Translated 9.1.6-9.1.15

astrosander edited Oct 17, 2024, 12:22 AM

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		+	<title>A square frame with current is located in a uniform magnetic field of induction B. The frame mass is m, and the current in it is I. Determine the frequency of free oscillations of the frame around the OO^\prime axis.</title>
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		+	<span><img src="../../img/book.png"></span><span>Savchenko Solutions</span>
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		+	<p class="author">
		+	Solutions of Savchenko Problems in Physics <br>
		+	<i><b>knowledge must be free</b></i>
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		+	<h3 id="back-link"><a href="../../#9.1">$\leftarrow$Back</a></h3>
		+
		+	<h3> Statement </h3>
		+	<p>
		+	$9.1.8^*.$ A square frame with current is located in a uniform magnetic field of induction $B$. The frame mass is $m$, and the current in it is $I$. Determine the frequency of free oscillations of the frame around the $OO^\prime$ axis.
		+	</p>
		+	<center>
		+	<figure>
		+	<img src="https://savchenkosolutions.com/9/9.1.8/statement.png"
		+	loading="lazy" width="200" />
		+	<figcaption>
		+	For problem $9.1.8^*$
		+	</figcaption>
		+	</figure>
		+	</center>
		+	<p>
		+	</p>
		+
		+	<h3>Solution</h3>
		+	<p>
		+	Let's consider a small deviation from equilibrium by an angle $\alpha$: </p><p>
		+	The moment acting on the frame from the magnetic field:
		+	$$ M = BIS\cdot \sin\alpha = BIa^2 \sin\alpha $$
		+	where $a$ is the side of the frame</p><p>
		+	gravity does not create a moment as it passes through the point of rotation</p><p>
		+	From the basic equation of rotational motion:
		+	$$ J\ddot{\alpha} = \sum\pm{M_i} $$
		+	for our case:
		+	$$ J\ddot{\alpha} = -BIa^2 \sin\alpha\quad (1) $$
		+	now we will find the moment of inertia $J$ of the frame during the observed oscillations</p><p>
		+	The moment of inertia of the upper and lower sides of the frame is
		+	$$ J_1=J_2 = \frac{m}{4}\cdot \left(\frac{a}{2}\right)^2 = \frac{m a^2}{16}\quad (2) $$
		+	where $a/2$ is the distance to this side.</p><p>
		+	Moment of inertia of the lateral sides:
		+	$$ J_3 = J_4 = \frac{1}{12} \cdot \frac{m}{4}\cdot a^2 = \frac{m a^2}{48}\quad (3) $$
		+	We find the moment of inertia $J$ by substituting $(2)$ and $(3)$
		+	$$ J = J_1 + J_2 + J_3 + J_4 $$
		+
		+	$$ J = 2\cdot \frac{m a^2}{16} + 2 \cdot \frac{m a^2}{48} = \frac{m a^2}{6} $$
		+	then equation $(1)$ will look like:
		+	$$ \frac{m a^2}{6}\ddot{\alpha} = -BIa^2 \sin\alpha $$
		+
		+	$$ \ddot{\alpha} = -\frac{6BI}{m} \alpha (t) $$
		+	Solving the equation of harmonic oscillations, we find the oscillation frequency
		+	$$ \boxed{\omega = \sqrt{\frac{6BI}{m}}} $$
		+
		+	</p>
		+
		+	<h4>Answer</h4>
		+	<p>
		+	$$\omega = \sqrt{\frac{6BI}{m}}$$
		+	</p>
		+	<p style="text-align: right; font-style: italic; font-size: 14;">
		+	Andrei Yersh<br>
		+	</p>
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			<meta name="keywords" content="Savchenko Problems in Physics, Savchenko solutions, physics problems, physics olympiad preparation, IPhO, Jaan Kalda">
			<meta name="description" content="A square frame with current is located in a uniform magnetic field of induction B. The frame mass is m, and the current in it is I. Determine the frequency of free oscillations of the frame around the OO^\prime axis.">
			<meta name="author" content="Aliaksandr Melnichenka">
			<meta name="date" content="2023-10" scheme="YYYY-MM">
			<meta property="og:title" content="A square frame with current is located in a uniform magnetic field of induction B. The frame mass is m, and the current in it is I. Determine the frequency of free oscillations of the frame around the OO^\prime axis.">
			<meta property="og:image" content="img/logo.png">
			<meta property="og:description" content="A square frame with current is located in a uniform magnetic field of induction B. The frame mass is m, and the current in it is I. Determine the frequency of free oscillations of the frame around the OO^\prime axis.">
			<meta name="yandex-verification" content="6cfda41f74038368">
			<title>A square frame with current is located in a uniform magnetic field of induction B. The frame mass is m, and the current in it is I. Determine the frequency of free oscillations of the frame around the OO^\prime axis.</title>
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			<i><b>knowledge must be free</b></i>
			</p>
			</header>

			<h3 id="back-link"><a href="../../#9.1">$\leftarrow$Back</a></h3>

			<h3> Statement </h3>
			<p>
			$9.1.8^*.$ A square frame with current is located in a uniform magnetic field of induction $B$. The frame mass is $m$, and the current in it is $I$. Determine the frequency of free oscillations of the frame around the $OO^\prime$ axis.
			</p>
			<center>
			<figure>
			<img src="https://savchenkosolutions.com/9/9.1.8/statement.png"
			loading="lazy" width="200" />
			<figcaption>
			For problem $9.1.8^*$
			</figcaption>
			</figure>
			</center>
			<p>
			</p>

			<h3>Solution</h3>
			<p>
			Let's consider a small deviation from equilibrium by an angle $\alpha$: </p><p>
			The moment acting on the frame from the magnetic field:
			$$ M = BIS\cdot \sin\alpha = BIa^2 \sin\alpha $$
			where $a$ is the side of the frame</p><p>
			gravity does not create a moment as it passes through the point of rotation</p><p>
			From the basic equation of rotational motion:
			$$ J\ddot{\alpha} = \sum\pm{M_i} $$
			for our case:
			$$ J\ddot{\alpha} = -BIa^2 \sin\alpha\quad (1) $$
			now we will find the moment of inertia $J$ of the frame during the observed oscillations</p><p>
			The moment of inertia of the upper and lower sides of the frame is
			$$ J_1=J_2 = \frac{m}{4}\cdot \left(\frac{a}{2}\right)^2 = \frac{m a^2}{16}\quad (2) $$
			where $a/2$ is the distance to this side.</p><p>
			Moment of inertia of the lateral sides:
			$$ J_3 = J_4 = \frac{1}{12} \cdot \frac{m}{4}\cdot a^2 = \frac{m a^2}{48}\quad (3) $$
			We find the moment of inertia $J$ by substituting $(2)$ and $(3)$
			$$ J = J_1 + J_2 + J_3 + J_4 $$

			$$ J = 2\cdot \frac{m a^2}{16} + 2 \cdot \frac{m a^2}{48} = \frac{m a^2}{6} $$
			then equation $(1)$ will look like:
			$$ \frac{m a^2}{6}\ddot{\alpha} = -BIa^2 \sin\alpha $$

			$$ \ddot{\alpha} = -\frac{6BI}{m} \alpha (t) $$
			Solving the equation of harmonic oscillations, we find the oscillation frequency
			$$ \boxed{\omega = \sqrt{\frac{6BI}{m}}} $$

			</p>

			<h4>Answer</h4>
			<p>
			$$\omega = \sqrt{\frac{6BI}{m}}$$
			</p>
			<p style="text-align: right; font-style: italic; font-size: 14;">
			Andrei Yersh<br>
			</p>


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