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.

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Solutions of Savchenko Problems in Physics
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    <h3 id="back-link"><a href="/#9.1">$\leftarrow$Back</a></h3>

    <h3> Statement </h3>
    <p>
        $9.1.15^*.$ 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.
    </p>

    <h3>Solution</h3>
    <p>
        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:

$$F_B = T\alpha = IRB\cdot \alpha\Rightarrow \boxed{B = \frac{T}{RI}}$$

    </p>

    <h4>Answer</h4>
    <p>
        $$B = \frac{F}{RI}$$
    </p>
	<p style="text-align: right; font-style: italic; font-size: 14;">
      Andrei Yersh<br>
    </p>


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