Question

Suppose that a particle accelerator is used to move two beams of particles in opposite directions. In a particular region, electrons move to the right at 7130 m/sand protons move to the left at 2583 m/s. The particles are evenly spaced with 0.0288 m between electrons and 0.0747 m between protons. Assuming that there are no collisions and that the interactions between the particles are negligible, what is the magnitude of the average current in this region?

Answer 1

Total current,
`I=4.51* 10^(-14)\ A`

Step-by-step explanation:

Velocity of electrons is 7130 m/s and particles are evenly spaced with 0.0288 m between electrons. We can find no of electrons passing per second as follows :

`n_e=(7130\ m/s)/(0.0288\ m)\\\\n_e=247569.44`

Velocity of protons is 2583 m/s and particles are evenly spaced with 0.0747 m between electrons. We can find no of protons passing per second as follows :

`n_p=(2583 \ m/s)/(0.0747 \ m)\\\\n_p=34578.31`

Total current in this region is equal to sum of current due to electrons and current due to protons.

`I=n_e* e+n_p* e\\\\I=e(n_e+n_p)\\\\I=1.6* 10^(-19)* (247569.44+34578.31)\\\\I=4.51* 10^(-14)\ A`

Hence, this is the required solution.