Question

Given that B = curl A, use the divergence theorem to show that Bndσ\oint \mathbf{B} \cdot \mathbf{n} d \sigma over any closed surface is zero.

Solution

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Answered 1 year ago
Answered 1 year ago
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Using the divergence theorem: vBndσ=v(B)dτ\oint_{\partial v} \boldsymbol{B}\cdot\boldsymbol{n}d\sigma=\int_{v}(\boldsymbol{\nabla}\cdot\boldsymbol{B})d\tau, but B=×A\boldsymbol{B}=\boldsymbol{\nabla}\times \boldsymbol{A}, then:

vBndσ=v(×A)dτ\oint_{\partial v} \boldsymbol{B}\cdot\boldsymbol{n}d\sigma=\int_{v}\boldsymbol{\nabla}\cdot(\boldsymbol{\nabla}\times \boldsymbol{A})d\tau

but (×A)=0\boldsymbol{\nabla}\cdot(\boldsymbol{\nabla}\times \boldsymbol{A})=0 everywhere, so:

vBndσ=0\oint_{\partial v} \boldsymbol{B}\cdot\boldsymbol{n}d\sigma=0

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