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Given is a vector potential

$$A(\vec{x},t)= \frac{\mu_0}{4\pi}\frac{\vec{m}\times\vec{x}}{|\vec{x}|^3}$$

Now I want to calculate the magnetic induction $\vec B$: $$\vec{B} = \nabla\times{\vec{A}} = \frac{\mu_0}{4\pi}\left(\nabla\times (\vec{m}\times\vec{x})\frac{1}{|\vec{x}|^3}+\nabla\left(\frac{1}{|\vec{x}|^3}\right)\times(\vec{m}\times\vec{x})\right)$$

Can someone please explain me how to get the grad term in this equation?

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  • $\begingroup$ You can find out yourself by writing out the expression in components. $\endgroup$
    – my2cts
    Jun 8, 2020 at 22:02

1 Answer 1

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You have a scalar $\frac{\mu_0}{4\pi|\vec{x}|^3}\equiv k$ multiplied by the cross product of two vectors $\vec{m}\times\vec{x}\equiv \vec{z}$. There is a vector identity:

$$\nabla\times(k\vec{z})=k(\nabla\times\vec{z})+(\nabla k)\times\vec{z}$$

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