Magnetostatics

Step 1: Set up geometry

Place the loop in the $xy$-plane centered at the origin. A current element at angle $\phi'$ is $d\boldsymbol{\ell}' = R\,d\phi'(-\sin\phi'\,\hat{x}+\cos\phi'\,\hat{y})$. The field point is on the axis: $\mathbf{r} = z\hat{z}$.

Step 2: Compute the separation vector

The vector from source to field point is $\mathbf{r}-\mathbf{r}' = -R\cos\phi'\,\hat{x}-R\sin\phi'\,\hat{y}+z\hat{z}$ with magnitude $|\mathbf{r}-\mathbf{r}'| = \sqrt{R^2+z^2}$.

Step 3: Evaluate the cross product

Computing $d\boldsymbol{\ell}'\times(\mathbf{r}-\mathbf{r}')$, the $x$ and $y$ components integrate to zero by symmetry. The $z$-component gives $R^2 d\phi'$.

Step 4: Integrate over the loop

$$B_z = \frac{\mu_0 I}{4\pi}\int_0^{2\pi}\frac{R^2\,d\phi'}{(R^2+z^2)^{3/2}} = \frac{\mu_0 I R^2}{2(R^2+z^2)^{3/2}}$$

Step 5: Limiting cases

At the center ($z=0$): $B_z = \mu_0 I/(2R)$. Far from the loop ($z \gg R$): $B_z \approx \mu_0 m/(2\pi z^3)$ where $m = I\pi R^2$ is the magnetic dipole moment, confirming the dipole approximation.

Step 1: Write B in terms of a vector identity

The Biot-Savart integrand involves $\mathbf{J}\times\hat{\mathscr{r}}/\mathscr{r}^2$. Using $\hat{\mathscr{r}}/\mathscr{r}^2 = -\nabla(1/\mathscr{r})$:

$$\mathbf{B} = \frac{\mu_0}{4\pi}\int \mathbf{J}(\mathbf{r}')\times\nabla\!\left(\frac{1}{|\mathbf{r}-\mathbf{r}'|}\right)d^3r' = \frac{\mu_0}{4\pi}\nabla\times\int\frac{\mathbf{J}(\mathbf{r}')}{|\mathbf{r}-\mathbf{r}'|}\,d^3r'$$

Step 2: Take the curl of B

Apply $\nabla\times(\nabla\times\mathbf{A}) = \nabla(\nabla\cdot\mathbf{A})-\nabla^2\mathbf{A}$ to the integral expression. The Laplacian of $1/|\mathbf{r}-\mathbf{r}'|$ gives $-4\pi\delta^3(\mathbf{r}-\mathbf{r}')$.

Step 3: The divergence term vanishes for steady currents

The $\nabla(\nabla\cdot\mathbf{A})$ term involves $\nabla'\cdot\mathbf{J}$, which vanishes by the continuity equation for steady currents. Therefore:

$$\nabla\times\mathbf{B} = \frac{\mu_0}{4\pi}\int \mathbf{J}(\mathbf{r}')\,4\pi\delta^3(\mathbf{r}-\mathbf{r}')\,d^3r'$$

Step 4: Evaluate using the delta function

$$\boxed{\nabla\times\mathbf{B} = \mu_0\mathbf{J}}$$

This is Ampere's law in differential form.

Step 5: Integral form via Stokes' theorem

Integrating over a surface $\mathcal{S}$ bounded by loop $\mathcal{C}$ and applying Stokes' theorem:

$$\oint_\mathcal{C}\mathbf{B}\cdot d\boldsymbol{\ell} = \int_\mathcal{S}(\nabla\times\mathbf{B})\cdot d\mathbf{a} = \mu_0\int_\mathcal{S}\mathbf{J}\cdot d\mathbf{a} = \mu_0 I_{\text{enc}}$$

Step 1: Substitute B = curl A into Ampere's law

$$\nabla\times(\nabla\times\mathbf{A}) = \mu_0\mathbf{J}$$

Step 2: Apply the vector identity

$$\nabla(\nabla\cdot\mathbf{A}) - \nabla^2\mathbf{A} = \mu_0\mathbf{J}$$

Step 3: Choose the Coulomb gauge

Set $\nabla\cdot\mathbf{A} = 0$. This is always achievable: if $\nabla\cdot\mathbf{A}' \neq 0$, perform a gauge transformation with $\nabla^2\Lambda = -\nabla\cdot\mathbf{A}'$.

Step 4: Obtain the vector Poisson equation

$$\boxed{\nabla^2\mathbf{A} = -\mu_0\mathbf{J}}$$

Each Cartesian component satisfies a scalar Poisson equation.

Step 5: Solution by analogy with electrostatics

By analogy with $\nabla^2\Phi = -\rho/\epsilon_0$ having solution $\Phi = (1/4\pi\epsilon_0)\int\rho/|\mathbf{r}-\mathbf{r}'|\,d^3r'$, we immediately write the solution above.

Step 1: Start from the general definition

For a filamentary loop, $\mathbf{m} = \frac{1}{2}\oint \mathbf{r}'\times(I\,d\boldsymbol{\ell}') = \frac{I}{2}\oint\mathbf{r}'\times d\boldsymbol{\ell}'$.

Step 2: Apply the vector identity for loop integrals

The integral $\frac{1}{2}\oint\mathbf{r}'\times d\boldsymbol{\ell}'$ equals the vector area $\mathbf{a}$ of the loop (by Stokes' theorem applied to the identity function).

Step 3: For a planar circular loop

With $\mathbf{r}' = R(\cos\phi'\hat{x}+\sin\phi'\hat{y})$ and $d\boldsymbol{\ell}' = R(-\sin\phi'\hat{x}+\cos\phi'\hat{y})d\phi'$:

$$\mathbf{r}'\times d\boldsymbol{\ell}' = R^2 d\phi'\,\hat{z}$$

Step 4: Integrate

$$\mathbf{m} = \frac{I}{2}\int_0^{2\pi}R^2\,d\phi'\,\hat{z} = I\pi R^2\hat{z} = IA\hat{z}$$

Step 5: General result

$$\boxed{\mathbf{m} = NIA\hat{n}}$$

For $N$ turns, the direction $\hat{n}$ follows the right-hand rule. This result holds for any planar loop shape, not just circular.

Step 1: Normal component from Gauss's law for B

Apply $\nabla\cdot\mathbf{B}=0$ to a thin pillbox of area $A$ and vanishing thickness straddling the interface:

$$\oint\mathbf{B}\cdot d\mathbf{a} = B_\perp^{\text{above}}A - B_\perp^{\text{below}}A = 0$$

Step 2: Normal component is continuous

$$B_\perp^{\text{above}} = B_\perp^{\text{below}}$$

Step 3: Tangential component from Ampere's law

Apply $\oint\mathbf{B}\cdot d\boldsymbol{\ell} = \mu_0 I_{\text{enc}}$ to a narrow rectangular loop of width $\ell$ perpendicular to $\mathbf{K}$:

$$B_\parallel^{\text{above}}\ell - B_\parallel^{\text{below}}\ell = \mu_0 K \ell$$

Step 4: Tangential discontinuity

$$B_\parallel^{\text{above}} - B_\parallel^{\text{below}} = \mu_0 K$$

Step 5: Compact vector form

In full vector notation: $$\boxed{\hat{n}\times(\mathbf{B}^{\text{above}} - \mathbf{B}^{\text{below}}) = \mu_0\mathbf{K}}$$

This is the magnetic analog of the surface charge boundary condition in electrostatics. The vector potential $\mathbf{A}$ itself is continuous across any interface.