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Is it possible to introduce magnetic monopoles without breaking $∇ · B = 0$?

It is not possible. To understand why, you need to visualise what vector equations really mean. $\Delta B = 0: Take any small volume. 'Number of field lines' entering this volume will be equal to '...
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6 votes
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Is it possible to introduce magnetic monopoles without breaking $∇ · B = 0$?

No, it is not possible. If you have a magnetic monopole then, by definition, there is some closed Gaussian surface $S$ over which $$\oint_S \vec B \cdot d\vec S \ne 0$$ which by implies that $$\int_V \...
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2 votes

Is it possible to introduce magnetic monopoles without breaking $∇ · B = 0$?

"Net magnetic charge is zero everywhere" "Magnetic current is non zero" Can be done using $$\rho_{m} = -\vec{P} \cdot \nabla \delta^3(r)$$ Where $$\frac{d\vec{P}}{dt} ≠ 0$$ Here, $\...
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-1 votes

Maximum $B$-field at the surface of a permanent magnet

If you stay at the surface of the magnet, your $r=R$ the radius of the magnet. If you want to maximize $B$, you need to be careful, because you have a second term that isn’t aligned so you could have ...
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11 votes

Why are magnetic monopoles "incompatible" with quantum mechanics?

For your first question, on the link between quantum mechanics and potentials is that currently, there is no formulation of QM without these potentials and only in terms of the gauge invariant EM ...
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32 votes
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Why are magnetic monopoles "incompatible" with quantum mechanics?

The argument here is supposed to go something like this: Classical electrodynamics, as long as you don't insist on a Lagrangian formulation, is fully described by Maxwell's equations in terms of the ...
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8 votes

Why are magnetic monopoles "incompatible" with quantum mechanics?

You are correct. This really has nothing to do with quantum mechanics. The existence of magnetic monopoles would require a change in Maxwell's equations (i.e., the divergence of the magnetic field ...
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3 votes

Why are magnetic monopoles "incompatible" with quantum mechanics?

Magnetic monopoles are incompatible with $\nabla \cdot B=0$ hence with $B = \nabla \times A$. They are incompatible with the Maxwell equations and with the electromagnetic potential. The ...
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