Worked examples — Ampere's circuital law — magnetostatic form

This is a companion to the parent note. Here we do not learn what the law is again — we stress-test it. We march through every kind of situation a problem can hand you: currents pointing one way, then the other; loops that enclose the current, then loops that don't; the field right at the centre ($s\to 0$), far away ($s\to \infty$), and at the exact boundary of a wire; a real-world cable; and an exam trap with two nested currents.

Before any symbols appear: let me name the few we reuse everywhere, in plain words.

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The scenario matrix

Every Ampère's-law problem lives in one of these cells. The goal of this page is to leave no cell untouched.

Cell	What makes it special	Covered by
A. Sign / direction	current up vs. down; does circulation flip sign?	Ex 1
B. Loop encloses current	full turn, $\oint d\varphi =2\pi$	Ex 1, 2
C. Loop does NOT enclose	angle returns, $\oint d\varphi =0$, circulation $=0$	Ex 3
D. Degenerate: point on axis	$s\to 0$ limit, is $B$ finite?	Ex 4
E. Boundary case	thick wire exactly at $s=R$, inside-formula meets outside-formula	Ex 4
F. Limiting: far field	$s\to \infty$, $B\to 0$	Ex 4
G. Cancellation trap	two opposite currents, net $I_{\text{enc}}=0$ but $B\ne 0$	Ex 5
H. Real-world word problem	coaxial cable, current inside + return outside	Ex 6
I. Exam twist: nested sheets/solenoid	superposition of two enclosed currents	Ex 7

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Example 1 — Sign & direction (Cell A, B)

Recall Forecast

Guess: does reversing the current flip the sign of the circulation, or the magnitude, or both?

Steps (a):

1. The right-hand rule: point your thumb out of the page (current direction), fingers curl counter-clockwise. Why this step? This tells us $\vec{B}$ points along our chosen walk direction, so every $\vec{B}\cdot d\vec{l}$ term is positive.
2. Apply the law directly: $\oint \vec{B}\cdot d\vec{l}={\mu }_0{I}_{\text{enc}}={\mu }_0(5)$. Why this step? The loop encloses the full current (Cell B), so $I_{\text{enc}}=I=5\text{A}$.
3. Compute: $\oint \vec{B}\cdot d\vec{l}=(4\pi \times 10^{-7})(5)=2\pi \times 10^{-6}\text{T⋅m}\approx 6.28\times 10^{-6}\text{T⋅m}$.

Steps (b): 4. Current now into the page. By the sign convention, current piercing the surface against the loop's right-hand normal is counted negative: $I_{\text{enc}}=-5\text{A}$. Why this step? The right-hand rule for the loop (walk counter-clockwise ⇒ normal points out of page) now disagrees with the current's direction. 5. So $\oint \vec{B}\cdot d\vec{l}=-2\pi \times 10^{-6}\text{T⋅m}$ — same magnitude, opposite sign.

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Example 2 — Loop encloses current, non-circular shape (Cell B)

Recall Forecast

Does the strange shape change the answer? Guess before reading.

Steps:

1. Recall from the parent's Step 3: in polar coordinates $\oint \vec{B}\cdot d\vec{l}=\frac{{\mu }_{0}I}{2\pi }\oint d\varphi$. Why this step? $\vec{B}$ has no radial component, so only the angular sweep survives — the shape of the path is irrelevant, only how much angle it sweeps.
2. A loop that wraps once around the wire sweeps $\oint d\varphi =2\pi$. Why this step? You return to your starting angle after exactly one full turn.
3. Therefore $\oint \vec{B}\cdot d\vec{l}=\frac{{\mu }_0(5)}{2\pi }(2\pi )={\mu }_0(5)=2\pi \times 10^{-6}\text{T⋅m}$.

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Example 3 — Loop does NOT enclose the wire (Cell C)

Recall Forecast

Guess two things: (1) the circulation value, (2) whether $\vec{B}=0$ on the loop.

Steps:

1. $I_{\text{enc}}=0$: no current pierces any surface bounded by $C$. Why this step? The wire is outside — you can stretch a surface across $C$ that the wire never punctures.
2. Therefore ${\oint }_C\vec{B}\cdot d\vec{l}={\mu }_0(0)=0$.
3. But $\vec{B}\ne 0$ on the loop! Look at the figure: near the wire $\vec{B}$ is strong, and it threads into the loop on the near edge and out on the far edge. Why this step? The positive contributions (where $\vec{B}$ points along the walk) exactly cancel the negative ones — this is Cell G's flavour but with a single external wire.

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Example 4 — Degenerate, boundary & limiting behaviour (Cells D, E, F)

Recall Forecast

Sketch $B$ vs $s$ in your head. Where is $B$ biggest? Is it finite on the axis?

