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Cyclotron deuterons, flux density and energy

Short answer

From the cyclotron resonance condition f = qB/(2πm), flux density B = 2πmf/q = 1.3 T. The exit speed v = 2πfR = 2.0×10⁷ m/s gives kinetic energy K = ½mv² = 6.7×10⁻¹³ J ≈ 4.2 MeV.

Deuterons in a cyclotron describe a circle of radius 0.32 m just before emerging from dees. The frequency of the applied emf's is 10 MHz. Find the flux density of the magnetic field and the energy of deuterons emerging out of the cyclotron. (mass of deuterons = 3.32 × 10⁻²⁷ kg.)

Given: R = 0.32 m, f = 10 MHz = 10⁷ Hz, m = 3.32 × 10⁻²⁷ kg. The deuteron has charge q = +e = 1.6 × 10⁻¹⁹ C.

Flux density

The cyclotron resonance condition is f = qB/(2πm).

B = 2πmf/q = (2π × 3.32 × 10⁻²⁷ × 10⁷) / (1.6 × 10⁻¹⁹) B = (2.086 × 10⁻¹⁹)/(1.6 × 10⁻¹⁹) = 1.30 T

Answer: B ≈ 1.3 T

Energy of emerging deuterons

  1. Speed at the exit: v = 2πfR = 2π × 10⁷ × 0.32 = 2.01 × 10⁷ m/s
  2. K = ½ m v² = ½ × 3.32 × 10⁻²⁷ × (2.01 × 10⁷)²
  3. K = 6.71 × 10⁻¹³ J
  4. In electron volts: K = 6.71 × 10⁻¹³ / 1.6 × 10⁻¹⁹ = 4.19 × 10⁶ eV

Check using K = q²B²R²/(2m) gives the same value.

Answer: K ≈ 6.7 × 10⁻¹³ J ≈ 4.2 MeV (v ≈ 2.0 × 10⁷ m/s, about 7% of the speed of light, so relativistic effects are small).

✓ Answer checked by Claude Opus 5.5 (high effort)

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