Always true, rarely a shortcut

For any closed path, called an Amperian loop,

$$\oint \vec{B}\cdot d\vec{\ell} = \mu_0 I_{\text{enc}},$$

where $I_{\text{enc}}$ is the current threading the surface bounded by that loop. The vector $\vec{B}$ in the integrand is the total magnetic field at each point on the loop, produced by every current in the universe, not only the enclosed ones.

So what does an outside current do? It contributes field at every point of the loop, and its contributions to the circulation cancel out around the trip: the loop passes with the field for part of the path and against it for another part, and the two exactly balance. Zero net circulation is not zero field. Deleting a neighboring wire's field because the loop does not enclose it is the same mistake as claiming a charge outside a Gaussian surface produces no field on it.

The practical consequence: a nearby wire cannot change the right-hand side of the equation, but it can absolutely destroy the symmetry that made the left-hand side solvable. The law survives; the shortcut does not.

The step everybody writes,

$$\oint \vec{B}\cdot d\vec{\ell} = B\,(2\pi r),$$

requires two separate things to be true along the entire loop: $\vec{B}$ has the same magnitude at every point of it, and $\vec{B}$ is everywhere tangent to it. If either fails, $B$ is inside the integral and cannot come out.

Three geometries deliver both, and they are the three the exam uses:

1. Infinite straight wire (or the interior of a thick one): circular loop concentric with the wire gives $B = \mu_0 I_{\text{enc}}/(2\pi r)$.
2. Ideal solenoid: rectangular loop with one side inside parallel to the axis, one side outside where $B \approx 0$, and two sides perpendicular to $\vec{B}$ contributing nothing, giving $B = \mu_0 n I$ with $n$ turns per unit length.
3. Toroid: circular loop inside the windings gives $B = \mu_0 N I/(2\pi r)$, which is not uniform across the cross-section.

Draw a circle around a pair of wires, or around a finite wire segment, and the field varies from point to point on that circle. The circulation is still $\mu_0 I_{\text{enc}}$, which is a true statement and a useless one, because one equation cannot recover a function of position. For the pointwise field there, superpose the individual wire fields as vectors instead.

$I_{\text{enc}}$ is not "add up the currents nearby". Counting it is three rules applied in order.

1. Sign. Curl the right-hand fingers along the direction you traverse the loop. A current piercing along the thumb counts positive; one piercing against it counts negative. Two equal wires with opposite currents inside one loop give $I_{\text{enc}} = 0$, and therefore zero circulation, even though $B$ is nonzero everywhere on the loop.
2. Multiplicity. A wire counts once per piercing. A solenoid's windings pierce a rectangular loop of length $L$ a total of $nL$ times, so $I_{\text{enc}} = nLI$. Missing the turn count is how $B = \mu_0 n I$ collapses to $\mu_0 I/(2\pi r)$.
3. Fraction. Inside a thick wire of radius $R$ carrying $I$ with uniform current density, a loop of radius $r < R$ encloses only the current passing through the AREA it bounds: $I_{\text{enc}} = I r^2/R^2$. Then $B = \mu_0 I r/(2\pi R^2)$, rising linearly from zero at the axis to $\mu_0 I/(2\pi R)$ at the surface, then falling as $1/r$ outside.

The area rule is where a plausible shortcut goes wrong: prorating by $r/R$ instead of $r^2/R^2$ gives a field that is constant inside the wire, which is not what a uniform current density produces. Use $I_{\text{enc}} = \int \vec{J}\cdot d\vec{A}$ and let the geometry do the work.

A useful check on rule 1: outside a toroid, any circular loop encloses $N$ turns going one way and $N$ coming back, so $I_{\text{enc}} = 0$ and the external field of an ideal toroid is zero.

A flat loop of $N$ turns, area $A$, carrying current $I$, has magnetic dipole moment

$$\vec{\mu} = NIA\,\hat{n},$$

where $\hat{n}$ is the loop's normal, found by curling the right-hand fingers along the current. In a uniform field the forces $I\vec{L}\times\vec{B}$ on opposite sides of the loop are equal and opposite and cancel in pairs, so the net force is zero. What survives is a torque:

$$\vec{\tau} = \vec{\mu}\times\vec{B}, \qquad \tau = NIAB\sin\theta, \qquad U = -\vec{\mu}\cdot\vec{B},$$

with $\theta$ the angle between $\vec{\mu}$ and $\vec{B}$. Read the geometry carefully, because the sine has caught generations of students:

1. Maximum torque at $\theta = 90^\circ$: $\vec{\mu}$ perpendicular to $\vec{B}$, which means the loop's plane contains the field.
2. Zero torque, stable at $\theta = 0$: $\vec{\mu}$ aligned with $\vec{B}$, which means the loop's plane is perpendicular to the field. This is where a free loop settles, and it is the compass needle's rest position.
3. Zero torque, unstable at $\theta = 180^\circ$: $\vec{\mu}$ antiparallel to $\vec{B}$.

Translating the loop toward the magnet requires a nonuniform field, where the forces on the two sides no longer match. That is why a magnet attracts iron, whose aligned atomic dipoles sit in the strongly nonuniform field near the pole, and why nothing translates in the idealized uniform-field problems on this exam.