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Question: An insulating pipe of cross-section area 'A' contains an electrolyte which has two types of ions, th...

An insulating pipe of cross-section area 'A' contains an electrolyte which has two types of ions, their charges being e- e and +2e+ 2e . A potential difference applied between the ends of the pipe results in the drifting of the two types of ions, having drift speed =v(ve)= v( - ve) ion and v4(+ve)\dfrac{v}{4}( + ve) ion . Both ions have the same number per unit volume =n= n . The current flowing through the pipe is.
(A) nevA2nev\dfrac{A}{2}
(B) nevA4nev\dfrac{A}{4}
(C) 5nevA25nev\dfrac{A}{2}
(D) 4nevA24nev\dfrac{A}{2}

Explanation

Solution

First find the total current in the pipe due to both the given charges then use the formula to find the current using drift velocity, area of cross-section, charge, and number of ions per unit volume. We will get our required current in the pipe.
I=neA×VdI = neA \times {V_d}
Where, nn is the number of ions per unit volume. AA is the area of cross-section. ee is the charge. Vd{V_d} is the drift velocity.

Complete Step By Step Answer:
We have been given a pipe of cross section ‘A’ inside which there is an electrolyte which has two charges e- e and +2e+ 2e their respective drift velocity are, v(ve)v( - ve) and v4(+ve)\dfrac{v}{4}( + ve) .
The total current flowing through the pipe will be due to both the charges e- e and +2e+ 2e
Itotal=I+2eIe\Rightarrow {I_{total}} = {I_{ + 2e}} - {I_{ - e}}
I+2e{I_{ + 2e}} ​ is due to the positive ion moving along the field.
Ie{I_{ - e}} is due to the negative ion moving opposite to the field.
The equation for the relationship between current and drift velocity is
I=neA×VdI = neA \times {V_d}
Where, nn is the number of ions per unit volume. AA is the area of cross-section. ee is the charge. Vd{V_d} is the drift velocity.
Since both ions have the same number per unit volume , then the current will be
Itotal=n×(2e)×A×v4(n×(e)×A×(v))\Rightarrow {I_{total}} = n \times (2e) \times A \times \dfrac{v}{4} - \left( {n \times ( - e) \times A \times ( - v)} \right)
Itotal=nevA2\Rightarrow {I_{total}} = nev\dfrac{A}{2}
Hence option A) nevA2nev\dfrac{A}{2} is the correct option.

Note:
The average velocity gained by free electrons in a conductor under the effect of an electric field applied across the conductor is known as drift velocity. If an electric field is applied to a solution, the positive ions will migrate inside the solution (following the field), whereas the negative ions will move in the opposite direction.