239 words
1 minute
静电静磁
2025-11-17
Electrodynamics
Electrodynamics
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Study
∇⋅(R⃗R3)=4πδ(R⃗)\nabla \cdot (\frac{\vec{R}}{R^3}) = 4\pi \delta(\vec{R}) ∇(p⃗⋅r⃗)=p⃗\nabla (\vec{p} \cdot \vec{r}) = \vec{p} φ=14πϵ0p⃗⋅r⃗r3\varphi = \frac{1}{4\pi\epsilon_0} \frac{\vec{p} \cdot \vec{r}}{r^3} E⃗=−14πϵ0∇(p⃗⋅r⃗r3)\vec{E} = -\frac{1}{4\pi\epsilon_0} \nabla (\frac{\vec{p} \cdot \vec{r}}{r^3}) =−14πϵ0(p⃗r3+(p⃗⋅r⃗)(−3r^r4))= -\frac{1}{4\pi\epsilon_0} (\frac{\vec{p}}{r^3} + (\vec{p} \cdot \vec{r}) (-3\frac{\hat{r}}{r^4})) =−14πϵ0p⃗−3(p⃗⋅r^)r^r3= -\frac{1}{4\pi\epsilon_0} \frac{\vec{p} - 3(\vec{p} \cdot \hat{r})\hat{r}}{r^3} dF⃗12=μ04πj⃗1dτ1×(j⃗2dτ2×r^12)r122dF⃗12≠dF⃗21d\vec{F}_{12} = \frac{\mu_0}{4\pi} \frac{\vec{j}_1 d\tau_1 \times (\vec{j}_2 d\tau_2 \times \hat{r}_{12})}{r_{12}^2} \quad d\vec{F}_{12} \neq d\vec{F}_{21} B⃗=μ04π∫j⃗dτ′×r⃗r2r⃗r2=−∇(1r)\vec{B} = \frac{\mu_0}{4\pi} \int \frac{\vec{j}d\tau' \times \vec{r}}{r^2} \quad \frac{\vec{r}}{r^2} = -\nabla(\frac{1}{r}) B⃗=μ04π∫∇(1r)×j⃗(r⃗′)dτ′\vec{B} = \frac{\mu_0}{4\pi} \int \nabla(\frac{1}{r}) \times \vec{j}(\vec{r}') d\tau' =μ04π∫[∇×(j⃗(r⃗′)r)−∇×j⃗(r⃗′)r]dτ′= \frac{\mu_0}{4\pi} \int \left[ \nabla \times (\frac{\vec{j}(\vec{r}')}{r}) - \frac{\nabla \times \vec{j}(\vec{r}')}{r} \right] d\tau' =∇×(μ04π∫j⃗(r⃗′)rdτ′)A⃗(r⃗)= \nabla \times \left( \frac{\mu_0}{4\pi} \int \frac{\vec{j}(\vec{r}')}{r} d\tau' \right) \quad \vec{A}(\vec{r}) a⃗×(b⃗×c⃗)=b⃗(a⃗⋅c⃗)−c⃗(a⃗⋅b⃗)\vec{a} \times (\vec{b} \times \vec{c}) = \vec{b}(\vec{a} \cdot \vec{c}) - \vec{c}(\vec{a} \cdot \vec{b}) ∇×B⃗=∇×(∇×A⃗)=∇(∇⋅A⃗)−∇2A⃗\nabla \times \vec{B} = \nabla \times (\nabla \times \vec{A}) = \nabla(\nabla \cdot \vec{A}) - \nabla^2 \vec{A} ∇⋅A⃗=μ04π∇⋅∫j⃗(r⃗′)Rdτ′=μ04π∫j⃗(r⃗′)⋅∇(1R)dτ′\nabla \cdot \vec{A} = \frac{\mu_0}{4\pi} \nabla \cdot \int \frac{\vec{j}(\vec{r}')}{R} d\tau' = \frac{\mu_0}{4\pi} \int \vec{j}(\vec{r}') \cdot \nabla(\frac{1}{R}) d\tau' =−μ04π∫j⃗(r⃗′)⋅∇′(1R)dτ′=−μ04π∫(∇′⋅(j⃗(r⃗′)R)−1R∇′⋅j⃗(r⃗′)⏟0)dτ′= -\frac{\mu_0}{4\pi} \int \vec{j}(\vec{r}') \cdot \nabla'(\frac{1}{R}) d\tau' = -\frac{\mu_0}{4\pi} \int \left( \nabla' \cdot (\frac{\vec{j}(\vec{r}')}{R}) - \frac{1}{R} \underbrace{\nabla' \cdot \vec{j}(\vec{r}')}_{0} \right) d\tau' −∇2A⃗=−μ04π∇2∫j⃗(r⃗′)dτ′R=∫μ04π(−4πδ(r⃗−r⃗′))j⃗(r⃗′)dτ′-\nabla^2 \vec{A} = -\frac{\mu_0}{4\pi} \nabla^2 \int \frac{\vec{j}(\vec{r}') d\tau'}{R} = \int \frac{\mu_0}{4\pi} (-4\pi \delta(\vec{r}-\vec{r}')) \vec{j}(\vec{r}') d\tau' =μ0j⃗(r⃗)= \mu_0 \vec{j}(\vec{r}) ⇒∇×B⃗=μ0j⃗(r⃗)\Rightarrow \nabla \times \vec{B} = \mu_0 \vec{j}(\vec{r})

对于磁偶极子

A⃗=μ0I4π∮dl⃗R=μ0I4π∮dl⃗∣r⃗−r⃗′∣≈μ0I4πr3∫(r⃗⋅r⃗′)dl⃗′\vec{A} = \frac{\mu_0 I}{4\pi} \oint \frac{d\vec{l}}{R} = \frac{\mu_0 I}{4\pi} \oint \frac{d\vec{l}}{|\vec{r}-\vec{r}'|} \approx \frac{\mu_0 I}{4\pi r^3} \int (\vec{r} \cdot \vec{r}') d\vec{l}' =μ0I4πr3⋅12∮(r⃗′×dl⃗′)×r⃗= \frac{\mu_0 I}{4\pi r^3} \cdot \frac{1}{2} \oint (\vec{r}' \times d\vec{l}') \times \vec{r} B⃗=μ04π∇×(m⃗×r⃗r3)=μ04π((∇⋅r⃗r3)m⃗−(m⃗⋅∇)r⃗r3)\vec{B} = \frac{\mu_0}{4\pi} \nabla \times (\frac{\vec{m} \times \vec{r}}{r^3}) = \frac{\mu_0}{4\pi} \left( (\nabla \cdot \frac{\vec{r}}{r^3})\vec{m} - (\vec{m} \cdot \nabla)\frac{\vec{r}}{r^3} \right) =−μ04π(m⃗⋅∇)r⃗r3=−μ04πm⃗−3(m⃗⋅r^)r^r3= -\frac{\mu_0}{4\pi} (\vec{m} \cdot \nabla) \frac{\vec{r}}{r^3} = -\frac{\mu_0}{4\pi} \frac{\vec{m} - 3(\vec{m} \cdot \hat{r})\hat{r}}{r^3}
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