Vector Potential Preview

Physics Electricity and Magnetism • Magnetic Fields and Sources

Written by STEM Calculators Team Published January 17, 2026 Updated February 24, 2026

6. Vector Potential Preview

Introduces the vector potential $ \mathbf{A} $ such that $\mathbf{B}=\nabla\times\mathbf{A}$ . This tool computes $ \mathbf{A} $ and verifies the curl for simple, symmetric cases (wire + ideal solenoid), and visualizes what the vectors mean.

Units: meters (m), amperes (A), tesla (T). Vector potential units: $ \mathrm{T\cdot m} $. Inputs accept 1e-3, pi, sqrt(2), sin(), cos(), tan(), ln(), log(), abs(). Use * for multiplication.

Inputs

Geometry:

Wire inputs

Current $I$ (A):

Wire is along the $z$-axis; field is azimuthal $\hat{\boldsymbol{\phi}}$.

Current direction:

Reverses $\hat{\boldsymbol{\phi}}$ direction by the right-hand rule.

Reference $r_0$ (m):

Sets the “zero” of $\ln(r/r_0)$. It does not change $\mathbf{B}$.

Add constant shift to $A_z$:

Adding a constant to $A_z$ does not change $\nabla\times\mathbf{A}$.

Constant $C$ (T·m):

Used only if shift is ON.

Solenoid inputs (ideal, infinite)

Turns density $n$ (turns/m):

Ideal infinite solenoid gives $B_0=\mu_0 n I$ inside.

Current $I$ (A):

This $I$ is the current in each turn.

Solenoid radius $R$ (m):

Inside region: $r\le R$. Outside: $r>R$.

Field direction (inside):

Sets the sign of $B_0$ and the sense of $\hat{\boldsymbol{\phi}}$ for $\mathbf{A}$.

Vector potential form:

Continuous form avoids a jump at $r=R$. (Ideal model: $B$ inside, ~0 outside.)

Observation point

$x$ (m):

In the cross-section plane.

$y$ (m):

Radius is $r=\sqrt{x^2+y^2}$.

$z$ (m):

Wire/solenoid are infinite ⇒ results do not depend on $z$ (kept for consistency).

Visual controls

Arrow scale (visual): Show guides:

Pan/zoom: drag to pan • mouse wheel / trackpad / pinch to zoom • “Reset view” restores default.
Play animates the point around a circle at fixed radius to show how $\hat{\boldsymbol{\phi}}$ rotates.

Ready

Steps

Enter values and click Solve.

Vector visual

Cross-section view ($x$-$y$ plane): axis, $P$, $\mathbf{r}$, $\mathbf{A}$, and $\mathbf{B}$

What the vectors represent

Source axis (wire or solenoid axis = z) at the origin
Solenoid boundary ($r=R$) (solenoid mode only)
$ \mathbf{r} $: from axis to observation point $P$
$ \mathbf{A} $: wire mode uses $A_z$ (⊙/⊗); solenoid mode uses $A_\phi$ (tangential arrow)
$ \mathbf{B}=\nabla\times\mathbf{A} $: wire mode shows azimuthal arrow; solenoid mode shows $B_z$ as ⊙/⊗ (inside) or 0 (outside)
$P$: observation point

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Theory

Vector potential definition

The vector potential $\mathbf{A}$ is defined so that the magnetic field is its curl:

\[ \mathbf{B}=\nabla\times\mathbf{A}. \]

This is possible because $\nabla\cdot\mathbf{B}=0$ (no magnetic monopoles). The choice of $\mathbf{A}$ is not unique:

\[ \mathbf{A}\;\to\;\mathbf{A}+\nabla\chi \quad\Longrightarrow\quad \nabla\times\mathbf{A}\ \text{unchanged}. \]

This is called a gauge transformation. Adding a constant vector to $\mathbf{A}$ also does not change $\nabla\times\mathbf{A}$.

