Describe an experiment to show that a force acts on a current-carrying conductor in a magnetic field, including the effect of reversing: (a) the current (b) the direction of the field

4.5.4 Force on a Current‑Carrying Conductor

Learning Objective

Students will be able to:

1. Describe an experiment that shows a magnetic force on a straight conductor placed in a uniform magnetic field.
2. Explain how the direction of the force changes when
   

     (a) the current is reversed, and
   

     (b) the magnetic field direction is reversed.

3. Relate the result to the force on a moving charged particle and to Fleming’s left‑hand rule.

Key Terminology (AO1)

Theory (AO2)

• The magnetic force on a straight conductor of length L carrying a current I in a uniform magnetic field B is given by the vector equation

  \(\displaystyle \mathbf{F}=I\,\mathbf{L}\times\mathbf{B}\)

  where L is a vector directed along the direction of conventional current (not merely the scalar length).

• Scalar (magnitude) form:

  \(\displaystyle F = I L B \sin\theta\)

  \(\theta\) = angle between the direction of current (L) and the magnetic field (B).

• When the conductor is perpendicular to the field (\(\theta=90^{\circ}\)), \(\displaystyle F = I L B\). This shows the proportionality

  \(F\propto I,\;F\propto L,\;F\propto B\)

• Units: I (A), L (m), B (T), therefore F (N).

Fleming’s Left‑Hand Rule (Motor Rule)

Relation to a Moving Charged Particle

A single charge q moving with velocity v in the same field experiences

\(\displaystyle \mathbf{F}=q\,\mathbf{v}\times\mathbf{B}\)

The left‑hand rule for a current‑carrying wire is the macroscopic analogue of the right‑hand rule for a positively charged particle.

Apparatus

• U‑shaped magnet (or a pair of bar magnets) giving a uniform field in the gap.
• Straight insulated copper wire, length ≈ 0.20 m.
• Low‑friction pivot or lightweight wooden plank allowing the wire to swing freely in a vertical plane.
• DC power supply with ammeter (0–5 A range) and a double‑pole switch that can reverse the current.
• Ruler or graduated scale, protractor (optional for angle work).
• Clamp or stand to hold the magnet securely.

Experimental Setup (Diagram)

Procedure (Numbered for Syllabus Alignment)

1. Secure the U‑magnet on a stable surface with the poles facing each other.
2. Place the straight wire across the gap so that it is perpendicular to the field lines.
3. Mount the wire on the pivot; it must be free to swing up or down without obstruction.
4. Connect the circuit: power supply → ammeter → wire → double‑pole switch → power supply.
5. Close the switch in the “forward” position. Record the current I from the ammeter.
6. Observe and measure the vertical deflection d (mm) of the wire (or the angular deflection \(\alpha\) with a protractor).
7. Reverse the current by flipping the double‑pole switch. Record the new direction of deflection.
8. Return the current to its original direction, then rotate the magnet 180° so that the magnetic field direction is reversed. Record the deflection.
9. Repeat steps 5–8 for at least three different current values (e.g., 0.5 A, 1.0 A, 1.5 A). For each current, perform the three combinations:
   

     – original field & forward current
   

     – original field & reversed current
   

     – reversed field & forward current.

10. Optional (angle experiment): Tilt the wire by a known angle \(\theta\) (30°, 45°, 60°) to the field and repeat a single current value to verify the \(\sin\theta\) dependence.

Observations Table

Data Handling (AO2)

• Plot deflection \(d\) (or \(\alpha\)) against current \(I\) for a given field direction. The graph should be a straight line through the origin, confirming \(F\propto I\) (and therefore \(d\propto I\)).
• If the optional angle work is done, plot \(F\) (derived from torque) versus \(\sin\theta\); the slope should equal \(I L B\).
• Use the slope of the linear fit to estimate the product \(L B\) and compare with the known magnet field strength.

Sample Calculation – Converting Deflection to Force

Assume the wire is attached to a light horizontal beam that pivots at its centre. The beam has a torsional spring constant \(k = 0.02\;\text{N·m rad}^{-1}\).

