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B Magnetic field outside of a solenoid conceptual

  1. May 6, 2016 #1
    How come the magnetic field outside a solenoid is practically zero?
    I've read reasons along the lines of:

    -The magnetic field cancels out on the outside.
    Of course the net force cancels out, but what if you have an object placed on just one spot? The force on that object clearly is not 0 because it is closer to one side of the solenoid than the other.

    -The field lines spread outside the solenoid so much, that the density goes to zero as the solenoid gets longer.
    This seems like it only happens in cases that say you start out with a solenoid of fixed length and current, then extend it out to a very large number. If you increase the current at the same time you are stretching the coil, the magnetic flux density will remain somewhat constant won't it? Now what if you don't stretch the solenoid at all? What if you had a solenoid of infinite length to begin with- is the field still 0 outside?

    Another question: Why is the field outside nearly zero at all? Each current running through each section will contribute to a magnetic field outside...
  2. jcsd
  3. May 6, 2016 #2

    Paul Colby

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    Gold Member

    A linear solenoid should have a field similar to a bar magnet. While the field is largest near the "poles" it's hardly 0 since magnetic field must close on themselves.
  4. May 9, 2016 #3
    Just consider the field produced by two small elements of the loop situated at the two opposite ends of a diameter.
    See how the fields add in the center of the loop and somewhere outside, in a point on the same diameter.
    This will show you why the field is weaker outside.
  5. May 20, 2016 #4
    You are right, this is not obious

    Regarding the first statement, we know it is right but we want to know why. As for the second statement it is only a qualitative one and does not prove anything

    One way we prove the correct result is this:
    - Apply Biort-Savart to four differential elements, two of them situated obove your position and the other two below it. The two elementes must be situated on a circle perpendicular to the cylinder axis and simetrically situated from your point of sight
    - The above analysis yields that the the field is paralell to the cylinder's axis
    - Using Ampere's Law to an appropiate contour inside and outside the cylinder yields the the field mus be constant. The values outside and inside could be different however.
    - Using Ampere´s law to an appropiate contour one part of it lying inside and the other outside we get:$$B(outside)-B(inside)=-\mu_0NI$$
    - By integration of Biot-Savart Low calculate B on the axis of the cylinder which gives : B(inside)=##\mu_0NI##
    - The last two results render B(outside)=0
    Last edited: May 20, 2016
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