31  The Motor Effect

31.1 Syllabus inquiry question

  • Under what circumstances is a force produced on a current-carrying conductor in a magnetic field?
Feynman Insight

From The Feynman Lectures on Physics, Vol II, Chapter 13:

When you pass current through a wire in a magnetic field, the magnetic force on the moving charges is transmitted to the wire itself. This is the motor effect—the foundation of all electric motors.

31.2 Learning Objectives

  • Calculate force on a current-carrying conductor: \(F = BIl\sin\theta\)
  • Apply the right-hand rule to determine force direction
  • Analyse force between parallel current-carrying wires
  • Define the ampere from the force between conductors

31.3 Content

31.3.1 Force on a Current-Carrying Conductor

When a current-carrying wire is placed in a magnetic field, it experiences a force:

\[F = BIl\sin\theta\]

where: - \(F\) = force on conductor (N) - \(B\) = magnetic field strength (T) - \(I\) = current (A) - \(l\) = length of conductor in field (m) - \(\theta\) = angle between current direction and field

Key Conditions
  • Maximum force when current is perpendicular to field (\(\theta = 90°\))
  • Zero force when current is parallel to field (\(\theta = 0°\) or 180°)
  • Force direction is perpendicular to both I and B

31.3.2 Interactive: Force on a Current-Carrying Wire

Visualise the motor effect force on a wire in a magnetic field:

31.3.3 The Right-Hand Rule

To find the force direction, use the right-hand palm rule:

  1. Point fingers in the direction of the magnetic field (B)
  2. Bend fingers in the direction of conventional current (I)
  3. Thumb points in the direction of the force (F)

Alternative: FBI rule with right hand: - First finger → Field (B) - Second finger → Current (I) - Thumb → Force/thrust (F)

31.3.4 Force Between Parallel Wires

Two parallel wires carrying currents exert forces on each other:

\[\frac{F}{l} = \frac{\mu_0}{2\pi}\frac{I_1 I_2}{r}\]

where: - \(F/l\) = force per unit length (N/m) - \(\mu_0 = 4\pi \times 10^{-7}\) T·m/A (permeability of free space) - \(I_1, I_2\) = currents in each wire (A) - \(r\) = separation between wires (m)

Attraction vs Repulsion
  • Parallel currents attract (currents in same direction)
  • Antiparallel currents repel (currents in opposite directions)

31.3.5 Interactive: Parallel Wires

31.3.6 Definition of the Ampere

The ampere is defined based on the force between parallel wires:

One ampere is the constant current which, if maintained in two straight parallel conductors of infinite length, placed 1 metre apart in vacuum, produces a force of exactly \(2 \times 10^{-7}\) N per metre of length.

Physical Basis

This definition connects the unit of current directly to the measurable force between wires, making it a fundamental SI unit.

31.3.7 Understanding Why Parallel Currents Attract

Consider two wires with currents in the same direction:

  1. Wire 1 creates a circular magnetic field around itself
  2. At the location of Wire 2, this field points in a specific direction
  3. The current in Wire 2, in this external field, experiences a force toward Wire 1
  4. By Newton’s third law, Wire 1 experiences an equal force toward Wire 2

For opposite currents, the field direction reverses, and the force becomes repulsive.

31.4 Worked Examples

31.4.1 Example 1: Force on a conductor

A wire of length 0.12 m carries 3.0 A perpendicular to a 0.40 T magnetic field. Find the force on the wire.

Solution:

  1. Use \(F = BIl\sin\theta\) with \(\theta = 90°\)

  2. \(F = 0.40 \times 3.0 \times 0.12 \times 1\)

  3. \(F = 0.144\ \text{N}\)

31.4.2 Example 2: Force at an angle

A wire carrying 2.5 A makes an angle of 30° with a 0.50 T field. The wire length in the field is 0.08 m. Find the force.

Solution:

  1. Use \(F = BIl\sin\theta\)

  2. \(F = 0.50 \times 2.5 \times 0.08 \times \sin(30°)\)

  3. \(F = 0.50 \times 2.5 \times 0.08 \times 0.5 = 0.050\ \text{N}\)

31.4.3 Example 3: Force between parallel wires

Two parallel wires 5.0 cm apart carry currents of 2.0 A and 3.0 A in the same direction. Find the force per metre between them.

