Experiment: Factors affecting the force on a current-carrying conductor

Aim

To investigate how the magnitude of the force on a current-carrying conductor in a magnetic field depends on the current.

Variables

• Manipulated: Current in the conductor, I
• Responding: Force on the conductor (shown by the change in the balance reading)
• Constant: The magnetic field strength and the length of the conductor in the field

Apparatus & materials

• Electronic balance
• Magnadur magnets on a steel yoke
• Stiff copper wire fixed horizontally
• Power supply
• Rheostat
• Ammeter
• Switch
• Connecting wires
• Retort stand

Procedure

1. Place the magnet assembly on the electronic balance and set the balance reading to zero (tare).
2. Clamp the stiff copper wire horizontally between the magnet poles, connected in series with the ammeter, rheostat and power supply.
3. Switch on and adjust the rheostat until the current is I = 1.0 A; record the change in the balance reading.
4. Increase the current to I = 2.0, 3.0 and 4.0 A, recording the balance reading each time.
5. Convert each balance reading (a mass) to a force using F = mg.
6. Record the current I and the force F.

Tabulating results

Record the current I in A and the change in the balance reading (a mass) in g or kg, then the calculated force F = mg in N. Keep the decimal places consistent.

The graph

Plot F (y-axis) against I (x-axis). A straight line through the origin shows the force is directly proportional to the current.

Analysis

The force is given by F = BIl. When B and l are constant, F is directly proportional to I, and the gradient of the F against I graph equals Bl.

Electromagnetism · Graph skills

Sample results and what they show

Here is a set of example readings, with the balance tared to zero before the current is switched on (label them as example data):

• I / A = 1.0, balance change m = 0.51 g, F = mg = 0.0050 N
• I / A = 2.0, m = 1.02 g, F = 0.0100 N
• I / A = 3.0, m = 1.53 g, F = 0.0150 N
• I / A = 4.0, m = 2.04 g, F = 0.0200 N

The balance reads a small mass change because the downward push on the magnets is the reaction to the upward force on the wire. As the current doubles from 1.0 A to 2.0 A the force also doubles, and trebling the current trebles the force.

The force is directly proportional to the current, which is exactly what F = BIl predicts when the field B and the length l in the field are kept constant.

Reading the graph and finding the answer

Plot F on the y-axis against I on the x-axis. The expected shape is a straight line passing through the origin, because F = BIl gives F directly proportional to I when B and l are constant.

The gradient of that line equals Bl.

Draw a large triangle on the best-fit line using two well-separated points, for example (1.0 A, 0.0050 N) and (4.0 A, 0.0200 N):

gradient = (0.0200 − 0.0050) N ÷ (4.0 − 1.0) A = 0.0150 N ÷ 3.0 A = 0.0050 N A⁻¹.

This gradient is Bl, in units of N A⁻¹ (which is the same as T m). Before that, each force value comes from the balance reading, for example F = mg = 0.00051 kg × 9.81 m s⁻² = 0.0050 N. If the length in the field is known, say l = 0.050 m, then B = 0.0050 ÷ 0.050 = 0.10 T.

Marks examiners look for

Precautions that protect the result: tare the balance to zero before each set of readings, switch the current on only briefly for each reading so the wire does not heat up, and keep the field strength B and the length l in the field constant while you change only the current.

For the Paper-3 science-process skills:

• Identify the manipulated variable (current I), the responding variable (force F) and the controlled variables (B and l).
• Convert each balance reading to a force with F = mg, showing the unit N.
• Draw a best-fit straight line through the origin with points scattered evenly.
• Use a large triangle for the gradient and give its unit, N A⁻¹.
• Conclude that F is directly proportional to I, supported by the straight line through the origin.

Source: DSKP KSSM Physics Form 4 and 5 (Versi English) (Bahagian Pembangunan Kurikulum (BPK), KPM)