Problem 82 - Fundamentals of Physics

Question

A linear accelerator produces a pulsed beam of electrons. The pulse current is \(0.50 \mathrm{~A}\), and the pulse duration is \(0.10 \mu \mathrm{s}\). (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 an energy of \(50 \mathrm{MeV}\), what are the (c) average power and (d) peak power of the accelerator?

Step-by-Step Solution

Answer

(a) About \(3.125 \times 10^{11}\) electrons per pulse. (b) Average current is \(2.5 \times 10^{-5}\) A. (c) Average power is 125 W. (d) Peak power is 2500 W.

Step 1: Calculate the charge per pulse

The charge per pulse produced by the accelerator can be found using the formula: \[ Q = I imes t \] where \( I = 0.50 \, \text{A} \) is the current and \( t = 0.10 \times 10^{-6} \, \text{s} \) is the pulse duration.Substitute the values:\[ Q = 0.50 \, \text{A} \times 0.10 \times 10^{-6} \, \text{s} = 5 \times 10^{-8} \, \text{C} \]

Step 2: Find the number of electrons per pulse

To determine how many electrons are accelerated per pulse, divide the total charge per pulse by the charge of a single electron. The charge of an electron is \( e = 1.6 \times 10^{-19} \, \text{C} \).\[ n = \frac{Q}{e} = \frac{5 \times 10^{-8} \, \text{C}}{1.6 \times 10^{-19} \, \text{C/electron}} = 3.125 \times 10^{11} \text{ electrons} \]

Step 3: Calculate the average current

The average current can be calculated by multiplying the pulse current by the fraction of the time the accelerator is active. Since it operates at 500 pulses per second, each pulse time is \( 0.10 \times 10^{-6} \, \text{s} \).\[ \text{Duty factor} = \text{pulse duration} \times \text{frequency} = 0.10 \times 10^{-6} \, \text{s} \times 500 \, \text{pulses/s} = 5 \times 10^{-5} \]\[ I_{\text{avg}} = I_{\text{pulse}} \times \text{Duty factor} = 0.50 \, \text{A} \times 5 \times 10^{-5} = 2.5 \times 10^{-5} \, \text{A} \]

Step 4: Calculate the average power

The energy per electron is given as 50 MeV. First, convert this energy to joules:\[ 1 \text{ MeV} = 1.6 \times 10^{-13} \, \text{J} \]\[ E = 50 \, \text{MeV} \times 1.6 \times 10^{-13} \, \text{J/MeV} = 8 \times 10^{-12} \, \text{J} \]The average power can be found using:\[ P_{\text{avg}} = n \times E \times \text{pulse frequency} = 3.125 \times 10^{11} \times 8 \times 10^{-12} \, \text{J} \times 500 \, \text{pulses/s} \]\[ P_{\text{avg}} = 125 \, \text{W} \]

Step 5: Calculate the peak power

Peak power is the power per pulse, which can be calculated using:\[ P_{\text{peak}} = n \times E = 3.125 \times 10^{11} \times 8 \times 10^{-12} \, \text{J} \]\[ P_{\text{peak}} = 2500 \, \text{W} \]

Key Concepts

Current Calculation

Understanding current calculation in a linear accelerator involves figuring out how many electrons are moving through the accelerator in each pulse. Here, we're given a pulse current of 0.50 Amperes and a pulse duration of 0.10 microseconds. Current, measured in Amperes (A), represents the rate of flow of electric charge. It essentially tells us how much charge is flowing per second.

To calculate the total charge (\(Q\)) per pulse, we use the formula:

• \(Q = I \times t\)

where \( I \) is the pulse current and \( t \) is the pulse duration. By substituting the given values, we get \(Q = 5 \times 10^{-8} C\). This calculation is crucial for understanding the overall number of electrons and their movement, or flow, within the accelerator.

Average Power

Average power in the context of a linear accelerator is the energy delivered over a period, divided by that time span. It's an essential concept as it helps in determining the operational efficiency of the accelerator. Given that each electron is accelerated to an energy of 50 MeV, the energy needs to be converted into Joules: 50 MeV equals \(8 \times 10^{-12} \,\text{J}\).

To find the average power, we multiply the energy per electron by the number of electrons and the pulse frequency:

• \(P_{\text{avg}} = n \times E \times \text{pulse frequency}\)

where \(n = 3.125 \times 10^{11}\) is the number of electrons per pulse and the pulse frequency is 500 pulses per second. Thus, the average power is calculated to be 125 watts. This means the accelerator provides 125 watts of energy on average every second.

Peak Power

Peak power represents the power generated during a single pulse of the linear accelerator. This is a critical measure because it reflects the energy capacity and efficiency of the accelerator in its most active state. During the pulse, the maximum power transferred to the electron beam is termed as the peak power.

It is calculated similarly to average power, but only considers a single pulse:

• \(P_{\text{peak}} = n \times E \)

By substituting the known values of \( n = 3.125 \times 10^{11}\) electrons and \( E = 8 \times 10^{-12} \,\text{J}\) per electron, we find the peak power to be 2500 watts. This implies that during each pulse, the accelerator operates at a maximum power of 2500 watts.

Electron Charge

The electron charge is a fundamental physical constant used in various calculations involving electronics and atomic particles. It provides the basis for calculating the number of electrons in a given charge and is crucial for quantifying interactions in electric and magnetic fields.

The charge of an electron is \( e = 1.6 \times 10^{-19} \,\text{C} \) (Coulombs). In the problem of the linear accelerator, knowing this charge allows us to calculate how many electrons are in the pulsed charge. By dividing the total charge per pulse by the electron charge:

• \(n = \frac{Q}{e} = \frac{5 \times 10^{-8} \,\text{C}}{1.6 \times 10^{-19} \,\text{C/electron}}\)

we find there are \(3.125 \times 10^{11}\) electrons per pulse. Understanding this charge and its applications is essential in physics and engineering, particularly in fields dealing with electricity and magnetism.