Problem 5: Let B = (B1(x, y), B2(x, y), 0) be a divergence-free vector, ∇ · B = 0. Find a vector function A~ = (0, 0, A(x, y)), such that ∇ × A~= B, and ∇ · A = 0.

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Problem 5: Let B = (B1(x, y), B2(x, y), 0) be a divergence-free vector, ∇ · B = 0. Find a vector function A~ = (0, 0, A(x, y)), such that ∇ × A~= B, and ∇ · A = 0.

Mathematics
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this is a special case of Helmholtz decomposition. we have that $$\vec B=(B_1(x,y),B_2(x,y),0)$$ and we see its divergence is: $$\frac{\partial B_1}{\partial x}+\frac{\partial B_2}{\partial y}=0\\\frac{\partial B_1}{\partial x}=-\frac{\partial B_2}{\partial y}$$ okay, now consider that we want a vector field \(\vec A=(0,0,A(x,y))\) such that $$\nabla \times \vec A=\left(-\frac{\partial A}{\partial y},\frac{\partial A}{\partial x},0\right)=B$$ so it follows we have $$B_1=-\frac{\partial A}{\partial y}\\B_2=\frac{\partial A}{\partial x}$$ and the earlier equation tells us $$-\frac{\partial^2 A}{\partial x\,\partial y}=-\frac{\partial^2 A}{\partial y\,\partial x}$$
so we can just take: $$A(x,y)=-\int B_1\, dy+f(x)\\A(x,y)=\int B_2\, dx+g(y)$$ and combine these into a single solution of the form $$A(x,y)=-\int B_1 dy+\int B_2\, dx+C$$
we call \(A\) a vector potential of \(B\): https://en.wikipedia.org/wiki/Vector_potential

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