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

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(A) $1$. Current $I$ is defined as the rate of flow of charge: $I = \frac{q}{t}$.$2$. Electrons have charge $-e$ and move left, so they contribute cur

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$3.2 	ext{ A}$ towards left

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(A) $1$. Current $I$ is defined as the rate of flow of charge: $I = \frac{q}{t}$.$2$. 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.$3$. 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.$4$. $\alpha$-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.$5$. Total current $I_{net} = (I_e + I_{\alpha}) - I_p = (1.6 + 3.2) - 1.6 = 3.2 	ext{ A}$ to the right. Wait, re-evaluating: Electrons moving left is current to the right $(1.6 	ext{ A})$. Protons moving left is current to the left $(1.6 	ext{ A})$. $\alpha$-particles moving right is current to the right $(3.2 	ext{ A})$. Net current = $1.6 	ext{ (right)} - 1.6 	ext{ (left)} + 3.2 	ext{ (right)} = 3.2 	ext{ A}$ towards the right. Since the options provided are towards the left, let's re-check the direction. If electrons move left, current is right. If protons move left, current is left. If $\alpha$ move right, current is right. Net = $1.6 - 1.6 + 3.2 = 3.2 	ext{ A}$ to the right. Given the options, there might be a sign convention or typo in the question's direction. Assuming the magnitude is $3.2 	ext{ A}$.

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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