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B1=A*sin(z)+C*cos(y)

B2=B*sin(x)+A*cos(z)

B3=C*sin(y)+B*cos(x)

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- Thread starter Nakul Aggarwal
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- #1

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B1=A*sin(z)+C*cos(y)

B2=B*sin(x)+A*cos(z)

B3=C*sin(y)+B*cos(x)

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Given ## \nabla \times \vec{B}=k \, \vec{B} ##, this differential equation does have an integral solution. Sometimes there are additional homogeneous solutions that need to be added, but it is basically of the Biot-Savart form: Maxwell's equation ## \nabla \times \vec{B} =\mu_o \vec{J} ## has the Biot-Savart solution: ## \vec{B}(\vec{x})=\frac{\mu_o}{4 \pi} \int \frac{\vec{J}(\vec{x'}) \times (\vec{x}-\vec{x'})}{|\vec{x}-\vec{x'}|^3} \, d^3x' ##. ## \\ ## If your last equation in the previous post is correct, it should be possible to at least show that the solution is consistent, i.e. if you replace ## \mu_o \vec{J} ## by ## k \vec{B} ##, with the form as provided, you should get consistency on both sides. ## \\ ## Edit: The better way would be to simply put in the solution of ## \nabla \times \vec{B}=k \, \vec{B} ##, and solve for ## k ##. The ## A ## term satisfies it for ## k=1 ## , and I believe the ## B ## and ## C ## do also. Starting with ## \nabla \times \vec{B}=\vec{B} ##, it really is only necessary to show that your solution satisfies the differential equation. It is unnecessary to derive anything.

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