Question

The relation between current density j and drift velocity $ {{v}_{d}} $ is?
  $ A.J=ne{{v}_{d}} $
  $ B.J=\dfrac{ne}{{{v}_{d}}} $
  $ C.J=\dfrac{{{v}_{d}}e}{n} $
  $ D.J=nev_{d}^{2} $

Answer 1

Hint: Knowledge of charge transport phenomenon in conductors is vital to solve this problem. Making a circuit diagram of the charge transport phenomenon will assist in solving this problem easily. Drift velocity $ '{{v}_{d}}' $ is given by, $ {{v}_{d}}=\dfrac{l}{t} $ , where ‘l’ is the length of the conductor and ‘t’ is the time taken to cover that distance. The total charge in the conductor will be, $ Q=n.A.l.e $ , where ‘n’ is the electron density of the conductor. Hence, Current will be, $ I=\dfrac{n.A.l.e}{t}=n.e.A.{{v}_{d}} $ . The current density ‘J’ is current flowing per unit cross sectional area. That is, $ J=\dfrac{I}{A}. $

Complete step-by-step answer:
Let’s make a diagram of current flowing through a conductor to get an idea for the charge transport phenomenon in a conductor.

We will consider the length of the conductor to be ‘l’ and the cross sectional area of the conductor to be ‘A’. Further, the free electron density of the conductor is ‘n’ electrons per meter cube. The conductor contains multiple cations and multiple electrons as shown in the diagram above.
When a potential difference ‘V’ is applied as shown in the diagram, an electric field ‘E’ will be produced across the conductor from the positive terminal towards the negative terminal. The free electrons of the conductor travel in the opposite direction of the applied electric field. Hence, the electrons travel from the negative end of the conductor (as marked in the diagram) towards the positive end of the conductor (as marked in the diagram). The velocity with which these electrons travel is known as drift velocity of the electrons or just drift velocity of the conductor.
The actual definition of drift velocity $ '{{v}_{d}}' $ is as follows: The average velocity with which the electrons drift towards the positive end of the conductor under the influence of an applied external electric field.
  $ \therefore {{v}_{d}}=\dfrac{l}{t} $ , that is the length of the conductor taken to cover that distance.
We have free electron density as n electrons/ $ {{m}^{3}}. $ Therefore, total number of electrons in the conductor is ‘N’: $ N=n.A.l $
Therefore the total charge in the conductor will be, $ Q=N.e=n.A.l.e $
This implies, the amount of current flowing through the conductor will be $ I=\dfrac{Q}{t}=\dfrac{N.e}{t}=\dfrac{n.A.l.e}{t} $
Since, the drift velocity is: $ {{v}_{d}}=\dfrac{l}{t}\Rightarrow l={{v}_{d}}.t $
Substituting this value of length in the value of current, we get, $ I=n.e.A.{{v}_{d}} $
Further, the current density ‘J’ is current flowing per unit cross sectional area. That is, $ J=\dfrac{I}{A}. $
  $ J=\dfrac{n.e.A.{{v}_{d}}}{A}=n.e.{{v}_{d}} $
Hence, $ J=n.e.{{v}_{d}} $

Note: It’s important to know as to why; we call the movement of electrons in the conductor from the negative end towards the positive end in the presence of an externally applied electric field as drift velocity and not just velocity of the electrons. It is known as drift velocity, because in the presence of no externally applied electric field, the electrons would still be randomly moving at high velocities, however upon the application of the electric field, the electrons start to get aligned as per the electric field, and the these electrons start to gain speed due to the applied field itself. These electrons hit with other electrons to transfer the momentum being generated, hence, giving an overall picture of the electron drifting through the conductor, leading to the name drift velocity. This whole phenomenon is known as charge transport phenomenon.