Problem 22.11

22.11.
The magnetic force on each part of the ring is given by
$\vec{F_{m}}= I d\vec{\ell} \times \vec{B}$
$\vec{F_{m}}= I d\ell B$ in a direction radially inward, and above the ring by an angle $\theta$.
Therefore, when we add the contributions from all the segments of the ring, the radial components cancel perfectly, leaving only the components in the direction normal to the plane of the ring. (upward) These components are given by
$d F_{up} = I d \ell B \sin \theta$
$F_{up} = I (2 \pi r) (B) \sin \theta$
So, the net force acting on the ring from the magnetic field produced by the magnet is given by:
$\vec{F_{m}} = 2\pi IrB \sin \theta$ upward