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written 6.5 years ago by | • modified 5.4 years ago |

Gauss law for static electric field: Gaussian surface can be constructed for the magnetic field as shown in figure below

The gaussian surface, even those that cut through the bar magnet do not enclose net magnetic charge because every cut through the magnet gives a piece having both north and south pole. Therefore, magnetic form of gauss law is written as

$$\Phi_B=\oint\vec{B}.\vec{ds}=0$$

This means that the net flux of the magnetic field through any closed surface is zero. This is called the gauss’s law for magnetic field in integral form.

Using divergence theorem the magnetic gauss law can be written as

$$\oint_S\vec{B}.\vec{ds}=\oint_S\left( \vec{\nabla}.\vec{B}\right)dv=0$$

Or $ \vec{\nabla}.\vec{B}=0$

This is called the differential or point form of magnetic gauss law

2.Ampere’s Circuital law states that “the line integral of magnetic field intensity H around a closed path is exactly equal to the direct current enclosed by that path.” The mathematical representation of Ampere’s law is

$$\vec{H}.\vec{dl}=I\tag{1}$$

The law is very useful to determine $\vec{H}$ when the current distribution is symmetrical.

Since, $\vec{B}=\mu\vec{H}$

Equation (1) can be written as

$$ \oint\vec{B}.\vec{dl}=\mu_0I$$

This is called the integral form of Ampere’s circuital law

Replacing $I=\int_s\vec{J}.\vec{ds}$ where $\vec{j}$ is current density and $S$ is the surface area bounded by the path of integration of $\vec{H}$, we can write $$\vec{H}.\vec{dl}=\int_S\vec{J}.\vec{ds} $$

Using stokes theorem, this can be written as $$\oint_S\left( \vec{\nabla}\times\vec{H}\right).\vec{ds}=\int_S\vec{J}.\vec{ds} $$

Hence, $\vec{\nabla}\times\vec{H}=\vec{J}$

Since, $\vec{B}=\mu_0\vec{H}$ in free space, we can write

$$ \vec{\nabla}\times\vec{B}=\mu_0\vec{J}$$

This is called the differential or point form of ampere’s circuital law.