📂Electrodynamics

Bound Current Density and the Vector Magnetic Field Created by a Magnetized Object

Explanation¹

Suppose there is an object magnetized by an external magnetic field. This object will have a magnetization density $\mathbf{M}$, and this magnetization density will generate a new magnetic field. The vector potential created by a magnetic dipole is as follows.

$$\mathbf{A} (\mathbf{r}) = \dfrac{\mu_{0}}{4\pi}\dfrac{\mathbf{m} \times \crH }{\cR ^2}$$

Since magnetization density is the dipole moment per unit volume, $\mathbf{M}=\dfrac{\mathbf{m}}{d\tau}$. By substituting this into the above equation and integrating over the entire volume, the vector potential created by the magnetized object is as follows.

$$\mathbf{A} (\mathbf{r}) = \dfrac{\mu_{0}}{4\pi} \int\dfrac{ \mathbf{M}(\mathbf{r}^{\prime}) \times \crH }{\cR ^2} d \tau^{\prime}$$

Given that the gradient of the separation vector is $\nabla^{\prime} \dfrac{1}{\cR} = \dfrac{\crH}{\cR^2}$,

$$\mathbf{A} (\mathbf{r}) = \dfrac{\mu_{0}}{4\pi} \int \mathbf{M}(\mathbf{r}^{\prime}) \times \nabla^{\prime}\dfrac{1}{\cR} d \tau^{\prime}$$

The product rule involving the del operator

$$\nabla \times (f \mathbf{A})=(\nabla f)\times \mathbf{A} + f(\nabla \times A) \implies \mathbf{A} \times (\nabla f) = f(\nabla \times \mathbf{A}) -\nabla \times ( f\mathbf{A})$$

Applying the product rule to the cross product within the integral gives the following.

$$\mathbf{A} (\mathbf{r}) = \dfrac{\mu_{0}}{4\pi} \left[ \int \frac{1}{\cR}\left[\nabla^{\prime} \times \mathbf{M} (\mathbf{r}^{\prime}) \right] d \tau^{\prime} - \int \nabla^{\prime} \times \dfrac{\mathbf{M}(\mathbf{r^{\prime}})}{\cR}d \tau^{\prime}\right]$$

Here, Gauss’s theorem is used to swap the second term.

Gauss’s theorem (Divergence theorem)

$$\int_\mathcal{V} \nabla \cdot \mathbf{ F} dV = \oint _\mathcal{S} \mathbf{F} \cdot d \mathbf{S}$$

Then the equation becomes as below.

$$\mathbf{A} (\mathbf{r}) = \dfrac{\mu_{0}}{4\pi} \left[ \int \frac{1}{\cR} \left[\nabla^{\prime} \times \mathbf{M} (\mathbf{r}^{\prime}) \right] d \tau^{\prime} + \oint \dfrac{\mathbf{M}(\mathbf{r^{\prime}})}{\cR} \times d \mathbf{a}^{\prime}\right]$$

Here, the first term can be considered as the potential created by volume current density $\mathbf{J}_{b}$. The subscript $b$ stands for bound.

$$\mathbf{J}_{b}=\nabla \times \mathbf{M}$$

The second term can be seen as the potential created by surface current density $\mathbf{K}_{b}$.

$$\mathbf{K}_{b}=\mathbf{M} \times \hat{\mathbf{n}}$$

Here, $\hat{\mathbf{n}}$ is the unit normal vector perpendicular to each surface. Now, representing the vector potential in terms of bound current density, we get the following.

$$\mathbf{A} (\mathbf{r})=\dfrac{\mu_{0}}{4\pi} \int_\mathcal{V} \dfrac{\mathbf{J}_{b}(\mathbf{r}^{\prime})}{\cR}d\tau^{\prime}+\dfrac{\mu_{0}}{4\pi}\oint_\mathcal{S}\dfrac{\mathbf{K}_{b} (\mathbf{r}^{\prime})}{\cR}da^{\prime}$$

Therefore, the vector potential created by the magnetized object is the same as the potential created by the object’s internal volume current density $\mathbf{J}_{b}=\nabla \times \mathbf{M}$ and the surface current density $\mathbf{K}_{b}=\mathbf{M} \times \hat{\mathbf{n}}$ on its surface. This is analogous to defining the bound charges $\rho_{b}$ and $\sigma_{b}$ and the potential created by a polarized object.