Motor Effect

Lessons › 5 Electricity and Magnetism › 5.3 ElectromagnetismPrevious lessonNext lesson

5.3.4 Motor Effect

By the end of this lesson you can:

  • Calculate the force on a current-carrying wire in a magnetic field
  • Use the right-hand rule to find the direction of the force
  • Explain the factors that affect the force
Find your course: NSW Year 11NSW Year 12: Module 6, IQ2QLD Unit 3: ElectromagnetismVIC Unit 3: Fields and motionAP Physics 2: Unit 12IB: D.3 Motion in electromagnetic fields
🎥 Watch📝 Practise🕹️ Explore🎯 Exam ready🍎 Teachers
🎥Watch
Learn the idea: a short explanation, then a check.

Because a current-carrying conductor generates its own magnetic field, placing it within an external magnetic field causes the two magnetic fields to interact. As a result, the conductor experiences a mechanical force—a phenomenon known as the motor effect.

The Mathematical Model

The magnitude of the force (F) exerted on a straight wire depends on four key factors: the strength of the magnetic field (B), the electric current (I), the length of the conductor inside the field (l), and the angle (θ) between the wire and the field lines:

F = BIl sin(θ)

Where:

  • F is the magnetic force in Newtons (N)
  • B is the magnetic flux density in Teslas (T)
  • I is the conventional current in Amperes (A)
  • l is the length of the wire within the field in meters (m)
  • θ is the angle between the wire and the magnetic field vectors

The direction of the resulting force is always perpendicular to both the direction of the current and the magnetic field lines, as determined by the Right-Hand Palm Rule (or Fleming's Left-Hand Rule). Maximum force occurs when the wire is perpendicular to the field (θ = 90°), whereas no force is exerted if the wire is parallel to the field (θ = 0°).

Start by watching the introductory video below to see the motor effect in action.

📝Practise
Worked examples first, then problems to try, then worksheets.
Some problems
Download the sheet to practise using the formula F = Bilsinθ
Motor effect worksheet
🎥Watch
Learn the idea: a short explanation, then a check.

The force between two wires

The Force Between Parallel Conductors

Since any current-carrying conductor generates its own magnetic field, placing a second current-carrying wire parallel to it creates a mutual magnetic interaction. Each conductor finds itself immersed in the magnetic field produced by the other, causing both to experience equal and opposite magnetic forces in accordance with Newton's Third Law.

Quantitative Model

The magnitude of the force per unit length (F⁄l) exerted between two parallel conductors carrying currents I1 and I2 separated by a distance r is given by:

F⁄l = μ0 I1 I2⁄2πr    &text{or}    F⁄l = k · I1 I2⁄r

Where:

  • F/l is the force per unit length in Newtons per meter (N/m)
  • I1 and I2 are the currents in each conductor in Amperes (A)
  • r is the distance between the parallel wires in meters (m)
  • μ0 is the magnetic permeability of free space (4π × 10-7 T·m/A)
  • k is the magnetic constant (μ0⁄2π = 2 × 10-7 N/A2)

Direction of Force (Attraction vs. Repulsion)

Applying the Right-Hand Grip Rule and Right-Hand Palm Rule reveals two fundamental interaction rules:

  • Parallel Currents (Same Direction): The wires attract each other.
  • Anti-parallel Currents (Opposite Directions): The wires repel each other.

Watch the video below to explore how these field interactions are modeled and how this relationship historically defined the SI unit of electric current—the Ampere.

📝Practise
Worked examples first, then problems to try, then worksheets.
A sample problem
Picture
Check your understanding

Check Your Understanding: Force Between Current-Carrying Wires

1. Two long, straight, parallel wires carry currents in the same direction. What is the force between the wires?
2. Two parallel wires carry steady currents. The wires are moved further apart, and nothing else is changed. What happens to the size of the force between them?

Practice Problems: Force Between Parallel Current-Carrying Wires

For two long parallel wires, the force per metre is F / L = μ0I1I2 / (2πd), where μ0 = 4π × 10-7 T m A-1. Currents in the same direction attract. Currents in opposite directions repel.

1. Two long, straight, parallel wires are 0.10 m apart. They carry currents of 5.0 A and 8.0 A in the same direction. What is the force per metre between the wires, and do the wires attract or repel?

