Solenoids & Toroids

A single wire's magnetic field is a whisper — concentric circles that die off as $1/d$ , a few microtesla at arm's length even for a hefty current. Yet an MRI scanner holds a field a hundred thousand times Earth's steady over a volume big enough for a human body, and the humble relay clicking inside a car indicator does its job with a field thousands of times stronger than the bare wire could manage. The trick in both machines is the same, and it is geometrical: take that whispering wire and wind it. Stack hundreds of turns side by side and each loop's field reinforces its neighbours' down the bore while cancelling them outside. The result is the solenoid — physics' field factory — with a uniform, controllable magnetic field inside and almost nothing outside. Bend the solenoid around until it bites its own tail and you get the toroid, a doughnut that goes one better: its field never leaves the doughnut at all. That total confinement is why the most expensive magnets ever built — the coils of fusion reactors — are toroids. This module gives you both devices at exam grade: $B = \mu_0 n I$ for the solenoid and $B = \mu_0 N I / 2\pi s$ for the toroid, where each formula comes from, the $n$ -versus- $N$ bookkeeping that costs more marks than any other slip in this chapter, and the design logic that makes these two shapes the workhorses of everything from doorbells to tokamaks.

Key takeaways

A solenoid is loops in a row: inside, the turns' fields reinforce into a nearly uniform axial field; outside, they cancel to almost nothing.
The formula: $B = \mu_0 n I$ inside a long solenoid, where $n = N/L$ is turns per metre — not the total turn count $N$ . This is the single most-tested conversion in the chapter.
The field is the same everywhere inside (wall to wall, not just on the axis) and doesn't care about the solenoid's radius. At the very ends it drops to about half.
A toroid is a solenoid bent into a doughnut: $B = \mu_0 N I / 2\pi s$ inside the winding ( $s$ measured from the doughnut's central axis), and exactly zero outside — total confinement.
An iron core multiplies the field by the relative permeability $\mu_r$ (hundreds to thousands) — but iron saturates near 2 T, so the multiplication is not unlimited.
KEY IDEA: winding a wire into a solenoid trades the wire's weak, spread-out circles for a strong, uniform field concentrated inside the coil — $B = \mu_0 n I$ , set only by turns per metre and current.

The core insight

Inside a long solenoid the magnetic field is uniform, axial, and equal to $B = \mu_0 n I$ — independent of the radius and of where you stand in the bore — while a toroid confines the field $B = \mu_0 N I/2\pi s$ entirely inside its doughnut.

Where this lives today

The two formulas of this module are quietly priced into two of the most expensive machines humanity builds. An MRI scanner is a superconducting solenoid asking equation (1) for 1.5 T: that demands $nI = B/\mu_0 \approx 1.2\times10^{6}$ ampere-turns per metre — say a thousand turns per metre carrying over a thousand amperes each, which is why the winding must be superconducting and bathed in liquid helium. The patient lies along the axis precisely because the deep interior is where the field is uniform; the bore length exists to push the "ends" (and their half-strength fringing) away from the imaging volume. The toroid's zero-leak theorem, meanwhile, is the architecture of magnetic-confinement fusion: ITER's toroidal field system is equation (3) at monster scale — 5.3 T at the plasma axis, of order $10^{8}$ ampere-turns — with the built-in $1/s$ gradient from the worked example as a first-order design constraint. Closer to home, the ring-shaped toroidal-core inductors on every laptop motherboard use the same confinement so one component's field doesn't whisper into its neighbours.

The physical picture

Start with what you already own. Fields of Straight Wires & Circular Loops gives a single circular loop a field that threads through its centre and fountains around the outside — a tiny bar magnet made of current. Now imagine sliding a second loop next to the first, carrying current the same way around. Along the shared axis, the two fields point the same direction: they add. In the space between the wires outside the coil, one loop's returning field runs against its neighbour's: they fight. Keep stacking — ten, a hundred, a thousand turns — and the fight outside ends in near-total cancellation while the cooperation inside builds a set of straight, parallel, evenly spaced field lines running down the bore like rails.

