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I'm confused about two things : the illustration and the "vector" in magnetic field . I didn't understand why the potential created by magnetic field got vector . I know about Helmholtz theorem , but it's unclear for me since we know that potential is scalar unit but here it's vector . Also I don't get why there is no illustration of magnetic potential in real life except Aharonov-Bohm effect which I think it's too complex !

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    $\begingroup$ To what illustration are you referring? $\endgroup$
    – J. Murray
    Commented Jun 5, 2020 at 16:48
  • $\begingroup$ @J.murray I meant Something that represent A in real life $\endgroup$ Commented Jun 7, 2020 at 4:39

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In electrostatics, the electric field is curl-free: $\nabla\times\mathbf{E}=0$. Any curl-free field can be described as the gradient of a scalar function. This scalar is related to the electric potential (namely, $\mathbf{E}=-\nabla\phi$).

But the magnetic field is not curl-free: $\nabla\times \mathbf{B}=\mu_0\mathbf{j}$ for current density $\mathbf{j}$, so we cannot define a scalar potential in the same way. However, the magnetic field is divergence-free: $\nabla\cdot\mathbf{B}=0$. This makes the magnetic field a solenoidal field, which means it can be described as the curl of some other vector field, which we call the vector potential: $\nabla\times \mathbf{A}=\mathbf{B}$.

So we have a vector potential for two reasons:

  • A scalar potential cannot be used to describe a vector field with nonzero curl, and
  • Gauss's Law for Magnetism, which says that the magnetic field is divergence-free.

The effect of the vector potential is extremely evident in real life! In particular, whenever you're not dealing only with electrostatics, the vector potential gives you a very real and visible contribution to the electric field:

$$\mathbf{E}=-\nabla\phi-\frac{\partial \mathbf{A}}{\partial t}$$

So whenever you have moving charges, changing currents, or changing fields, the electric field depends explicitly on how the vector potential is changing. You can see this in many situations: eddy currents, electric motors, generators, etc.

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  • $\begingroup$ But you defined magnetic potential as vector unit and it's not a vector unit ! At least we should name A something else not potential . $\endgroup$ Commented Jun 7, 2020 at 4:36
  • $\begingroup$ @AlirezaOe In physics, we use the word "potential" to refer to a quantity which you differentiate to obtain some other physical quantity. If it's potential energy, then you differentiate it to get force. The scalar and vector potentials give you the electric and magnetic fields. In thermodynamics, we differentiate thermodynamic potentials to obtain pressure, temperature, etc and the list goes on. $\endgroup$
    – J. Murray
    Commented Jun 7, 2020 at 4:41
  • $\begingroup$ @AlirezaOe What do you mean by "it's not a vector unit"? The same units can apply to both vector and scalar quantities: speed (a scalar) and velocity (a vector) both have units of m/s, for example. $\endgroup$ Commented Jun 7, 2020 at 18:53
  • $\begingroup$ @AlirezaOe Anyway, the units of the magnetic vector potential are perfectly sensible: just like the electric potential is the energy per unit charge, the magnetic vector potential is the momentum per unit charge. $\endgroup$ Commented Jun 7, 2020 at 18:56
  • $\begingroup$ @J.Murray So it's not write to say "potential" is scalar unit . Actually for example "potential energy" is scalar for example , right ? $\endgroup$ Commented Jun 8, 2020 at 19:17
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Because one important feature of the magnetic field (and one of the Maxwell laws) is: $\nabla.\mathbf B = 0$

And it can be easily shown that for any vector $V$:
$\nabla.(\nabla \times \mathbf V) = 0$

So, it is possible to define a vector potential so that its curl is $\mathbf B$.

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  • $\begingroup$ But potential is scalar and you defined it as vector now ! We should name A something else . $\endgroup$ Commented Jun 7, 2020 at 4:37

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