= Solution
Let $\rho_q=3q/(4\pi a^3)$ be the uniform <charge density> and take $\boldsymbol\omega=\omega\widehat{\mathbf z}$. Rigid rotation produces <current density> $\mathbf J=\rho_q\boldsymbol\omega\times\mathbf r$. The <magnetic dipole moment> is
$$
\mathbf m_s=\frac12\int\mathbf r\times\mathbf J\,d^3r
=\frac{\rho_q}{2}\int\left[r^2\boldsymbol\omega-\mathbf r(\mathbf r\cdot\boldsymbol\omega)\right]d^3r.
$$
<Spherical symmetry> gives $\int r_i r_j\,d^3r=\delta_{ij}\int r^2d^3r/3$, while $\int r^2d^3r=4\pi a^5/5$. Therefore
$$
\boxed{\mathbf m_s=\frac{qa^2}{5}\boldsymbol\omega.}
$$
It is parallel to the <angular velocity> for positive $q$ and antiparallel for negative $q$. The fully signed magnitude statement is $|\mathbf m_s|=|q|\,|\omega|a^2/5$; the printed magnitude assumes positive $q$ and positive angular speed.
<Axisymmetry> permits a purely azimuthal <magnetic vector potential>, $\mathbf A=f(r)\sin\theta\,\widehat{\boldsymbol\varphi}$, with $A_r=A_\theta=0$. For the source's <associated Legendre function> convention, $P_1^1(\cos\theta)=+\sin\theta$, since its definition omits the Condon–Shortley phase. The given differential operator reduces to
$$
\widehat F[f(r)\sin\theta]=\left(f''+\frac2r f'-\frac{2f}{r^2}\right)\sin\theta.
$$
Inside the sphere, $J_\varphi=\rho_q\omega r\sin\theta$. Since the radial operator maps $r^3$ to $10r$, the particular coefficient is $\alpha_0=-\mu_0\rho_q\omega/10$. Regularity at the origin and decay at infinity give
$$
f_{\rm in}=Ar-\frac{\mu_0\rho_q\omega}{10}r^3,\qquad
f_{\rm out}=\frac C{r^2}.
$$
There is no surface sheet <electric current>, so both normal and tangential <magnetic field> components are continuous at $r=a$. Using
$$
B_r=\frac{2f(r)}r\cos\theta,\qquad
B_\theta=-\frac{f(r)+rf'(r)}r\sin\theta,
$$
this is equivalent to continuity of $f$ and $f'$. The two equations give $A=\mu_0\rho_q\omega a^2/6$ and $C=\mu_0\rho_q\omega a^5/15$. Thus
$$
\boxed{\mathbf A_{\rm in}=\frac{\mu_0\rho_q}{30}(5a^2-3r^2)(\boldsymbol\omega\times\mathbf r),\quad r<a,}
$$
$$
\boxed{\mathbf A_{\rm out}=\frac{\mu_0}{4\pi}\frac{\mathbf m_s\times\mathbf r}{r^3},\quad r>a.}
$$
In the homogeneous expansion, $\alpha_1=A$, $\beta_1=C$ in their respective interior and exterior regions; all higher angular orders vanish. Indeed, matching the regular $r^\ell$ and decaying $r^{-\ell-1}$ terms with zero source at $\ell>1$ forces both coefficients to zero. The outside <magnetic field> is exactly the <magnetic dipole field> of $\mathbf m_s$, not merely its leading far-field approximation. For a positive moment along $z$, $A_\varphi=\mu_0m_s\sin\theta/(4\pi r^2)$, which is the printed cross-product magnitude expression.
For the <flux coupling of a rotating charged sphere to a circular loop>, take a loop of radius $b>a$ in the equatorial plane and use <Stokes theorem> on the regular full <vector potential>, evaluated around the rim. Its linked <magnetic flux> is
$$
\Phi_b=\oint\mathbf A\cdot d\boldsymbol\ell
=2\pi b\frac{\mu_0m_s}{4\pi b^2}=\frac{\mu_0m_s}{2b}.
$$
With $m_s=\pi a^2I_0$, the <mutual inductance> defined using this effective <electric current> is
$$
\boxed{M=\frac{\Phi_b}{I_0}=\frac{\mu_0\pi a^2}{2b}.}
$$
Integrating an ideal <magnetic dipole field> through the entire disk would incorrectly use its singular expression inside the actual sphere; the rim calculation avoids that error.
For the prescribed time dependence, the quasistatic <Faraday's law of induction> result and the purely resistive loop approximation give
$$
\boxed{I_b(t)=-\frac1R\frac{d\Phi_b}{dt}=-\frac{\mu_0qa^2\omega_0\nu}{10bR}\cos(\nu t).}
$$
Positive loop <electric current> is defined to produce a positive $z$ <magnetic dipole moment>, so the minus sign is <Lenz's law>. This assumes negligible retardation and $\nu L_b\ll R$, where $L_b$ is the loop's <self-inductance>. If that is not negligible, the governing equation is instead $L_b\dot I_b+RI_b=-\dot\Phi_b$; $R$ alone is insufficient for the exact dynamic response.
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