Why is the magnetic field not a conservative field?

Question

1. Why is the magnetic field not a conservative field?

2. If it were conservative, which equation in the theory of electromagnetism would change/be canceled?

Answer 1

A field is non-conservative if its field lines form loops. This is because if a field line forms a loop a charge can travel along that field line and back to its original position. The problem is if we have some field $f$ and charge $q$ the field exerts a force:

$$ \mathbf{F} = \mathbf{f} q $$

so in travelling round the loop the force does some non-zero work on the charge:

$$ W = \int \mathbf{F} \cdot d\ell $$

This means when the charge has got back to its original position work has been done on it so its potential energy must have changed, and that means its PE at its starting point cannot have a unique value.

We can write the condition for a field to be non-conservative as:

$$ \nabla \times \mathbf{f} \ne 0 $$

that is, the curl of the field must be non-zero.

The reason why magnetic fields are always non-conservative is that field lines can only start or end on a charge. However magnetic charges (magnetic monopoles) do not exist so a magnetic field line cannot have a start or an end so it must always form a loop.

We should note that electric fields can be non-conservative as well. In fact electric generators create energy by creating non-conservative electric fields. However while electric can be loops they don't have to be. Electric charges do exist so an electric field line can start on a positive charge and end on a negative charge and not form a loop.

To answer your question (2): the magnetic field can be conservative only if magnetic charges exist, and if so this modifies the Maxwell's equations for the divergence of the magnetic field and the curl of the electric field. The changes needed are described in the answers to Reformulation of Maxwell's equations in the event of finding a magnetic monopole.

Answer 2

Conservative fields have a scalar potential i.e. if we denote the field by $\vec{F}$, and it's conservative, it can be written as $$\vec{F} = -\nabla \Phi(\vec{x}),$$ where $\Phi$ is a scalar function of coordinates. Magnetic field $\vec{B}$ (in contrast to Newtonian gravity or the $E$-field in electrostatics) doesn't have this property. It has what we call a vector potential $\vec{A}$ that satisfies

$$\vec{B} = \nabla \times \vec{A},$$

but not a scalar potential. Another way of looking at conservative fields is the integral form. Using Stokes' theorem we can get that for a conservative field $\vec{F} = -\nabla \Phi$ it holds that $$\oint \vec{F} \cdot d\vec{\ell} = 0.$$ If we consider $\vec{F}$ being a force, this basically means that if you go around in a closed loop, and end up at the same point, you don't do any mechanical work (as $dW = \vec{F} \cdot d\vec{\ell}$).

On condition that the magnetic field was conservative, it would be in direct contradiction to Ampère's circuital law which says that $$\oint_{\partial\Sigma} \vec{B} \cdot d\vec{\ell} = \mu_0I + \frac{1}{c^2}\int_\Sigma \frac{\partial \vec{E}}{\partial t} \cdot d\vec{S} \neq 0.$$