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I have the following expression of oscillating dipole, $$d = \vec{d}_0\sin(\omega t).$$ Find the strength of the magnetic field, $$\vec{H} = \,?$$
My solution at the moment: $$\vec{A} = \frac{\vec{d}(t-\frac{r}{c})}{r c}$$ $$\varphi = \frac{\vec{d}(t-\frac{r}{c})\cdot\vec{r}}{r^3}$$ $$\vec{H}=\operatorname{rot}\vec{A}$$ $$\vec{E} = -\frac{1}{c}\frac{\partial \vec{A}}{\partial t}-\operatorname{grad}\varphi$$ $$\vec{A} = \frac{\vec{d}_0\sin\left[\omega(t-\frac{r}{c})\right]}{rc} $$ $$ \vec{H} = \operatorname{rot}{\vec{A}}= \begin{bmatrix} \vec{\imath} & \vec{\jmath} & \vec{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ A_x & A_y & A_z \end{bmatrix} = \vec{\imath}\left(\frac{\partial}{\partial y}A_z-\frac{\partial}{\partial z}A_y\right) + \vec{\jmath}\left(\frac{\partial}{\partial z}A_x-\frac{\partial}{\partial x}A_z\right) + \vec{k}\left(\frac{\partial}{\partial x}A_y-\frac{\partial}{\partial y}A_x\right) = (*) $$
How do we find $A_x$, $A_y$, and $A_z$, as we don’t have any $x$, $y$, or $z$ here? Am I missing vector above any of $r$? If yes, where exactly and what to do next?
My tutor once told me that $r$ is something like a square root of sum $x^2+y^2+z^2$? Is that true?
