Question bank — Magnetic field of straight wire, circular loop, solenoid, toroid

True — $B={\mu }_{0}I/2\pi r$ blows up as $r\to 0$, so shrinking $r$ or raising $I$ gives arbitrarily large field near an idealised line wire.

False — an Amperian circle drawn inside the metal, at radius $r$ from the axis, encloses only the fraction of current threading it; that fraction is $I{r}^2/R^2$, giving $B={\mu }_{0}Ir/2\pi R^2$, which grows straight up from zero at the centre to a peak at the surface (see figure).

False — it only fixes the total circulation (the running sum of $\vec{B}$ tangent to the loop), not $\vec{B}$ at each point. Genuine counterexample: outside a solenoid, an Amperian loop can enclose zero net current yet sit in a region where a small stray field still exists — the plus and minus contributions of $\vec{B}$ around the loop merely cancel to zero sum while $\vec{B}$ itself is nonzero.

True — and here is the why: each turn adds its own little push of field down the tube, and the pushes stack. Twice as many turns in the same length means twice as many stacked pushes per metre, so the steady breeze inside is twice as strong. Symbolically $n=N/L$ doubles, and $B={\mu }_{0}nI$ follows.

False — the loop's field is only known cleanly on its axis; off-axis and in the plane it is complicated and non-uniform, strongest right at the wire.

True (ideally) — for a perfectly wound toroid $B=0$ in the central hole and outside; only the small leakage from finite winding pitch escapes.

True — $\vec{B}$ is linear in $I$ in every formula here, so flipping the current's sign flips $\vec{B}$ by the right-hand rule.

False — both fields point the same way at the midpoint (grip rule), so they add, not cancel; they cancel far outside, not between.

Error — no Amperian path around the axis has constant $B$ parallel to $d\vec{l}$, so $B$ cannot be pulled out; you must integrate with Biot-Savart Law.

Error — $n=N/L$ contains no radius; the ideal long-solenoid field is genuinely radius-independent, so making it fatter changes nothing.

Error — $r$ is the distance from the centre of the doughnut to the field point (inside the core), not the thickness of the wire or the core's minor radius.

Error — at the mouth the field drops to exactly $\frac{1}{2}{\mu }_{0}nI$. Why: deep inside, a slice of coil has wire on both sides feeding field into it; at the very end, only half the coil remains (all on one side), so it delivers half the push. The field lines fan out and weaken there — see the flaring lines in the figure.

Error — for the infinite wire the ends run to $\pm 90^{\circ }$, so ${\theta }_1={\theta }_2=90^{\circ }$ and $\sin =1$, recovering ${\mu }_{0}I/2\pi r$; setting them to $0$ would wrongly give zero.

Error — forgot $N$; each turn contributes, so the centre field is $N{\mu }_{0}I/2a$.

Error — the element law is $1/r^2$, but after integrating the whole infinite wire the result is $1/r$; only the point-charge (or point element) case is $1/r^2$.

For the wire the Amperian path is a circle of circumference $2\pi r$, and $B$ divides by that whole length. For the loop centre you instead add up $dB={\mu }_{0}Idl/4\pi a^2$ from every element of the ring; summing $dl$ over the circumference gives $2\pi a$, and $\frac{2\pi a}{4\pi a^2}=\frac{1}{2a}$ — the $\pi$ cancels, leaving the clean ${\mu }_{0}I/2a$. See the ring geometry in the figure.

Pick two current elements at exactly opposite ends of the ring. Each produces a $d\vec{B}$ that leans partly along the axis and partly sideways (toward the axis). Their sideways parts point in exactly opposite directions and are equal in size, so they cancel pair by pair; only the axial parts, which point the same way, survive and add. The figure shows one such opposite pair with the sideways arrows annihilating.

The return field from all the loops spreads over a huge region, and an Amperian loop outside encloses zero net current, so the circulation vanishes; combined with symmetry this forces $B\approx 0$.

Its Amperian circle has circumference $2\pi r$ that varies across the core, so $B\propto 1/r$ — larger at the inner edge (small $r$), smaller at the outer edge; only a thin core looks uniform.

Magnetic field circulates around the current rather than pointing away from it, and the cross product is exactly the operation that turns "direction of flow" plus "direction to the point" into that perpendicular, circulating direction.

A finite wire lacks the full translational symmetry, so $B$ is not constant around any circle and cannot leave the integral — you must fall back on Biot-Savart Law with the $\sin {\theta }_1+\sin {\theta }_2$ result.

Its field threads through the hole and loops back around outside exactly like a dipole, giving it a north and south face — the basis of Bar magnet and magnetic dipole.

Undefined/singular for an idealised line wire ($B\to \infty$); for a real wire of finite radius $R$, $B\to 0$ at the true centre where enclosed current is zero.

Zero — an Amperian circle in the hole encloses no current, so $\oint \vec{B}\cdot d\vec{l}=0$ and by symmetry $B=0$ there.

$B\to {\mu }_{0}I{a}^2/2x^3$, the $1/x^3$ dipole falloff — the loop looks like a point magnetic dipole from far off (see Bar magnet and magnetic dipole).

The interior field ${\mu }_{0}nI\to 0$; spreading the same turns over infinite length dilutes them to nothing.

Every formula gives $B=0$ — no moving charge, no magnetic field, regardless of how cleverly the wire is bent.

The $B={\mu }_{0}nI$ approximation breaks down; with one turn you have a single loop and must use the loop formula ${\mu }_{0}I/2a$, not the long-solenoid result.

The axial formula does not apply; the field there is genuinely off-axis, curls back opposite to the interior direction, and requires full Biot-Savart Law integration — no simple closed form.