Deduce Maxwell's second equation in differential and integral form. (asked in MU Dec 2025)

Answer: \(\nabla\cdot\vec B=0\) (differential) and \(\oint_S\vec B\cdot d\vec S=0\) (integral): the net magnetic flux through any closed surface is zero.

Basis – Gauss's law for magnetism. Magnetic field lines are always closed loops: they come out of the N-pole and return to the S-pole, and continue inside the magnet. Isolated magnetic poles (monopoles) do not exist. So for any closed surface, every field line that enters it also leaves it.

Integral form. The total magnetic flux leaving a closed surface \(S\) is therefore zero:

\[\oint_S\vec B\cdot d\vec S=0\qquad\ldots(1)\]

Differential form. By Gauss's divergence theorem, for the volume \(V\) enclosed by \(S\):

\[\oint_S\vec B\cdot d\vec S=\int_V(\nabla\cdot\vec B)\,dV\]

From (1), \(\int_V(\nabla\cdot\vec B)\,dV=0\). Since this holds for every volume, however small, the integrand must vanish:

\[\boxed{\nabla\cdot\vec B=0}\]

Significance: the magnetic field is solenoidal (divergence-free); there is no "magnetic charge" analogous to \(\rho\) in \(\nabla\cdot\vec D=\rho\). It also allows \(\vec B\) to be written as the curl of a vector potential, \(\vec B=\nabla\times\vec A\), since \(\nabla\cdot(\nabla\times\vec A)=0\).