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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 4910 m/s and protons move to the left at 2583 m/s. The particles are evenly spaced with 0.0758 m between electrons and 0.0577 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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Answer:

The answer is "[tex]17.28 \times 10^{-14}\ A[/tex]"

Explanation:

Calculating the number of electrons passing in per second:

[tex]\to n_e = \frac{4910}{0.0758} = 6.4 \times 10^5[/tex]

Calculating the current in electron:

[tex]\to I_e = 6.4 \times 10^5 \times 1.6 \times 10^{-19} = 10.24 \times 10^{-14}\ A[/tex]

Calculating the number of protons passing in per second:

[tex]\to n_p = \frac{2583}{0.0577} = 4.4 \times 10^{4}[/tex]

Calculating the current in proton:

[tex]\to I_p = 4.4 \times 10^{4} \times 1.6 x 10^{-19} = 7.04 \times 10^{-14} \ A[/tex]

Calculating the total current:

[tex]\to I = I_p + I_e = 17.28 \times 10^{-14}\ A[/tex]

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