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Magnetic flux through loop inside of solenoid

  1. Oct 17, 2008 #1
    1. The problem statement, all variables and given/known data

    A solenoid has 1800 turns of wire and is 125 cm long and 10.0 cm in diameter; see Figure P.14 (attached). A circular wire loop of diameter 5.0 cm lies along the axis of the solenoid near the middle of its length as shown.
    (a) If the current in the solenoid initially is 4.0 A, find the magnetic flux through the smaller loop.
    (b) If the current in the solenoid is switched off and falls to zero in 3.0 s, calculate the average value of the emf induced in the smaller loop.

    2. Relevant equations
    1. B(inside solenoid) = [tex]\mu[/tex]0*n*I, where n=number turns, I= current
    2. [tex]\Phi[/tex]magnetic=[tex]\int[/tex][tex]\stackrel{\rightarrow}{B}[/tex]*[tex]\stackrel{\rightarrow}{dA}[/tex]

    3. The attempt at a solution

    [tex]\Phi[/tex]magnetic=[tex]\int[/tex][tex]\stackrel{\rightarrow}{B}[/tex]*[tex]\stackrel{\rightarrow}{dA}[/tex] = BA = B*[tex]\pi[/tex]*r2 (this did not work, and as I did not get the first part, I had no idea how to even approach the second part...thanks for any advice).

    Attached Files:

  2. jcsd
  3. Oct 18, 2008 #2

    Doc Al

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    Staff: Mentor

    Why do you say it "didn't work"? Show exactly what you did.
  4. Oct 18, 2008 #3
    My apologies.

    So B(inside solenoid) = [tex]\mu[/tex]0*n*I = (1.2566 x 10-6)*(1800)*(4.0A) = .009048 T

    Area loop = [tex]\pi[/tex]*r2 = [tex]\pi[/tex]* (6.25 x 10-4) = .0019635 m2

    so, if [tex]\Phi[/tex]= BA = (.009048 T) * (.0019635 m2) = 1.7766 x 10-5

    but this answer was marked incorrect.
  5. Oct 19, 2008 #4

    Doc Al

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    Staff: Mentor

    In the equation B = μ*n*I, n is the number of turns per unit length, not just the number of turns.
    Last edited: Oct 19, 2008
  6. Oct 20, 2008 #5
    So to find the flux through the smaller loop do you first find the flux through the solenoid using its area and then somehow use the smaller loop's area to find its flux?
  7. Oct 20, 2008 #6

    Doc Al

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    Staff: Mentor

    You find the field within the solenoid (which is given by B = μ*n*I), then use it to calculate the flux through the smaller loop's area.
  8. Oct 20, 2008 #7
    Thanks that worked; I tried that before, but used the diameter instead radius. woops

    Part B is giving me trouble as well.

    induced emf = -NA * dB/dt

    =-1800*(.025)^2 * pi * (-4/3)
  9. Oct 20, 2008 #8
    Wrong equation, never mind.

    induced emf = -L * dI/dt

    L= flux / I
  10. Oct 20, 2008 #9
    I don't think you have to use an equation; it's asking for the average emf and

    emf = -dflux /dt

    and dflux= flux2 - flux1 , but flux2 = 0 because the current goes to 0 (stated in problem)

    and dt is given in the problem
  11. Oct 20, 2008 #10
    Your right; you can get the same solution both ways,
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