$9.1.15^*.$ A conducting ring is placed in a magnetic field perpendicular to its plane. Current $I$ circulates through the ring. If the wire of the ring can withstand a breaking load $F$, then at what magnetic field induction will the ring break? The radius of the ring is $R$. Neglect the effect of the magnetic field created by the current $I$ on the ring.
The force acting on a small section of length $\alpha R$ from the magnetic field is equal to $$ F_B = IRB \cdot \alpha $$ This force is balanced by both tension forces: $$ F_B = 2T\cdot \cos\left(\frac{\pi}{2} - \frac{\alpha}{2}\right) = 2T\cdot \sin\frac{\alpha}{2} $$ In the approximation of a small angle $\alpha$ we get $\sin\frac{\alpha}{2} \approx \frac{\alpha}{2}$ then: $$ F_B = T\alpha = IRB\cdot \alpha\Rightarrow \boxed{B = \frac{T}{RI}} $$
$$B = \frac{F}{RI}$$