Question

An insulating pipe of cross-section area $A$ contains an electrolyte which has two types of ions: their charges being $-e$ 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 $=v(-veion)$ and $\frac{v}{4}(+veion)$. Both ions have the same number per unit volume $=n$. The current flowing through the pipe is$\frac{x}{y}\left(nevA\right)$. Find $(x-y)$.

Answer 1

Step 1: Current due to the ions:

Given,

Drift speed of ion having $-e$ charge

Drift speed of ion having $+2e$ charge

Carrier density

We know, $I=neA{v}_d$

So, for negative charged ion,

Current $i_n=neAv$

Since $–ve$ charges are drifting towards left , current $i_n$ will be towards right

For positive charged ion, current $i_p=n\left(2e\right)A\frac{v}{4}$

Since positive charges are drifting towards right, current $i_p$ will be towards right.

Step 2: Calculation of force $F_2$

Now, net current $=i_n+i_p$ ( both currents in same direction i.e. towards right )

$=\left(neAv\right)+ne\frac{A}{2}v$

$=\frac{3}{2}neAv$

So, comparing with $\frac{x}{y}\left(nevA\right),x=3,y=2$

Hence, value of $x-y=1$