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Field inside a finite solenoid - At the ends and in the middle

  1. Oct 3, 2012 #1

    AGNuke

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    A solenoid of length 0.4 m and diameter 0.6 m consists of a single layer of 1000 turns of fine wire carrying a current of 5 x 10-3 A. Calculate the magnetic field strength on the axis at the middle and at the ends of the solenoid.

    I know the field inside a long solenoid, B = μNI, which is unfortunately not applicable here. I tried it, but the answer was not matching, by a long distance.

    Answers:
    1.) π/√13 × 10-5 T
    2.) 2π × 10-6 T
     
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  3. Oct 9, 2012 #2

    TSny

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    Do you know the formula for the magnetic field produced by a circular ring of current for any point on the axis of the ring? If so, what would be a natural way to break the solenoid into infinitesimal parts and integrate?
     
  4. Oct 9, 2012 #3

    AGNuke

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    Here's what I attempted a week ago. But its illogical in just the last step.

    Let this solenoid of radius R, having N turns per unit length. If we take an elemental ring of width dx, the magnetic field on a given point can be given by Biot-Savart's special case of current carrying loop. Here, [tex]\mathrm{d}B=\frac{\mu _0 I N\mathrm{d}x R^2}{2(R^2+x^2)^{\frac{3}{2}}}[/tex]

    Sorry if I didn't mentioned, "x" is the distance of point from the centre of element loop and "X" which is given instead, is the distance of the point from one end of the solenoid.

    We can see that from the figure, we can substitute x = R cot θ. Hence, dx = -R cosec2 θ dθ. Using these facts in our formula,[tex]\mathrm{d}B = -\frac{\mu _0 INR^3\mathrm{cosec}^2\theta \mathrm{d}\theta}{2(R^2 + R^2\mathrm{cot}^2\theta)^{\frac{3}{2}}}[/tex][tex]\int \mathrm{d}B =-\frac{\mu _0IN}{2} \int \mathrm{sin}\theta \mathrm{d}\theta[/tex][tex]B=\frac{\mu _0NI}{2}[\mathrm{cos}\theta][/tex]

    Now the thing that I don't understand is the limit which I instinctively took. I tried "setting" the limits in my answer and I got the relation if I use the limit -θ2 → θ1. Can you explain why did I do so. The only explanation I could come up with is symmetry, that is, the field will be same for the point at the distance "X" from the right side too, as for the left side.
     
  5. Oct 9, 2012 #4

    TSny

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    For part (a) you want the field at the center of the solenoid. Then, in your figure,how would the magnitude of θ1 compare to θ2?

    Also note that if you measure θ as shown in your figure so that it starts at θ1 and then increases clockwise, what would be the value for the upper limit of θ?
     
    Last edited: Oct 9, 2012
  6. Oct 9, 2012 #5

    AGNuke

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    for part (a), I think θ1 = θ2. This is one of the reasons to make my final expression as [tex]B=\frac{\mu _0NI}{2}(\mathrm{cos}\theta_1+\mathrm{cos}\theta_2)[/tex]

    As for the upper limit of θ, I interpreted it as [itex]\pi - \theta_1-\theta_2[/itex]
     
  7. Oct 9, 2012 #6

    TSny

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    That's not quite right. Remember, you are measuring θ clockwise from the horizontal. See the attached figure for θ at the upper limit.
     

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  8. Oct 10, 2012 #7

    AGNuke

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    So you are saying that I should integrate it from θ1 to π - θ2. Which would return the formula stated above. So, thanks for stating where I was wrong.
     
  9. Oct 10, 2012 #8

    TSny

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    That looks right. You are going to get an overall negative value for B. You might think about whether or not that's ok. Would your original integral in terms of x be negative? It wasn't clear to me where you were taking the origin for x and in which direction you were taking x to be positive.
     
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