11.1.23∗. A metal ring is placed vertically in the gravity field. A metal rod of length $L$ and mass $m$ is pivotally fixed in the center of the ring and touches it with the other end. A uniform magnetic field of induction $B$ is perpendicular to the ring plane. By what law is it necessary to change the current in the rod so that the rod rotates evenly with an angular velocity $\omega$, if at the initial moment the rod was in the upper position? Ignore the friction.
Solution
For problem $11.1.23$
The angle between the rod and the vertical (shown in green) is $\omega t$ at time $t$ after the initial moment. If there is current $I$ flowing radially outward through the rod, then the magnetic force on the rod is $F=IBL$. The angular speed is constant when counterclockwise moment due to $F$ equals clockwise moment due to gravity $mg$. Thus,
