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Physics

Question

derive ohm's law in microscopic form​

2 Answer

  • Answer:

    Let us consider a conductor of length L having cross-sectional 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 :))

  • 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

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