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Homework Help: Magnetic Flux through a loop:

  1. Sep 29, 2009 #1
    1. The problem statement, all variables and given/known data

    A single-turn square loop of side L is centered on the axis of a long solenoid. In addition, the plane of the square loop is perpendicular to the axis of the solenoid. The solenoid has 1350 turns per meter and a diameter of 5.55 cm, and carries a current of 2.20A

    2. Relevant equations
    Magnetic Field (B) of a solenoid: B = [tex]\mu_{0}[/tex]*(N/L)*I
    Where [tex]\mu_{0} = 4pi * 10^{-7} T*m/A[/tex] , N/L is loops per unit length, and I is current

    [tex]\Phi = B*A*Cos\theta[/tex]

    3. The attempt at a solution
    I tried doing this: I use the Magnetic Field of a solenoid formula to find the magnetic field through the solenoid - because I know the magnetic field within the axis of the solenoid is constant, I thought I could just find the value of that constant B and then find the flux using the other formula for the different values of L (area) - and for areas larger than the circle of the solenoid - I could take the area of the flux as the area of the circle.

    However, when I put my answer in, it tells me that the answer must be expressed in terms of L (I don't know if that's supposed to be the side length of the square or if it's supposed to be the length of the solenoid).

    Last edited: Sep 29, 2009
  2. jcsd
  3. Sep 29, 2009 #2


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    Homework Helper

    Take the area of the loop as L^2 and find the flux through it.
  4. Sep 29, 2009 #3
    Oh, sorry: I forgot a bit of information:

    There are three parts to the question:

    A) Find the magnetic flux through the loop when L = 2.35 cm.

    B) Find the magnetic flux through the loop when L = 5.55×10−2 cm

    C) Find the magnetic flux through the loop when L = 12.5 cm

    That's what I was getting confused about, since it gives you L - why is it asking you to put it in terms of L - or does it mean put it in terms of the length of the solenoid?

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