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Units help

  1. Nov 6, 2008 #1

    tony873004

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    Gold Member

    Help with units
    The book gives the following forumla
    [tex]E_0 = \sqrt {2\mu _0 c\left\langle S \right\rangle } [/tex]

    And when it uses it in an example, the units go from:
    [tex]\sqrt {\left( {{\rm{H/m}}} \right)\left( {{\rm{m/s}}} \right)\left( {{\rm{W/m}}^{\rm{2}} } \right)} [/tex]
    to [tex]V/m[/tex], with no intermediate steps. My two efforts fall short:

    Attempt #1
    [tex]\sqrt {\left( {{\rm{H/m}}} \right)\left( {{\rm{m/s}}} \right)\left( {{\rm{W/m}}^{\rm{2}} } \right)} = \sqrt {\frac{{{\rm{HmW}}}}{{{\rm{msmm}}}}} = \sqrt {\frac{{{\rm{HW}}}}{{{\rm{m}}^{\rm{2}} {\rm{s}}}}} [/tex]
    Does a Henry - Watt divided by a second equal a Volt squared? If so, = V/m

    Attempt #2 using N/A^2 instead of H/m for units of mu0
    [tex]\begin{array}{l}
    \sqrt {\left( {{\rm{N/A}}^{\rm{2}} } \right)\left( {{\rm{/s}}} \right)\left( {{\rm{W/}}{\rm{m}}} \right)} = \\
    \\
    \sqrt {\frac{{{\rm{NW}}}}{{{\rm{A}}^{\rm{2}} {\rm{s}} \cdot {\rm{m}}}}} = \sqrt {\frac{{{\rm{kgmJ}}}}{{{\rm{A}}^{\rm{2}} {\rm{s}} \cdot {\rm{ms}}^{\rm{2}} {\rm{s}}}}} = \sqrt {\frac{{{\rm{kg - m - kg - m - m}}}}{{\left( {\frac{{\rm{C}}}{{\rm{s}}}} \right)^{\rm{2}} {\rm{s - m - s}}^{\rm{2}} {\rm{s - s}}^{\rm{2}} }}} = \sqrt {\frac{{{\rm{kg - m - kg - m - }}{\rm{ - }}}}{{{\rm{C}}^{\rm{2}} {\rm{s - }}{\rm{ - s}}^{\rm{2}} {\rm{ - s - }}}}} \\
    \\
    = \sqrt {\frac{{{\rm{kg - m - kg - m}}}}{{{\rm{C}}^{\rm{2}} {\rm{s - s}}^{\rm{2}} {\rm{ - s}}}}} = \sqrt {\frac{{{\rm{kg - m - kg - m}}}}{{\left( {{\rm{FV}}} \right)^{\rm{2}} {\rm{s - s}}^{\rm{2}} {\rm{ - s}}}}} = \sqrt {\frac{{{\rm{kg - m - kg - m}}}}{{{\rm{F}}^{\rm{2}} {\rm{V}}^{\rm{2}} {\rm{ - s - s}}^{\rm{2}} {\rm{ - s}}}}} = \sqrt {\frac{{{\rm{NW}}}}{{{\rm{A}}^{\rm{2}} {\rm{s - m}}}}} = \sqrt {\frac{{{\rm{kg - }}{\rm{ - mJ}}}}{{{\rm{A}}^{\rm{2}} {\rm{s}}{\rm{ - s - s}}^{\rm{2}} }}} = \\
    \\
    \sqrt {\frac{{{\rm{kg - mJ}}}}{{{\rm{A}}^{\rm{2}} {\rm{s - s - s}}^{\rm{2}} }}} \\
    \end{array}[/tex]

    This is not beginning to resemble [tex]\sqrt {\frac{{{\rm{V}}^{\rm{2}} }}{{{\rm{m}}^2 }}} [/tex]
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Nov 6, 2008 #2

    D H

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    Henries have the same units as do volt-seconds/ampere. Watts are of course volt-amperes. Thus (H/m)·(m/s)·(W/m2) is equivalent to (V·s)/(A·m)·(m/s)·(V·A/m2) = (V/A)·(V·A/m2) = (V/m)2
     
  4. Nov 6, 2008 #3
    An equation that may come in handy:

    L*dI/dt = V.

    Since we're doing a dimensional analysis, we can ignore the (dimensionless) d's"

    LI/t = V.

    Multiplying both sides by V, LIV/t = V^2.

    Power = P = IV, so substituting this in we get LP/t = V^2.

    The unit of L is H (inductance is measured in Henrys), the unit of P is W (power is measured in watts) and the unit of t is s, so it is indeed true that a volt squared = henry times a watt (or a henry-watt) divided by a second.
     
  5. Nov 6, 2008 #4

    tony873004

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    Thanks, DH and Joeyar. I doubt I was required to justify this step since the book doesn't, but I just had to know :)
     
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