Question

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 \(\underline{\hspace{3cm}}\). 

(Given: \(e = 1.6 \times 10^{-19} \, \text{C}\))

• A. 3.2 A towards left
• B. 3.2 A towards right
• C. 1.6 A towards left
• D. 1.6 A towards right

Hint:

Conventional current is always in the direction of positive charge flow and opposite to the direction of negative charge flow. Using a fixed sign convention (e.g., right = positive, left = negative) for all vector quantities prevents summation errors.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
Conventional electric current is defined as the flow of positive charge per unit time. We must calculate the individual current contributions from each type of charge carrier and find their vector sum.
Step 3: Detailed Explanation:
The rate of flow for each particle type is given as \( n = 10^{19} \text{ particles per second} \).
1. Electrons: These have a negative charge (\( -e \)). Moving to the left, they contribute to a conventional current directed towards the right.
\( I_e = n \times e = 10^{19} \times 1.6 \times 10^{-19} = 1.6 \text{ A (Right)} \)
2. Protons: These have a positive charge (\( +e \)). Moving to the left, they contribute to a conventional current directed towards the left.
\( I_p = n \times e = 10^{19} \times 1.6 \times 10^{-19} = 1.6 \text{ A (Left)} \)
3. \(\alpha\)-particles: These have a positive charge (\( +2e \)). Moving to the right, they contribute to a conventional current directed towards the right.
\( I_{\alpha} = n \times 2e = 10^{19} \times 2 \times 1.6 \times 10^{-19} = 3.2 \text{ A (Right)} \)
Resultant Current Calculation:
Assigning "right" as the positive direction:
\( I_{\text{net}} = I_e (\text{right}) + I_p (\text{left}) + I_{\alpha} (\text{right}) \)
\( I_{\text{net}} = 1.6 + (-1.6) + 3.2 = 3.2 \text{ A (Right)} \)
Step 4: Final Answer:
The resulting electric current is 3.2 A towards the right.