Question

During the 4.0 mina 5.0 Acurrent is set up in a wire, how many (a) coulombs pass through any cross section across the wire’s width? (b) electrons pass through any cross section across the wire’s width?

Short answer

1. The number of coulombs passing through any cross section across the wire’s width is$1.2\times {10}^3\mathrm{C}$
2. The number of electrons passing through any cross section across the wire’s width is$7.5\times {10}^{21}$ .

Step-by-step solution

The given data

1. Time of current flow, t = 4.0 min 0r 240 s
2. Current passing through the area, i = 5.0 A

Understanding the concept of the flow of current

We can use the formula of current to find the number of charges by using the given values of time and current. Then, by dividing this charge by the magnitude of the charge on a single electron, we can find the number of electrons that pass through any cross-section across the wire’s width.

Formulae:

The current passing through the cross-sectional area in terms of charge, i = q/t …(i)

The number of electrons passing through the area, N = q/e …(ii)

(a) Calculation of the value of charge

Using the given data in equation (i), the amount of coulombs or the charge passing through any cross-section can be given as follows:

$\begin{array}{l}\mathrm{q}=\mathrm{it}\\ =\left(5.0\mathrm{A}\right)\left(240\mathrm{s}\right)\\ =1.2\times {10}^3\mathrm{C}\end{array}$

Therefore, the number of coulombs (charges) that pass through any cross section across the wire’s width is$1.2\times {10}^3\mathrm{C}$

(b) Calculation of the number of electrons

For the amount of charge passing through the area, the number of electrons passing can be calculated using equation (ii) as follows:

(is the magnitude of the charge of an electron,$\mathrm{e}=1.6\times {10}^{-19}\mathrm{C}$ )

$$\begin{gathered} \mathrm{N}=\frac{1.2\times {10}^3\mathrm{C}}{1.6\times {10}^{-19}\mathrm{C}} \\ =7.5\times {10}^{21} \end{gathered}$$

Therefore, the number of electrons that pass through any cross section across the wire’s width is$7.5\times {10}^{21}$ .