∇ → × E → = − ∂ B → ∂ t \overrightarrow{\nabla}\times\overrightarrow{E}=-\frac{\partial \overrightarrow{B}}{\partial t } ∇ × E = − ∂ t ∂ B ∇ → × E → = − ∂ B → ∂ t \overrightarrow{\nabla}\times\overrightarrow{E}=-\frac{\partial \overrightarrow{B}}{\partial t } ∇ × E = − ∂ t ∂ B ∇ → × E → = − ∂ B → ∂ t \overrightarrow{\nabla}\times\overrightarrow{E}=-\frac{\partial \overrightarrow{B}}{\partial t } ∇ × E = − ∂ t ∂ B ∇ → × E → = − ∂ B → ∂ t \overrightarrow{\nabla}\times\overrightarrow{E}=-\frac{\partial \overrightarrow{B}}{\partial t } ∇ × E = − ∂ t ∂ B
∇ → × B → = μ 0 ϵ 0 ∂ E → ∂ t \overrightarrow{\nabla}\times\overrightarrow{B}=\mu_0\epsilon_0\frac{\partial \overrightarrow{E}}{\partial t } ∇ × B = μ 0 ϵ 0 ∂ t ∂ E
∇ → × B → = μ 0 ϵ 0 ∂ E → ∂ t \overrightarrow{\nabla}\times\overrightarrow{B}=\mu_0\epsilon_0\frac{\partial \overrightarrow{E}}{\partial t } ∇ × B = μ 0 ϵ 0 ∂ t ∂ E ∇ → × B → = μ 0 ϵ 0 ∂ E → ∂ t \overrightarrow{\nabla}\times\overrightarrow{B}=\mu_0\epsilon_0\frac{\partial \overrightarrow{E}}{\partial t } ∇ × B = μ 0 ϵ 0 ∂ t ∂ E ∇ → × B → = μ 0 ϵ 0 ∂ E → ∂ t \overrightarrow{\nabla}\times\overrightarrow{B}=\mu_0\epsilon_0\frac{\partial \overrightarrow{E}}{\partial t } ∇ × B = μ 0 ϵ 0 ∂ t ∂ E
∇ → × E → ( x , t ) i → = ∣ i → j → k → ∂ ∂ x ∂ ∂ y ∂ ∂ z E → ( x , t ) 0 0 ∣ = ∂ E ∂ x j → \overrightarrow{\nabla}\times\overrightarrow{E}(x,t)\overrightarrow{i}=\begin{vmatrix} \overrightarrow{i} & \overrightarrow{j} & \overrightarrow{k}\\ \frac{\partial }{\partial x } & \frac{\partial }{\partial y } & \frac{\partial }{\partial z } \\ \overrightarrow{E}(x,t) &0 & 0 \end{vmatrix}=\frac{\partial E}{\partial x }\overrightarrow{j} ∇ × E ( x , t ) i = ∣ ∣ i ∂ x ∂ E ( x , t ) j ∂ y ∂ 0 k ∂ z ∂ 0 ∣ ∣ = ∂ x ∂ E j
∇ → × B → ( x , t ) j → = ∣ i → j → k → ∂ ∂ x ∂ ∂ y ∂ ∂ z 0 B → ( x , t ) 0 ∣ = − ∂ B ∂ x i → \overrightarrow{\nabla}\times\overrightarrow{B}(x,t)\overrightarrow{j}=\begin{vmatrix} \overrightarrow{i} & \overrightarrow{j} & \overrightarrow{k}\\ \frac{\partial }{\partial x } & \frac{\partial }{\partial y } & \frac{\partial }{\partial z } \\ 0 & \overrightarrow{B}(x,t) & 0 \end{vmatrix}=-\frac{\partial B}{\partial x }\overrightarrow{i} ∇ × B ( x , t ) j = ∣ ∣ i ∂ x ∂ 0 j ∂ y ∂ B ( x , t ) k ∂ z ∂ 0 ∣ ∣ = − ∂ x ∂ B i
∂ E ∂ x = − ∂ B ∂ t \frac{\partial E}{\partial x }=-\frac{\partial B}{\partial t } ∂ x ∂ E = − ∂ t ∂ B
∂ B ∂ x = − μ 0 ϵ 0 ∂ E ∂ t \frac{\partial B}{\partial x }=-\mu_0\epsilon_0\frac{\partial E}{\partial t } ∂ x ∂ B = − μ 0 ϵ 0 ∂ t ∂ E
∂ 2 E ∂ x 2 = − ∂ ∂ x ∂ B ∂ t = − ∂ ∂ t ∂ B ∂ x = − ∂ ∂ t ( − μ 0 ϵ 0 ∂ E ∂ t ) = μ 0 ϵ 0 ∂ 2 E ∂ t 2 \frac{\partial^2 E}{\partial x^2 }=-\frac{\partial}{\partial x }\frac{\partial B}{\partial t }=-\frac{\partial}{\partial t }\frac{\partial B}{\partial x }=-\frac{\partial}{\partial t }(-\mu_0\epsilon_0\frac{\partial E}{\partial t })=\mu_0\epsilon_0\frac{\partial^2 E}{\partial t^2 } ∂ x 2 ∂ 2 E = − ∂ x ∂ ∂ t ∂ B = − ∂ t ∂ ∂ x ∂ B = − ∂ t ∂ ( − μ 0 ϵ 0 ∂ t ∂ E ) = μ 0 ϵ 0 ∂ t 2 ∂ 2 E
∂ 2 E ∂ x 2 = μ 0 ϵ 0 ∂ 2 E ∂ t 2 \frac{\partial^2 E}{\partial x^2 }=\mu_0\epsilon_0\frac{\partial^2 E}{\partial t^2 } ∂ x 2 ∂ 2 E = μ 0 ϵ 0 ∂ t 2 ∂ 2 E
#340153 on Dec 2023
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