Deep Dive: Derivation & Example

This page details the mathematical transformation from the dipole potential integral to the concept of Bound Currents.

1. Deriving Bound Currents

To understand where the integral comes from, we must first visualize the coordinate system. We are summing up the contributions of infinitesimal dipoles $\bfM d\tau'$ located at source points $\vecrp$.

Step 1: The Potential Integral

The vector potential $\bfA$ at observation point $\vecr$ is the sum (integral) of the potentials produced by dipoles in the volume:

$$ \bfA(\vecr) = \frac{\mu_0}{4\pi} \int \frac{\bfM(\vecrp) \times \rhat}{r^2} \dtau $$

$\vecr$ : Observation point.

$\vecrp$ : Source point inside the material.

$r$ : Separation distance $|\vecr - \vecrp|$.

$\rhat$ : Unit vector pointing from $\vecrp$ to $\vecr$.

Step 2: Vector Calculus Identity

We use the identity $\nabla' (1/r) = \rhat/r^2$ to rewrite the integrand:

$$ \frac{\bfM \times \rhat}{r^2} = \bfM \times \left( \nabla' \frac{1}{r} \right) $$

Using the product rule $\nabla' \times (f\mathbf{v}) = f(\nabla' \times \mathbf{v}) - \mathbf{v} \times \nabla' f$, we expand:

$$ \bfM \times \nabla' \left(\frac{1}{r}\right) = \frac{1}{r} (\nabla' \times \bfM) - \nabla' \times \left( \frac{\bfM}{r} \right) $$

Step 3: Resulting Bound Currents

Substituting this back into the integral, we use the Divergence Theorem on the second term to convert it to a surface integral:

$$ \bfA(\vecr) = \frac{\mu_0}{4\pi} \int_V \frac{\nabla' \times \bfM}{r} \dtau + \frac{\mu_0}{4\pi} \oint_S \frac{\bfM \times \nhat}{r} da' $$

Comparing this to the standard potentials for currents, we identify the terms:

\bfJ_b = \nabla \times \bfM $$ $$ \bfK_b = \bfM \times \nhat

2. Example: Uniformly Magnetized Sphere

💡 The Strategy

Instead of trying to evaluate the complex integral for $\bfA(\vecr)$ directly, we use the result derived above.

Calculate the bound currents $\bfJ_b$ and $\bfK_b$.
Treat these currents as real physical sources.
Calculate $\bfB$ using standard Ampere's Law or Biot-Savart methods.

Problem: Find the magnetic field of a sphere of radius $R$ with uniform magnetization $\bfM = M \zhat$.

Step 1: Calculate Bound Currents

Choosing the $z$ axis along the direction of $\bfM$ ($\bfM = M\zhat$):

$$ \bfJ_b = \nabla \times \bfM = 0 $$

Since $\bfM$ is constant. Now for the surface current $\bfK_b$, with $\nhat = \rhat$:

$$ \bfK_b = \bfM \times \nhat = (M\zhat) \times \rhat = M \sin\theta \phihat $$

Step 2: Analogy to Rotating Shell

Recall that a rotating spherical shell of uniform surface charge $\sigma$ has a surface current density:

$$ \bfK = \sigma \mathbf{v} = \sigma \omega R \sin\theta \phihat $$

Comparing the two expressions, we see they are identical if we make the identification $\sigma R \omega \rightarrow M$.

Step 3: The Result

Therefore, the field of a uniformly magnetized sphere is identical to that of a spinning spherical shell. Concluding from that known problem:

\text{Inside } (r < R): \quad \bfB = \frac{2}{3} \mu_0 \bfM $$ $$ \text{Outside } (r > R): \quad \bfB \text{ is a perfect dipole field with } \mathbf{m} = \frac{4}{3}\pi R^3 \bfM

Notice that the internal field is uniform, like the electric field inside a uniformly polarized sphere ($\bf{E} = -\frac{1}{3\epsilon_0}\bfP$), although the coefficients ($2/3$ vs $-1/3$) are different.

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