Constant magnetic field > The magnetic field from a current distributed over a surface of space > Problem 9.3.24

Statement

$9.3.24.$ A long cylindrical iron rod of radius $r$ is magnetized in a magnetic field perpendicular to the rod axis. Magnetic moment of the rod volume unit $M$. How does the magnetic field induction depend on $x$ at distances much smaller than the rod length?

Solution

Let's cut our cylinder into plates along its axis, the thickness of such a plate is $h = Rsin\varphi d\varphi$, where $\varphi$ is the azimuth angle, and calculate the current density through the arc $Rd\varphi$:

$$i R d\varphi = M R sin\varphi d \varphi$$
$$i = M sin \varphi$$

The harmonic law is already familiar to us, so the field inside the cylinder is homogeneous and equal to:

$$B_{in} = \frac{\mu_0 M}{2}$$

(see task $9.3.22$)

Now we use the result of the same problem to find the equivalent current density through the cross section and use the limit transition, considering the field from the cylinder as a superposition of two cylinders at a distance of $a\to 0$ with $j\to \inf$:
$$\frac{\mu_0 j a}{2} = \frac{\mu_0 M}{2}$$
$$j = \frac{M}{a}$$
The resulting field at a distance of x:
$$B_{out} = \frac{\mu_0 j r^2}{2x}-\frac{\mu_0 j r^2}{2(x+a)}$$
$$B_{out} = \frac{\mu_0 M}{2}(\frac{r}{x})^2$$

Answer

$$B_{in} = \frac{\mu_0 M}{2}$$
$$B_{out} = \frac{\mu_0 M}{2}(\frac{r}{x})^2$$