- #1

theodoros.mihos

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- 38

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## ?

$$ B(z) = \frac{\mu_0I}{2}\frac{r^2}{(r^2+z^2)^{3/2}} $$

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## ?

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