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Current and current density

  1. Jun 30, 2015 #1
    we know
    i = dq/qt = enLA/dt = en(Vd)(A)
    standard derivation i en(Vd)(A)
    where e is charge on electron
    n is electrons per unit vol.
    Vd is drift velocity of electrons
    A is area of cross section of conductor

    SO my doubt is if we have a conical conductor or some kind of conductor whose CSA is not constant , will we have variable current across its length according to i = en(Vd)A


    I also want to ask what is current density ??
    Precisely i want to know - current density(j) = di/dA
    here what is ''di'' signifying ???
     
  2. jcsd
  3. Jun 30, 2015 #2

    mfb

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    No, current is constant, your drift velocity won't be constant.

    d/dA is a notation for a derivative with respect to A. di/dA means the derivative of i with respect to A.
     
  4. Jul 2, 2015 #3
    i = dq/dt
    so for a conductor of length L, maintained at constant potential difference, we know that q = enAL
    where e is charge on electron
    n is electrons per unit vol.
    A is area of cross section of conductor
    SO
    in q = enAL what is changing with time so that we come across rate of change of charge
     
  5. Jul 3, 2015 #4

    vanhees71

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    It is very important in physics to define the quantities properly. In electrodynamics you have to strictly distinguish between a current (a scalar) and the current density (a vector field). The more fundamental notion is the current density ##\vec{j}(t,\vec{x})##. It's easier to understand, writing it in more microscopic (classical) quantities. First you have the charge density (a scalar) ##\rho(t,\vec{x})##. It gives the electric charge per unit volume around ##\vec{x}## at each time ##t##. In the same way at time ##t## you have the velocity field ##\vec{v}(t,\vec{x})## of the charges, i.e., the velocity of the charged particles at time ##t## that are located at ##\vec{x}##. Then the current density is given by
    $$\vec{j}(t,\vec{x})=\rho(t,\vec{x}) \vec{v}(t,\vec{x}).$$
    To get its meaning take a surface ##A## with surface-normal vector elements ##\mathrm{d}^2 \vec{A}##. Then the charge running through this surface within a time ##t## is given by
    $$\mathrm{d} Q=\mathrm{d} t \int_{A} \mathrm{d}^2 \vec{A} \cdot \vec{j},$$
    because ##\mathrm{d} V=\mathrm{d} t \int_A \mathrm{d}^2 \vec{A} \cdot \vec{v}## is the volume swept out in the infinitesimal time interval ##\mathrm{d} t## by the particles running through the surface, and the charge contained in this volume is ##\mathrm{d} V \rho##.

    Now the current is by definition the charge per unit of time running through the surface ##A##. So you have
    $$i(t)=\frac{\mathrm{d} Q}{\mathrm{d} t} = \int_A \mathrm{d}^2 \vec{A} \cdot \vec{j}.$$

    It's important to keep in mind that a current only makes sense when you specify the surface and its orientation, i.e., the arbitrarily chosen direction of the surface normal element vectors.

    In the usual applications, of course, you define the current as those running through the complete cross sectional area of the wire, and you only have to be careful with the orientation.
     
    Last edited: Jul 3, 2015
  6. Jul 3, 2015 #5
    You mean to say j is property of a material
    j⃗ (t,x⃗ )=ρ(t,x⃗ )v⃗ (t,x⃗ )
    and only having defined that we can get to current.
    Is it what you are trying to say ??
     
  7. Jul 3, 2015 #6
    Further you said that

    dQ=dt ∫dA⃗ ⋅j⃗

    so bringing dt to the LHS we get current. and you are saying that only those electrons from the stream of electrons contribute to current which are along the direction of dA our area vector ??

    AND
    here
    j⃗ (t,x⃗ )=ρ(t,x⃗ )v⃗ (t,x⃗ )
    is ''v'' drift speed of electrons ??
     
  8. Jul 3, 2015 #7
    have I correctly understood what you meant ??
     
  9. Jul 3, 2015 #8

    vanhees71

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    Of course, that's not the only way to describe a current or current density, but it is the most fundamental one. For the usual case of electrons in a wire, ##\vec{v}(t,\vec{x})## is the drift velocity as a function of time and position of the electrons within the wire.
     
