Question

In a straight conductor of uniform cross-section charge \[q\] is flowing for time \[t\] . Let \[s\] be the specific charge of an electron. The momentum of all the free electrons per unit length of the conductor, due to their drift velocity only is:
A. $\dfrac{q}{{ts}}$
B. ${\left( {\dfrac{q}{{ts}}} \right)^2}$
C. $\sqrt {\dfrac{q}{{ts}}} $
D. $qts$

Answer 1

Hint:To answer the issue and determine the momentum of all free electrons per unit length of the wire due to their drift velocity, we will utilise the drift velocity formula and arrive at our final conclusion.

Formula used:
\[I = {\text{ }}nAe{V_d}\]
Where, where \[n\] is the number of electrons per unit volume, \[A\] is the wire's cross-sectional area, \[{V_d}\] is the electron’s drift velocity, and \[e\] is the electron’s charge.

Complete step by step answer:
A drift velocity is the average velocity obtained by charged particles in a medium, such as electrons, due to an electric field in physics. In general, an electron propagates at the Fermi velocity at random in a conductor, resulting in an average velocity of zero. An electric field adds a modest net flow in one direction to this random motion; this is the drift.Now coming to the question, In order to answer it we will apply the formula for drift velocity i.e.
\[I = {\text{ }}nAe{V_d}\]

We're looking for momentum per unit of wire length. (i.e.) \[m{V_d}\] , where \[m\] is the total mass of all electrons in a unit length of wire.Now, we know that;
$I = \dfrac{q}{t} = nAe{V_d}$
And, \[m = {\text{ }}n \times \left( {A \times 1} \right) \times ({m_e})\]
Where \[n\] = number of electrons per unit volume, \[A \times 1\] = volume of unit length cross section, and \[{m_e}\] = mass of one electron.

Therefore,
\[{\text{ }}m{V_{d\;}}\; = {\text{ }}(nA{V_d}) \times {m_{e\;}} \\
\Rightarrow {\text{ }}m{V_{d\;}} = \dfrac{{(q \times {m_e})}}{{\left( {t \times e} \right)}}\]
Due to the fact that \[\dfrac{e}{{{m_e}}}\] = specific charge = \[s\] ,
\[m{V_{d{\text{ }}\;}} = {\text{ }}\dfrac{q}{{\left( {t \times s} \right)}} \\
\therefore m{V_{d{\text{ }}\;}}= \dfrac{q}{{ts}}\]
Therefore, the momentum of all the free electrons per unit length of the conductor is \[\dfrac{q}{{ts}}\].

Therefore, the correct option is A.

Note:When dealing with a macroscopic (normal, everyday life) wire, the drift velocity is independent of the wire's length or cross sectional area. However, if the wire is too short, say, comparable to the typical distance a charge carrier travels before colliding, it may begin to rely on the wire length, but a wire will not be so short for all practical uses.