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Energy and force in a solenoid with multiple wirings

  1. Mar 14, 2014 #1
    1. The problem statement, all variables and given/known data
    A solenoid has length L = 2.0m, radius R = 28.0cm, field strength B = 2.7T. The field is maintained by superconducting wires, with cross section 2mm x 2mm, wound with a 2mm spacing between windings and 20 layers of winding. Use the result that the current is I = 215A.

    a. If the coil were made of room-temperature copper (resistivity ρ = 1.680×10-8 Ωm), how much power would be consumed in maintaining this current?
    b. The magnetic field strength falls, from its full value in the innermost layer of wires to zero in the outermost layer, so on average the wires feel half the field strength present inside the solenoid. Compute the force per unit area, pushing outward on the wires.
    c. Find the ratio of the energy density stored in the field and the force per unit area acting on the wires.


    2. Relevant equations
    -


    3. The attempt at a solution
    I have calculated the energy density stored, which should be 2.90x106J/m3. However, I can't seem to find the answer to a. I'm using the equation [itex]R = \frac{pL}{A}[/itex] and then [itex]P = I^{2}R[/itex]. For the resistance, I'm taking the length to be [itex]L = 2\pi r[/itex] and the area to be the cross-sectional area of the wires. Then, the current is the result obtained. However, I'm not sure what's wrong with my reasoning.

    On the other hand, for question b., I got that the force per unit area should be given by [itex]F = \frac{BIl}{2A}[/itex], where the length would be the length given by [itex]L = 2\pi r[/itex] and the area should be the product [itex]A = 2\pi rw[/itex], where w is one of the sides which is 2mm. However, this result as well is wrong. I really can't visualize how can multiple winding solenoids work. From this, then the result to c. should come easily.

    Thank you for your time and patience.
     
    Last edited: Mar 14, 2014
  2. jcsd
  3. Mar 14, 2014 #2

    NascentOxygen

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    Staff: Mentor

    L=2Pi.r sounds like the length of wire per single turn. How many turns in the total solenoid winding?
     
  4. Mar 14, 2014 #3

    It would be N = 1000 turns as we divide the total length by the spacing. I used this and i stilll get a wrong anser for a. (Using it as the total length of wire).
     
  5. Mar 14, 2014 #4

    NascentOxygen

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    Staff: Mentor

    I don't make it 1000 turns.
     
  6. Mar 14, 2014 #5
    What do you mean? Is the result different?
     
  7. Mar 14, 2014 #6

    NascentOxygen

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    Staff: Mentor

    Yes, I calculate a different number.
     
  8. Mar 14, 2014 #7
    P = (I^2)(resistivity)l / A

    A = 0.002^2
    n = LW/0.002
    l = 2pi*r*n

    for a, that is what you should be using, I believe
     
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