Given: I1 = 5.0 A, I2 = 8.0 A, d = 0.10 m

Formula: F / L = μ0I1I2 / (2πd)  (μ0 / 2π = 2.0 × 10-7 T m A-1)

Calculation: F / L = (2.0 × 10-7 × 5.0 × 8.0) / 0.10 = 8.0 × 10-5 N/m

Direction: The currents are in the same direction, so the wires attract.

Answer: 8.0 × 10-5 N/m, attraction
2. Two parallel wires, 20 cm apart, each carry a current of 15 A in opposite directions. What is the magnitude of the force on a 2.5 m length of one of the wires, and is it attractive or repulsive?

Given: I1 = I2 = 15 A, d = 20 cm = 0.20 m, L = 2.5 m

Step 1 (Force per metre): F / L = (2.0 × 10-7 × 15 × 15) / 0.20 = 2.25 × 10-4 N/m

Step 2 (Force on 2.5 m): F = (F / L) × L = 2.25 × 10-4 × 2.5 = 5.6 × 10-4 N

Direction: The currents are in opposite directions, so the wires repel.

Answer: 5.6 × 10-4 N, repulsion
3. A wire carries a current of 12 A. A second wire runs parallel to it, 5.0 cm away. The force per metre between the wires is 4.8 × 10-5 N/m. What current flows in the second wire?

Given: I1 = 12 A, d = 5.0 cm = 0.050 m, F / L = 4.8 × 10-5 N/m

Rearrange: I2 = (F / L) × d / (2.0 × 10-7 × I1)

Calculation: I2 = (4.8 × 10-5 × 0.050) / (2.0 × 10-7 × 12) = (2.4 × 10-6) / (2.4 × 10-6) = 1.0 A

Answer: 1.0 A
4. Two parallel wires carry currents of 20 A and 30 A in the same direction. The force per metre between them is 1.2 × 10-3 N/m. How far apart are the wires?

Given: I1 = 20 A, I2 = 30 A, F / L = 1.2 × 10-3 N/m

Rearrange: d = (2.0 × 10-7 × I1 × I2) / (F / L)

Calculation: d = (2.0 × 10-7 × 20 × 30) / (1.2 × 10-3) = (1.2 × 10-4) / (1.2 × 10-3) = 0.10 m

Answer: 0.10 m (10 cm)
5. A long horizontal wire carries a current of 50 A. A second, lighter wire lies parallel to it, 2.0 cm above, carrying current in the opposite direction. The second wire has a mass of 1.0 g per metre, and it floats in the air because the magnetic force balances its weight. What current does the second wire carry? (g = 9.8 m/s2)

Given: I1 = 50 A, d = 2.0 cm = 0.020 m, mass per metre = 1.0 × 10-3 kg/m

Step 1 (Balance the forces): The currents are opposite, so the wires repel and the upward magnetic force balances the weight. For each metre, F / L = (m / L)g = 1.0 × 10-3 × 9.8 = 9.8 × 10-3 N/m

Step 2 (Find the current): I2 = (F / L) × d / (2.0 × 10-7 × I1) = (9.8 × 10-3 × 0.020) / (2.0 × 10-7 × 50) = (1.96 × 10-4) / (1.0 × 10-5) = 19.6 A

Answer: about 20 A (19.6 A), in the opposite direction to the 50 A current

Two applications


Here are two applications that rely on the motor effect to work.
Useful to view, to consolidate your understanding.
A galvanometer is a device that responds to current. Its designed in such away that in recorded deflection is proportional to the current input.
This is useful, as it allows it to be calibrated so that it can act as an ammeter, or voltmeter
Classic speakers vibrate a cone, which in turn generates sound.
They use an alternating current and the motor effect to do this
An important application of the motor effect is the working of a DC motor. This will addressed in the next lesson
🎯Exam ready
Check you can do it under exam conditions.
 

How well do you know the motor effect?

Hopefully you have understood the concepts if you get here.
​​Test your understanding of motor effect. So do the Quiz and try to get full marks

Then check your understanding if necessary with the video
🍎For teachers
Ideas for using this lesson in class, and where to find the answers.
The motor effect, the foundational concept in all electric motors, is easy to demonstrate with just a few simple items.

​ In this next video I not only demonstrate it qualitatively, I also show how to collect data to establish the magnetic fields of the magnets that are used in this demonstration A simple yet effective demo teachers can use the classroom to help their students
A current balance is a tool used in school to examine the motor effect. Its basically a swing which experiences a force upward on one side due to the motor effect, and a weigh on the other which counter balances it.



​