Biot-Savart Law & Ampere's Law

Step 1: Write B in terms of J using Biot-Savart

$$\mathbf{B}(\mathbf{r}) = \frac{\mu_0}{4\pi}\int \frac{\mathbf{J}(\mathbf{r}') \times \hat{\mathscr{r}}}{\mathscr{r}^2}\,d\tau' = \frac{\mu_0}{4\pi}\int \mathbf{J}(\mathbf{r}') \times \nabla\!\left(\frac{-1}{\mathscr{r}}\right)d\tau'$$

where we used the identity $\hat{\mathscr{r}}/\mathscr{r}^2 = -\nabla(1/\mathscr{r})$.

Step 2: Take the curl of B

Using the vector identity $\nabla \times (\mathbf{J} \times \nabla f) = \mathbf{J}(\nabla^2 f) - (\mathbf{J}\cdot\nabla)\nabla f$ (where $\mathbf{J}$ depends on $\mathbf{r}'$, not $\mathbf{r}$):

$$\nabla \times \mathbf{B} = \frac{\mu_0}{4\pi}\int\left[\mathbf{J}(\mathbf{r}')\,\nabla^2\!\left(\frac{1}{\mathscr{r}}\right) - (\mathbf{J}(\mathbf{r}')\cdot\nabla)\nabla\!\left(\frac{1}{\mathscr{r}}\right)\right]d\tau'$$

Step 3: Apply the Laplacian identity

The key identity is $\nabla^2(1/\mathscr{r}) = -4\pi\delta^3(\boldsymbol{\mathscr{r}})$. The first term becomes:

$$\frac{\mu_0}{4\pi}\int \mathbf{J}(\mathbf{r}')\bigl[-4\pi\delta^3(\mathbf{r}-\mathbf{r}')\bigr]\,d\tau' = -\mu_0\,\mathbf{J}(\mathbf{r})\cdot(-1) = \mu_0\,\mathbf{J}(\mathbf{r})$$

Step 4: Show the second integral vanishes for steady currents

The second term can be rewritten using integration by parts. For magnetostatics, $\nabla' \cdot \mathbf{J} = 0$ (continuity equation with $\partial\rho/\partial t = 0$). This ensures the second integral vanishes:

$$\int (\mathbf{J}\cdot\nabla)\nabla\!\left(\frac{1}{\mathscr{r}}\right)d\tau' = 0 \quad \text{(for steady currents)}$$

Step 5: Obtain the differential form of Ampere's law

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

Step 6: Convert to integral form using Stokes' theorem

Integrate both sides over a surface $S$ bounded by closed loop $C$ and apply Stokes' theorem to the left side:

$$\int_S (\nabla \times \mathbf{B})\cdot d\mathbf{a} = \oint_C \mathbf{B}\cdot d\boldsymbol{\ell} = \mu_0 \int_S \mathbf{J}\cdot d\mathbf{a} = \mu_0 I_{\text{enc}}$$

Step 7: Final integral form

$$\boxed{\oint_C \mathbf{B}\cdot d\boldsymbol{\ell} = \mu_0 I_{\text{enc}}}$$

where $I_{\text{enc}}$ is the total current passing through the surface bounded by the Amperian loop $C$.

Step 1: Symmetry arguments

By the translational symmetry along the axis and rotational symmetry about it, $\mathbf{B}$ can only depend on the distance $s$ from the axis and can only point in the $\hat{z}$ direction (along the axis). Using the right-hand rule: $\mathbf{B} = B(s)\,\hat{z}$.

Step 2: Show B = 0 outside the solenoid

Choose a rectangular Amperian loop entirely outside the solenoid, with two sides parallel to the axis (at distances $s_1 > s_2 > R$). The perpendicular sides contribute nothing (B is along $\hat{z}$). No current is enclosed, so:

$$B(s_1)L - B(s_2)L = \mu_0 \cdot 0 \implies B(s_1) = B(s_2)$$

Since $B \to 0$ as $s \to \infty$, we conclude $B = 0$ everywhere outside.

Step 3: Apply Ampere's law to a loop straddling the solenoid wall

Choose a rectangular Amperian loop with one side (length $L$) inside the solenoid and one side outside. The two short sides are perpendicular to $\hat{z}$ and contribute nothing:

$$\oint \mathbf{B}\cdot d\boldsymbol{\ell} = B_{\text{inside}}\cdot L + 0 \cdot L + 0 + 0 = B_{\text{inside}} \cdot L$$

Step 4: Count the enclosed current

The loop of length $L$ encloses $nL$ turns, each carrying current $I$:

$$I_{\text{enc}} = nLI$$

Step 5: Equate and solve

$$B_{\text{inside}} \cdot L = \mu_0 n L I$$

$$\boxed{B = \mu_0 n I \quad \text{(inside)}, \qquad B = 0 \quad \text{(outside)}}$$

The field inside is uniform and independent of position -- a remarkable result. This makes solenoids essential for creating uniform magnetic fields in laboratories.

Step 1: Write B in a convenient form

Express the Biot-Savart law using the identity $\hat{\mathscr{r}}/\mathscr{r}^2 = -\nabla(1/\mathscr{r})$:

$$\mathbf{B}(\mathbf{r}) = \frac{\mu_0}{4\pi}\int \mathbf{J}(\mathbf{r}') \times \left[-\nabla\!\left(\frac{1}{\mathscr{r}}\right)\right]d\tau' = -\frac{\mu_0}{4\pi}\int \mathbf{J}(\mathbf{r}') \times \nabla\!\left(\frac{1}{\mathscr{r}}\right)d\tau'$$

Step 2: Rewrite as the curl of an integral

Using the vector identity $\mathbf{J} \times \nabla f = -\nabla \times (f\mathbf{J})$ (when $\mathbf{J}$ does not depend on $\mathbf{r}$):

$$\mathbf{B} = \frac{\mu_0}{4\pi}\nabla \times \int \frac{\mathbf{J}(\mathbf{r}')}{\mathscr{r}}\,d\tau'$$

That is, $\mathbf{B} = \nabla \times \mathbf{A}$ where $\mathbf{A} = \frac{\mu_0}{4\pi}\int \frac{\mathbf{J}(\mathbf{r}')}{\mathscr{r}}\,d\tau'$.

Step 3: Take the divergence

Now compute $\nabla \cdot \mathbf{B}$:

$$\nabla \cdot \mathbf{B} = \nabla \cdot (\nabla \times \mathbf{A})$$

Step 4: Apply the vector calculus identity

The divergence of any curl is identically zero -- this is a fundamental theorem of vector calculus. For any well-behaved vector field $\mathbf{F}$:

$$\nabla \cdot (\nabla \times \mathbf{F}) = 0 \quad \text{(always, identically)}$$

Step 5: Conclude

$$\boxed{\nabla \cdot \mathbf{B} = 0}$$

This holds for any current distribution whatsoever. In integral form (by the divergence theorem): $\oint \mathbf{B}\cdot d\mathbf{a} = 0$. The net magnetic flux through any closed surface is always zero -- there are no magnetic monopoles. Unlike electric field lines that begin and end on charges, magnetic field lines always form closed loops.