Правка разделов «Statement», «Solution», «Answer»

jzmicer правка от 28 июн. 2026 г., 18:55

правка #19353 предыдущая #19352 ← раньше

en/11.6.3.md +27 −3

		@@ -1,11 +1,35 @@
			### Statement

		−	$11.6.3.$ [Insert the problem statement]
		+	$11.6.3.$ What is the electric displacement flux through the area bounded by a closed loop if, when this flux decreases uniformly to zero over $1\ \mu s$, a circulation of magnetic induction of $0.001\ \text{T}\cdot\text{m}$ is induced in the loop?

			### Solution

		−	1
		+	The relation between the circulation of the magnetic induction around the loop and the electric displacement flux through the surface "stretched" over this loop is given by the equation:
		+	$$
		+	\oint \vec B \, d\vec l = \mu_0 \varepsilon_0 \frac{d}{dt} \left( \int \vec E \, d\vec S \right).
		+	$$
		+	Or
		+	$$
		+	C_B = \mu_0 \varepsilon_0 \frac{d\Phi_E}{dt},
		+	$$
		+	where $C_B$ is the circulation of the magnetic induction, and $\Phi_E$ is the electric displacement flux.

		+	According to the condition ($t_0 = 1\ \mu s$),
		+	$$
		+	\Phi_E(t) = kt + b, \qquad \Phi_E(t_0) = 0 \ \Rightarrow\ \Phi_E(t) = kt - k t_0.
		+	$$
		+	Then, taking the time derivative:
		+	$$
		+	C_B = \mu_0 \varepsilon_0 k,
		+	$$
		+	$$
		+	\Phi_E(t) = \frac{C_B}{\mu_0 \varepsilon_0} (t - t_0) = C_B c^2 (t - t_0).
		+	$$
		+	$$
		+	\Phi_E(0) = 1 \cdot 10^{-3}\ (\text{T}\cdot\text{m}) \cdot (3 \cdot 10^8\ (\text{m/s}))^2 \cdot 1 \cdot 10^{-6}\ (\text{s}) = 9 \cdot 10^7\ (\text{V}\cdot\text{m}).
		+	$$
		+	Savchenko's answer contains a misprint.
		+
			#### Answer

		−	[Insert a concise answer or boxed result]
		+	$\Phi_E = 9 \cdot 10^7\ (\text{V}\cdot\text{m})$

	@@ -1,11 +1,35 @@
	### Statement		### Statement

	$11.6.3.$ [Insert the problem statement]		$11.6.3.$ What is the electric displacement flux through the area bounded by a closed loop if, when this flux decreases uniformly to zero over $1\ \mu s$, a circulation of magnetic induction of $0.001\ \text{T}\cdot\text{m}$ is induced in the loop?

	### Solution		### Solution

	1		The relation between the circulation of the magnetic induction around the loop and the electric displacement flux through the surface "stretched" over this loop is given by the equation:
			$$
			\oint \vec B \, d\vec l = \mu_0 \varepsilon_0 \frac{d}{dt} \left( \int \vec E \, d\vec S \right).
			$$
			Or
			$$
			C_B = \mu_0 \varepsilon_0 \frac{d\Phi_E}{dt},
			$$
			where $C_B$ is the circulation of the magnetic induction, and $\Phi_E$ is the electric displacement flux.

			According to the condition ($t_0 = 1\ \mu s$),
			$$
			\Phi_E(t) = kt + b, \qquad \Phi_E(t_0) = 0 \ \Rightarrow\ \Phi_E(t) = kt - k t_0.
			$$
			Then, taking the time derivative:
			$$
			C_B = \mu_0 \varepsilon_0 k,
			$$
			$$
			\Phi_E(t) = \frac{C_B}{\mu_0 \varepsilon_0} (t - t_0) = C_B c^2 (t - t_0).
			$$
			$$
			\Phi_E(0) = 1 \cdot 10^{-3}\ (\text{T}\cdot\text{m}) \cdot (3 \cdot 10^8\ (\text{m/s}))^2 \cdot 1 \cdot 10^{-6}\ (\text{s}) = 9 \cdot 10^7\ (\text{V}\cdot\text{m}).
			$$
			Savchenko's answer contains a misprint.

	#### Answer		#### Answer

	[Insert a concise answer or boxed result]		$\Phi_E = 9 \cdot 10^7\ (\text{V}\cdot\text{m})$