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Measuring a Coil of Wire With Ohmmeter and Balance

  1. Apr 18, 2015 #1
    1. The problem statement, all variables and given/known data

    A coil of fine copper wire has a resistance of 6.2 -Ω and a total mass is 14.4 g. What is the diameter d of the wire and what is its length L? (The density of copper is 8.96 g/cm3.)
    Hint Given: •Model the wire as a long cylinder of diameter d and length L. Write one relation for its mass, another for its resistance--from these two relations, you can determine the two desired quantities.

    2. Relevant equations

    R=p*L/A where R = Resistance, p = resistivity coefficient, L = length, A = Area

    In this case resistivity coefficient of copper is given as p = 1.68*10^-8 Ohms/m

    3. The attempt at a solution

    Using the equation R=p*L/A I know the value of R is 6.2 Ohms and the value of p is 1.68*10^-8 Ohms/m
    I know that to find the diameter and length of the cylinder, I need to find the Area.
    To find the area, my method is: p/R * L = A
    But I only know p and R, so I'm not sure how to solve for the length and the area.
    Would I just use v = m/d where v = volume and m/d = mass/density ?
    If I do that:
    (14.4g)/(8.96g/cm^3) = 1.607142 (volume)
    What do I do once I get the volume? Do I convert the volume measurements since they're in cm^3?

    Maybe this is more of a math question but I'm just stuck at that part so I would appreciate some help.
     
  2. jcsd
  3. Apr 18, 2015 #2

    Merlin3189

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    Gold Member

    Yes, just a math problem: two unknowns need two equations before you can solve them.
    One equation gives you A/L = k or A=kL (from resistance) and the other gives you AL = v because Area x Length =Volume.
    You can combine the two equations to eliminate either A or L, then solve for the other. Then use the value you find, put into either equation and solve for the other.

    You need to be careful because,
    1- you are mixing units (m and cm): resistivity is given in Ohm metres and density in grammes per cubic centimetre. Maybe you should decide what units to work in and convert the constants? Otherwise be careful!
    2- you wrote, "the value of p is 1.68*10^-8 Ohms/m" but the units of resistivity are Ohm metres not Ohms per metre

    Maths example:
    If you know P= 5 Q and P x Q = 10
    then you can use the first in the second to say
    5 Q x Q = 10 so
    5 Q2 = 10 so
    Q2 = 2 so
    Q = sqrt(2) = 1.414 and
    P = 5 Q = 5 x 1.414 = 7.07
    Check P x Q = 7.07 x 1.414 = 9.996 = 10 to 3sf
     
  4. Apr 18, 2015 #3
    So if I do this:

    (1.68*10^-8) = L/A
    and multiply both sides by A then
    A(1.68*10^-8)=L

    then substitute that into the AL=V

    A(A*1.68*10^-8A)=1.6071 (calculated volume from given density and mass)
    and calculating for A, I get
    A=457.346

    ?
     
  5. Apr 18, 2015 #4

    Merlin3189

    User Avatar
    Gold Member

    Originally you said,
    So your work now,
    does not look quite right, does it?
    You should have
    R = (p*L)/A so 6.2 = ( 1.68^-8 * L) / A etc

    yes, provided you are sure the units are compatible.

    Finally,
    well the ? is appropriate, because you have not given any units!
    But it was wrong because of the first mistake and you did well to realise it could not be correct.

    I think you are nearly there now.
     
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