Year 12 Physics Module 6: Electromagnetism IQ4: Applications Lessons 16–20 45 min

Checkpoint 4: Applications of Electromagnetism

Test your understanding of AC induction motors, DC motors, mass spectrometers, eddy currents, and power transmission. Covers Lessons 16–20.

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Lesson 16

AC Induction Motors and Generators

  • Rotating stator field induces rotor currents
  • Squirrel cage rotor; slip is necessary
  • Synchronous speed: $n_s = 120f/p$
Lesson 17

DC Motors in Depth

  • Back emf and its effect on current
  • Stall current vs running current
  • Motor efficiency calculations
Lesson 18

Mass Spectrometers

  • Velocity selector: $v = E/B$
  • Magnetic analysis: $r = mv/(qB)$
  • Separation by mass-to-charge ratio
Lessons 19–20

Eddy Currents and Power Transmission

  • Eddy currents: applications and losses
  • High-voltage transmission: $P_{loss} = I^2 R$
  • Transformers in the grid
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Essential Formulae — Applications

$n_s = \\dfrac{120f}{p}$Synchronous speed (RPM)
$I = \\dfrac{V - \\varepsilon_{back}}{R}$DC motor current
$r = \\dfrac{mv}{qB}$Radius in mass spectrometer
$v = \\dfrac{E}{B}$Velocity selector
$P_{loss} = I^2 R$Transmission line loss
Key Terms
SlipDifference between synchronous and rotor speed in induction motors
Back emfInduced voltage opposing applied voltage; proportional to speed
Mass spectrometerDevice separating ions by mass using electric and magnetic fields
Eddy currentInduced current in conductors; useful for braking, unwanted in cores
LaminationThin insulated layers reducing eddy current losses in cores
Step-up transformerIncreases voltage for efficient power transmission
Multiple Choice — 10 Questions

1. A 4-pole induction motor on 50 Hz has synchronous speed:

A3000 RPM
B1500 RPM ($n_s = 120 \\times 50 / 4$)
C1000 RPM
D750 RPM

2. In an AC induction motor, the rotor currents are produced by:

ADirect connection to the stator
BElectromagnetic induction from the rotating stator field
CA battery in the rotor
DFriction

3. A DC motor connected to 24 V has coil resistance 4.0 ohms. At full speed, back emf is 20 V. The running current is:

A6.0 A
B1.0 A ($I = (24-20)/4.0$)
C5.0 A
D1.2 A

4. A velocity selector has E = 2000 V/m and B = 0.10 T. The speed of ions passing through undeflected is:

A2.0 x 10^3 m/s
B2.0 x 10^4 m/s ($v = E/B = 2000/0.10$)
C2.0 x 10^5 m/s
D5.0 x 10^-5 m/s

5. In a mass spectrometer, ions with larger mass-to-charge ratio:

AHave smaller radius
BHave larger radius of curvature
CTravel faster through the selector
DExperience less magnetic force

6. Magnetic braking is smooth and wear-free because:

AIt uses friction pads
BThere is no physical contact between components
CIt uses hydraulic fluid
DIt relies on air resistance

7. 400 MW transmitted at 400 kV through lines with R = 4.0 ohms. Power loss is:

A40 MW
B4.0 MW ($I = 10^6$ A; $P = (10^6)^2 \\times 4$ wait, $I = 400\\times10^6/400\\times10^3 = 1000$ A; $P = 1000^2 \\times 4 = 4\\times10^6$ W)
C16 MW
D0.4 MW

8. Transformer cores are laminated to:

AIncrease magnetic field strength
BReduce eddy current energy losses
CMake them lighter
DImprove electrical insulation

9. A DC motor draws more current when loaded because:

ABack emf decreases when speed drops
BCoil resistance decreases
CApplied voltage increases
DMagnetic field strengthens

10. A step-up transformer with Np = 100, Ns = 500, Vp = 20 V produces:

A4 V
B100 V ($V_s = 20 \\times 500/100$)
C20 V
D500 V
Short Answer — 2 Questions

1. (4 marks) A DC motor has coil resistance 2.5 ohms and is connected to 15 V. At full speed, back emf is 12 V.

  • Calculate the current at startup and at full speed. (2 marks)
  • Explain why a starting resistor might be used in a large DC motor. (2 marks)

2. (4 marks) A power station generates 600 MW at 25 kV. This is stepped up to 500 kV for transmission through lines with total resistance 2.5 ohms.

  • Calculate the current in the transmission lines. (1 mark)
  • Calculate the power loss in the lines. (2 marks)
  • Calculate the percentage of power lost. (1 mark)
Answers

Multiple Choice: 1-B, 2-B, 3-B, 4-B, 5-B, 6-B, 7-B, 8-B, 9-A, 10-B

Short Answer 1: (a) I_start = 15/2.5 = 6.0 A; I_run = (15-12)/2.5 = 1.2 A. (b) At startup, back emf is zero and current is V/R, which can be 5-10 times the running current. This surge can overheat coils and damage the commutator. A starting resistor limits the initial current to a safe value.

Short Answer 2: (a) I = P/V = 600x10^6 / 500x10^3 = 1200 A. (b) P_loss = I^2 R = 1200^2 x 2.5 = 3.6x10^6 W = 3.6 MW. (c) % loss = (3.6/600) x 100 = 0.6%.