A coil of radius R was bent along its diameter at a right angle and placed in a uniform magnetic field of induction B so that one of the planes of the coil was located at an angle \alpha, the other at an angle \frac{\pi}{2}-\alpha to the direction of induction B. The current in the coil is I. Determine the moment of forces acting on the coil.

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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>
        $9.1.12.$ A coil of radius $R$ was bent along its diameter at a right angle and placed in a uniform magnetic field of induction B so that one of the planes of the coil was located at an angle $\alpha$, the other at an angle $\frac{\pi}{2}-\alpha$ to the direction of induction $B$. The current in the coil is $I$. Determine the moment of forces acting on the coil.

    <h3>Solution</h3>
    <p>
        Let's split the circuit into two circuits with the same current, but in different planes.</p><p>

Let's find the moment of forces for each of them:
$$M_1 = \frac{1}{2}\pi R^2IB\cdot \sin\alpha$$
$$M_2 = \frac{1}{2}\pi R^2IB\cdot \sin\left(\frac{\pi}{2} -\alpha\right) = \frac{1}{2}\pi R^2IB\cdot \cos\alpha$$
Total moment of forces
$$\boxed{M = M_1 + M_2 = \frac{1}{2}\pi R^2IB\cdot (\sin\alpha + \cos\alpha )}$$

    </p>

    <h4>Answer</h4>
    <p>
        $$M = \frac{1}{2}\pi R^2IB\cdot (\sin\alpha + \cos\alpha )$$
    </p>
	<p style="text-align: right; font-style: italic; font-size: 14;">
      Andrei Yersh<br>
    </p>


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