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# 1.6 - current in the-z-direction I e = ∆ Q ∆ t = 3 ×...

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Problem 1.6 A certain cross section lies in the x y plane. If 3 × 10 20 electrons go through the cross section in the z -direction in 4 seconds, and simultaneously, 1 . 5 × 10 20 protons go through the same cross section in the negative z -direction, what is the magnitude and direction of the current flowing through the cross section? Solution: Negatively charged electrons moving along + z -direction constitute a
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Unformatted text preview: current in the-z-direction: I e = ∆ Q ∆ t = 3 × 10 20 × 1 . 6 × 10-19 4 = 12 A, along-z-direction . Positively charged protons moving along-z-direction constitute a current in the-z-direction: I p = ∆ Q ∆ t = 1 . 5 × 10 20 × 1 . 6 × 10-19 4 = 6 A, along-z-direction . Total net current is: I = I e + I p = 12 + 6 = 18 A, along-z-direction....
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