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For a thin wire carrying current $I$, the given magnetic vector potential reduces to $\mathbf{A}=({\mu }_{0}I/4\pi ){\oint }_{C}d{\mathbf{r}}^{\prime }/|\mathbf{r}-{\mathbf{r}}^{\prime }|$. Taking its curl and using ${\nabla }_{\mathbf{r}}(1/|\mathbf{r}-{\mathbf{r}}^{\prime }|)=-(\mathbf{r}-{\mathbf{r}}^{\prime })/|\mathbf{r}-{\mathbf{r}}^{\prime }{|}^3$ gives the Biot-Savart lawThis is exactly the sign convention in the printed form using ${\mathbf{r}}^{\prime }-\mathbf{r}$.

Let $\mathbf{v}=(\mathbf{r}-{\mathbf{r}}^{\prime })/|\mathbf{r}-{\mathbf{r}}^{\prime }{|}^3$. Away from the wire, $\nabla \cdot \mathbf{v}=0$. The supplied divergence and curl of a cross product therefore gives $\nabla \times (d{\mathbf{r}}^{\prime }\times \mathbf{v})=-(d{\mathbf{r}}^{\prime }\cdot {\nabla }_{\mathbf{r}})\mathbf{v}$. But ${\nabla }_{\mathbf{r}}\mathbf{v}=-{\nabla }_{{\mathbf{r}}^{\prime }}\mathbf{v}$, so the integrand is the total differential of $\mathbf{v}$ along the source loop. Its closed-loop integral is zero. Thus $\nabla \times \mathbf{B}=0$ at every point outside $C$, as demanded by the current-free magnetostatic Ampère's law.

For $r^{\prime }=f(\theta )>0$, $z^{\prime }=0$, and counterclockwise current, $d{\mathbf{r}}^{\prime }=(f^{\prime }{\mathbf{e}}_r+f{\mathbf{e}}_{\theta })d\theta$. At the origin, $\mathbf{v}=-{\mathbf{e}}_r/f^2$, soFor the ellipse expressed relative to a focus, $1/f=(1-e\cos \theta )/\ell$. The cosine integrates to zero, givingHere $\ell$ is the semilatus rectum. Reversing the current reverses the field direction.

Use the usual setting of a continuously differentiable solenoidal vector field defined on all radial segments from the origin, including the origin. More generally an open star-shaped set containing zero suffices. Differentiation under the integral sign givesApply divergence and curl of a cross product with the second field $x$, using $\nabla \cdot x=3$ and $(D\cdot \nabla )x=D$:The scaling identity in part (a) now converts the last expression to a one-variable derivative:Continuity at the origin ensures the lower endpoint is zero. Hence the radial vector potential of a solenoidal vector field isThe order of the cross product matters: $x\times D$ would produce $-B$. This construction is not valid on every punctured or multiply connected domain. For example $B=x/|x{|}^3$ is solenoidal away from zero, but its radial integral diverges at zero and it has nonzero flux through a surrounding sphere. The stated formula therefore implicitly needs the regular radial-domain hypothesis, rather than just local zero divergence.

In a region with no current, the Ampère-Maxwell equation and Faraday's law areThereforewhere the last line uses the divergence and curl of a cross product. Thus Poynting's theorem takes the formThe vector $\mathbf{S}$ is the Poynting vector.

In the inertial frame, rigid corotation gives $\mathbf{u}=\mathbf{\Omega }\times \mathbf{r}$. The ideal magnetohydrodynamics condition is the motional electric fieldApply Gauss's law in Gaussian units and use the divergence and curl of a cross product:The exterior field is produced by currents inside the star, so $\nabla \times {\mathbf{B}}_0=0$ there, while rotation with constant angular velocity has $\nabla \times \mathbf{u}=2\mathbf{\Omega }$. Hence the required Goldreich-Julian charge density isHere $n_q$ denotes electric charge per unit volume, as in the question. If each carrier has charge $q$, its signed number density is $n_q/q$.

Use the Levi-Civita symbol and the product rule. First,For the second identity, the contractiongivesIn vector notation this isThese are the divergence and curl of a cross product identities.