Question

An insulating pipe of cross-section area $A$ contains an electrolyte which has two types of ions and their charges being $-2e$ and $+2e$. A potential difference applied between the ends of the pipe result in the drifting of the two types of ions, having drift speed $v\text{(-ve ion)}$ and $v/4\text{(+ ve ion)}$. Both ions have the same number per unit volume $n$. The current flowing through the pipe is

• A. $\frac{neAv}{2}$
• B. $\frac{neAv}{4}$
• C. $\frac{5neAv}{2}$
• D. $\frac{3neAv}{2}$

Answer 1

The correct option is C $\frac{5neAv}{2}$

Time rate of flow of charge through a cross-sectional area is called current.

Relation between current and drift velocity is given by

$I=nqA{v}_d$

$n\to$ no. of free charge particles per unit volume
$q\to$ charge of each free particle
$I\to$ charge flow per unit time

For the given electrolyte the electrical neutrality of the solution will be maintained, since both the $+ve$ and $-ve$ ions have same magnitude of charge and number of ions per unit volume$\left(n\right)$ is also same for both.

$I_1=n\left(2e\right)Av\text{and}{I}_2=n\left(2e\right)A\left(v/4\right)$
Thus, $I_1=2neAv\text{and}{I}_2=\frac{neAv}{2}$

The direction of current is along the direction of flow of positive charge or opposite to the direction of flow of negative charge.

In our case opposite charges are moving in opposite direction therefore the value of their equivalent current will add.

$\Rightarrow I=I_1+I_2$

$\text{or},I=\frac{5neAv}{2}$
Hence, option $\left(C\right)$ is the correct choice.