derive ohm's law in microscopic form
Question
2 Answer

1. User Answers roshankr1000
Answer:
Let us consider a conductor of length L having crosssectional area A. Let i is the current passing through through the conductor and R is the resistance of the conductor and V is the potential difference applied across the two ends of the conductor then by ohm's law :
V=iR ........(1)
Since the resistance of a conductor with resistivity r, length L and area A is given by :
R=rL/A
Putting above value of R in equation (1) :
V=irL/A .........(2)
If E is the electric field intensity across the ends of conductor and V is the potential difference then the relation between field and potential difference is
V=EL .........(3)
Comparing equation (2) and equation (3) :
EL=irL/A
E=ir/A
E=(i/A)r .........(4)
Since by definition of current density J :
J=i/A
Putting this value in equation (4) :
E=Jr
Since r is resistivity and conductivity c is the the reciprocal of resistivity i.e
c=1/r
=> r=1/c
Hence;
E=J/c
=> J=cE .........(5)
Equation (5) is the Microscopic Form of Ohm's Law in terms of conductivity, electric field intensity and current density.
Hope so it will help you :))

2. User Answers Anonym
Answer:
Explanation:
Ohm's Law :
It states that , on a constant temperature , the current I , flowing through the electric circuit , is directly proportional to the potential difference V , across its ends.
V ∝ I
I ∝ V
V/ I = R , constant of proportionality
R = Resistance [ opposes flow of current ]
V = IR
R = V/I
I = V / R