$9.1.13^*.$ A coil with a current flowing through its turns stands vertically on a plane. The total weight of the coil is $P$, the number of turns is $n$, the radius is $R$, and the current in the turns is $I$. At what induction of a uniform magnetic field directed horizontally will the coil overturn under the action of this field?
Moment of gravity for the coil relative to the edge of the coil: $$ M_{mg} = R\cdot mg = PR $$ Moment of magnetic field forces: $$ M_B = n\cdot BSI = n\cdot \pi R^2BI $$ In the equilibrium condition, the total moment of forces, taking into account the signs, is equal to zero $$ n\cdot \pi R^2BI = PR \Rightarrow \boxed{ B = \frac{P}{n\pi RI}} $$
$$B = \frac{P}{n\pi RI}$$