Q82P - Fundamentals Of Physics

Question

A linear accelerator produces a pulsed beam of electrons. The pulse current is 0.50 A, and the pulse duration is 0.10 ms. (a) How many electrons are accelerated per pulse? (b) What is the average current for a machine operating at 500 pulses/s ? If the electrons are accelerated to energy of 50 MeV, what are the (c) average power and (d) peak power of the accelerator?

Step-by-Step Solution

Answer

1. The number of accelerated electrons per pulse is $3.1\times {10}^{11}$ .
2. The average current for a machine operating at 500 pulses/s is $25\times {10}^{-6}\mathrm{A}$ .
3. The average power of the accelerator is 1.3kW.
4. The peak power of the accelerator is 25MW.

Step 1: The given data

1. The pulse current is / = $0.50\mathrm{A}$ 
2. The duration of the pulse is $\mathrm{t}=0.10\mathrm{\mu s} \mathrm{or} 0.10\times {10}^{-6}\mathrm{s}$ .
3. The energy of the accelerated electron is $\mathrm{E}=50\mathrm{MeV} \mathrm{or} 50\times 1.6\times {10}^{-19}\mathrm{J}$  

Step 2: Understanding the concept of current and power

We can use the concept of current and power. The current is the rate of flow of charge and power is the rate of energy of the accelerated electron.

Formulae:

The current flowing through the area, $I=\frac{q}{t}$                                             (i)

The average current flow in the charge flow, $I_{avg}=q\times N/s$                     (ii)

The power or the rate of energy transfer during work done, $P=\frac{W}{t}$        (iii)

The total charge on the body, q = ne                                                        (iv)

The potential energy of a system, U = qV                                             (v)

The electric power due to potential difference, P = VI                              (vi)

Step 3: a) Calculation of the number of accelerated electrons per pulse

Charge of electron, $\mathrm{e}=1.6\times {10}^{-19}\mathrm{C}$ .

Using the given data and equations (i) and (iv), we can get the value of the number of the accelerated electrons per pulse as follows:

$$\begin{gathered} n=\frac{It}{e} \\ =\frac{0.50A\times 0.10\times {10}^{-6}s}{1.6\times {10}^{-19}C} \\ =3.1\times {10}^{11} \end{gathered}$$ 

Hence, the value of the number electrons is $3.1\times {10}^{11}$ .

Step 4: b) Calculation of the average current for the operating machine

The expression of the average current is the product of the charge of pulse   and number of pulse per second N/s , then using the given data in equations (i) and (iii) as follows:

 $$\begin{gathered} {\mathrm{I}}_{\mathrm{avg}}=\mathrm{It}\times \mathrm{N}/\mathrm{s} \\ =0.50\mathrm{A}\times 0.10\times {10}^{-6}\mathrm{s}\times 500\frac{\mathrm{pulses}}{\mathrm{s}} \\ =25\times {10}^{-6}\mathrm{A} \end{gathered}$$

Hence, the value of the average current is $25\times {10}^{-6}\mathrm{A}$ .

Step 5: c) Calculation of the average power on the accelerator

 Accelerating potential difference can be found using the given energy, 

$\mathrm{E}=50\mathrm{MeV} \mathrm{or} 50\times 1.6\times {10}^{19 }\mathrm{J}$ . 

Now, for the conservation case, we can get the kinetic energy of the system as equal to the potential energy of the system. Thus, the electric potential of the electron can be calculated using equation (v) as follows:

$$\begin{gathered} \mathrm{K}=\mathrm{eV} \\ =\frac{\mathrm{K}}{\mathrm{e}} \\ =\frac{50\times {10}^6\times {10}^{-19}}{1.6\times {10}^{-19}} \\ =50\times {10}^6\mathrm{V} \end{gathered}$$ 

Average power of the electron system can be calculated using the given data in equation (vi) as follows:

$$\begin{gathered} \mathrm{P}=25\times {10}^{-6}\times 50\times {10}^6 \\ =1.25\times {10}^3\mathrm{W} \\ \approx 1.3\mathrm{kW} \end{gathered}$$

Hence, the value of the electric power is 1.3 kW.

Step 6: d) Calculation of the peak power of the accelerator

The peak power is the ratio of the energy of the pulse to the duration of the pulse. Thus, using the given data in equation (iii) the power can be given as follows:

$$\begin{gathered} \mathrm{P}=\frac{3.1\times {10}^{11}\times 50\times {10}^6\times 1.6\times {10}^{19 }\mathrm{J}}{0.10\times {10}^{-6}\mathrm{s}} \\ =25\times {10}^6\mathrm{W} \\ =25\mathrm{MW} \end{gathered}$$ 

Hence, the value of the peak power is 25MW.