Recent content by Bright Liu

##(\nabla\times\vec B) \times \vec B=\nabla \cdot (\vec B\vec B -\frac 1 2B^2\mathcal I)-(\nabla \cdot \vec B)\vec B## ##\mathcal I## is the unit tensor

I think you are right and the answer is wrong

I think he has defined this node as ground node

Maybe you can solve it like this （In China we called it Current distribution method ）

Mathematics is important in physics I think

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