Translated 9.2.10-9.2.17

astrosander правка от 17 окт. 2024 г., 19:43

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		+	<title>Determine the magnetic field induction of a rectangular frame a \times b with a current I at a point A located at a distance r much greater than the linear dimensions of the frame. The radius vector \vec{r} forms an angle \alpha with the plane of the frame.</title>
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		+	<span><img src="../../img/book.png"></span><span>Savchenko Solutions</span>
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		+	<h3 id="back-link"><a href="../../#9.2">$\leftarrow$Back</a></h3>
		+
		+	<h3> Statement </h3>
		+	<p>
		+	$9.2.17^*.$ Determine the magnetic field induction of a rectangular frame $a \times b$ with a current $I$ at a point $A$ located at a distance $r$ much greater than the linear dimensions of the frame. The radius vector $\vec{r}$ forms an angle $\alpha$ with the plane of the frame.
		+	</p>
		+
		+	<h3>Solution</h3>
		+	<p>
		+
		+	</p>
		+	<center>
		+	<figure>
		+	<img src="https://savchenkosolutions.com/9/9.2.17/9.2.17_1.png"
		+	loading="lazy" width="350" />
		+	<figcaption>
		+	Direction of total magnetic induction
		+	</figcaption>
		+	</figure>
		+	</center>
		+	<p>
		+	From symmetry, it follows that the magnetic field induction will be present only in the plane defined by the angle $\alpha$, since the components of the wire of length $a$ will compensate each other</p><p>
		+	Let's write out the cosines of the angles that $\vec{r}$ makes with wires of length $b$
		+	let the angle $\alpha_1$ be for the nearest wire, and the angle $\alpha_2$ be for the farthest one, then </p><p>
		+	$$ \cos\alpha_1 = \frac{{2r\cos\alpha} - a}{2r} = \cos\alpha - \frac{a}{2r} $$
		+
		+	$$ \cos\alpha_2 = \frac{{2r\cos\alpha} + a}{2r} = \cos\alpha + \frac{a}{2r} $$
		+	<b><i>Theoretical digression</i></b>
		+
		+	</p>
		+	<center>
		+	<figure>
		+	<img src="https://savchenkosolutions.com/9/9.2.17/9.2.17_2.png"
		+	loading="lazy" width="140" />
		+	<figcaption>
		+	Magnetic induction created by a wire of length $l$
		+	</figcaption>
		+	</figure>
		+	</center>
		+	<p>
		+	A magnetic wire of length $l$ creates a magnetic field equal to
		+	$$ B=\frac{\mu_0I}{2\pi r}\cos\alpha_0 $$
		+	From the geometry of the problem, we obtain
		+	$$ \cos\alpha_0 = \frac{b/2}{\sqrt{r^2+b^2/4}} = \frac{b}{2r}\left(1+\frac{b^2}{4r^2}\right)^{1/2} $$
		+	Using the approximation $(1+x)^\alpha\approx1+\alpha x$, we transform to the first order of smallness
		+	$$ \cos\alpha_0 = \frac{b}{2r}\left(1-\frac{b^2}{8r^2}\right)\approx\frac{b}{2r} $$
		+	A more detailed derivation of this formula is on pages 76-78 <a href="http://belpho.org/books/Magnitostatika.pdf">"Magnetostatics. Electrodynamics. Theory of Oscillations"</a> — A.I. Slobodanyuk
		+	From where we get
		+	$$ \boxed{B_0=B_1=B_2=\frac{\mu_0Ib}{4\pi r^2}} $$
		+	NO: we used an approximation that will later give us an answer with an accuracy of up to first order</p><p>
		+	From the <a href="#drawing">drawing</a> it follows that the total magnetic induction of the magnetic field $\vec{B}$ will be equal to the vector sum of the components $\vec{B}_1$ and $\vec{B}_2$
		+	$$ \vec{B}=\vec{B}_1+\vec{B}_2 $$
		+	Then the vector modulus can be found using the cosine law.
		+	$$ B = \sqrt{B_1^2+B_2^2-2B_1B_2\cos\Delta\alpha} $$
		+
		+	$$ B = B_0\sqrt{2-2\cos\Delta\alpha} = 2B_0\sin\frac{\Delta\alpha}{2}\quad(1) $$
		+	Considering the smallness of the angle $\Delta \alpha$,
		+	$$ \sin\frac{\Delta\alpha}{2} \approx \frac{\Delta\alpha}{2}=\frac{a}{2r} $$
		+	We substitute $(1)$:
		+	$$ B = 2B_0\sin\frac{\Delta\alpha}{2} = B_0 \Delta\alpha $$
		+	We obtain in approximation to the first order:
		+	$$ \boxed{B = \frac{\mu_0Ib}{4\pi r^2}\frac{a}{r} = \frac{\mu_0M}{4\pi r^3}} $$
		+	If, when solving the problem, you managed to get an answer up to the second order, be the first to send the solution to <a href="mailto:alex@savchenkosolutions.com" target="_blank">alex@savchenkosolutions.com</a>
		+	</p>
		+
		+	<h4>Answer</h4>
		+	<p>
		+	$$B=\mu_0M\sqrt{1+3\sin^2\alpha} /(4\pi r^3), ~M=Iab$$
		+	</p>
		+
		+	<p style="text-align: right; font-style: italic; font-size: 14;">
		+	Aliaksandr Melnichenka<br>
		+	Andrei Yersh<br>
		+	</p>
		+
		+
		+	<footer class="row container">
		+	<br>
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			<i><b>knowledge must be free</b></i>
			</p>
			</header>

