In an electrolytic solution, 2×10^18 positive ions, each of charge +e…

The correct answer is Option (2) → 0.8 A, towards the right

In an electrolytic solution both positive and negative ions carry the current. A negative charge moving to the left has the same effect as an equal positive charge moving to the right. So the two flows add up to a current towards the right.

Taking the right as the forward direction, the positive charge that crosses the section in one second is:

    \( \begin{aligned} q_+ &= 2\times10^{18}\times1.6\times10^{-19}\,\mathrm{C} \\ &= 0.32\,\mathrm{C} \end{aligned} \)

The negative ions move to the left, that is, backwards. So the negative charge carried in the forward direction in one second is:

    \( \begin{aligned} q_- &= -\,3\times10^{18}\times1.6\times10^{-19}\,\mathrm{C} \\ &= -0.48\,\mathrm{C} \end{aligned} \)

The net charge flowing forward per second, which is the current, is:

    \( \begin{aligned} I &= \frac{q_+-q_-}{t} \\ &= \frac{0.32\,\mathrm{C}+0.48\,\mathrm{C}}{1\,\mathrm{s}} \\ &= 0.8\,\mathrm{A} \end{aligned} \)

The value is positive, so the current is towards the right, the direction in which the positive ions move.

NCERT (2026-27):
Electric Current – “Imagine a small area held normal to the direction of flow of charges. Both the positive and the negative charges may flow forward and backward across the area. In a given time interval t, let q₊ be the net amount (i.e., forward minus backward) of positive charge that flows in the forward direction across the area. Similarly, let q_– be the net amount of negative charge flowing across the area in the forward direction. The net amount of charge flowing across the area in the forward direction in the time interval t, then, is q = q₊– q_–. This is proportional to t for steady current and the quotient”

    \( \displaystyle I=\frac{q}{t} \qquad (3.1) \)

“is defined to be the current across the area in the forward direction.”