Physics, 22.04.2020 03:26 tory74

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 6020 m/s and protons move to the left at 1681 m/s. The particles are evenly spaced with 0.0476 m between electrons and 0.0662 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 from: hawszanderpelv2v

$2.429783984\times 10^{-14}\ A$

Explanation:

Velocity of electron = 6020 m/s

Velocity of proton = 1681 m/s

Electron space = 0.0476 m

Proton space = 0.0662 m

e = Charge of particle = $1.6\times 10^{-19}\ C$

Number of electrons passing per second

$n_e=\dfrac{6020}{0.0476}\\\Rightarrow n_e=126470.588$

Number of protons passing per second

$n_p=\dfrac{1681}{0.0662}\\\Rightarrow n_p=25392.749$

Current due to electrons

$I_e=n_ee\\\Rightarrow I_e=126470.588\times 1.6\times 10^{-19}\\\Rightarrow I_e=2.0235\times 10^{-14}\ A$

Current due to protons

$I_p=n_pe\\\Rightarrow I_p=25392.749\times 1.6\times 10^{-19}\\\Rightarrow I_p=4.06283984\times 10^{-15}\ A$

Total current

$I=2.0235\times 10^{-14}+4.06283984\times 10^{-15}\\\Rightarrow I=2.429783984\times 10^{-14}\ A$

The average current is $2.429783984\times 10^{-14}\ A$

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Suppose that a particle accelerator is used to move two beams of particles in opposite directions. I...

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