Measuring a Coil of Wire With Ohmmeter and Balance

[Angie K.]

Homework Statement

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.

Homework 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

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.

 

[Merlin3189]

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

 

[Angie K.]

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

?

 

[Merlin3189]

Originally you said,

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

So your work now,

So if I do this: (1.68*10^-8) = L/A

does not look quite right, does it?
You should have
R = (p*L)/A so 6.2 = ( 1.68^-8 * L) / A etc

then substitute that into the AL=V

yes, provided you are sure the units are compatible.

Finally,

and calculating for A, I get A=457.346 ?

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 realize it could not be correct.

I think you are nearly there now.