Правка разделов «Statement», «Solution», «Answer»
en/11.6.1.md
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| @@ -1,10 +1,12 @@ | |||
| ### Statement | |||
| − | $11.6.1.$ [Insert the problem statement] | ||
| + | $11.6.1.$ According to the law of electromagnetic induction, an alternating magnetic field generates a vortex electric field. Similarly, an alternating electric field generates a vortex magnetic field, but when the electric field changes, the direction of the vector $B$ forms a right screw with the direction of the vector $dE/dt$. The coefficient of proportionality in the GHS, which connects these fields, is the same in both phenomena. Using this property of the electromagnetic field, determine in the GHS and in the SI the dependence of the circulation of the induction of a magnetic field along a closed circuit on the rate of change of the flow of electric displacement through this circuit. | ||
| + |  | ||
| + | |||
| ### Solution | |||
| − | In the GHS, the | ||
| + | In the GHS, the FLI 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: | |||
| @@ -16,13 +18,19 @@Solution | |||
| $$B_{GHS} = B_{SI} \cdot\sqrt{\frac{4\pi}{\mu_0}}$$ | |||
| − | $$ | ||
| + | $$\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}$$ | |||
| Savchenko apparently made a typo in his answer to SI and wrote a connection with the flow of the electric field and not with the displacement flow. | |||
| #### Answer | |||
| − | [Insert a concise answer or boxed result] | ||
| + | In GHS: | ||
| + | |||
| + | $$C_B = \frac{1}{c} \frac{d \Phi_D}{dt}$$ | ||
| + | |||
| + | In SI: | ||
| + | |||
| + | $$C_B = \mu_0 \frac{d \Phi_D}{dt}$$ | ||
| @@ -1,10 +1,12 @@ | |||
| ### Statement | ### Statement | ||
| $11.6.1.$ [Insert the problem statement] | $11.6.1.$ According to the law of electromagnetic induction, an alternating magnetic field generates a vortex electric field. Similarly, an alternating electric field generates a vortex magnetic field, but when the electric field changes, the direction of the vector $B$ forms a right screw with the direction of the vector $dE/dt$. The coefficient of proportionality in the GHS, which connects these fields, is the same in both phenomena. Using this property of the electromagnetic field, determine in the GHS and in the SI the dependence of the circulation of the induction of a magnetic field along a closed circuit on the rate of change of the flow of electric displacement through this circuit. | ||
|  | |||
| ### Solution | ### Solution | ||
| In the GHS, the |
In the GHS, the FLI for a vortex electric field is written: | ||
| $$C_E = - \frac{1}{c} \frac{d \Phi_B}{dt}$$ | $$C_E = - \frac{1}{c} \frac{d \Phi_B}{dt}$$ | ||
| The right screw, unlike the left screw, means a change of sign before the coefficient, so in the GHS the law for the vortex magnetic field looks like: | The right screw, unlike the left screw, means a change of sign before the coefficient, so in the GHS the law for the vortex magnetic field looks like: | ||
| $$C_B = \frac{1}{c} \frac{d \Phi_D}{dt}$$ | $$C_B = \frac{1}{c} \frac{d \Phi_D}{dt}$$ | ||
| In GHS, we have a displacement flow equal to the field flow. To convert to SI, we will use the standard ratios: | In GHS, we have a displacement flow equal to the field flow. To convert to SI, we will use the standard ratios: | ||
| @@ -16,13 +18,19 @@Solution | |||
| $$B_{GHS} = B_{SI} \cdot\sqrt{\frac{4\pi}{\mu_0}}$$ | $$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 $\frac{\Phi_D}{\varepsilon_0}$: | After substituting this into the GHS equation and replacing the electric field flux with $\frac{\Phi_D}{\varepsilon_0}$: | ||
| $$C_B = \mu_0 \frac{d \Phi_D}{dt}$$ | $$C_B = \mu_0 \frac{d \Phi_D}{dt}$$ | ||
| 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. | 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 | #### Answer | ||
| [Insert a concise answer or boxed result] | In GHS: | ||
| $$C_B = \frac{1}{c} \frac{d \Phi_D}{dt}$$ | |||
| In SI: | |||
| $$C_B = \mu_0 \frac{d \Phi_D}{dt}$$ | |||