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?

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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>
        $9.1.13^*.$ 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?
    </p>

    <h3>Solution</h3>
    <p>
        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
$$n\cdot \pi R^2BI = PR \Rightarrow \boxed{ B = \frac{P}{n\pi RI}}$$

    </p>

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

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