Question

Verify the Gauss’s law for magnetic field of a point dipole of dipole moment $M$ at the origin for the surface which is a sphere of radius $R.$

Answer 1

Gauss' Law for magnetism applies to the magnetic flux through a closed surface. In this case the area vector points out from the surface. Because magnetic field lines are continuous loops, all closed surfaces have as many magnetic field lines going in as coming out. Hence, the net magnetic flux through a closed surface is zero.

Net flux $\varphi =\int B.dA=0$



Now, magnetic moment of dipole at origin $O$ is along $z-axis.$ Let $P$ be a point at distance $r$ from $O$ and $OP$ makes an angle $\theta$ with $z-axis.$ Component of $M$ along $OP=M\cos \theta$

Now, the magnetic field induction at $P$ due to dipole of moment $M\cos \theta$ is

$B=\frac{{\mu }_0}{4\pi }\frac{2M\cos \theta }{r^3}\hat{r}$

$r$ is the radius of sphere with centre at $O$ lying in $yz-plane.$ Take an elementary area $dS$ at the surface at $P$. Then,

$dS=r\left(r\sin \theta d\theta \right)\hat{r}=r^2\sin \theta d\theta \hat{r}$

$B.ds=\frac{{\mu }_0}{4\pi }\frac{2M\cos \theta }{r^3}\hat{r}\left(r^2\sin \theta d\theta \hat{r}\right)$

$=\frac{{\mu }_0}{4\pi }\frac{M}{r}{\int }_0^{2\pi }2\sin \theta .\cos \theta d\theta$

$=\frac{{\mu }_0}{4\pi }\frac{M}{r}{\int }_0^{2\pi }\sin 2\theta d\theta$

$=-\frac{{\mu }_0}{4\pi }\frac{M}{2r}{\left(\frac{-\cos 2\theta }{2}\right)}_0^{2\pi }$

$=-\frac{{\mu }_0}{4\pi }\frac{M}{2r}[\cos 4\pi -\cos 0]$

$=-\frac{{\mu }_0}{4\pi }\frac{M}{2r}[1-1]$

$=0$

Final Answer: $0$