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Physics 1251 HW8 Question 4

0 votes
2,196 views
I need help with part B!

For a research project, a student needs a solenoid that produces an interior magnetic field of 0.0100 T. She decides to use a current of 1.00 A and a wire 0.500 mm in diameter. She winds the solenoid in layers on an insulating form 1.00 cm in diameter and 19.0 cm long.

~~Determine the number of layers of wire needed. (Round your answer up to the nearest integer.)

The answer to this is 4 layers

~~Determine the total length of the wire. (Use the integer number of layers and the average layer diameter.)

I need help with this part.  Since each layer will have a different diameter, do we just average the 4 diameters?  Doing this I got an average diameter of 1.075 cm.  Don't we just multiply the total number of turns by the circumference (pi*0.01075) to get the length needed?  I did it this way and the answer I got was wrong.  Can someone let me know how they solved this?
asked Mar 14, 2014 in Intro College Physics by Biomed94 (1,080 points)

1 Answer

0 votes
b)

$$L_{wire}= n_{layers} \cdot n_{turns per layer} \cdot d_{avg} \cdot \pi $$

Where \( n_{layers} \) was found in part (a), \(n_{turns per layer} = L_{solenoid}/d_{wire}\), and the average diameter needs some consideration:

The diameter of the first layer should be taken from the CENTER of the wire. This means that instead of 0.0100m we will have a layer with diameter 0.0105m. This is because we add half a diameter of wire on both sides of our circle. Then each additional layer adds 0.0010m...half from one side of the circle, half from the other.

\(n_{layers} = 4\)
\(n_{turns per layer} = 380 \)

\( d_{avg} = \frac{\sum\limits_{i=1}^{n_layers} (0.01005+0.0010(i-1))}{n_{layers}} =0.0120\)

 
Like this, I found \( L_{wire} = 57.3 m \)
answered Mar 14, 2014 by jcf1559 (2,140 points)
edited Mar 17, 2014 by jcf1559
Can you show me the numbers you are plugging in to get that answer?
I have:
Number of layers=4
Number of turns per layer=380

L=4*380*0.01075*π
=51.3 m

Which is pretty different than what you got, can you let me know your approach?
updated the original answer...you are correct. When I entered numbers into my equation I used 10.0 cm for the length of the solenoid...that is fixed now.

If 51.3m is the answer you had originally, I would suggest only carrying three sig figs throughout the problem (including the average diameter) and you will end up with a slightly different result of 51.6m


Actually, now that I think about it I bet I know the issue:

The average diameter should be from the CENTER of the wire, not the inner edge. This effectively increases the diameter of each layer by the diameter of one wire (half a diameter on one side and half a diameter on the other). This would mean our  \(d_{avg} = 0.01125 m\) resulting in \(L_{wire}= 53.7m\).
So I checked a few other sources, and webassign finally took 57.3 m as the acceptable answer...don't ask me how that works haha.
Ok I should not attempt to help when i'm half awake...the issue is related to what i said before with looking at the average diameter at the center of the wire in each layer. This make sthe first one 1.05 cm, then 1.15cm, 1.25 cm, and 1.35 cm averaging at 1.20cm. this will yield the correct 57.3m anwery
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