Steps (D) — axis, $s\to 0$:

1. Inside formula (parent Ex 2): $B_{\text{in}}=\frac{{\mu }_{0}Is}{2\pi R^2}$. Why this step? Only the current inside radius $s$ threads the loop, and that fraction is $s^2/R^2$.
2. Set $s=0$: $B=0$. Why this step? No current is enclosed by a zero-radius loop — so the field on the axis is exactly zero, finite and well-behaved. (Contrast: a thin wire would blow up, but real wires don't.)

Steps (E) — surface, $s=R$ from both sides: 3. From inside: $B_{\text{in}}(R)=\frac{{\mu }_{0}IR}{2\pi R^2}=\frac{{\mu }_{0}I}{2\pi R}$. 4. From outside: $B_{\text{out}}(R)=\frac{{\mu }_{0}I}{2\pi R}$ (full current enclosed). Why this step? At the boundary both formulas must agree — the field is continuous, no sudden jump. Plug $R=0.01$: $B(R)=\frac{(4\pi \times 10^{-7})(5)}{2\pi (0.01)}=1.0\times 10^{-4}\text{T}$. This is the maximum — the peak of the graph.

Steps (F) — far field, $s\to \infty$: 5. Outside: $B_{\text{out}}=\frac{{\mu }_{0}I}{2\pi s}$. At $s=0.10$: $B=\frac{(4\pi \times 10^{-7})(5)}{2\pi (0.10)}=1.0\times 10^{-5}\text{T}$. 6. As $s\to \infty$, $B\to 0$. Why this step? The $1/s$ decay means the field fades but never abruptly stops — consistent with an infinite straight wire.

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Example 5 — The cancellation trap (Cell G)

Recall Forecast

Net enclosed current is zero — so is the field zero? Careful.

Steps:

1. $I_{\text{enc}}=(+7)+(-7)=0$. Why this step? Out-of-page counts positive (right-hand normal), into-page negative; they cancel in the sum.
2. $\oint \vec{B}\cdot d\vec{l}={\mu }_0(0)=0$.
3. Yet $\vec{B}\ne 0$ anywhere on the loop: each wire produces its own circular field, and at a general point these two fields do not cancel (they point in different directions with different magnitudes). Why this step? The law constrains the integral, not the pointwise field.

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Example 6 — Real-world word problem: coaxial cable (Cell H)

Recall Forecast

Why do we shield cables this way? Guess the field outside the cable before computing.

Steps (i) — between conductors, $s=1\text{mm}$:

1. Loop of radius $s=1\text{mm}$ encloses only the inner conductor: $I_{\text{enc}}=3\text{A}$. Why this step? The return current at $b=2\text{mm}$ is outside this loop, so it doesn't count.
2. $B=\frac{{\mu }_{0}I}{2\pi s}=\frac{(4\pi \times 10^{-7})(3)}{2\pi (0.001)}=6.0\times 10^{-4}\text{T}$.

Steps (ii) — outside everything, $s=3\text{mm}$: 3. Loop now encloses both currents: $I_{\text{enc}}=(+3)+(-3)=0$. Why this step? The go and return currents are both inside — they cancel. 4. $B=\frac{{\mu }_0(0)}{2\pi s}=0$.

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Example 7 — Exam twist: nested current sheets / solenoid superposition (Cell I)

Recall Forecast

Inside the innermost region, both solenoids contribute. Will the fields add or subtract?

Steps (a) — deep inside the inner solenoid:

1. Each solenoid alone gives $B={\mu }_{0}nI$ inside it (parent Ex 3). Why this step? The rectangular Amperian loop picks up ${\mu }_{0}nLI$ and $BL={\mu }_{0}nLI\Rightarrow B={\mu }_{0}nI$.
2. By superposition, both fields are present in the inner region; they point oppositely, so subtract: $B={\mu }_{0}I(n_1-n_2)$. Why this step? The full field is the vector sum; opposite windings mean opposite signs.
3. $B=(4\pi \times 10^{-7})(2)(1000-400)=(4\pi \times 10^{-7})(2)(600)=1.508\times 10^{-3}\text{T}$.

Steps (b) — in the gap (outside inner, inside outer): 4. Here only the outer solenoid's field is present (the inner solenoid's field is ~zero outside itself). $B={\mu }_0{n}_{2}I=(4\pi \times 10^{-7})(400)(2)=1.005\times 10^{-3}\text{T}$, pointing the outer solenoid's way. Why this step? An Amperian loop in the gap encloses only outer-solenoid turns.

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Wrap-up

Recall Which cell did each example cover?

Ex 1 :::: A (sign) + B (encloses) Ex 2 :::: B (shape-independence) Ex 3 :::: C (no enclosure, $\oint =0$, $\vec{B}\ne 0$) Ex 4 :::: D (axis $s\to 0$) + E (boundary $s=R$) + F (far field) Ex 5 :::: G (cancellation trap) Ex 6 :::: H (real-world coax) Ex 7 :::: I (nested solenoid superposition)

Related: Biot–Savart law · Solenoid and toroid fields · Gauss's law for magnetism · Stokes' theorem · Displacement current · Maxwell's equations.