  10. Jul 3, 2015 #9
    j is property of a material ??
     
  11. Jul 3, 2015 #10
    j = di/dA definition i found in my textbook

    but for a conductor maintained at constant potential difference i should stay the same
    then j = 0 ??
     
  12. Jul 3, 2015 #11

    vanhees71

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    If you have a wire at constant potential difference (after some short transient time) you have a direct current because of the static electric field.

    Microscopically this is explained as follows due to the constant electric field along the wire on the electrons acts the electric force ##\vec{F}_{\text{el}}=-e E \vec{e}_{\text{wire}}## as well as a friction force ##\vec{F}_{\text{fr}}=-\gamma v \vec{e}_{\text{wire}}##, where I assume that the wire is so thin that we can assume the electrons move always in direction of the wire. The equation of motion for the electrons thus reads (in non-relativistic approximation, which is very accurate for everyday house-hold currents)
    $$m \ddot{x} =m \dot{v}=-\gamma v -e E.$$
    These are all components in direction of ##\vec{e}_{\text{wire}}##.

    The general solution is
    $$v(t)=(v_0+e E/\gamma)\exp(-\gamma t/m)-e E/\gamma.$$
    For ##t \gg m/\gamma## we have ##v=-e E/\gamma=\text{const}##. In this stationary state the current density is
    $$\vec{j}=-n e \vec{v}=n e^2/\gamma \vec{E}= \sigma \vec{E},$$
    where ##n## is the number density of conduction electrons in the material and ##\sigma=n e^2/\gamma## the electric conductivity. Thus you get just Ohm's Law.

    You get a more famliar form by first integrating over the cross-sectional area of the wire (normal vector in direction of ##\vec{E}_{\text{wire}}##). Then the voltage between the end of the wire is given by ##U=l E##, where ##l## is the length of the wire. You thus get for the current
    $$i=\sigma A E=\sigma A U/l =U/R \; \Rightarrow R=\frac{l}{\sigma A},$$
    which is the usual Ohm law in circuit theory.
     
  13. Jul 6, 2015 #12
    can you put this in a simpler way !!
     
  14. Jul 12, 2015 #13

    vanhees71

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    It's difficult to make this more simple. Where is your problem with this treatment? It's known as the Drude model of metals. You can find it in textbooks on statistical mechanics. A very good source is

    A. Sommerfeld, Lectures on Theoretical Physics, vol. V

    Note that Sommerfeld was the one who first applied Fermi-Dirac statistics to this model, solving a related puzzle with it, namely why the electrons in the metal do not essentially contribute to the specific heat of the metal, while the naive treatment of conduction electrons is pretty successful in deriving Ohm's Law.
     
  15. Jul 13, 2015 #14
    ok its very fine
     
  16. Jul 13, 2015 #15
    But i have a doubt that
    you said j⃗ (t,x⃗ )=ρ(t,x⃗ )v⃗ (t,x⃗ )
    but as i said previously -

    i = dq/qt = enLA/dt = en(Vd)(A)
    standard derivation i=en(Vd)(A)
    where e is charge on electron
    n is electrons per unit vol.
    Vd is drift velocity of electrons
    A is area of cross section of conductor

    then j=en(Vd)
    so how do you correlate both the definitions ????
     
  17. Jul 13, 2015 #16

    Dale

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    ##\rho=en## and ##\vec{v}=Vd## so the correlation is pretty direct.
     
  18. Jul 14, 2015 #17

    vanhees71

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    I don't understand the 2nd formula. What's ##d##? The charge density is ##\rho=en## (now ##e<0## for electrons) and ##\vec{v}## is the drift-velocity field of the electrons. Then ##\vec{j}=\rho \vec{v}=e n \vec{v}##.
     
  19. Jul 14, 2015 #18

    atyy

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    The current is constant means di/dt = 0, where t is time.

    However, di/dA means the current per area, which is non-zero. If di/dA is constant, then the total current is just the A times di/dA.
     
  20. Jul 14, 2015 #19
    I don't understand the 2nd formula. What's d?[/QUOTE]
    Vd is drift velocity
     
  21. Jul 15, 2015 #20

    vanhees71

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    Then you should write ##\vec{v}_d##. It's important to use clear symbols!
     
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