a) The metal ring broke when the current in the ring was I_0. They made exactly the same ring, but from a material whose tensile strength was ten times greater. What current will break the new ring? b) What current will break the new ring, made from this stronger material, if all the dimensions of the new ring are twice the dimensions of the old one?

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    <h3 id="back-link"><a href="/#9.2">$\leftarrow$Back</a></h3>

    <h3> Statement </h3>
    <p>
        $9.2.15.$ a) The metal ring broke when the current in the ring was $I_0$. They made exactly the same ring, but from a material whose tensile strength was ten times greater. What current will break the new ring? </p><p>

b) What current will break the new ring, made from this stronger material, if all the dimensions of the new ring are twice the dimensions of the old one?

    <h3>Solution</h3>
    <p>
        a) Having carried out calculations similar to those in the problem <a href="/9.1.15">9.1.15</a>, we obtain that the limit force $F = IBR$, and since $B \sim \frac{I}{R}$, then $F \sim I^2$ <br><br>

Then it is obvious that when $F$ increases by 10 times, the current $I$ only needs to be increased by $\sqrt{10}$ times
$$\boxed{I = I_0\sqrt{10}}$$
b) When the linear dimensions increase, the resistance increases, which means the current strength decreases; here the linear dimensions are increased by 2 times, which means that to achieve the same strength, a current twice as large is needed:
$$\boxed{I = 2I_0\sqrt{10}}$$

    </p>

    <h4>Answer</h4>
    <p>
        $$I = I_0\sqrt{10}; \quad I = 2I_0\sqrt{10}$$
    </p>
	<p style="text-align: right; font-style: italic; font-size: 14;">
      Andrei Yersh<br>
    </p>


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