Why we calculate B of solenoid with this way?

1. Jun 7, 2015

theodoros.mihos

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. Jun 7, 2015

Simon Bridge

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
3. Jun 8, 2015

theodoros.mihos

I think this is the point I need. Thanks.

4. Jun 8, 2015

No worries.