A triangular wire frame with current can rotate about a horizontal axis OO^\prime passing through the vertex of the triangle. The mass of a unit length of wire is \rho, the current in the frame is I. The frame is in a magnetic field of induction B directed along the gravitational field. Determine the angle of deviation of the plane of the triangle from the vertical.

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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>
        $9.1.9.$ A triangular wire frame with current can rotate about a horizontal axis $OO^\prime$ passing through the vertex of the triangle. The mass of a unit length of wire is $\rho$, the current in the frame is $I$. The frame is in a magnetic field of induction $B$ directed along the gravitational field. Determine the angle of deviation of the plane of the triangle from the vertical.

    <h3>Solution</h3>
    <p>
        To solve this, we will use the fact that $\vec{N} = \vec{M} \times \vec{B}$ is valid for any contour $S$ with magnetic moment $M$ (for the proof, see <a href="/9.1.10">9.1.10</a>)</p><p>

The moment of magnetic field forces
$$N_{B} = |\vec{M} \times \vec{B}| = ISB \cos\alpha\quad(1)$$
The center of mass will be at the intersection point of the medians, the distance from this point to the suspension point:
$$h_C=\frac{2h}{3}$$
Moment of gravity
$$M_{mg} = 3a\rho g\cdot \frac{2}{3}h \sin\alpha\quad(2)$$
where $a$ is the side of the triangle, and its area $S = \frac{1}{2}ah$
We substitute the result into $(2)$
$$M_{mg}= 4S\rho g\cdot \sin\alpha\quad(3)$$
We equate the obtained moments from $(1)$ and $(3)$:
$$ISB\cdot \cos\alpha = 4S\rho g \cdot \sin\alpha\Rightarrow \boxed{\tan\alpha = \frac{IB}{4\rho g}}$$

    </p>

    <h4>Answer</h4>
    <p>
        $$\tan\alpha = \frac{IB}{4\rho g}$$
    </p>
	<p style="text-align: right; font-style: italic; font-size: 14;">
      Andrei Yersh<br>
    </p>


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