Figure

Solenoid in cross-section — turns reinforce into uniform rails inside, cancel outside

Two facts about that interior field deserve to feel surprising, because both are exam favourites. First, it is genuinely uniform — the same strength halfway to the wall as on the axis, wall to wall across the whole bore. Second, it is completely blind to the radius: a fat solenoid and a skinny one with the same turns per metre and the same current produce exactly the same interior field. Nothing about a single loop suggests either fact; they emerge only when you make the coil long, and the clean argument for both comes from applying Ampère's Circuital Law — the tool the previous module built — to one shrewdly chosen rectangle. The solenoid also settles which end is "north": face the coil, and if the current circulates anticlockwise, field lines emerge toward you — a north pole. Clockwise: south. A solenoid is a bar magnet you can switch off — the observation that launched the electromagnet in 1824.

The solenoid formula, earned

The claim to be earned: inside a long solenoid with $n$ turns per metre carrying current $I$ ,

B = \mu_0 n I \qquad (1)

where: $\mu_0 = 4\pi\times10^{-7}\ \mathrm{T\,m/A}$ is the permeability of free space, $n = N/L$ is the number of turns per metre of length (a 1000-turn solenoid half a metre long has $n = 2000\ \mathrm{m^{-1}}$ ), and $I$ is the current in the wire. Read the formula physically: more turns packed into each metre means more current sheets stacked per metre of bore, and each ampere in each turn contributes its share. Nothing else appears — not the radius, not the position inside, not the total length. At $n = 2000\ \mathrm{m^{-1}}$ and $I = 3.0\ \mathrm{A}$ , equation (1) gives $7.5\ \mathrm{mT}$ — about 150 times Earth's field, from bench-top hardware.

A scope note before the derivation, because the boards genuinely split here. CBSE, ISC, JEE-Main, JEE-Advanced, NEET and AP Physics 2 all expect equation (1) as a working formula. IB (SL and HL), IGCSE and CIE A-Level ask only for the field pattern — sketching and interpreting the solenoid's field lines, bar-magnet resemblance included — with no $B = \mu_0 n I$ calculation on the paper. If you sit one of the pattern-only boards, own the picture above and the pole rule; the formula below is enrichment that will make the picture unforgettable.

Derivation AThe solenoid field from one rectangular loop

We want equation (1), and with it the two surprises — uniformity across the bore and independence of radius — using Ampère's Circuital Law, whose statement and justification belong to the previous module (circulation around a closed loop equals $\mu_0$ times the threaded current). The strategy: for an ideal long solenoid the field inside is axial and the field outside is negligible, so we choose a rectangular loop with one long side inside the bore and one outside, and let the law convert "current threading the rectangle" into "field times length." The thing to watch is how the length of the rectangle cancels — that cancellation is the uniformity claim.

Step 1. Set up the rectangle. Take a rectangle of length $\ell$ with one long side inside the solenoid, parallel to the axis, and the other long side outside; the two short sides pierce the winding. For the ideal long solenoid, the field inside is axial (the stacked-loop picture: sideways components of neighbouring turns cancel) and the field outside is effectively zero (the return flux spreads through all of space).

Step 2. Walk the loop and add up the circulation. Along the inside long side, the field $B$ runs parallel to our path: contribution $B\ell$ . Along the outside long side: field is zero, contribution zero. Along the two short sides: the field (where it exists) is axial, perpendicular to these legs, so they contribute nothing.

\text{circulation} = B\ell

Why the short legs die: circulation only counts the field component along your direction of travel, and an axial field has no component along a leg that runs radially.

