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I have two questions.

Firstly, how is the equation solved after taking the curl of A(r) here. How did they arrive at the final expression for B(r). I tried taking the product rule for curl, but got two terms zero, as m doesn't depend on r, but couldn't get the expression given here by solving further.

$$\mathbf{A(r)} = \frac{\mu_0}{4\pi r^3}\mathbf{m\times r}$$

$$\mathbf{B(r)} = \mathbf{\nabla \times A}=\frac{\mu_0}{4\pi r^3}\left( \frac{3\mathbf{r(m\cdot r)}}{r^2} - \mathbf{m}\right)$$

Secondly, how and why the vector potential described in the above link different from one of the "retarded potentials" shown here. They're both used to find B(r), right?

$$\mathbf{A}(\mathbf{r},t) = \frac{\mu_0}{4\pi}\int \frac{\mathbf{J}(\mathbf{r'},t_r)}{|\mathbf{r-r'}|} d^3\mathbf{r'}$$

where $t_t = t - \frac 1 c |\mathbf{r-r'}|$

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A straightforward way to find $\mathbf{B} = \nabla\times\mathbf{A}$ given the expression for the vector potential of a magnetic dipole is using Einstein's tensor notation, in which the cross product and curl operator are written as

$$ \mathbf{L} = \mathbf{M\times N} \rightarrow L_i = \varepsilon_{ijk}M_jN_k,\\ \mathbf{L} = \mathbf{\nabla\times M} \rightarrow L_i = \varepsilon_{ijk}\frac{\partial}{\partial x_j}M_k. $$

In this notation, your equations can be rewritten as

$$ A_i = \frac{\mu_0}{4\pi}\frac{\varepsilon_{ijk}m_jx_k}{r^3},\\ B_i = \varepsilon_{ijk}\frac{\partial}{\partial x_j}A_k. $$ Then, by substituting the first one on the second one, $$ B_g = \varepsilon_{ghi}\varepsilon_{ijk}\frac{\partial}{\partial x_h}\left(\frac{m_jx_k}{r^3}\right). $$ The Levi-Civita symbol remains invariant under cyclic permutations, so $\varepsilon_{igh} = \varepsilon_{ghi}$, and we may then use the identity which relates it with the Kronecker-delta

$$ \varepsilon_{igh}\varepsilon_{ijk} = \delta_{gj}\delta_{nk} - \delta_{gk}\delta_{hj}. $$ This yields $$ B_g = m_g\frac{\partial}{\partial x_h}\left(\frac{x_h}{r^3}\right) - m_h\frac{\partial}{\partial x_h}\left(\frac{x_g}{r^3}\right). $$ In vector notation, this is written as $$ \mathbf{B} = \mathbf{m}\nabla\cdot\left(\frac{\mathbf{x}}{r^3}\right) - (\mathbf{m}\cdot\nabla)\left(\frac{\mathbf{x}}{r^3}\right). $$ To continue the calculation using Einstein's notation, we only require the results $$ \frac{\partial}{\partial x_i}x_i = \nabla\cdot\mathbf{x} = 3, \\ \frac{\partial}{\partial x_j}x_i = \nabla\mathbf{x} = \delta_{ij},\\ \frac{\partial}{\partial x_i}r = \nabla r = \frac{\mathbf{x}}{r} = \frac{x_i}{r}. $$ With these and application of the chain-rule, we readily obtain $$ B_g = \frac{3m_hx_hx_g}{r^5} - \frac{m_g}{r^3} $$ which in tensor notation is the desired result $$ \mathbf{B} = \frac{3\mathbf{x}(\mathbf{m}\cdot\mathbf{x})}{r^5} - \frac{\mathbf{m}}{r^3}. $$ As for the reason the two expressions you provide for the vector potential are different, this is because they are caused by different sources. The first one due to a magnetic pole corresponds to an infinitesimal, constant, magnetic dipole, which can be related to a constant electric current through the equation $$ \mathbf{m} = \frac{1}{2}\int\mathbf{x}'\times\mathbf{J}(\mathbf{x}')d^3x' $$ which you may find in the chapter on magnetostatics of the book "Classical electrodynamics" by Jackson. The expression due to a retarded potential is derived from consideration of a time-dependent current density as a source. However, you may verify with this last expression that they are the same for the case of a time-independent current.

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