Three wires meet at a junction. Wire 1 has a current of 0.40 A into the junction. The current of wire 2 is 0.59 A out of the junction. How many electrons per second move past a point in wire 3?

To determine the current in wire 3, we apply Kirchhoff's junction rule, which states that the sum of currents entering a junction equals the sum of currents leaving it. The current entering (wire 1) is 0.40 A, and the current leaving (wire 2) is 0.59 A. Thus, the current in wire 3 is 0.40 A - 0.59 A = -0.19 A, indicating it flows into the junction. To find the number of electrons per second, we use the formula ( I = n \cdot e ), where ( I ) is the current, ( n ) is the number of electrons, and ( e ) (the charge of an electron) is approximately ( 1.6 \times 10^{-19} ) C. Thus, ( n = \frac{0.19 , \text{A}}{1.6 \times 10^{-19} , \text{C/electron}} \approx 1.19 \times 10^{18} ) electrons per second flow past a point in wire 3.