Answer
64.8k+ views
Hint: Current flows when there is flow of charges. The amount of charge flowing will give you the current Also, the current is the total charge flowing with respect to time, so be careful while counting charges;
Complete solution:
Here, in this question the time is given to us. It is 1 second so we need to see how much charge flows in 1 second;
Here the 8 ampere current in the helium gas tube is due to the positive helium ions moving towards right and also because of the n electrons moving towards left.
We can mathematically write it as;
$8A = \dfrac{{40 \times {{10}^{18}} \times {q_{H{e^ + }}}C}}{{1\sec }}$$ + \dfrac{{n \times {q_{electron}}}}{{1\sec }}$ (equation:1)
Here ${q_{H{e^ + }}}$ is the magnitude of charge of a positive helium ion.
Hence, ${q_{H{e^ + }}} = 1.6 \times {10^{ - 19}}C$.
Also, ${q_{electron}}$ is the magnitude of charge of an electron.
So, ${q_{electron}} = 1.6 \times {10^{ - 19}}C$.
Thus, now we substitute these values in equation 1;
$8 = 40 \times {10^{18}} \times 1.6 \times {10^{ - 19}} + n \times 1.6 \times {10^{ - 19}}$
Solving this equation for n will give us the value of n;
$8 = \left( {40 \times {{10}^{18}} + n} \right)1.6 \times {10^{ - 19}}$
$40 \times {10^{18}} + n = \dfrac{8}{{1.6 \times {{10}^{ - 19}}}} = 50 \times {10^{18}}$
Hence,$n = \left( {50 - 40} \right){10^{18}} = 10 \times {10^{18}}$
Hence, option A is correct.
Note:(1) The current through a conductor is equal to charge flowing per second through its cross section.
(2) For conductors like the gas tube the positive and negative charge both were flowing and so the net current is the sum of the current due to flow of positive charge per second and the flow of negative charge per second.
(3) For conductors like the metal wire the positive charge is not mobile so we take into account the flow of negative charge only.
Complete solution:
Here, in this question the time is given to us. It is 1 second so we need to see how much charge flows in 1 second;
Here the 8 ampere current in the helium gas tube is due to the positive helium ions moving towards right and also because of the n electrons moving towards left.
We can mathematically write it as;
$8A = \dfrac{{40 \times {{10}^{18}} \times {q_{H{e^ + }}}C}}{{1\sec }}$$ + \dfrac{{n \times {q_{electron}}}}{{1\sec }}$ (equation:1)
Here ${q_{H{e^ + }}}$ is the magnitude of charge of a positive helium ion.
Hence, ${q_{H{e^ + }}} = 1.6 \times {10^{ - 19}}C$.
Also, ${q_{electron}}$ is the magnitude of charge of an electron.
So, ${q_{electron}} = 1.6 \times {10^{ - 19}}C$.
Thus, now we substitute these values in equation 1;
$8 = 40 \times {10^{18}} \times 1.6 \times {10^{ - 19}} + n \times 1.6 \times {10^{ - 19}}$
Solving this equation for n will give us the value of n;
$8 = \left( {40 \times {{10}^{18}} + n} \right)1.6 \times {10^{ - 19}}$
$40 \times {10^{18}} + n = \dfrac{8}{{1.6 \times {{10}^{ - 19}}}} = 50 \times {10^{18}}$
Hence,$n = \left( {50 - 40} \right){10^{18}} = 10 \times {10^{18}}$
Hence, option A is correct.
Note:(1) The current through a conductor is equal to charge flowing per second through its cross section.
(2) For conductors like the gas tube the positive and negative charge both were flowing and so the net current is the sum of the current due to flow of positive charge per second and the flow of negative charge per second.
(3) For conductors like the metal wire the positive charge is not mobile so we take into account the flow of negative charge only.
Recently Updated Pages
Write a composition in approximately 450 500 words class 10 english JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Arrange the sentences P Q R between S1 and S5 such class 10 english JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
What is the common property of the oxides CONO and class 10 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
What happens when dilute hydrochloric acid is added class 10 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
If four points A63B 35C4 2 and Dx3x are given in such class 10 maths JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
The area of square inscribed in a circle of diameter class 10 maths JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Other Pages
A boat takes 2 hours to go 8 km and come back to a class 11 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Electric field due to uniformly charged sphere class 12 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
In the ground state an element has 13 electrons in class 11 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
According to classical free electron theory A There class 11 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Differentiate between homogeneous and heterogeneous class 12 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Excluding stoppages the speed of a bus is 54 kmph and class 11 maths JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)