Step 3. Count the threaded current. The rectangle's face is pierced by every turn that lies within the length $\ell$ — that is $n\ell$ turns, each carrying $I$ , all punching through the same way:

I_{\text{threaded}} = n \ell I

Step 4. Apply the law and cancel:

B\ell = \mu_0\, n\ell I \;\;\Rightarrow\;\; B = \mu_0 n I

Why this proves uniformity and radius-blindness: nowhere did we say how far inside the bore the inner leg sat, or how wide the solenoid was — slide the inner leg anywhere inside and the same equation returns the same $B$ . Every interior point sees $\mu_0 n I$ ; every exterior point sees zero. The solenoid is magnetism's answer to the parallel-plate capacitor: all field inside, uniform, none outside. *Verified: $n = 2000\ \mathrm{m^{-1}}$ , $I = 3.0\ \mathrm{A}$ gives $B = 4\pi\times10^{-7}\times2000\times3.0 = 7.54\times10^{-3}\ \mathrm{T}$ , recomputed; units $\mathrm{(T\,m/A)(m^{-1})(A)} = \mathrm{T}$ ✓; matches the NCERT Class-12 result and the loop-stacking route in Fields of Straight Wires & Circular Loops.*

Honesty about the ends, because examiners are honest about them too. Equation (1) is the deep interior value of a long solenoid. At the exact centre of an end face the field falls to almost exactly half — picture an infinite solenoid as two half-infinite ones joined at your location: each half supplies half the field, and at an end face one half is simply missing. In practice "long" arrives fast: a 50 cm solenoid of 2 cm radius already reaches 99.7% of $\mu_0 n I$ at its centre. Board answers assume the ideal value inside; the half-at-the-ends fact is a favourite one-marker and the fringing field at the mouth is what lets a solenoid pull in iron — a story that belongs to the field-gradient physics later in the course. One upgrade turns the solenoid into an industrial electromagnet: fill the bore with soft iron. The iron's own atomic magnets line up with the coil's field and add to it, multiplying $B$ by the relative permeability $\mu_r$ :

B = \mu_r \mu_0 n I \qquad \text{(iron core, before saturation)} \qquad (2)

where: $\mu_r$ is a dimensionless multiplier — several hundred to several thousand for soft iron. Read it with a physicist's caution: the multiplication is spectacular but not unlimited, because iron saturates near 2 T — once essentially all its atomic magnets are aligned, pushing more current buys almost nothing. A naive $\mu_r = 1200$ applied to a 6.3 mT air-core field predicts 7.5 T; real iron quietly tops out around 2 T and the "prediction" is fiction beyond that. State the saturation ceiling and you sound like an examiner's favourite student.

The toroid: a solenoid that ate its tail

Bend a solenoid into a circle and join the ends. The field lines, which ran straight down the bore, now close into perfect circles that live entirely inside the doughnut. Applying Ampère's Circuital Law to one of those circles — a circular path of radius $s$ measured from the doughnut's central axis (the axis of the hole, not the centre of the tube) — is even quicker than the rectangle.

Figure

Toroid from above — the field circulates inside the doughnut and is zero outside

Go deeper · Derivation B · 1 minThe toroid field and why it cannot leak

Derivation BThe toroid field and why it cannot leak

We want the field inside a toroid of $N$ total turns carrying current $I$ , and the celebrated bonus — that the field outside is exactly zero. The strategy: by symmetry the field lines are circles centred on the toroid's central axis, so we choose the Amperian path to be one of those circles; the circulation is then just $B$ times the circumference, and everything reduces to counting how many turns thread the circle. Watch the counting — it is where the confinement comes from.

Step 1. Choose a circle of radius $s$ that runs inside the doughnut's tube. By the rotational symmetry of the winding, $B$ has the same magnitude everywhere on this circle and points along it, so the circulation is $B \times 2\pi s$ .

Step 2. Count the threaded turns. The flat disc spanned by our circle is pierced once by each of the $N$ turns — every turn's inner leg passes up (say) through the disc, and its outer leg lies beyond the circle. Threaded current: $NI$ .

