Define electric current density and write its relation with the current. Derive the relation between current density and electric field intensity.

Explanation

Electric Current Density (J)
The current density at a point in a conductor is the ratio of the current at that point in the conductor to the area of cross-section of the conductor at that point. If a currentI is distributed uniformly over the cross-section A of a conductor, then the current density at that point is
`J=(I)/(A)`
It is characteristic property of a point inside the conductor. It is a vector quantity. Its direction at a point is the direction of flow of motion of positive charge at that point.
The unit of current density J is ampere per metre squar or `A//m^(2)` in SI system. The dimension of J is `AL^(-2)`.
Now, consider a conductor of length l and A be the uniform area of cross-section.
`therefore` Volume of conductor = Al
If the conductor contains free electrons n per unit volume.
Then, number of free electrons in the conductor = nAl
If e is the charge of an electron, then total charge on all free electrons in the conductor is given by
`q=nAl e....(i)`
Time taken by the free electrons to cross the length of the conductor,
`t=(l)/(v_(d))...(ii)`
Since, current is the rate of flow of the charge through conductor.
`I=(q)/(t)`
Using Eqs. (i) and (ii), we get
`I=(nAl e)/(l//v_(d))impliesI=n eAv_(d)...(iii)`
Eq. (iii) give the relation between the current flowing through the conductor and drift velocity of the electron.
Putting the value of `v_(d)=((eEtau)/(m))` in Eq. (iii), we get
`I=(An e^(3)tauE)/(m)`
Also, current density, `J=(I)/(A)`
`J=(n eAv_(d))/(A)impliesJ=n ev_(d)orJ=(n e^(2)tauE)/(m)`