The discussion clarifies that the magnetic vector potential, denoted as ## \vec A ##, is not uniquely determined for a given magnetic field ## \vec B ##. While the magnetic field has a unique solution, there are infinitely many equivalent magnetic vector potentials related by the equation ## \vec {A'}=\vec A+\vec \nabla \phi ##, where ## \phi ## is a scalar field. This means that although the magnetic field is uniquely defined, the vector potential can vary infinitely while still producing the same magnetic field.
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Shyan said:For any magnetic field ## \vec B ##, there are an infinite number of equivalent magnetic vector potentials##\vec A##(##\vec B=\vec \nabla\times\vec A ##), related by ## \vec {A'}=\vec A+\vec \nabla \phi ## for some scalar field ## \phi ##. So the magnetic vector potential of a magnetic field is not uniquely determined.
As I said, all those infinite number of vector potentials are equivalent, which means they all give the same magnetic field. That's because ## \vec \nabla \times \vec \nabla \phi=0 ## for any scalar field ## \phi ##. So ## \vec{A'}=\vec A+\vec\nabla \phi \Rightarrow \vec\nabla\times\vec {A'}=\vec\nabla\times\vec A ##.fricke said:thank you very much for the reply!
As what you have explained, magnetic vector potential is not uniquely determined since there are an infinite number of equivalent vector potential A, for magnetic field B. But does it mean B also not uniquely determined since we could choose any scalar field of A?
Shyan said:As I said, all those infinite number of vector potentials are equivalent, which means they all give the same magnetic field. That's because ## \vec \nabla \times \vec \nabla \phi=0 ## for any scalar field ## \phi ##. So ## \vec{A'}=\vec A+\vec\nabla \phi \Rightarrow \vec\nabla\times\vec {A'}=\vec\nabla\times\vec A ##.