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Does the electric current have a direction?

  1. Mar 1, 2009 #1
    We know that the electrical current is scalar. Also we know that a scalar hasn't got a direction but electric current has got a direction. I've confused! Please help me..
     
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  3. Mar 1, 2009 #2

    ZapperZ

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    How do you know that electric current is a scalar?

    Zz.
     
  4. Mar 1, 2009 #3
    Electric current does have a direction, and, as such, is a vector, and not a scalar.
     
  5. Mar 1, 2009 #4
    Electic current does have a direction, the problem is the frequent appearance of the equation:

    [tex]i = \frac{dq}{dt} [/tex]

    which does not really tell exactly what current is; this equation only gives you the magnitude, not the direction. I prefer to define the surface current density:

    [tex]\vec{j} = \rho \vec{v}[/tex]

    in terms of the charge density and the velocity. Then we have:

    [tex]\vec{i} = A \vec{j} [/tex]

    where A is the area.
     
  6. Mar 1, 2009 #5
    Last edited by a moderator: May 4, 2017
  7. Mar 1, 2009 #6

    jtbell

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    The current density [itex]\vec J[/itex] is a vector, but the current I through a given surface is a scalar, as can be seen from the relationship between the two:

    [tex]I = \int{\vec J \cdot d \vec a}[/tex]

    When you're calculating e.g. the magnetic force on a current-carrying wire, the directionality of the current is properly associated with the length of the wire rather than with the current itself:

    [tex]\vec F = I \vec l \times \vec B[/tex]

    for a straight wire segment and uniform [itex]\vec B[/itex], or

    [tex]\vec F = I \int {d \vec l \times \vec B}[/tex]

    otherwise. This assumes that [itex]\vec B[/itex] doesn't vary significantly over the cross-section of the wire. If it does, then you have to calculate the force by using the current density and integrating over the volume of the wire:

    [tex]\vec F = \int {(\vec J \times \vec B) dV}[/tex]
     
    Last edited: Mar 1, 2009
  8. Mar 1, 2009 #7

    jtbell

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    You have to allow for the area not being perpendicular to the current. If [itex]\vec J[/itex] is uniform, then you can use

    [tex]I = \vec J \cdot \vec A[/tex]

    where the direction of [itex]\vec A[/itex] is perpendicular to the surface. If [itex]\vec J[/itex] is not uniform, then you have to integrate.
     
  9. Mar 1, 2009 #8
    I don't understand what [itex]\vec B[/itex] stands for.
     
  10. Mar 1, 2009 #9

    Vanadium 50

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    Magnetic field.
     
  11. Mar 1, 2009 #10
    Current is scalar.

    Current (in amperes) is the amount of charge that passes through a point on a conductor every second. It is just a number. An "ampere" is a scalar quantity.

    But, and this might make it confusing, but the electrons do flow in a given direction.
    The electrical current does have a direction, but that information isn't contained in the unit "ampere".

    Does that make sense at all? I could try to explain it better...
     
    Last edited by a moderator: May 4, 2017
  12. Mar 1, 2009 #11
    The units only ever measure magnitude, not direction.

    It doesn't even make sense to have vector units.
     
  13. Mar 2, 2009 #12
    All you are very helpful. Thanks for replies.
     
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