Answer
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Hint:The cumulative velocity of all the free flowing electrons in the conductor is known as the conductor’s drift velocity of electrons. The unit of drift velocity is $m{s^{ - 1}}$. Drift velocity is produced in an applied electric field.
Complete step by step answer:
Drift velocity is equal to the mobility of the charge carriers and the applied electric field intensity $E$. The holes shift towards the negative part of the battery and the electrons shift towards the positive part.
This combined influence of charge carrier movement forms a current known as the drift current.
The drift current density is the current flowing across a square centimetre at an angle of ninety degree to the flow path due to charge carriers such as free electrons and holes
Here the volume of the conductor is taken as $AL$, where $A = $ area and $L = $ length of the conductor.
Let the number of electrons per unit volume be $n$ .
So, the complete number of electrons that are free $ = nAL$.
The cumulative charge on all the free electrons in the conductor if the charge of free electrons is $e$ then, $q = ALne$
If the constant potential difference between the ends of the conductor applied is $V$, then the electric field set up around the conductor is given by $E = \dfrac{V}{L}$.
Due to the field $E = \dfrac{V}{L}$, the free electrons in the conductor begin to drift with a velocity ${v_d}$ towards the positive part of the conductor.
Time taken by the free electrons to get from one end of the conductor to the other end is given by
$t = \dfrac{L}{{{v_d}}}$
Hence, current is the given by
$
I = \dfrac{q}{t} = \dfrac{{ALne}}{{\dfrac{L}{{{v_d}}}}} = neA{v_d} \\
\therefore I = neA{v_d} \\
$
Note:The equations should be kept in mind along with the variables. Here everyone gets confused between the variables of the equations as they tend to appear the same. The equation for time taken and electric field is of utmost importance.
Complete step by step answer:
Drift velocity is equal to the mobility of the charge carriers and the applied electric field intensity $E$. The holes shift towards the negative part of the battery and the electrons shift towards the positive part.
This combined influence of charge carrier movement forms a current known as the drift current.
The drift current density is the current flowing across a square centimetre at an angle of ninety degree to the flow path due to charge carriers such as free electrons and holes
Here the volume of the conductor is taken as $AL$, where $A = $ area and $L = $ length of the conductor.
Let the number of electrons per unit volume be $n$ .
So, the complete number of electrons that are free $ = nAL$.
The cumulative charge on all the free electrons in the conductor if the charge of free electrons is $e$ then, $q = ALne$
If the constant potential difference between the ends of the conductor applied is $V$, then the electric field set up around the conductor is given by $E = \dfrac{V}{L}$.
Due to the field $E = \dfrac{V}{L}$, the free electrons in the conductor begin to drift with a velocity ${v_d}$ towards the positive part of the conductor.
Time taken by the free electrons to get from one end of the conductor to the other end is given by
$t = \dfrac{L}{{{v_d}}}$
Hence, current is the given by
$
I = \dfrac{q}{t} = \dfrac{{ALne}}{{\dfrac{L}{{{v_d}}}}} = neA{v_d} \\
\therefore I = neA{v_d} \\
$
Note:The equations should be kept in mind along with the variables. Here everyone gets confused between the variables of the equations as they tend to appear the same. The equation for time taken and electric field is of utmost importance.
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