Visual walkthrough — Ampere's circuital law — magnetostatic form

Step 0 — What we are even trying to explain

Two symbols we will earn, not assume:

The quantity we chase is the circulation ${\oint }_C\underset{\text{field arrow here}}{\underbrace{\vec{B}}}\cdot \underset{\text{tiny step here}}{\underbrace{d\vec{l}}}$ The dot $\cdot$ between them is the dot product: it keeps only the part of $\vec{B}$ that lies along $d\vec{l}$. We define that next, because it is the one piece of maths this whole page depends on.

Step 1 — WHY a dot product? "How much along the path?"

WHY this tool and not another? We asked "how much does the field push me along my walk?" The word "along" is projection, and projection is what $\cos \alpha$ computes. No other operation answers that exact question. That is why the dot product — not the cross product, not plain multiplication — is the right instrument here.

PICTURE: the orange step $d\vec{l}$ and the blue field $\vec{B}$ meet at angle $\alpha$; the green dashed segment is the shadow $|\vec{B}|\cos \alpha$ — the only part that counts.

Step 2 — The one input fact: field of a straight wire

Before we can add anything up, we need to know what $\vec{B}$ actually is near a wire. This comes from the Biot–Savart law (the previous chapter), which we take as given here.

WHY only $s$? By symmetry the wire looks identical from every angle around it and from every point along it. So $\vec{B}$ cannot depend on which angle or where along the wire — only on how far out, $s$. And it must wrap in circles (right-hand rule: thumb along $I$, fingers curl the way $\vec{B}$ points).

PICTURE: wire pointing out of the page (the dot), field arrows forming concentric circles; inner circles have longer arrows (stronger $B$), outer circles shorter — the $1/s$ falloff you can see.

Step 3 — Add it up on the friendly loop (a circle)

Because $\vec{B}\parallel d\vec{l}$ everywhere, $\alpha =0$ and $\cos \alpha =1$, so each tiny contribution is just $B|d\vec{l}|$: ${\oint }_C\vec{B}\cdot d\vec{l}={\oint }_{C}Bdl=B{\oint }_{C}dl$

• $B$ came out of the sum because at fixed radius $s$ it is the same number at every point (Step 2).
• ${\oint }_{C}dl$ is just "add up all the tiny lengths around the circle" = the circumference $=2\pi s$.

${\oint }_C\vec{B}\cdot d\vec{l}=\underset{B}{\underbrace{\frac{{\mu }_{0}I}{2\pi s}}}\times \underset{\text{circumference}}{\underbrace{2\pi s}}={\mu }_{0}I$

PICTURE: the blue circle loop with orange $d\vec{l}$ steps, each sitting on top of a blue $\vec{B}$ arrow — perfectly aligned all the way round.

Step 4 — WHY any wiggly loop gives the same answer

Now deform the loop into any shape you like (still enclosing the wire). Describe each step $d\vec{l}$ by two independent moves: a radial part (in/out from the wire) and an angular part (around the wire).

${\oint }_C\vec{B}\cdot d\vec{l}={\oint }_C\frac{{\mu }_{0}I}{2\pi s}(\underset{\text{sideways arc}}{\underbrace{sd\varphi }})=\frac{{\mu }_{0}I}{2\pi }{\oint }_{C}d\varphi$

• Radial moves contribute $0$: $\vec{B}$ points purely sideways (angular), so it is perpendicular to any in/out step — dot product zero (Step 1, $\alpha =90^{\circ }$).
• The sideways part of a step at radius $s$ covering angle $d\varphi$ has length $sd\varphi$. The $s$ here cancels the $1/s$ in $B$ again — same secret, general loop.
• ${\oint }_{C}d\varphi$ = "total angle swept as you go around once."

PICTURE: a lumpy loop; each step split into a green radial piece (killed by the dot product) and an orange sideways arc (all that survives).

Step 5 — The edge case: loop that does NOT enclose the wire

PICTURE: wire outside the loop; on the near arc $\vec{B}$ helps your walk (green, positive), on the far arc it opposes it (red, negative). Their sums cancel: circulation $=0$ even though $\vec{B}\ne 0$ anywhere.

Step 6 — Superpose: the law for any bundle of currents

Real setups have many wires. But circulation adds: if $\vec{B}={\vec{B}}_1+{\vec{B}}_2+\dots$ from several wires, then ${\oint }_C\vec{B}\cdot d\vec{l}={\oint }_C{\vec{B}}_1\cdot d\vec{l}+{\oint }_C{\vec{B}}_2\cdot d\vec{l}+\dots$ Each term is ${\mu }_0{I}_k$ if wire $k$ is enclosed, and $0$ if not. Adding only the enclosed ones:

PICTURE: three wires; a loop encloses two of them ($+I$ and $-I$, note the sign from direction) and misses the third. Only the enclosed ones enter $I_{\text{enc}}$; the outside one is greyed out.

The one-picture summary

Everything on this page compressed: the circular field ($B\propto 1/s$), a wiggly loop whose steps split into killed radial pieces and surviving sideways pieces, the $s$-cancellation, and the verdict — ${\mu }_{0}I$ if you loop the wire, $0$ if you don't.

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