Question

Given that $\mathbf{B}\mathbf{}\mathbf{=}\mathbf{}\mathbf{curl}\mathbf{}\mathbf{A}$, use the divergence theorem to show that over any closed surface is zero.

Short answer

The solution of the integrals is ${\text{∮}}_{\mathrm{\partial }\mathrm{V}}\mathrm{B}\times \mathrm{nd\sigma }=0$.

Step-by-step solution

Given Information.

The given integral is$\text{∮}\mathrm{B}\times \mathrm{nd}\sigma .$

Definition of Divergence’s Theorem.

Divergence theorem, often known as Gauss' theorem or Ostrogradsky's theorem, is a theorem that connects the flow of a vector field across a closed surface to the field's divergence in the volume enclosed. According to this theorem, the surface integral of a vector field over a closed surface, also known as the flux through the surface, equals the volume integral of the divergence over the region inside the surface.

Apply Divergence Theorem.

Use the divergence theorem.

${\text{∮}}_a\mathrm{B}\times \mathrm{n}d\sigma ={\int }_1 (\mathrm{\nabla }\times \mathrm{B})d{\tau }_{\mathrm{I}}$

$B=\mathrm{\nabla }\times A$

Then, the equation is ${\text{∮}}_{\mathrm{\partial }v}\mathrm{B}\times \mathrm{nd}\sigma ={\int }_V \mathrm{\nabla }\times (\mathrm{\nabla }\times \mathrm{A})d\tau$true.

${\text{∮}}_{\mathrm{\partial }v}Bnd\sigma ={\int }_v \mathrm{\nabla }(\mathrm{\nabla }\times A)d\tau$

But, $\mathrm{\nabla }\times (\mathrm{\nabla }\times A)=0$everywhere .

${\text{∮}}_{\mathrm{\partial }v}\mathrm{B}\times \mathrm{n}d\sigma =0$

Hence, the solution of the integrals is ${\text{∮}}_{\mathrm{\partial }v}\mathrm{B}\times \mathrm{nd}\sigma =0$.