Step 3. Apply Ampère's Circuital Law:

B\,(2\pi s) = \mu_0 N I \;\;\Rightarrow\;\; B = \frac{\mu_0 N I}{2\pi s} \qquad (3)

Why the field is stronger at the inner wall: the same $NI$ drives every interior circle, but an inner circle is shorter — smaller $2\pi s$ , bigger quotient. A toroid's field carries a built-in $1/s$ gradient no winding trick removes.

Step 4. Now take a circle outside the doughnut, beyond the outer wall. Its disc is pierced by every turn twice — once by the turn's inner leg going up, once by the outer leg coming down. The two crossings cancel, turn by turn: net threaded current zero, so $B \times 2\pi s = 0$ and the exterior field vanishes. A circle inside the hole threads nothing at all: zero again. The toroid does not leak. Verified: $N = 1500$ , $I = 2.0\ \mathrm{A}$ , $s = 0.25\ \mathrm{m}$ : $B = (4\pi\times10^{-7}\times1500\times2.0)/(2\pi\times0.25) = 2.4\ \mathrm{mT}$ , recomputed; the thin-toroid limit below closes the loop with equation (1).

The consistency check worth doing once in your life: for a thin toroid (tube radius much smaller than the ring radius $R_0$ ), every interior point has $s \approx R_0$ , and the turns-per-metre along the ring is $n = N/2\pi R_0$ . Substitute into (3): $B = \mu_0 N I / 2\pi R_0 = \mu_0 n I$ — the solenoid formula, recovered. A toroid is a solenoid that ate its tail, and the two formulas are one formula in two costumes.

Limiting cases

case	result	meaning
deep inside a long solenoid	$B = \mu_0 n I$	uniform, radius-blind — the working formula
centre of an end face	$B \approx \tfrac12 \mu_0 n I$	half the rings are missing — classic one-marker
short fat coil	formula fails	it's just a stack of loops; no "long solenoid" shortcut
iron core, below saturation	$B = \mu_r\mu_0 n I$	multiplied hundreds-fold; ceiling ≈ 2 T
toroid, inner wall vs outer	$B \propto 1/s$	stronger inboard; not uniform
toroid, outside or in the hole	$B = 0$ exactly	every outside path nets zero threaded current
thin toroid, $s \approx R_0$	$B \to \mu_0 n I$	the bent solenoid closes the circle

Common trapPlugging in the total turns N instead of turns per metre n

The costliest slip in this chapter, and it is a units slip, not a physics slip. Equation (1) wants $n$ , the turn count per metre; problems hand you $N$ , the total count, plus a length in centimetres — two conversions in one line. "A solenoid of 800 turns and length 40 cm at 2.5 A" means $n = 800/0.40 = 2000\ \mathrm{m^{-1}}$ and $B = 6.3\ \mathrm{mT}$ ; grabbing $n = 800$ inflates the answer by ×0.4 wrongly ( $2.5\ \mathrm{mT}$ ), and using $L = 40$ deflates it a hundredfold. The toroid formula plays the opposite trick: equation (3) wants the total $N$ — no length ever appears, because $2\pi s$ is the length. Solenoid: per metre. Toroid: total. Say it twice before every problem.

Worked examples

Board level — the standard solenoid, plus its honest end (CBSE/ISC pattern)

A solenoid 50 cm long has 1000 turns and carries a current of 3.0 A. Find (a) the magnetic field deep inside, (b) the approximate field at the centre of one end face, and (c) the field if a soft-iron core of relative permeability 400 is inserted (comment on your answer).

Step 1. Convert to turns per metre first, before touching the formula: $n = N/L = 1000/0.50 = 2000\ \mathrm{turns/m}$ .

Step 2. Apply equation (1): $B = \mu_0 n I = (4\pi\times10^{-7}) \times 2000 \times 3.0$ . Keep $4\pi$ symbolic: $4\pi\times 6000 \times10^{-7} = 24\pi\times10^{-4}$ , so $\boxed{B = 7.5\ \mathrm{mT}}$ (7.54 mT to three figures) — directed along the axis, sense given by curling your right hand with the current.