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

			<h3> Statement </h3>
			<p>
			$9.2.17^*.$ Determine the magnetic field induction of a rectangular frame $a \times b$ with a current $I$ at a point $A$ located at a distance $r$ much greater than the linear dimensions of the frame. The radius vector $\vec{r}$ forms an angle $\alpha$ with the plane of the frame.
			</p>

			<h3>Solution</h3>
			<p>

			</p>
			<center>
			<figure>
			<img src="https://savchenkosolutions.com/9/9.2.17/9.2.17_1.png"
			loading="lazy" width="350" />
			<figcaption>
			Direction of total magnetic induction
			</figcaption>
			</figure>
			</center>
			<p>
			From symmetry, it follows that the magnetic field induction will be present only in the plane defined by the angle $\alpha$, since the components of the wire of length $a$ will compensate each other</p><p>
			Let's write out the cosines of the angles that $\vec{r}$ makes with wires of length $b$
			let the angle $\alpha_1$ be for the nearest wire, and the angle $\alpha_2$ be for the farthest one, then </p><p>
			$$ \cos\alpha_1 = \frac{{2r\cos\alpha} - a}{2r} = \cos\alpha - \frac{a}{2r} $$

			$$ \cos\alpha_2 = \frac{{2r\cos\alpha} + a}{2r} = \cos\alpha + \frac{a}{2r} $$
			<b><i>Theoretical digression</i></b>

			</p>
			<center>
			<figure>
			<img src="https://savchenkosolutions.com/9/9.2.17/9.2.17_2.png"
			loading="lazy" width="140" />
			<figcaption>
			Magnetic induction created by a wire of length $l$
			</figcaption>
			</figure>
			</center>
			<p>
			A magnetic wire of length $l$ creates a magnetic field equal to
			$$ B=\frac{\mu_0I}{2\pi r}\cos\alpha_0 $$
			From the geometry of the problem, we obtain
			$$ \cos\alpha_0 = \frac{b/2}{\sqrt{r^2+b^2/4}} = \frac{b}{2r}\left(1+\frac{b^2}{4r^2}\right)^{1/2} $$
			Using the approximation $(1+x)^\alpha\approx1+\alpha x$, we transform to the first order of smallness
			$$ \cos\alpha_0 = \frac{b}{2r}\left(1-\frac{b^2}{8r^2}\right)\approx\frac{b}{2r} $$
			A more detailed derivation of this formula is on pages 76-78 <a href="http://belpho.org/books/Magnitostatika.pdf">"Magnetostatics. Electrodynamics. Theory of Oscillations"</a> — A.I. Slobodanyuk
			From where we get
			$$ \boxed{B_0=B_1=B_2=\frac{\mu_0Ib}{4\pi r^2}} $$
			NO: we used an approximation that will later give us an answer with an accuracy of up to first order</p><p>
			From the <a href="#drawing">drawing</a> it follows that the total magnetic induction of the magnetic field $\vec{B}$ will be equal to the vector sum of the components $\vec{B}_1$ and $\vec{B}_2$
			$$ \vec{B}=\vec{B}_1+\vec{B}_2 $$
			Then the vector modulus can be found using the cosine law.
			$$ B = \sqrt{B_1^2+B_2^2-2B_1B_2\cos\Delta\alpha} $$

			$$ B = B_0\sqrt{2-2\cos\Delta\alpha} = 2B_0\sin\frac{\Delta\alpha}{2}\quad(1) $$
			Considering the smallness of the angle $\Delta \alpha$,
			$$ \sin\frac{\Delta\alpha}{2} \approx \frac{\Delta\alpha}{2}=\frac{a}{2r} $$
			We substitute $(1)$:
			$$ B = 2B_0\sin\frac{\Delta\alpha}{2} = B_0 \Delta\alpha $$
			We obtain in approximation to the first order:
			$$ \boxed{B = \frac{\mu_0Ib}{4\pi r^2}\frac{a}{r} = \frac{\mu_0M}{4\pi r^3}} $$
			If, when solving the problem, you managed to get an answer up to the second order, be the first to send the solution to <a href="mailto:alex@savchenkosolutions.com" target="_blank">alex@savchenkosolutions.com</a>
			</p>

			<h4>Answer</h4>
			<p>
			$$B=\mu_0M\sqrt{1+3\sin^2\alpha} /(4\pi r^3), ~M=Iab$$
			</p>

			<p style="text-align: right; font-style: italic; font-size: 14;">
			Aliaksandr Melnichenka<br>
			Andrei Yersh<br>
			</p>


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