id: phys-ex-03a
type: calculation
difficulty: hard
points: 20
related_lesson: phys-03
question: |
  A burst-mode coil has ε = 60 J/m. To reach L = 1.5 m in a 200 μs pulse,
  what power is required? Is this realistic for a burst-mode Tesla coil?

hints:
  - "Use growth rate equation: dL/dt = P/ε"
  - "Rearrange: P = ε × dL/dt"
  - "Calculate dL/dt = L/T for the pulse duration"
  - "Consider typical DRSSTC power levels"

solution:
  steps:
    - "Calculate growth rate needed:"
    - "dL/dt = L / T = 1.5 m / (200×10⁻⁶ s) = 7,500 m/s"
    - "Calculate required power:"
    - "P = ε × dL/dt"
    - "P = 60 J/m × 7,500 m/s"
    - "P = 450,000 W = 450 kW"
    - "Analysis of realism:"
    - "Energy per pulse: E = P × T = 450 kW × 200 μs = 90 J"
    - "For primary: C = 0.5 μF, need V² = 2E/C = 360,000, so V ≈ 600 V"
    - "Peak power: 450 kW is high but achievable for large DRSSTC"
    - "Conclusion: Challenging but realistic for large coil"

  answer: "450"
  unit: "kW"
  energy_per_pulse: "90"
  realistic: "yes, but requires large DRSSTC"
  tolerance: 5.0

explanation: |
  Growing 1.5 m in just 200 μs requires extremely high instantaneous power
  (450 kW). However, the total energy per pulse is only 90 J, which is achievable
  with a 0.5 μF primary capacitor charged to 600 V. This high power/short duration
  trade-off is characteristic of burst mode operation. The high ε = 60 J/m reflects
  inefficiency (branching, radiation) in burst mode. A QCW coil with ε = 10 J/m
  would need only 75 kW for the same growth rate, or could grow the same length
  with less power over a longer time.

related_concepts: ["energy-per-meter", "growth-rate", "burst-mode", "power-requirements"]