Step 3. End face: half the deep-interior value, $\boxed{B_{\text{end}} \approx 3.8\ \mathrm{mT}}$ . One sentence of justification earns the mark: "an end face sees only half the stack of current turns, so the field is half."

Step 4. Iron core: $B = \mu_r \mu_0 n I = 400 \times 7.54\ \mathrm{mT} = 3.0\ \mathrm{T}$ by the formula — but soft iron saturates near 2 T, so the realistic field caps out around $\boxed{2\ \mathrm{T}}$ . Writing the formula value and the saturation caveat is exactly the "comment" the question is fishing for.

Competitive level — toroid anatomy and the bent-solenoid check (JEE-Main pattern)

A toroid has an inner radius of 24 cm and an outer radius of 26 cm, is wound with 1500 turns, and carries 2.0 A. Find (a) the field inside the hole of the toroid, (b) the field at the mean radius of the core, (c) the field outside the toroid, and (d) show the mean-radius answer agrees with treating the toroid as a bent solenoid.

Step 1. Parts (a) and (c) are free marks if you know the confinement theorem: inside the hole, an Amperian circle threads no current — $\boxed{B = 0}$ ; outside the body, every turn threads twice with opposite signs — $\boxed{B = 0}$ . Nonzero field exists only within the winding, 24 cm $\le s \le$ 26 cm.

Step 2. Mean radius $s = 25\ \mathrm{cm} = 0.25\ \mathrm{m}$ . Equation (3): $B = \mu_0 N I/2\pi s = (4\pi\times10^{-7}\times1500\times2.0)/(2\pi\times0.25)$ . The $\pi$ 's partially cancel: $2\times10^{-7}\times 3000/0.25 = \boxed{2.4\ \mathrm{mT}}$ . (For interest: 2.5 mT at the inner wall, 2.3 mT at the outer — the built-in $1/s$ gradient.)

Step 3. Bent-solenoid check: the ring circumference is $2\pi\times0.25 = 1.571\ \mathrm{m}$ , so $n = 1500/1.571 = 955\ \mathrm{turns/m}$ , and $\mu_0 n I = 4\pi\times10^{-7}\times955\times2.0 = 2.4\ \mathrm{mT}$ ✓ — the same number by the other formula, which is the intended "show that." Timing note: 3 minutes. Parts (a) and (c) should take ten seconds total — if you find yourself computing anything for them, you've missed the confinement theorem, which is the point of the question. The real arithmetic lives only in Step 2.

Exam traps

The n-versus-N swap. Solenoid formula takes turns per metre; toroid formula takes total turns. Every board tests this, usually by handing you $N$ and a length in centimetres. Convert $L$ to metres, then divide. (Full anatomy in the warning box above.)

Centimetres left unconverted. $L = 40\ \mathrm{cm}$ must become 0.40 m before $n = N/L$ . An unconverted length inflates $n$ — and your field — a hundredfold, an error size examiners specifically scan for.

Toroid radius measured from the wrong place. The $s$ in $B = \mu_0 NI/2\pi s$ runs from the toroid's central axis (the middle of the hole), not from the centre of the tube's cross-section. Given inner and outer radii, the mean of those two is the usual evaluation point.

"Field outside a solenoid is zero" stated as exact. Ideal-solenoid zero is an idealisation; a real, finite solenoid leaks a weak return field (that's how it attracts a paperclip at its mouth). For an ideal toroid, though, exterior zero is exact — know which claim is which.

Iron-core answers above saturation. $\mu_r \mu_0 n I$ predictions beyond about 2 T are unphysical for iron; say so. NEET and JEE both like conceptual options exploiting this.

