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Homework Help: Magnetic Field on an Axial point within a Solenoid

  1. May 12, 2010 #1
    1. The problem statement, all variables and given/known data

    Thin Shell Solenoid

    2. Relevant equations
    General Case Formula Found:


    3. The attempt at a solution

    plugged in variables and got .00062919475 T

    4 other attempts were made using trivial solutions.
    Even tried finding the energy density due to the field and then the magnetic field from the volume of the center to that point.

    I think the basic concept is all the Field contributions over the integral of the solenoid to that point.

    Looked over old homework and cant find a problem to reference this with.

    Please help!!

    Ps. First post :D
    Last edited by a moderator: Apr 25, 2017
  2. jcsd
  3. May 12, 2010 #2
    Well, who even knows if I got it right plugging into my calculator, but I got 1.086*10^-5 T. Check to make sure your units are all in SI. One thing I wasn't sure about was if it meant .0716m from the end of the solenoid (how I interpreted it), or .0716m from the very center of the solenoid, still inside. If it's the latter case then of course you are wrong because the formula you're using wouldn't work, but I think it's the former case. Try again, being super duper careful with your numbers.
  4. May 12, 2010 #3
    In this problem I think it is the latter case, being the point is still inside the solenoid from the center, but I do agree that the problem is ambiguous as to which case it may be. (Would it have hurt them to put a diagram?)

    That being the case, I do not understand why the formula I am using would not work? In describing x1 and x2 it says the distance between the point and the ends, which is what i used.

    From my experience in this course I'm pretty sure the answer lies within some kind of integral of the magnetic field. Any Input on how I could find the magnetic field inside the solenoid?
  5. May 12, 2010 #4
    Ok so after a bit more investigating, I found what I did wrong. For the Magnetic field inside of the solenoid the equation you must use is:

    B = [(Uo)*N*I / (2l) ] * [sin(02) + sin(01)]

    sin01 = d1/sqrt(d1^2 + r^2)

    sin02 = d2/sqrt(d2^2 + r^2)

    Uo = 4*pi*10^-7 Tm/A
    N = number of turns
    I = current in Amps
    l = length in meters
    d1 & d2 = distance from ends of solenoid to point within in meters)
    r = radius of the solenoid

    Now, If anyone wants to help me understand or clarify, the reason for this is that basically we are trying to add the contributions of electric field from both ends of the solenoid to the point within the solenoid. Is this the correct understanding?
  6. May 13, 2010 #5
    Yeah, that's more or less the reason. Really it's just how you do your integrations, and the first formula you were using did its integrations, with its own set of coordinates, for the field outside of the solenoid. For the field inside you would have to use different coordinates. If you really want to see the difference, and you're good with your vector calculus, try the integrations for yourself. If you just want to set them up I could check them. You could even do a general formula that covers both cases, though the math could be quite nasty.
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