Magnetic field of a rotating sphere

The charges move on circles aroud the z axis with the velocity $\vec v = 2 \pi f r_\perp \vec e_\varphi$ , where $r_\perp = r \sin \theta$ denotes the distance from the ration axis. The currents density yields $\vec j = \rho \vec v = \rho \omega r \sin \theta \vec e_\phi$ with the charge desinty $\rho = \frac{Q}{\frac{4}{3} \pi R^3}$ of the sphere. The magnetic field at the point $\vec r_0$ results according to the Biot-Savart law: $\vec H(\vec r_0) = \frac{1}{4 \pi} \int \frac{\vec j(\vec r) \times (\vec r_0 - \vec r)}{|\vec r_0 - \vec r|^3} dV$ First, we calculate the cross product $\begin{aligned} \vec j(\vec r) \times (\vec r_0 - \vec r) &= -\rho \omega r \sin \theta \left( \begin{array}{c} -\sin \phi \\ \cos \phi \\ 0 \end{array} \right) \times \left( \begin{array}{c} r \cos \phi \sin \theta \\ r \sin \phi \sin \theta \\ r \cos \theta \end{array} \right) \\ &= -\rho \omega r^2 \sin \theta \left( \begin{array}{c} \cos \phi \cos \theta \\ \sin \phi \cos \theta \\ - \sin \theta \end{array} \right) \end{aligned}$ Now we can calculate the magnetic field by volume integration in spherical coordinates: $\begin{aligned} \vec H(0) &= - \frac{1}{4 \pi} \int_0^{2 \pi} \int_{0}^\pi \int_0^R \frac{\rho \omega r^2 \sin \theta}{r^3} \left( \begin{array}{c} \cos \phi \cos \theta \\ \sin \phi \cos \theta \\ - \sin \theta \end{array} \right) r^2 \sin \theta dr d\theta d\phi \\ &= - \frac{\rho \omega}{4 \pi} \int_{0}^\pi \int_0^R r \sin^2 \theta \left( \begin{array}{c} 0 \\ 0 \\ - 2 \pi \sin \theta \end{array} \right) dr d\theta \\ &= \frac{\rho \omega}{2} \int_0^R r dr \int_{0}^\pi \sin^3 \theta d\theta \, \vec e_z \\ &= \frac{Q \omega}{\frac{8}{3} \pi R^3} \cdot \frac{R^2}{2} \cdot \frac{4}{3}\,\vec e_z \\ &= \frac{Q f}{2 R} \,\vec e_z \end{aligned}$ The absolute value of the magnetic field thus results $|\vec H| = \frac{Q f}{2 R} = \frac{1 \cdot 100}{2 \cdot 0.1} \,\frac{\text{A}}{\text{m}} = 500 \,\frac{\text{A}}{\text{m}}$