In a conducting region, 10^{19} electrons and | vedclass

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Question

(A) Current $I$ is defined as the rate of flow of charge: $I = \frac{q}{t}$.Electrons have charge $-e$ and move left, so they contribute current $I_e

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$3.2 	ext{ A}$ towards right

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(A) Current $I$ is defined as the rate of flow of charge: $I = \frac{q}{t}$.Electrons have charge $-e$ and move left, so they contribute current $I_e = \frac{10^{19} 	imes e}{1} = 1.6 	ext{ A}$ to the right.Protons have charge $+e$ and move left, so they contribute current $I_p = \frac{10^{19} 	imes e}{1} = 1.6 	ext{ A}$ to the left.$\alpha$-particles have charge $+2e$ and move right, so they contribute current $I_{\alpha} = \frac{10^{19} 	imes 2e}{1} = 3.2 	ext{ A}$ to the right.Total current $I_{net} = I_e - I_p + I_{\alpha} = 1.6 - 1.6 + 3.2 = 3.2 	ext{ A}$ towards the right.

In a conducting region, $10^{19}$ electrons and $10^{19}$ protons move to the left, while $10^{19}$ $\alpha$-particles move to the right per second. The resulting electric current is $(e = 1.6 \times 10^{-19} \text{ C})$

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