Formula on a pattern-only board. If you sit IB, IGCSE or CIE A-Level: the examiner wants the field-line sketch (uniform, parallel, evenly spaced inside; spreading loops outside; bar-magnet resemblance; poles by the current-circulation rule). Time spent deriving $\mu_0 n I$ there earns nothing — the sketch and the pole rule earn everything.

Constants: use $\mu_0 = 4\pi\times10^{-7}\ \mathrm{T\,m/A}$ on every Indian board and keep the $4\pi$ symbolic as long as possible — solenoid and toroid answers almost always simplify to a clean multiple of $\pi$ or cancel it entirely.

Origins · 1 minWilliam Sturgeon and how this was discovered

William Sturgeon

1783–1850 · Woolwich, London

Sturgeon was a shoemaker's apprentice who ran away to join the army, taught himself physics from borrowed books between drills, and at forty — by then a lecturer at the Royal Military College — built the device this module is really about. In 1824 he took a varnished iron bar bent into a horseshoe, wound it with eighteen loose turns of bare copper wire, and connected a single battery cell: the seven-ounce contraption lifted nine pounds of iron, twenty times its own weight, and dropped it dead the instant the circuit opened. It was the first electromagnet — the solenoid-plus-core made useful — and the first machine that could switch a strong magnetic field on and off at will. Joseph Henry soon insulated the wire and packed hundreds of tight turns (raising $n$ , exactly as equation (1) prescribes) to lift three-quarters of a tonne. Sturgeon went on to build the first practical English electric motor — and died poor while the relay, the telegraph sounder and the doorbell built empires on his coil.

History

1820

— Schweigger and Poggendorff wind Ørsted's wire into multiple turns: the coil is born as a way to multiply a feeble effect

1824

— Sturgeon's varnished horseshoe electromagnet lifts twenty times its own weight from one battery cell

1831

— Joseph Henry's insulated, many-turn windings lift 750 pounds at Yale — turn density becomes the design variable

1913

— Kamerlingh Onnes proposes superconducting windings, opening the road to tesla-scale solenoids

1977

— First human MRI scans run inside large-bore superconducting solenoids

2020

— ITER's eighteen D-shaped toroidal field coils begin assembly: the no-leak doughnut at fusion scale

Going further

Texts. NCERT Physics Class 12, Ch. 4 (Moving Charges and Magnetism, §4.8) — the CBSE/NEET canonical treatment · H C Verma, Concepts of Physics II, Ch. 35 · CIE 9702 syllabus topic 20.4 (field-pattern scope) · Griffiths, Introduction to Electrodynamics §5.3 for the full symmetry arguments (UG deep dive).

Provenance. Standard results per NCERT Ch. 4 and Griffiths §5.3. A verifier should recompute: $n = 1000/0.50 = 2000$ , $B = 4\pi\times10^{-7}\times2000\times3.0 = 7.54\ \mathrm{mT}$ , end $\approx 3.8\ \mathrm{mT}$ ; iron core $400\times7.54\ \mathrm{mT} = 3.0\ \mathrm{T}$ capped by saturation ≈ 2 T; toroid $N=1500$ , $I=2.0$ : $B = 2\times10^{-7}\times3000/0.25 = 2.4\ \mathrm{mT}$ (2.5 inner, 2.3 outer); bent-solenoid check $n = 1500/(2\pi\times0.25) = 955$ , $\mu_0 n I = 2.4\ \mathrm{mT}$ ; MRI $nI = 1.5/(4\pi\times10^{-7}) = 1.19\times10^{6}\ \mathrm{A\cdot turns/m}$ .

Prerequisites. Ampère's Circuital Law · Fields of Straight Wires & Circular Loops. Leads to. Parallel Currents & the Definition of the Ampere · Net Fields of Wire Systems & Zero-Field Points · Self-Inductance & Inductors (coming up in the induction chapter — the solenoid's other star role) · Ampère's Law (UG deep dive).

Why doesn't the solenoid's field depend on its radius? Surely a wider coil has its wires farther away.

