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?

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

1 Answer 1


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:

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.