1. Measured vertical displacement \(d = 8\;\text{mm}\); lever arm \(r = 0.10\;\text{m}\). Approximate angular deflection

   \(\displaystyle \alpha \approx \frac{d}{r}= \frac{8\times10^{-3}}{0.10}=0.08\;\text{rad}\).

2. Torque produced: \(\tau = k\alpha = 0.02\times0.08 = 1.6\times10^{-3}\;\text{N·m}\).
3. Force at the wire: \(F = \tau / r = \dfrac{1.6\times10^{-3}}{0.10}=1.6\times10^{-2}\;\text{N}\).
4. Theoretical force (with \(B = 0.40\;\text{T}\), \(L = 0.20\;\text{m}\), \(I = 1.0\;\text{A}\)):

   \(\displaystyle F_{\text{theory}} = I L B = 1.0\times0.20\times0.40 = 8.0\times10^{-2}\;\text{N}\).

5. The measured value is smaller because the simple torque balance neglects friction and the fact that the magnetic field is not perfectly uniform. Nevertheless, the linear relationship with \(I\) is evident.

Explanation of Results

• Reversing the current changes the direction of the vector \(\mathbf{L}\); the cross‑product \(\mathbf{L}\times\mathbf{B}\) therefore changes sign, so the wire deflects in the opposite direction.
• Reversing the magnetic field flips \(\mathbf{B}\); again the sign of \(\mathbf{L}\times\mathbf{B}\) reverses, giving an opposite deflection.
• The magnitude of the deflection grows linearly with the current, confirming \(F\propto I\). With constant \(L\) and \(B\), doubling \(I\) roughly doubles the deflection.
• When the wire is inclined at an angle \(\theta\) to the field, the measured force follows \(F = I L B \sin\theta\); a plot of \(F\) against \(\sin\theta\) should be a straight line through the origin.

Evaluation (AO3)

Consider the following sources of error and suggest how they could be minimised:

• Non‑uniform magnetic field – Use the central region of a well‑shaped U‑magnet or a calibrated Helmholtz coil.
• Friction at the pivot – Use a low‑friction knife‑edge or a lightweight air‑cushioned bearing.
• Current measurement tolerance – Read the ammeter at the centre of the scale and repeat measurements.
• Heating of the wire – Keep the current below the wire’s rating and limit the duration of each run.
• Parallax when reading deflection – Position the ruler directly in line with the wire and view from directly above.

Safety and Precautions

• Do not exceed the current rating of the copper wire (typically 2 A for 1 mm²) to avoid overheating.
• Keep the supply voltage low (≤ 5 V) to minimise shock risk.
• Secure the magnet firmly; it can attract nearby metal objects suddenly.
• Check all connections before switching on the supply to prevent sparks.
• Never touch the wire while current is flowing.

Extension / Enquiry Questions

1. Predict the direction of the force on a positively charged particle moving horizontally into the page through the same magnetic field. Use the right‑hand rule and compare with the conductor result.
2. How would the deflection change if the length of the conductor were doubled while keeping \(I\) and \(B\) constant? Explain using the formula.
3. Design a simple galvanometer using a coil of wire suspended in a magnetic field. Relate its operation to the principle demonstrated above.
4. (Optional) Perform the angle experiment, plot \(F\) versus \(\sin\theta\), and interpret the slope.

Summary

The experiment provides clear, observable evidence that a current‑carrying conductor experiences a magnetic force when placed in a uniform magnetic field. Reversing either the current or the magnetic field reverses the direction of the force, while the magnitude remains proportional to the product \(I L B\) (and to \(\sin\theta\) when the conductor is not perpendicular). Fleming’s left‑hand rule and the particle‑beam analogue give quick, reliable ways to predict the force direction. The quantitative relationship can be verified by measuring deflection for different currents, field directions, and angles, satisfying all requirements of Cambridge IGCSE/A‑Level syllabus 4.5.4.

Syllabus Alignment Table