What $\hat{\boldsymbol{\phi}}$ (phi-hat) means

In cylindrical coordinates $(\rho,\phi,z)$, \[ \rho=\sqrt{x^2+y^2}, \] and the unit vectors are:

$\hat{\boldsymbol{\rho}}$: points radially outward from the $z$-axis
$\hat{\boldsymbol{\phi}}$: points tangentially (azimuthally) around the $z$-axis
$\hat{\mathbf{z}}$: points along the $z$-axis

The magnetic field around a long straight wire is tangential: $\mathbf{B}\parallel\hat{\boldsymbol{\phi}}$.

Infinite straight wire (current along $z$)

For an infinite wire on the $z$-axis carrying current $I$, symmetry gives:

\[ \mathbf{B}(\rho)=\frac{\mu_0 I}{2\pi\rho}\,\hat{\boldsymbol{\phi}}. \]

One convenient gauge choice for the vector potential uses an axial component:

\[ \mathbf{A}(\rho)=A_z(\rho)\,\hat{\mathbf{z}}, \qquad A_z(\rho)=-\frac{\mu_0 I}{2\pi}\ln\!\left(\frac{\rho}{\rho_0}\right), \]

$\rho_0$ is an arbitrary reference radius (sets the “zero” of the logarithm). Different $\rho_0$ corresponds to a different gauge choice; $\mathbf{B}$ is unchanged.

Curl check (cylindrical coordinates): if $\mathbf{A}=A_z(\rho)\hat{\mathbf{z}}$, then

\[ \mathbf{B}=\nabla\times\mathbf{A} \;=\; -\frac{dA_z}{d\rho}\,\hat{\boldsymbol{\phi}} \;=\; \frac{\mu_0 I}{2\pi\rho}\,\hat{\boldsymbol{\phi}}. \]

Ideal solenoid (uniform $\mathbf{B}$ inside)

Inside an ideal long solenoid, the field is approximately uniform and axial:

\[ \mathbf{B}\approx \mu_0 n I\,\hat{\mathbf{z}} \quad (n=\text{turns per meter}). \]

A standard choice of vector potential for a uniform $\mathbf{B}=B_0\hat{\mathbf{z}}$ is the “symmetric gauge”:

\[ \mathbf{A}(\rho)=\frac{1}{2}B_0\,\rho\,\hat{\boldsymbol{\phi}} \quad\Longrightarrow\quad \nabla\times\mathbf{A}=B_0\,\hat{\mathbf{z}}. \]

For a finite-radius solenoid, the uniform-field approximation is best for $\rho$ well inside the coil.

Coulomb gauge (concept)

A common gauge condition is the Coulomb gauge:

\[ \nabla\cdot\mathbf{A}=0. \]

The calculator’s closed-form examples focus on symmetry; the gauge condition is mentioned to explain why $\mathbf{A}$ is not unique.

Frequently Asked Questions

What does the vector potential A represent in this calculator?

It is a vector field chosen so that the magnetic field is its curl: B = curl(A). The tool uses standard symmetric choices for A in a straight-wire case (Az) and a solenoid case (Aphi) to make the curl relationship easy to verify.

Why does changing r0 or adding a constant to Az not change the magnetic field?

For the wire model, Az(r) involves ln(r/r0), and changing r0 shifts Az by a constant amount. Adding a constant shift C to Az also changes A but does not change curl(A), so B is unchanged even though A itself is different.

What is the difference between the continuous and truncated solenoid Aphi options?

The continuous form uses Aphi = (B0 r)/2 inside and Aphi = (B0 R^2)/(2 r) outside, which avoids a jump at r = R. The truncated option forces Aphi = 0 for r > R to emphasize the idealized idea of B being inside the solenoid and approximately zero outside.

Why does the result not depend on z for the observation point?

Both the wire and the ideal infinite solenoid are modeled as infinite along the z-axis, so symmetry makes A and B depend only on the radial distance from the axis, not on z. The z input is included for consistency but does not change the computed values.