Magnetic vector potential

The magnetic vector potential $(\vec{A})$ is a vector field that serves as the potential for the magnetic field. The curl of the magnetic vector potential is the magnetic field.

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

The magnetic vector potential is preferred when working with the Lagrangian in classical mechanics and quantum mechanics.

The magnetic vector potential contributed by a length $d\vec{s}$ with current $I$ running through it is

$$d\vec{A} = \frac{\mu_0 I}{4\pi r} d\vec{s}.$$

What is the magnetic vector potential a distance $R$ from a long straight current element?

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Assume the wire is placed on the z-axis. Hence the distance from a differential element $d\vec{z}$ to a point in space is

$$r = \sqrt{ R^2 + z^2}.$$

Hence,

$$d\vec{A} = \frac{\mu_0 I}{4 \pi} \frac{dz}{\sqrt{ R^2 + z^2} } \hat{z}.$$

Now it is necessary to integrate over the length of the rod. Since the rod is arbitrarily said to be on the z-axis, it can be said, for simplicity, to extend from $z=0$ to $z=L.$ This simply requires multiplying by 2. Also, use the substitution $z = R \tan(\theta).$

$$\begin{aligned} \vec{A} &= 2 \int_0^L \frac{\mu_0 I}{4 \pi} \frac{dz}{\sqrt{ R^2 + z^2} } \hat{z} \\ &= \frac{\mu_0 I}{2 \pi} \hat{z} \int_0^\phi \sec(\theta) d\theta \\ &= \frac{\mu_0 I}{2 \pi} \ln \bigg( \frac{\sqrt{L^2 + R^2} + L }{R} \bigg) \hat{z} \end{aligned}$$

The magnetic field is the curl of the vector potential.

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

Find the magnetic field in a region with magnetic vector potential $\vec{A} = \sin(\theta)\hat{r} - r\hat{\theta}.$

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Since $\vec{A}$ is in spherical coordinates, use the spherical definition of the curl.

$$\begin{aligned} \vec{B} = \frac{1}{r\sin\theta} \left( \frac{\partial}{\partial \theta} \left(A_\varphi\sin\theta \right) - \frac{\partial A_\theta}{\partial \varphi} \right) &\hat{\mathbf r} \\ {}+ \frac{1}{r} \left( \frac{1}{\sin\theta} \frac{\partial A_r}{\partial \varphi} - \frac{\partial}{\partial r} \left( r A_\varphi \right) \right) &\hat{\boldsymbol \theta} \\ {}+ \frac{1}{r} \left( \frac{\partial}{\partial r} \left( r A_{\theta} \right) - \frac{\partial A_r}{\partial \theta} \right) &\hat{\boldsymbol \varphi} \end{aligned}$$

The only part that will survive for the given $\vec{A}$ is

$$\begin{aligned} \vec{B} &= \frac{1}{r} \big( \frac{\partial}{\partial r} (r A_\theta) - \frac{\partial A_r}{\partial \theta} \big) \hat{\varphi} \\ &= \frac{1}{r} \big( 2r - \cos(\theta) \big) \\ &= 2 - \frac{\cos(\theta)}{r} \end{aligned}$$

The partial derivative of the magnetic vector potential contributes partially to the induced electric field according to faraday's law.

$$\vec{E} = - \frac{\partial \vec{A} }{ \partial t}$$

Recall Faraday's law:

$$\nabla \times \vec{E} = - \frac{\partial \vec{B} }{\partial t}$$

$$\nabla \times \vec{E} = - \frac{\partial \big( \nabla \times \vec{A} \big) }{\partial t}$$

$$\nabla \times \vec{E} = - \nabla \times \frac{ \partial \vec{A} }{\partial t}$$

$$\rightarrow \vec{E} = - \frac{ \partial \vec{A} }{\partial t}$$