Two effects cancel exactly. Widen the coil and each turn's wire is indeed farther from an interior point — but each turn is also longer, and the geometry recruits its contribution more efficiently. The rectangle derivation shows the cancellation without any calculation: the inner leg of the rectangle can sit anywhere inside the bore, near the wall or on the axis, in a fat solenoid or a thin one, and the algebra $B\ell = \mu_0 n\ell I$ never notices. Whatever isn't in the equation can't be in the answer. The same logic explains the uniformity: no term says where inside you are, so every interior point must agree.

Is the field outside a real solenoid actually zero? My lab solenoid attracts a paperclip from outside the end.

Not exactly zero, and your paperclip has spotted why. First, real solenoids are finite: the field lines filling the bore must return through the outside space, so a weak, spread-out return field exists everywhere around the coil — strongest near the mouths, which is exactly where the paperclip jumps. Second, a real winding advances like a screw thread, so the coil carries a small net current along its length, wrapping a faint straight-wire field around the outside. "Zero outside" is exact only for the ideal infinitely long solenoid. The exam-safe phrasing: the exterior field is negligible compared with the interior field — near the middle, hundreds of times weaker.

Which formula do I use if the problem's solenoid is bent into a circle?

That's a toroid — use $B = \mu_0 NI/2\pi s$ with the total turn count and $s$ from the central axis. But check what the question wants: if the tube is thin compared with the ring (say a 2 cm tube on a 25 cm ring), the two formulas agree to about 4% and examiners often accept $\mu_0 n I$ with $n = N/2\pi R_0$ . The worked example's Step 3 is exactly this check. If inner and outer radii differ a lot, only equation (3) is honest — and the question probably wants the inner-vs-outer comparison too.

What actually makes the field inside stronger — more turns, more current, or more length?

More turns per metre and more current, in direct proportion each; more length changes nothing. This trips people because $N$ appears to help — but lengthen the coil while adding turns at the same density and $n$ is unchanged, so $B$ is too. The practical knobs: raise $I$ (limited by $I^2R$ heating — hence water-cooled or superconducting magnets), pack turns tighter (limited by wire thickness), or insert an iron core ( $\times\mu_r$ , capped by saturation near 2 T).

How can the toroid's field be exactly zero outside when a solenoid's is only approximately zero?

Closure. The solenoid's problem is its ends: interior field lines must exit and return through outside space, so a finite solenoid must leak. The toroid has no ends — its field lines close into circles that never leave the tube, and the Amperian counting argument (every outside path threads each turn twice, in and out) makes the cancellation a theorem rather than an approximation. The one honest footnote: a real toroid's winding creeps around the ring as it winds around the tube, leaving a tiny effective single-loop current around the doughnut; precision toroids cancel it with a return wire. On paper and on papers, exterior zero is exact.

Why does putting iron inside multiply the field so enormously?

Iron atoms are themselves tiny magnets (circulating and spinning electrons — the same loop-as-magnet idea coming up in Current Loops as Magnetic Dipoles). In unmagnetised iron they point every which way and cancel. The solenoid's modest field nudges them into alignment, and their combined contribution dwarfs the original: the coil supplies the instruction, the iron supplies most of the field. That's the meaning of $\mu_r$ in the hundreds or thousands. The catch is stock-taking: once nearly all the atomic magnets are aligned, there's nothing left to recruit — saturation, at roughly 2 T for iron — after which the core adds nothing and the coil is on its own again.

Do I need Ampère's law to answer solenoid questions, or can I just memorise B = μ0nI?

For the number, the formula suffices. But three recurring conceptual questions are unanswerable from the memorised formula alone: why the field is uniform across the bore, why the radius doesn't matter, and why the toroid's exterior is exactly zero. All three become one-line answers if you can sketch the rectangle or the circle and count threaded turns. JEE-Advanced in particular likes assertion-reason items where the formula is the assertion and the Amperian argument is the reason.