Constant magnetic field > The magnetic field from a current distributed over a surface of space > Problem 9.3.23

Statement

$9.3.23.$ The planes of the turns of a circular solenoid are inclined at an angle $\alpha$ to its axis. Current of the solenoid $I$, number of turns per unit of its length $n$, radius $R$. Determine the magnetic field induction inside sucha solenoid.

Solution

Consider our current as a superposition of longitudinal and transverse current, the induction inside the cylinder is determined by the linear current density, therefore, the longitudinal component of induction, which is created by the transverse component of the current, does not depend on its inclination (the number of turns with current crossing a unit length of any line on the surface of the cylinder parallel to its axis does not depend on the angle of inclination of the turns). Therefore, the longitudinal induction component will be as in a conventional solenoid:
$$B_{||} = \mu_0 n I$$

Now let's look at the perpendicular component, consider the current coil, let's have a coordinate system centered on the cylinder axis $(x, y, z)$, $$z = x ctg\alpha$$
Now let's move to a cylindrical coordinate system with the same origin: $(R,\varphi)$, where $\varphi$ is the azimuth angle:

$$z(\varphi) = R ctg \alpha cos \varphi$$

Now, to calculate the current density through a unit length of a coaxial circle on the surface of a cylinder, we calculate the ratio of the change in the z coordinate to the length of a piece of the circle:
$$\frac{dz}{Rd\varphi}=\frac{-R ctg\alpha sin \varphi d\varphi}{Rd\varphi}=-ctg\alpha sin \varphi$$
From here we find the current density distribution $i$:
$$i(\varphi) = - Inctg \alpha sin \varphi$$

It is an amazing coincidence that we are familiar with this distribution from the previous problem, and we know for sure that it creates a homogeneous field inside perpendicular to the axis of the cylinder. Using the result of the previous problem, we obtain the formula:
$$B_{\perp} = \frac{1}{2}\mu_0 I n ctg \alpha$$

If you look at Savchenko's answer, then he made at least a mathematical mistake, because at $\alpha = \pi/2$ the perpendicular component should reset to zero, but for him it goes to infinity.

Answer

$$B_{||} = \mu_0 n I$$
$$B_{\perp} = \frac{1}{2}\mu_0 I n ctg \alpha$$