Year 12 Physics Module 6: Electromagnetism IQ2: Motor Effect Lessons 7–10 45 min

Checkpoint 2: The Motor Effect

Test your understanding of the motor effect, forces between parallel conductors, torque on coils, DC motor design, and back emf. Covers Lessons 7–10.

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Lesson 7

The Motor Effect

  • $F = BIL$ on current-carrying conductor
  • Right-hand palm rule for direction
  • Force between parallel current-carrying wires
Lesson 8

Forces Between Parallel Conductors

  • $F/l = \\mu_0 I_1 I_2 / (2\\pi d)$
  • Attractive for same-direction currents
  • Repulsive for opposite-direction currents
Lesson 9

Torque on a Current-Carrying Coil

  • $\\tau = nBAI \\cos \\theta$
  • Split-ring commutator for DC motors
  • Radial magnetic field for smooth torque
Lesson 10

DC and AC Motors

  • Back emf: $\\varepsilon_{back} = NBA\\omega$
  • AC induction motor: rotating stator field
  • Squirrel cage rotor and slip
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Essential Formulae — Motor Effect

$F = BIL$Motor effect force (perpendicular)
$\\frac{F}{l} = \\frac{\\mu_0 I_1 I_2}{2\\pi d}$Force per unit length between parallel wires
$\\tau = nBAI \\cos \\theta$Torque on coil
$\\varepsilon_{back} = NBA\\omega$Back emf in rotating motor
$I = \\frac{V - \\varepsilon_{back}}{R}$Motor current with back emf
Key Terms
Motor effectForce on a current-carrying conductor in a magnetic field
CommutatorSplit-ring device that reverses current for continuous rotation
Back emfInduced voltage opposing supply voltage; increases with speed
Radial fieldMagnetic field always perpendicular to coil plane for smooth torque
Squirrel cageRotor with conducting bars shorted at ends; no brushes needed
SlipDifference between synchronous speed and rotor speed in induction motors
Multiple Choice — 10 Questions

1. A 0.50 m wire carries 4.0 A perpendicular to a uniform magnetic field of 0.20 T. The force on the wire is:

A0.20 N
B0.40 N ($F = BIL = 0.20 \\times 4.0 \\times 0.50$)
C0.80 N
D4.0 N

2. Two parallel wires carry currents in the same direction. The force between them is:

AAttractive
BRepulsive
CZero
DPerpendicular to both wires

3. A coil with 40 turns, area $3.0\\times10^{-3}$ m², carries 2.5 A in B = 0.40 T. Maximum torque is:

A0.040 N m
B0.12 N m ($\\tau = nBIA = 40 \\times 0.40 \\times 2.5 \\times 3.0\\times10^{-3}$)
C0.30 N m
DZero

4. In a DC motor, the split-ring commutator:

AIncreases the magnetic field
BReverses current every half-turn for continuous rotation
CReduces friction
DConverts AC to DC

5. A radial magnetic field in a DC motor is used to:

AKeep torque nearly constant throughout rotation
BReduce the current needed
CIncrease the number of turns
DReverse the magnetic field direction

6. A DC motor connected to 12 V has coil resistance 3.0 ohms. At full speed, back emf is 9.0 V. The running current is:

A4.0 A
B1.0 A ($I = (12-9)/3.0$)
C3.0 A
D7.0 A

7. The back emf of a DC motor is maximum when:

AThe motor is stalled
BThe motor is running at maximum speed
CThe current is maximum
DThe load is suddenly increased

8. An AC induction motor has no commutator because:

AThe AC current already alternates direction
BIt uses permanent magnets
CThe rotor is connected directly to power
DIt does not need to reverse torque

9. Two parallel wires 0.10 m apart carry currents I1 = 5.0 A and I2 = 3.0 A in opposite directions. The force per metre is:

A$1.5\\times10^{-5}$ N/m (attractive)
B$3.0\\times10^{-5}$ N/m (repulsive)
C$3.0\\times10^{-4}$ N/m
D$1.5\\times10^{-4}$ N/m

10. The torque on a coil is zero when:

AThe plane is parallel to the field
BThe plane is perpendicular to the field
CThe current is maximum
DThe coil has maximum area
Short Answer — 2 Questions

1. (4 marks) A DC motor has coil resistance 2.0 ohms and is connected to a 10 V supply. When running at full speed, the back emf is 8.0 V.

  • Calculate the current when the motor is running at full speed. (1 mark)
  • Calculate the current when the motor is stalled. (1 mark)
  • Explain why the stall current is much larger than the running current. (2 marks)

2. (4 marks) A rectangular coil with 80 turns and area $5.0\\times10^{-3}$ m² carries a current of 3.0 A in a uniform magnetic field of 0.25 T.

  • Calculate the maximum torque on the coil. (2 marks)
  • Explain why a DC motor uses a radial magnetic field rather than a uniform field. (2 marks)
Answers

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

Short Answer 1: (a) I = (10-8)/2.0 = 1.0 A. (b) I_stall = 10/2.0 = 5.0 A. (c) At stall, back emf = 0, so the full 10 V drives current through the 2.0 ohm coil. At full speed, back emf of 8 V opposes the applied voltage, leaving only 2 V to drive current.

Short Answer 2: (a) tau_max = nBIA = 80 x 0.25 x 3.0 x 5.0x10^-3 = 0.30 N m. (b) In a uniform field, torque varies as cos theta, dropping to zero when the coil is perpendicular to B. A radial field keeps B perpendicular to the coil plane at all positions, maintaining nearly constant torque and smooth rotation.