During the 3.7 min a 7.4 A current is set up in a wire, how many (a) coulombs and (b) electrons pass through any cross section of the wire's width?

Electric current is the circulation of electric charges in an electric circuit, while electric current intensity (I) is the amount of electricity or electric charge (Q) that circulates through a circuit in unit time (t).

Then, the intensity of electric current is expressed as:

[tex]I=\frac{Q}{t}[/tex]

Where:

I is the intensity expressed in Amps (A)
Q is the electric charge expressed in Coulombs (C)
t is the time expressed in seconds (s)

In this case:

I= 7.4 A
Q= ?
t= 3.7 min= 222 s (being 1 min= 60 s)

Replacing:

[tex]7.4 A=\frac{Q}{222 s}[/tex]

Solving:

Q= 7.4 A* 222 s

Q= 1,642.8 C

A Coulomb represents about 6.24*10¹⁸ electrons, so you can apply the following rule of three: if 1 C represents 6.24*10¹⁸ electrons, 222 C how many electrons does it represent?

[tex]electrons=\frac{222 C*6.24*10^{18} }{1C}[/tex]

electrons= 1.39*10²¹

1,642.8 coulombs and 1.39*10²¹ electrons pass through any cross section the width of the wire.