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Chapter 18: Problem 2

The current in a wire is \(0.500 \mathrm{A} .\) (a) How much charge flows through a cross section of the wire in 10.0 s? (b) How many electrons move through the same cross section in $10.0 \mathrm{s} ?$

Short Answer

Expert verified
Answer: The charge that flows through the cross-section of the wire is 5.00 C, and the number of electrons that pass through the cross-section in 10 seconds is approximately \(3.125 \times 10^{19}\) electrons.

Step by step solution

01

Determine the charge that passes through the cross section

We're given the current (I) and the time (t), so we can find the charge (Q) using the formula Q = I * t. Using the given values: \(I = 0.500 A\) \(t = 10.0 s\) Plugging these values into the formula, we get: \(Q = (0.500\,\text{A})(10.0\,\text{s})\) Multiplying the current and the time, we obtain the total charge: \(Q = 5.00\,C\)
02

Calculate the number of electrons

Now that we have the charge (Q), we can find the number of electrons (n) using the formula n = Q/e, where e is the elementary charge. The elementary charge is a constant value and it is approximately equal to: \(e = 1.6 \times 10^{-19}\,\text{C}\) Dividing the total charge by the elementary charge, we get the number of electrons: \(n = \frac{5.00\,\text{C}}{1.6 \times 10^{-19}\,\text{C}}\) Calculating the result gives us: \(n = 3.125 \times 10^{19}\) So, the number of electrons that pass through the cross-section in 10 seconds is approximately \(3.125 \times 10^{19}\).

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Most popular questions from this chapter

A galvanometer has a coil resistance of \(50.0 \Omega\). It is to be made into an ammeter with a full-scale deflection equal to \(10.0 \mathrm{A} .\) If the galvanometer deflects full scale for a current of \(0.250 \mathrm{mA},\) what size shunt resistor should be used?
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