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Question: In a Neon discharge tube \(2.9\times {{10}^{18}}N{{e}^{+}}\) ions move to the right each second, whi...

In a Neon discharge tube 2.9×1018Ne+2.9\times {{10}^{18}}N{{e}^{+}} ions move to the right each second, while 1.2×10181.2\times {{10}^{18}} electrons move to the left per sec, electron charge is 1.6×10−19C1.6\times {{10}^{-19}}C. The current in the discharge tube:-
A. 1A towards right
B. 0.66A towards right
C. 0.66A towards left
D. Zero

Explanation

Solution

Here, we are given two types of charges, among which one is positive and other negative. So, we have to find the current due both these charges separately. Then the net current due to them can be found by taking the sum by taking the sign of charges and their direction of movement under consideration.

Formula used:
Current,
I=nqtI=\dfrac{nq}{t}

Complete Step by step solution:
In the question, we are given a discharge tube in which there is flow of charges in either side. We are said that 2.9×1018Ne+2.9\times {{10}^{18}}N{{e}^{+}} ions and 1.2×10181.2\times {{10}^{18}} electrons are found to move in right and left directions in each second. The charge of electron is given as,
e=1.6×10−19Ce=1.6\times {{10}^{-19}}C

Let us recall the definition of electric current. Electric current is defined as the flow of charges per unit time. In that case, we have the flow of both positron and electron in the tube. Flow of electrons to the left per second is given by,

Ie−=ne−e−t{{I}_{{{e}^{-}}}}={{n}_{{{e}^{-}}}}\dfrac{{{e}^{-}}}{t}
⇒Ie−=1.2×1018×−1.6×10−191\Rightarrow {{I}_{{{e}^{-}}}}=1.2\times {{10}^{18}}\times \dfrac{-1.6\times {{10}^{-19}}}{1} ……………………………………… (1)
Ie+=ne+e+t{{I}_{{{e}^{+}}}}={{n}_{{{e}^{+}}}}\dfrac{{{e}^{+}}}{t}
⇒Ie+=2.9×1018×1.6×10−191\Rightarrow {{I}_{{{e}^{+}}}}=2.9\times {{10}^{18}}\times \dfrac{1.6\times {{10}^{-19}}}{1}………………………………………….. (2)

So the net current flow in the discharge tube will be the sum of equations (1) and (2),
I=Ie+−(−Ie−)=Ie++Ie−I={{I}_{{{e}^{+}}}}-\left( -{{I}_{{{e}^{-}}}} \right)={{I}_{{{e}^{+}}}}+{{I}_{{{e}^{-}}}}
⇒I=1.6×10−19(1.2×1018+2.9×1018)\Rightarrow I=1.6\times {{10}^{-19}}\left( 1.2\times {{10}^{18}}+2.9\times {{10}^{18}} \right)
⇒I=1.6×4.1×10−1\Rightarrow I=1.6\times 4.1\times {{10}^{-1}}
∴I=0.656A∼0.66A\therefore I=0.656A\sim 0.66A

Therefore, we found the net current in the discharge tube to be 0.66A. Now, we know that the direction of the current flow is opposite to that of the flow of electrons by convention. So the current in the discharge tube will flow to the right.

Hence, the correct option is B.

Note:
You may have noticed that we have simply added the current due to positive and negative charges to get the net current irrespective of both moving in opposite directions. Though it may appear to be so, we have considered both the direction as well as the sign of the charge. Direction of the net current is assigned as per convention.