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Magnetic field inside a solenoid

  1. May 28, 2015 #1
    Could someone check my calculations?
    1. The problem statement, all variables and given/known data
    Calculate intensity of magnetic field inside of solenoid, who is made from 1mm thick copper wire with 28 twines(n=28). Solenoid is 25 cm long, diameter - 4 cm, and it`s connected to 24V DC voltage.
    U=24V ; n=28 ; ρ for copper =0,0175 (ohm*mm2)/m
    2. Relevant equations
    H= i n / L
    I=U/R
    ρ= R*S/L
    R=ρ*L/S
    3. The attempt at a solution
    Perimeter of solenoid=2πr=2*3,14*0,02m=0,126m
    Length of solenoid=28 * 0,126 = 3,519 m
    S for wire = πr2 = 3,14 * 0,25*10-6 = 0,785*10-6
    Resistance R=ρ*L/S = 0,0175*3,519 / 0,785*10-6=0,078*107
    I=U/R=24/0,078*107 = 307,69*10-6
    H= In/L = 307,69*10-6*28/0,25 = 0,034 A/m
     
  2. jcsd
  3. May 28, 2015 #2

    Hesch

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    Repeated:

    Perimeter of solenoid=2πr=2*3,14*0,02m=0,126m
    Length of solenoid=28 * 0,126m = 3,519 m

    S for wire = πr2 = 3,14 * 0,25*10-6 = 0,785*10-6
    Resistance R=ρ*L/S = 0,0175*3,519 / 0,785*10-6=0,078*107
    I=U/R=24/0,078*107 = 307,69*10-6
    H= In/L = 307,69*10-6*28/0,25 = 0,034 A/m


    This is right and this is wrong, and I think it's because you are skipping the units. Then you don't know what you are doing.

    The resistance of a 3.519m long copper-wire with a diameter = 1mm will never be 780 kΩ.
     
    Last edited: May 28, 2015
  4. May 28, 2015 #3
    I tried it with all units but still don`t get the problem. Why calculations for wire area are wrong? I had to stick to mm2, instead of m2 or what?
     
  5. May 28, 2015 #4

    Hesch

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    Cross section area of wire = 785*10-9m2.

    In the next line you have:

    Resistance R=ρ*L/S = 0,0175*3,519 / 0,785*10-6=0,078*107

    Try to write this line with units included, and reduce the unit of the result.
     
  6. May 28, 2015 #5
    aargh, somehow at the and I get 34285,664 A*mm2/m and this sounds totally wrong :/ How should the units for final answer look like/how big numbers will be?
     
  7. May 28, 2015 #6

    Hesch

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    0,0175 (ohm*mm2)/m
    L = 3,519 m
    S =0,785*10-6 m2 = 0.785 mm2
    R=ρ*L/S = 0.0175 [Ωmm2/m] * 3.519 [m] / 0.785 [mm2] = 0.0784 Ω
     
  8. May 28, 2015 #7
    Thanks!
    So I=24V/0,078 ohm =306,122 A
    Final question - L in H=In/L stands for 3,519m, not 25 cm, right?
    So the answer is H=306,122 A * 28 /3,519m = 2435,753 A/m
     
  9. May 28, 2015 #8

    Hesch

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    No. (not quite)

    The formula: H = I * n / L is derived from Amperes law:

    circulationH⋅ds = N * I

    ( n / L ) expresses number of turns per length = 28 turns / 0,25 m. The calculated value for H is only valid at the center of the solenoid, so the formula could more exact be written:

    Hcenter = I * ΔN / ΔL
     
  10. May 28, 2015 #9
    but if I take L as 0,25 m, final answer is 34285,664 A/m, sounds huge
     
  11. May 28, 2015 #10

    Hesch

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    Yes, but N*I = 28*306A = 8568A spread out within 25cm is also a huge current-density. ( turn off the current after 0.1 s, or the copper will melt ).

    And, well, it will only be about 0.043T (magnetic induction).
     
    Last edited: May 28, 2015
  12. May 28, 2015 #11
    Thank you for big help!
     
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