Solution:

  1. Use \(\frac{F}{l} = \frac{\mu_0}{2\pi}\frac{I_1 I_2}{r}\)

  2. \(\frac{F}{l} = \frac{4\pi \times 10^{-7}}{2\pi} \times \frac{2.0 \times 3.0}{0.050}\)

  3. \(\frac{F}{l} = 2 \times 10^{-7} \times 120 = 2.4 \times 10^{-5}\ \text{N/m}\)

  4. Direction: attractive (same direction currents)

31.4.4 Example 4: Determining current from force

Two parallel wires 0.10 m apart experience an attractive force of \(3.0 \times 10^{-5}\) N per metre. One wire carries 5.0 A. Find the current in the other wire.

Solution:

  1. Rearrange: \(I_2 = \frac{F/l \times 2\pi r}{\mu_0 I_1}\)

  2. \(I_2 = \frac{3.0 \times 10^{-5} \times 2\pi \times 0.10}{4\pi \times 10^{-7} \times 5.0}\)

  3. \(I_2 = \frac{1.88 \times 10^{-5}}{6.28 \times 10^{-6}} = 3.0\ \text{A}\)

31.4.5 Example 5: Force direction using right-hand rule

A horizontal wire carries current toward the east in a magnetic field pointing vertically upward. Determine the direction of the force.

Solution:

  1. Current (I) → East

  2. Field (B) → Upward

  3. Apply right-hand rule: fingers point up (B), curl toward east (I)

  4. Thumb points South

  5. Force on wire is toward the South

31.5 Common Misconceptions

Common Misconceptions

  • Misconception: The force on a wire is along the direction of the current. Correction: The force is perpendicular to both the current and magnetic field.

  • Misconception: Opposite currents attract because opposites attract. Correction: This is not like electric charges. Parallel (same direction) currents attract; antiparallel (opposite direction) currents repel.

  • Misconception: Only the length of wire in the field matters, not the angle. Correction: The angle between current and field is crucial: \(F = BIl\sin\theta\).

  • Misconception: A stronger current always means a stronger force. Correction: True only if other factors (B, l, θ) remain constant.

  • Misconception: The motor effect requires permanent magnets. Correction: Any magnetic field works—from permanent magnets, electromagnets, or even Earth’s field.

31.6 Practice Questions

31.6.1 Easy (2 marks)

A wire of length 0.15 m carries 4.0 A perpendicular to a 0.30 T field. Calculate the force on the wire.

  • Use \(F = BIl\sin(90°) = BIl\) (1)
  • \(F = 0.30 \times 4.0 \times 0.15 = 0.18\) N (1)

Answer: 0.18 N

31.6.2 Medium (4 marks)

Two parallel wires separated by 4.0 cm carry currents of 6.0 A and 8.0 A. Calculate the force per metre between them and state whether it is attractive or repulsive if the currents are in opposite directions.

  • Use \(F/l = (\mu_0/2\pi)(I_1 I_2/r)\) correctly (1)
  • \(F/l = (4\pi \times 10^{-7}/2\pi)(6.0 \times 8.0/0.04)\) (1)
  • \(F/l = 2 \times 10^{-7} \times 1200 = 2.4 \times 10^{-4}\) N/m (1)
  • Repulsive (opposite direction currents) (1)

Answer: \(F/l = 2.4 \times 10^{-4}\) N/m, repulsive

31.6.3 Hard (5 marks)

A horizontal rectangular coil of 50 turns, with sides 0.08 m × 0.06 m, carries 0.25 A in a vertical magnetic field of 0.40 T. Calculate the maximum force on one side and explain why the net force on the coil is zero but there is a net torque.

  • Force on one side: \(F = BIln = 0.40 \times 0.25 \times 0.08 \times 50 = 0.40\) N (2)
  • Forces on opposite sides are equal and opposite → net force = 0 (1)
  • Forces create a couple (separated and opposite) → produces torque (1)
  • Torque causes rotation even though net force is zero (1)

Answer: \(F = 0.40\) N; net force is zero because opposite sides have opposite forces, but these create a couple producing rotation.

31.7 Multiple Choice Questions

Test your understanding with these interactive questions:

31.8 Summary

Key Takeaways
  • Force on conductor: \(F = BIl\sin\theta\) (maximum when θ = 90°)
  • Use right-hand rule: fingers B, curl to I, thumb = F direction
  • Parallel wires: \(F/l = (\mu_0/2\pi)(I_1 I_2/r)\)
  • Same direction currents → attractive force
  • Opposite direction currents → repulsive force
  • The ampere is defined by the force between parallel conductors

31.9 Self-Assessment

Check your understanding:

After studying this section, you should be able to: