Can you solve for the current directed to the right?

Question

In a given fluid, positive ions, each with three excess protons, move towards right at a steady rate of 4.5 x 1019 ions per minute and negative ions, each with two excess electron, move towards left at a steady rate of 2.5 x 1020 ions per minute. Determine the current directed to the right.

Answer 1

Total current is rate of flow of charge

The conventional is rate of flow of positive charge and opposite to the direction of flow of negative charge

In this case3 positive ions are moving to right at he rate of 4.5*10¹⁹ per min

So corresponding current to the right = 3*1.6*10^-19 *4.5*10¹⁹ / 60 Amp ( charge per sec)

Charge of each proton= charge of each electron = 1.6*10^-19 C

now for electrons moving to the left current is in the right direction

Current due to electrons = 2* 1.6*10^-19*2.5* 10²⁰/60 Amp

The currents being in the same directction are added

Hence net current is sum of the two currents ( right Direction)