Suppose that a particle accelerator is used to move two beams of particles in opposite directions. In a particular region, electrons move to

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?

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Adela 5 years 2021-07-15T15:21:37+00:00 1 Answers 76 views 0

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  1. thugiang
    0
    2021-07-15T15:23:30+00:00
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    Answer:

    Total current, I=4.51\times 10^{-14}\ A

    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=\dfrac{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=\dfrac{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\times e+n_p\times e\\\\I=e(n_e+n_p)\\\\I=1.6\times 10^{-19}\times (247569.44+34578.31)\\\\I=4.51\times 10^{-14}\ A

    Hence, this is the required solution.

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