Problem #0124 Electricity & Magnetism Sub-menu Problem #0126 Chapters Chapters

395.

Problem 32.51 (RHK)

An electron linear accelerator produces a pulsed beam of electrons. The pulse current is 485 mA and the pulse duration is 95.0 ns. (a) We have to find the number of electrons in each pulse; (b) average current for a machine operating at ; if the electrons are accelerated to an energy of , we have to find the values of the average and the peak power outputs of the accelerator.

Solution:             Click For PDF Version

(a)

In the linear accelerator the pulse current is 485 mA and the pulse duration is 95.0 ns. Therefore, the total charge contained in one pulse will be

Charge of one electron is Therefore, the number of electrons in one pulse will be

(b)

The average current for a machine operating at will be

(c)

The electrons are accelerated to an energy of . That is the accelerating potential is . The average output power of the accelerator will, therefore, be

(d)

The peak power of the accelerator will be