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.
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.
Velocity selection in crossed fields 17(m)
Question 1
Crossed fields have E = 6.0 × 10⁴ V m⁻¹ and B = 0.20 T. Find the undeflected speed and state the required force directions.
Check the model response
For opposite electric and magnetic forces, qE = qvB, so v = E/B = 3.0 × 10⁵ m s⁻¹. The selected beam direction and field orientations must make qE and qv × B antiparallel.
repair
2. Repair the common breaks
Use only the correction matching an error, then retry the corresponding diagnostic.
Velocity selection in crossed fields 17(m)
Check this idea
Misconception: Every particle passes undeflected through crossed fields.
Repair: Only particles with the geometry and speed v = E/B have opposing forces of equal magnitude.
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.
Velocity selection in crossed fields 17(m)
Model 1
Explain why a crossed-field selector transmits a narrow speed independent of charge magnitude and mass.
Check the model response
With forces opposed, |Q|E = B|Q|v, so v = E/B and |Q| cancels. Mass never enters the balance. Charge sign reverses both forces, leaving their opposition unchanged for the same geometry.
guided practice
4. Guided practice
Use each hint only to select the correct geometry, sine component, direction rule or force balance.
Velocity selection in crossed fields 17(m)
Question 1
E doubles and B halves. State the selected-speed factor.
Hint: Treat numerator and denominator changes separately.
Check the model response
v = E/B, so the factor is 2/(1/2) = 4.
independent practice
5. Independent practice
Solve without repair notes and state every long-wire, flat-coil, long-solenoid, uniform-field and perpendicular-motion assumption.
Velocity selection in crossed fields 17(m)
Question 1
State all geometric and idealising conditions behind v = E/B.
Check the model response
Uniform E and B must be mutually perpendicular and both perpendicular to the selected beam velocity, with electric and magnetic forces opposite. Other forces and collisions are neglected, and particles must traverse the common field region.
Practice exit check
6. Practice assessment
Use this as extra closed-book practice, then complete the separate recorded assessment in your plan.
Velocity selection in crossed fields 17(m)
Question 1
Ions pass undeflected through E = 4.5 × 10⁴ V m⁻¹ and B = 0.15 T. Find speed and explain why ions above it deflect magnetically more strongly.
Check the model response
v = E/B = 3.0 × 10⁵ m s⁻¹. At fixed E and B, electric force is qE while magnetic force qvB grows with speed, so above the selected speed the magnetic force dominates.
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.
Velocity selection in crossed fields 17(m)
Question 1
A selector transmits 2.0 × 10⁵ m s⁻¹ with B = 0.30 T. Find E.
Check the model response
E = vB = (2.0 × 10⁵)(0.30) = 6.0 × 10⁴ V m⁻¹.