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Why we calculate B of solenoid with this way?

  1. Jun 7, 2015 #1
    We see the magnetic field calculation by Biot-Savart law for a circular current I on some distance z on the vertical axis throw it centre as:
    $$ B(z) = \frac{\mu_0I}{2}\frac{r^2}{(r^2+z^2)^{3/2}} $$
    where r is the radius of circle.

    When we use summation of N rings lays on L length, like solenoid, we integrate as:
    $$ B(z) = \frac{\mu_0nIr^2}{2}\int_{-L/2}^{L/2}\frac{dz}{[r^2+(z-z_0)^2]^{3/2}} $$

    I think that ##B_i## for one ring gives to total ##B## an amount:
    $$ B_i = B(z_i) = \frac{\mu_0I}{2}\frac{r^2}{[r^2+(z_i-z_0)^2]^{3/2}} $$

    How we take the integral ##\int\,dz## ?
     
    Last edited: Jun 7, 2015
  2. jcsd
  3. Jun 7, 2015 #2

    Simon Bridge

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    How would you normally take the integral?
    (You may want to try a trig substitution.)

    Have you set up the integral correctly?
    When you take the definite integral over a variable, it goes away - so you cannot have a function like ##f(z)=\int_a^b g(z)\;dz## because that implies that f is a function of z.

    Usually you want to say something like - the ring between z=z' and z=z'+dz' contributes dB=something to the overall field at position z on the axis... then you put ##B=\int dB##. Then ##z'_i## is the position of the ith ring and you need to add them up.

    ##z_0## appears to be playing the role of ##z'## in your integral - so you need to integrate over that.

    Have you examined the physics carefully ? - usually a solenoid has a finite number of rings in length L.
    The integral implies that the rings are very thin.
     
    Last edited: Jun 7, 2015
  4. Jun 8, 2015 #3
    I think this is the point I need. Thanks.
     
  5. Jun 8, 2015 #4

    Simon Bridge

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    No worries.
     
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