
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 results in the drifting of the two types of ions, having drift speed $ = v( - ve) $ ion and $ \dfrac{v}{4}( + ve) $ ion . Both ions have the same number per unit volume $ = n $ . The current flowing through the pipe is.
(A) $ nev\dfrac{A}{2} $
(B) $ nev\dfrac{A}{4} $
(C) $ 5nev\dfrac{A}{2} $
(D) $ 4nev\dfrac{A}{2} $
Answer
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Hint: 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 \times {V_d} $
Where, $ n $ is the number of ions per unit volume. $ A $ is the area of cross-section. $ e $ is the charge. $ {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 $ and $ + 2e $ their respective drift velocity are, $ v( - ve) $ and $ \dfrac{v}{4}( + ve) $ .
The total current flowing through the pipe will be due to both the charges $ - e $ and $ + 2e $
$ \Rightarrow {I_{total}} = {I_{ + 2e}} - {I_{ - e}} $
$ {I_{ + 2e}} $ is due to the positive ion moving along the field.
$ {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 \times {V_d} $
Where, $ n $ is the number of ions per unit volume. $ A $ is the area of cross-section. $ e $ is the charge. $ {V_d} $ is the drift velocity.
Since both ions have the same number per unit volume , then the current will be
$ \Rightarrow {I_{total}} = n \times (2e) \times A \times \dfrac{v}{4} - \left( {n \times ( - e) \times A \times ( - v)} \right) $
$ \Rightarrow {I_{total}} = nev\dfrac{A}{2} $
Hence option A) $ nev\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.
$ I = neA \times {V_d} $
Where, $ n $ is the number of ions per unit volume. $ A $ is the area of cross-section. $ e $ is the charge. $ {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 $ and $ + 2e $ their respective drift velocity are, $ v( - ve) $ and $ \dfrac{v}{4}( + ve) $ .
The total current flowing through the pipe will be due to both the charges $ - e $ and $ + 2e $
$ \Rightarrow {I_{total}} = {I_{ + 2e}} - {I_{ - e}} $
$ {I_{ + 2e}} $ is due to the positive ion moving along the field.
$ {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 \times {V_d} $
Where, $ n $ is the number of ions per unit volume. $ A $ is the area of cross-section. $ e $ is the charge. $ {V_d} $ is the drift velocity.
Since both ions have the same number per unit volume , then the current will be
$ \Rightarrow {I_{total}} = n \times (2e) \times A \times \dfrac{v}{4} - \left( {n \times ( - e) \times A \times ( - v)} \right) $
$ \Rightarrow {I_{total}} = nev\dfrac{A}{2} $
Hence option A) $ nev\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.
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