Translated 9.2.10-9.2.17

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

правка #11454 GitHub ef227c0

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		+	<span><img src="../../img/book.png"></span><span>Savchenko Solutions</span>
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		+	<h3> Statement </h3>
		+	<p>
		+	$9.2.16.$ Determine the magnetic field induction on the axis of a circuit whose magnetic moment is $M$, at large distances $h$ in cases where the circuit is a circle, square, or regular triangle.
		+	</p>
		+
		+	<h3>Solution</h3>
		+	<p>
		+	For a circle, the magnetic moment
		+	$$ M = \pi IR^2 $$
		+	Magnetic field at the ring axis (see <a href="../9.2.10">9.2.10</a>)
		+	$$ B = \frac{\mu_0 I R^2}{2(R^2+h^2)^{3/2}} = \frac{\mu_0 M}{2\pi (R^2 + h^2)^{3/2}} \approx \frac{\mu_0 M}{2\pi h^3} $$
		+	For large distances the shape of the contour is obviously not important, then
		+	$$ B = \frac{\mu_0 M}{2\pi h^3} $$
		+	For the unbelievers, let us prove this: let us have a regular $n$ polygon, described around a circle of radius $r$, the side of the polygon $a$, its magnetic moment
		+	$$ M = \frac{1}{2}narI $$
		+	Let's find the projection of the induction of the side onto the axis:
		+	$$ B_i = \int dB = \int\limits_0^a\frac{\mu_0 I dx}{4\pi (h^2+r^2)} \cdot \frac{r}{\sqrt{h^2+r^2}} $$
		+
		+	$$ B_i = \frac{\mu_0 Ir}{4\pi}\int\limits_0^a \frac{dx}{(h^2 + r^2)^{3/2}} $$
		+	in approximation $r\ll h$ we get
		+	$$ B_i \approx \frac{\mu_0 Ir}{4\pi}\int\limits_0^a \frac{dx}{h^3} = \frac{\mu_0 Ira}{4\pi h^3} $$
		+	Summarizing $B_i$, I get:
		+	$$ B = \frac{\mu_0 Iran}{4\pi h^3} = \frac{\mu_0 2M}{4\pi h^3} \Rightarrow \boxed{B= \frac{\mu_0 M}{2\pi h^3}} $$
		+	And you didn't believe it!
		+	</p>
		+
		+	<h4>Answer</h4>
		+	<p>
		+	$$B = \frac{\mu_0 M}{2\pi h^3}$$
		+	</p>
		+	<p style="text-align: right; font-style: italic; font-size: 14;">
		+	Andrei Yersh<br>
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			</p>
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			<h3 id="back-link"><a href="../../#9.2">$\leftarrow$Back</a></h3>

			<h3> Statement </h3>
			<p>
			$9.2.16.$ Determine the magnetic field induction on the axis of a circuit whose magnetic moment is $M$, at large distances $h$ in cases where the circuit is a circle, square, or regular triangle.
			</p>

			<h3>Solution</h3>
			<p>
			For a circle, the magnetic moment
			$$ M = \pi IR^2 $$
			Magnetic field at the ring axis (see <a href="../9.2.10">9.2.10</a>)
			$$ B = \frac{\mu_0 I R^2}{2(R^2+h^2)^{3/2}} = \frac{\mu_0 M}{2\pi (R^2 + h^2)^{3/2}} \approx \frac{\mu_0 M}{2\pi h^3} $$
			For large distances the shape of the contour is obviously not important, then
			$$ B = \frac{\mu_0 M}{2\pi h^3} $$
			For the unbelievers, let us prove this: let us have a regular $n$ polygon, described around a circle of radius $r$, the side of the polygon $a$, its magnetic moment
			$$ M = \frac{1}{2}narI $$
			Let's find the projection of the induction of the side onto the axis:
			$$ B_i = \int dB = \int\limits_0^a\frac{\mu_0 I dx}{4\pi (h^2+r^2)} \cdot \frac{r}{\sqrt{h^2+r^2}} $$

			$$ B_i = \frac{\mu_0 Ir}{4\pi}\int\limits_0^a \frac{dx}{(h^2 + r^2)^{3/2}} $$
			in approximation $r\ll h$ we get
			$$ B_i \approx \frac{\mu_0 Ir}{4\pi}\int\limits_0^a \frac{dx}{h^3} = \frac{\mu_0 Ira}{4\pi h^3} $$
			Summarizing $B_i$, I get:
			$$ B = \frac{\mu_0 Iran}{4\pi h^3} = \frac{\mu_0 2M}{4\pi h^3} \Rightarrow \boxed{B= \frac{\mu_0 M}{2\pi h^3}} $$
			And you didn't believe it!
			</p>

			<h4>Answer</h4>
			<p>
			$$B = \frac{\mu_0 M}{2\pi h^3}$$
			</p>
			<p style="text-align: right; font-style: italic; font-size: 14;">
			Andrei Yersh<br>
			</p>


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