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What is the induced emf in one of the windings?

  1. Feb 10, 2013 #1
    1. A solenoid of length 3.64 cm and diameter 0.914 cm is wound with 170 turns per cm. If the current is decreasing at a rate of 36.3 A/s, what is the induced emf in one of the windings?


    this is what I did:
    Length of the solenoid l = 3.64cm
    = (3.64 cm)(10-2 m/ 1 cm)
    = 0.0364 m
    diameter d = 0.914 cm
    radius r = d/2
    = 0.914 cm /2
    = 0.457 cm
    = (0.457 cm)(10-2 m/ 1 cm)
    = 0.00457 m
    area A = πr2
    = (3.14)(0.00457 m)2
    = 6.55*10-5 m2
    the change of rate of current dI/dt = 36.3 A/s
    _______________________________________
    a)
    for one winding N = 1
    for one winding, n = N/1 cm
    = 1/10-2 m
    = 100 m
    per one meter of winding,
    n = 100
    the induced emf in one of the windings is
    ε = [μ0 n2A/l](dI/dt)
    = [(4π*10-7 T.m/A)(100)2(6.55*10-5 m2 )/(0.0364 m)](36.3 A/s)
    = 1.64*10-5 V <--------BUT THE ANSER IS WRONG...
    B. What is the induced emf in the entire solenoid?

    This is what I did:
    for entire solenoid, n = 170 turns/ 1 cm
    = 170 / 10-2 m
    = 17000 m
    per one meter, n = 17000
    the induced emf in the entire solenoid is
    ε = [μ0 n2A/l](dI/dt)
    = [(4π*10-7 T.m/A)(17000)2(6.55*10-5 m2 )/(0.0364 m)](36.3 A/s)
    but the answer was still wrong....
    could someone help me where did i do wrong, please \
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Feb 10, 2013 #2

    TSny

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    Looks like you are using the formula B = μo(N/L)I for the field inside a solenoid. What do N and L stand for here?
     
  4. Feb 11, 2013 #3
    N=how many turns
    L=length of the wire?!
     
  5. Feb 11, 2013 #4

    TSny

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    Gold Member

    N is the total number of turns and L is the total length of the solenoid. So, N/L is the number of turns per unit length, which you found to be 17000 turns per m.
     
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