Solution

Statement

$9.2.22.$ The magnetic field induction at the center of a thin steel disk of radius $R$ magnetized along its axis is equal to $B$. This disk is placed in a uniform magnetic field with induction $B_0$, which does not change the magnetic moment of the disk. How should the disk be oriented in this magnetic field so that the torque acting on it is maximum? What is this torque equal to?

Solution

Exactly the same disk was considered in problem 9.2.20. There, we concluded that it is equivalent to a circular loop of radius $R$ carrying a current $I$. The field at the center of such a loop is:

$$B=\frac{\mu_0I}{2R}$$

as was found in problem 9.2.10.

It is known that the torque acting on a current loop in a uniform magnetic field is:

$$\vec N=\vec p\times\vec B_0$$

Where $\vec p=I\vec S$ is the magnetic moment of a loop of area $S$ carrying a current $I$.

The magnetic moment is directed perpendicularly to the plane of the disk; therefore, since according to the rules of the cross product the maximum torque corresponds to $\vec p\perp\vec B_0$, the plane of the disk should be oriented parallel to the external field.

In this case, the magnetic moment is:

$$p=IS=\frac{2BR}{\mu_0}\cdot\pi R^2=\frac{2\pi BR^3}{\mu_0}$$

And the maximum torque (denoted as $N$ to distinguish it from the magnetic field $B$) is:

$$N=p B_0=\frac{2\pi BB_0R^3}{\mu_0}$$