Suppose that a particle accelerator is used to move two beam

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/s and protons move to the left at 1.230 × 103 m/s. The particles are evenly spaced with 0.0194 m between electrons and 0.0407 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?

Solution

Here,

distance between proton , dp = 0.0407 m

distance between electrons , de = 0.0194 m

speed of proton , vp = 1.230 *10^3 m/s

speed of electron , ve = 7130 m/s

Now ,

magnitude of average current = charge passing each second

magnitude of average current = e * (vp/dp + ve/de)

magnitude of average current = 1.602 *10^-19 *(1.23 *10^3/.0407 + 7130/.0194)

magnitude of average current = 6.37 *10^-14 A

the magnitude of average current is 6.37 *10^-14 A