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

Gauss's law of magnetism $\int \overrightarrow{\mathrm{B}} \cdot d \overrightarrow{\mathrm{S}}=0$.

Now, magnetic moment of dipole at origin "O" is along $z$-axis $\overrightarrow{\mathrm{M}}=\mathrm{M} \hat{k}$

Let $\mathrm{P}$ be a point at distance $r$ from $\mathrm{O}$ and OP makes an angle $\theta$ with $z$-axis component of $\overrightarrow{\mathrm{M}}$ along $\mathrm{OP}=\mathrm{M} \cos \theta$

Now, the magnetic field induction at P due to dipole of moment $\vec{M} \cos \theta$ is $\vec{B}=\frac{\mu_{0}}{4 \pi} \frac{2 M \cos \theta}{r^{3}} \hat{r}$

From the diagram, $r$ is the radius of sphere with centre at O lying in $y z$-plane. Take an elementary area $d \overrightarrow{\mathrm{S}}$ of the surface at P, then

$\therefore d \overrightarrow{\mathrm{S}}=r(r \sin \theta) \hat{r}=r^{2} \sin \theta d \theta \hat{r}$

$\left.\therefore \int\right] \overrightarrow{\mathrm{B}} \cdot d \overrightarrow{\mathrm{S}}=\int \frac{\mu_{0}}{4 \pi} \frac{2 \mathrm{M} \cos \theta}{r^{3}} \hat{r}\left(r^{2} \sin \theta d \theta\right) \hat{r}$

$=\frac{\mu_{0}}{4 \pi} \frac{\mathrm{M}}{r} \int_{0}^{2 \pi} 2 \sin \theta \cos \theta d \theta$

$=\frac{\mu_{0}}{4 \pi} \frac{\mathrm{M}}{r} \int_{0}^{2 \pi} \sin 2 \theta d \theta$

$=\frac{\mu_{0}}{4 \pi} \frac{\mathrm{M}}{r}\left(-\frac{\cos 2 \theta}{2}\right)_{0}^{2 \pi}$

$=\frac{\mu_{0}}{4 \pi} \frac{\mathrm{M}}{r}\left[\frac{\cos 2 \theta}{2}\right]_{0}^{2 \pi}(\because \cos (-0)=\cos 4)$

$=-\frac{\mu_{0}}{4 \pi} \frac{\mathrm{M}}{2 r}[\cos 4 \pi-\cos 0]$

$=\frac{\mu_{0} \mathrm{M}}{4 \pi(2 r)}[1-1]=0$