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Physics lab formula question - help please

  1. Nov 23, 2015 #1
    1. The problem statement, all variables and given/known data

    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
    2. Relevant equations


    3. 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 ^?
     
    Last edited by a moderator: Nov 23, 2015
  2. jcsd
  3. Nov 23, 2015 #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.
     
  4. Nov 24, 2015 #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.
     

    Attached Files:

  5. Nov 24, 2015 #4
    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..
     
  6. Nov 24, 2015 #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?
     
  7. Nov 24, 2015 #6
    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
     
  8. Nov 24, 2015 #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".
     
  9. Nov 24, 2015 #8
    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
     
  10. Nov 24, 2015 #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.
     
  11. Nov 24, 2015 #10
    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...
     
  12. Nov 24, 2015 #11

    andrevdh

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    M = amount of turns x current in coil x cross sectional area
     
  13. Nov 24, 2015 #12
    yes...which I show in the work above.....so are they saying put my "new value" A( .7090) in for my cross sectional^
     
  14. Nov 24, 2015 #13

    andrevdh

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    was the current about 30 mA?
     
  15. Nov 24, 2015 #14
    yeah 0.03 on the dot
     
  16. Nov 24, 2015 #15

    andrevdh

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    did you get the A (= 0.7090) fit parameter from logger pro by inserting n = -3?
     
  17. Nov 24, 2015 #16
  18. Nov 24, 2015 #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.
     
  19. Nov 24, 2015 #18
    That's why when I put in 0.7090, the results seem to be way off
     
  20. Nov 24, 2015 #19


    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
     
  21. Nov 24, 2015 #20
    those are kind've hard to read, but the magnetic field is about -.020mT
     
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