Новое решение

naz правка от 29 июн. 2026 г., 09:35

правка #19412 позже →

en/11.6.1.md +28 −0

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		+	### Statement
		+
		+	$11.6.1.$ [Insert the problem statement]
		+
		+	### Solution
		+
		+	In the GHS, the EMP law for a vortex electric field is written:
		+
		+	$$C_E = - \frac{1}{c} \frac{d \Phi_B}{dt}$$
		+
		+	The right screw, unlike the left screw, means a change of sign before the coefficient, so in the GHS the law for the vortex magnetic field looks like:
		+
		+	$$C_B = \frac{1}{c} \frac{d \Phi_D}{dt}$$
		+
		+	In GHS, we have a displacement flow equal to the field flow. To convert to SI, we will use the standard ratios:
		+
		+	$$B_{GHS} = B_{SI} \cdot\sqrt{\frac{4\pi}{\mu_0}}$$
		+
		+	$${\Phi_E}_{GHS} ={\Phi_E}_{SI} \cdot\sqrt{4\pi\varepsilon_0}$$
		+	After substituting this into the GHS equation and replacing the electric field flux with $\frac{\Phi_D}{\varepsilon_0}$:
		+
		+	$$C_B = \mu_0 \frac{d \Phi_D}{dt}$$
		+
		+	Savchenko apparently made a typo in his answer to SI and wrote a connection with the flow of the electric field and not with the displacement flow.
		+
		+	#### Answer
		+
		+	[Insert a concise answer or boxed result]

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			### Statement

			$11.6.1.$ [Insert the problem statement]

			### Solution

			In the GHS, the EMP law for a vortex electric field is written:

			$$C_E = - \frac{1}{c} \frac{d \Phi_B}{dt}$$

			The right screw, unlike the left screw, means a change of sign before the coefficient, so in the GHS the law for the vortex magnetic field looks like:

			$$C_B = \frac{1}{c} \frac{d \Phi_D}{dt}$$

			In GHS, we have a displacement flow equal to the field flow. To convert to SI, we will use the standard ratios:

			$$B_{GHS} = B_{SI} \cdot\sqrt{\frac{4\pi}{\mu_0}}$$

			$${\Phi_E}_{GHS} ={\Phi_E}_{SI} \cdot\sqrt{4\pi\varepsilon_0}$$
			After substituting this into the GHS equation and replacing the electric field flux with $\frac{\Phi_D}{\varepsilon_0}$:

			$$C_B = \mu_0 \frac{d \Phi_D}{dt}$$

			Savchenko apparently made a typo in his answer to SI and wrote a connection with the flow of the electric field and not with the displacement flow.

			#### Answer

			[Insert a concise answer or boxed result]