Source: High school physics (Chinese)
Problem
A linear accelerator produces an electron beam whose current is not steady but consists of pulses. Each pulse lasts $0.1$ $\mu$s with an average current (during the pulse) of $1.6$ A.
- How many electrons are accelerated in each pulse?
- If there are $1000$ pulses per second, what is the average current of the electron beam?
- If each electron gains an energy of $400$ MeV, what is the average power supplied to the accelerator?
- What is the average power during one pulse?
- What fraction of the total time does the accelerator actually spend accelerating particles? (This ratio is called the duty factor of the accelerator.)
(1) $N = 1.0 \times 10^{12}$ electrons; \quad (2) $\bar{I} = 1.6 \times 10^{-4}$ A; \quad (3) $\bar{P} = 64$ kW; \quad (4) $P_p = 640$ MW; \quad (5) duty factor $\eta = 10^{-4}$ (1 part in 10,000).
Let $\tau = 0.1$ $\mu$s $= 1.0 \times 10^{-7}$ s be the pulse duration, $I_p = 1.6$ A the in-pulse current, $f = 1000$ Hz the pulse repetition rate, $W = 400$ MeV the energy gained per electron, and $e = 1.6 \times 10^{-19}$ C.
(1) Electrons per pulse. Total charge per pulse divided by electron charge:
$$N = \frac{I_p \tau}{e} = \frac{1.6 \times 1.0 \times 10^{-7}}{1.6 \times 10^{-19}} = 1.0 \times 10^{12} \text{ electrons}$$(2) Average current. Total charge per second divided by 1 s:
$$\bar{I} = N e f = I_p \tau f = 1.6 \times 1.0 \times 10^{-7} \times 1000 = 1.6 \times 10^{-4} \text{ A}$$(3) Average power. Each accelerated electron receives $W$ of energy. With $N f$ electrons per second:
$$\bar{P} = N f W = \bar{I} \cdot \frac{W}{e}$$Equivalently, $W/e = 400$ MV is the equivalent accelerating voltage, so $\bar{P} = \bar{I} \times 4 \times 10^{8}$ V:
$$\bar{P} = 1.6 \times 10^{-4} \times 4 \times 10^{8} = 6.4 \times 10^{4} \text{ W} = 64 \text{ kW}$$(4) Power during a pulse. Same formula but with the in-pulse current $I_p$:
$$P_p = I_p \cdot \frac{W}{e} = 1.6 \times 4 \times 10^{8} = 6.4 \times 10^{8} \text{ W} = 640 \text{ MW}$$(5) Duty factor. Fraction of time the beam is on:
$$\eta = \tau f = 1.0 \times 10^{-7} \times 1000 = 1.0 \times 10^{-4} = \frac{1}{10\,000}$$Consistency check: $P_p / \bar{P} = 1/\eta = 10^{4}$, confirming average power is $10^{-4}$ of peak.