Question
Explain the Gauss’ law for magnetic fields.

Answer

i. Analogous to the Gauss' law for electric field, the Gauss' law for magnetism states that, the net magnetic flux $\left(\emptyset_{ B }\right)$ through a closed Gaussian surface is zero. $\emptyset_{ B }=\int \vec{B} \cdot \overrightarrow{d S}=0$
ii. Consider a bar magnet, a current carrying solenoid and an electric dipole. The magnetic field lines of these three are as shown in figures.
Image

iii. The areas (P) and (Q) are the cross – sections of three dimensional closed Gaussian surfaces. The Gaussian surface (P) does not include poles while the Gaussian surface (Q) includes N-pole of bar magnet, solenoid and the positive charge in case of electric dipole.

iv. The number of lines of force entering the surface (P) is equal to the number of lines of force leaving the surface. This can be observed in all the three cases.

v. However, Gaussian surface (Q) of bar magnet, enclose north pole. As, even thin slice of a bar magnet will have both north and south poles associated with it, the number of lines of Force entering surface (Q) are equal to the number of lines of force leaving the surface.

vi. For an electric dipole, the field lines begin from positive charge and end on negative charge. For a closed surface (Q), there is a net outward flux since it does include a net (positive) charge.
vii. Thus, according to the Gauss' law of electrostatics $\emptyset_{ E }=\int \vec{E} \cdot \overrightarrow{d S}=\frac{q}{\varepsilon_0}$, where $q$ is the positive charge enclosed.
viii. The situation is entirely different from magnetic lines of force. Gauss' law of magnetism can be written as $\varnothing_{ B }=\int \vec{B} \cdot \overrightarrow{d S}=0$
From this, one can conclude that for electrostatics, an isolated electric charge exists but an isolated magnetic pole does not exist.

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