Electromagnetic forces: currents, fields and beams

Key idea: Keep field production, conductor force and moving-charge force distinct, apply direction rules with charge sign, and compare electric and magnetic beam deflection before balancing crossed fields.

  • H2 Physics 9478 · 2027
  • Internally reviewed by MiniEducation Team
  • Recorded selected-response study loop available

Before you start: Current Electricity objective chainD.C. Circuits objective chainElectric Fields objective chain

By the end, you can

  • Represent and calculate magnetic fields produced by straight wires, flat coils and long solenoids.
  • Analyse conductor forces, define flux density, use a current balance and predict parallel-current interactions.
  • Calculate and direct magnetic force on moving positive or negative charges.
  • Compare charged-beam deflection in uniform electric and magnetic fields.
  • Explain and calculate crossed-field velocity selection.

Starting-point self-check

1. Check your starting point

Attempt all five groups without notes and mark the first geometry, direction, force, trajectory or balance decision you cannot justify. Use the recorded topic diagnostic above when you want scoring and a personalised repair plan.

Conductor force, flux density, current balance and parallel currents 17(e)–(i)

Question 1

A 0.12 m wire carries 4.0 A perpendicular to a 0.30 T field. Find the force and state how a current balance could measure B.

Check the model response

F = BIl = 0.144 N. Balance the magnetic force with a measured weight change Δmg; plotting Δmg against Il gives gradient B when the wire is perpendicular.

repair

2. Repair the common breaks

Use only the correction matching an error, then retry the corresponding diagnostic.

Conductor force, flux density, current balance and parallel currents 17(e)–(i)

Check this idea

Misconception: Magnetic force is parallel to current.

Repair: F = BIl sin θ acts perpendicular to both conventional current and B; use Fleming's left-hand rule for direction.

Check this idea

Misconception: Parallel currents in the same direction repel.

Repair: Same-direction currents attract; opposite-direction currents repel.

worked example

3. Follow five worked models

Follow how each solution fixes source geometry, conventional direction, charge sign, field uniformity or force opposition before calculating.

Conductor force, flux density, current balance and parallel currents 17(e)–(i)

Model 1

Two long parallel wires 5.0 cm apart carry 6.0 A in the same direction. Find force per unit length and explain its direction.

Check the model response

Field from one at the other is μ₀I/(2πd) = 2.4 × 10⁻⁵ T. F/l = BI = 1.44 × 10⁻⁴ N m⁻¹. The right-hand grip rule plus conductor-force direction gives attraction for currents in the same direction.

guided practice

4. Guided practice

Use each hint only to select the correct geometry, sine component, direction rule or force balance.

Conductor force, flux density, current balance and parallel currents 17(e)–(i)

Question 1

A wire turns from 90° to 30° relative to B at fixed B, I and l. State the force factor.

Hint: θ is the angle between current direction and field.

Check the model response

F ∝ sin θ, so the factor is sin30°/sin90° = 1/2.

independent practice

5. Independent practice

Solve without repair notes and state every long-wire, flat-coil, long-solenoid, uniform-field and perpendicular-motion assumption.

Conductor force, flux density, current balance and parallel currents 17(e)–(i)

Question 1

Define magnetic flux density operationally and explain how field superposition produces forces between parallel currents.

Check the model response

For a conductor perpendicular to B, flux density is force per unit current per unit length: B = F/(Il). Each wire produces a circular field at the other; applying F = BIl and the direction rule gives attraction for same-direction currents and repulsion for opposite currents.

Practice exit check

6. Practice assessment

Use this as extra closed-book practice, then complete the separate recorded assessment in your plan.

Conductor force, flux density, current balance and parallel currents 17(e)–(i)

Question 1

A current balance uses l = 0.080 m and I = 2.5 A; rebalancing needs 3.06 g. Find B using g = 9.81 m s⁻².

Check the model response

F = Δmg = 0.00306(9.81) = 0.0300 N. B = F/(Il) = 0.0300/[2.5(0.080)] = 0.150 T.

Re-test practice

7. Delayed re-test practice

Return after at least three days and solve these fresh contexts without reopening earlier responses. The recorded plan enforces the delay and uses a separate re-test family for selected-response skill-group evidence.

Conductor force, flux density, current balance and parallel currents 17(e)–(i)

Question 1

A perpendicular wire's current and active length both double. State the force factor.

Check the model response

F = BIl, so the force increases by 2 × 2 = 4.

Continue with established practice

Use the established seven-question structured set after the delayed re-test. It samples all five groups; the full three-geometry field-sketch requirement remains assessed in this chain, lessons and quiz.

Open Electromagnetic Forces structured practice