Verify Gauss's law for the magnetic field of a point dipole of dipole moment $\vec{m}$ at the origin for a surface which is a sphere of radius $R$.

(N/A) Gauss's law of magnetism states that the net magnetic flux through any closed surface is zero: $\oint \vec{B} \cdot d\vec{S} = 0$.
Consider a magnetic dipole of moment $\vec{m} = m \hat{k}$ placed at the origin $O$. The magnetic field $\vec{B}$ at a point $P$ on the surface of a sphere of radius $R$ (where $OP$ makes an angle $\theta$ with the $z$-axis) is given by:
$\vec{B} = \frac{\mu_0}{4\pi} \frac{m}{r^3} (2 \cos \theta \hat{r} + \sin \theta \hat{\theta})$.
The area element $d\vec{S}$ on the spherical surface is $d\vec{S} = R^2 \sin \theta d\theta d\phi \hat{r}$.
Now,calculate the flux $\Phi_B = \oint \vec{B} \cdot d\vec{S}$:
$\Phi_B = \int_0^{2\pi} \int_0^{\pi} \left( \frac{\mu_0}{4\pi} \frac{m}{R^3} (2 \cos \theta \hat{r} + \sin \theta \hat{\theta}) \right) \cdot (R^2 \sin \theta d\theta d\phi \hat{r})$.
Since $\hat{r} \cdot \hat{r} = 1$ and $\hat{\theta} \cdot \hat{r} = 0$:
$\Phi_B = \frac{\mu_0 m}{4\pi R} \int_0^{2\pi} d\phi \int_0^{\pi} 2 \cos \theta \sin \theta d\theta$.
$\Phi_B = \frac{\mu_0 m}{4\pi R} (2\pi) \int_0^{\pi} \sin(2\theta) d\theta$.
$\Phi_B = \frac{\mu_0 m}{2R} \left[ -\frac{\cos(2\theta)}{2} \right]_0^{\pi} = \frac{\mu_0 m}{4R} [-\cos(2\pi) + \cos(0)] = \frac{\mu_0 m}{4R} [-1 + 1] = 0$.
Thus,Gauss's law for magnetism is verified.