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## Main Question or Discussion Point

While I understand the mathematical definition of the magnetic vector potential field

For the (scalar) electric potential the matter seems rather simple. The dimensions of φ are energy per unit charge, and one can use it to easily find the amount of energy it takes to move a charged particle from one electric potential to another by subtracting the difference in the two potentials and multplying that by the charge.

This kind of approach does not work with magnetic potential, however. The dimensions of

To take a simple case, let

[tex]

\bf {A}=(0, 0, y)

[/tex]

then

[tex]

\bf {B} = \nabla \times \bf {A}

=(1,0,0).

[/tex]

A charged particle could move in a circular path around the x axis with constant speed v as long as r = mv/qB, which in this case is mv/q. The difference in its momentum as it moves between (0,0,r) and (0,0,-r) would be (0,-2mv,0), but the

So my question is, what is the

**A**( [tex] \bf {B} = \nabla \times \bf {A} [/tex] ), I don't have an intuitive grasp of its physical meaning.For the (scalar) electric potential the matter seems rather simple. The dimensions of φ are energy per unit charge, and one can use it to easily find the amount of energy it takes to move a charged particle from one electric potential to another by subtracting the difference in the two potentials and multplying that by the charge.

This kind of approach does not work with magnetic potential, however. The dimensions of

**A**are momentum per unit charge, but the change in a charged particle's momentum as it moves from one magnetic potential to another is*not*equal to its charge multiplied by the difference in the two vector potentials.To take a simple case, let

[tex]

\bf {A}=(0, 0, y)

[/tex]

then

[tex]

\bf {B} = \nabla \times \bf {A}

=(1,0,0).

[/tex]

A charged particle could move in a circular path around the x axis with constant speed v as long as r = mv/qB, which in this case is mv/q. The difference in its momentum as it moves between (0,0,r) and (0,0,-r) would be (0,-2mv,0), but the

**A**vector is (0,0,0) at both locations - it doesn't change at all.So my question is, what is the

*physical*meaning of the**A**field? What does it represent?
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