Physics lab formula question - help please

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Homework Statement



An ideal solenoid is expected to generate a dipole field that falls off quickly as you move away from the solenoid. The magnetic field at distance r along the axis of the solenoid is given by B = (Mo/ 2pie) (M/r^3) In this equation the parameter M is called the dipole moment and it is equal to M = NIA where N is the number of turns, and A
the cross-sectional area of the solenoid.
Calculate the value of the M , knowing that the number of turns in the solenoid you have is 1080 and the corss-sectional diameter is about 7.5 mm. Enter your measurements from the table above into the Logger Pro program and plot B vs r for each side of the solenoid. Then, perform a “variable power” fit of the form Y=AX^n with n set to -3 and identify the value of of the fit parameter A
Using the data you now have determine the value of the permeability of free space Mo


Be careful with the units

Homework Equations




The Attempt at a Solution



I have;

B= -.020mT = -.000020 T

2(p) = 6.28

M= 1.43x10^-3 Am^2

r = 20mm = 0.020m

fit parameter A = 0.7090

WHERE does the new A be placed ^?
 
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Answers and Replies

  • #2
marcusl
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From your equations for B and Y, you can see that A_fit = M_0 * M / (2*pi). You are supposed to find a value for M0.
 
  • #3
andrevdh
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It seems that the instructions are telling you to set the gradient of the power fit graph to the suggested value for A in Logger Pro.
 

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  • #4
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Yes I understand that, which I did. I found the Value for A with the new equation Y=Ax^-3 .......I just dont understand why I need that A when they gave me an A already for the formula..
 
  • #5
andrevdh
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Maybe they first want you to verify the power -3 with the supplied A
and then fit again with the power to obtain an experimental value for A?
 
  • #6
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Maybe they first want you to verify the power -3 with the supplied A
and then fit again with the power to obtain an experimental value for A?

the supplied A comes from the radius of 7.5mm ....solve for that A and you can solve M=NIA .....therefor I have B,r, and M and should be able to solve for Mo without anything else...I just dont get why they say with this new "data" you can now solve - which I thought I could have before without the new A
 
  • #7
andrevdh
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This is getting confusing, especially since there are talk of two As!
I get the cross sectional area of the solenoid as 4.42x10-05 m2.
I tend to agree with marcusl that you are suppose to use the obtained A form the power fit to calculate mu_zero - the permeability of free space.
This value of A from the power fit might be what they are referring to as the "new data".
 
  • #8
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I got the same cross sectional, i tried using the new A from the graph to solve for M and then solve for Mu_zero, however the answer is 7.4x10^-5, which is no where close to the actual value for Mo
 
  • #9
andrevdh
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I don't think you are suppose to "solve" for the magnetic dipole moment. You should calculate its value from the experimental values and then solve for mu_o using the fitted value of A.
 
  • #10
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Yeah I got that, by solve for M i meant find M (calculate M with the new A) and then solve for Mo....with the new formula....which gives that outrageous number still...
 
  • #11
andrevdh
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M = amount of turns x current in coil x cross sectional area
 
  • #12
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yes...which I show in the work above.....so are they saying put my "new value" A( .7090) in for my cross sectional^
 
  • #13
andrevdh
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was the current about 30 mA?
 
  • #14
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yeah 0.03 on the dot
 
  • #15
andrevdh
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did you get the A (= 0.7090) fit parameter from logger pro by inserting n = -3?
 
  • #17
andrevdh
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I assume you have a set of measurements where you measured the magnetic field at various distances from the central axis?
Could you maybe attach it here?
Another thing that is worrying me is that one gets two different probes. One for measuring the axial and another for the radial component of the magnetic field.
I am trying to find your formula in some literature, but have not been successful yet.
 
  • #18
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did you get the A (= 0.7090) fit parameter from logger pro by inserting n = -3?
That's why when I put in 0.7090, the results seem to be way off
 
  • #19
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I assume you have a set of measurements where you measured the magnetic field at various distances from the central axis?
Could you maybe attach it here?
Another thing that is worrying me is that one gets two different probes. One for measuring the axial and another for the radial component of the magnetic field.
I am trying to find your formula in some literature, but have not been successful yet.



B (T)

r(mm)


B(T)

r(mm)

-0.02

20


0.02

20

-0.019

21


0.019

21

-0.018

22


0.018

22

-0.017

23


0.017

23

-0.016

24


0.015

24

-0.015

25


0.014

25

-0.013

26


0.013

26

-0.011

27


0.012

27

-0.008

28


0.01

28

-0.007

29


0.009

29
 
  • #20
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those are kind've hard to read, but the magnetic field is about -.020mT
 
  • #21
andrevdh
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I'll analyze it. Going to take a while, but I will most likely get what you got for A.
 
  • #22
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After the value of A is found is where I get stuck completely...
 
  • #23
andrevdh
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That might be that the experiment was not carried out correctly or the instrument was not used correctly .... ?:)
 
  • #24
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Thats why I posted on here, hoping that wasnt the case haha!
 
  • #25
andrevdh
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I did not force it throught the origin. My fit gives A = 1.306 x 10-7 Tm3
which gives a value of 5.76 x 10-4 H/m for mu_o. Sorry that is how far I can help you.
 

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