Question

Answer the following question in detail.

Explain Gauss's law for magnetic fields.

Answer 1

1. Analogous to the Gauss’ law for the electric field, the Gauss’ law for magnetism states that the net magnetic flux (Φ_B) through a closed Gaussian surface is zero. Φ_B = `int vec"B".vec"dS"=0`
2. Consider a bar magnet, a current-carrying solenoid, and an electric dipole. The magnetic field lines of these three areas shown in figures.
   
            Bar magnet
   
   Current (I) carrying solenoid
   
        Electric dipole
3. 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 the N-pole of a bar magnet, solenoid, and the positive charge in case of an electric dipole.
4. 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 three cases.
5. However, the Gaussian surface (Q) of the bar magnet, enclose the north pole. As even a thin slice of a bar magnet will have both north and south poles associated with it, the number of lines of Force entering the surface (Q) is equal to the number of lines of force leaving the surface.
6. For an electric dipole, the field lines begin from a positive charge and end on a negative charge. For a closed surface (Q), there is a net outward flux since it does include a net (positive) charge.
7. Thus, according to Gauss’ law of electrostatics
   Φ_E = `int vec"E".vec"dS"="q"/ε_0`, where q is the positive charge enclosed.
8. The situation is entirely different from the magnetic lines of force. Gauss’ law of magnetism can be written as Φ_B = `int vec"B".vec"dS"=0`.
   From this, one can conclude that for electrostatics, an isolated electric charge exists but an isolated magnetic pole does not exist.