 kW of electrical power. To achieve this, a mechanical engine pulls the generator with 20 N·m of torque at 900 RPM. Calculate the efficiency.

Input Power: Pin = τω = 20 × (900 × 2π / 60) = 20 × 94.25 = 1885 W
Output Power: Pout = 1500 W
Efficiency: η = (Pout / Pin) × 100%
Solution: η = (1500 / 1885) × 100% = 79.6%