ARCHETYPE: Coils in Fields — Torque & Equilibrium

Hang a current-carrying coil in a uniform magnetic field and it does something a bar magnet on a string has always done: it twists. The field grabs the coil's magnetic moment and swings it toward alignment — and the moment you oppose that swing with anything, a spring, a hinge, a hanging weight, your own steadying hand, you have built the single most-recycled torque problem in Indian physics examinations. CBSE asks for the torque itself, two or three marks of clean substitution. JEE-Main hangs the coil from a hinge and asks for the equilibrium angle. JEE-Advanced attaches a spring, reverses the current mid-problem, and watches you decide which way everything snaps. The formula is not the challenge — $\tau = NIAB\sin\theta$ arrived fully derived in Current Loops as Magnetic Dipoles: Torque & Moment, built face-by-face from $F = BIL$ . The challenge is the bookkeeping: which angle is $\theta$ (from the normal, never the plane — half the wrong answers in this archetype die on that one word), which torque fights which, what work the field does versus what work you do, and what changes the instant a current reverses. This archetype drills exactly that bookkeeping: the master equations, the four equilibrium costumes examiners rotate — free coil, plane-angle trap, hinged coil against gravity, spring-held coil — and the work-to-rotate ladder that connects torque problems to energy problems in one line.

Key takeaways

A coil of $N$ turns, current $I$ , area $A$ in a uniform field $B$ feels zero net force but a torque $\tau = NIAB\sin\theta = mB\sin\theta$ , with $m = NIA$ and $\theta$ measured from the coil's normal to $\vec B$ .
If the question gives the angle $\alpha$ between the plane and the field, use $\theta = 90° - \alpha$ : torque $= NIAB\cos\alpha$ . Plane parallel to field ⟹ maximum torque; plane perpendicular ⟹ zero.
Equilibrium problems are torque ledgers: magnetic torque $mB\sin\theta$ (or $mB\cos\theta$ , geometry decides) balanced against spring torque $kx\cdot(\text{arm})$ or gravity torque $Mg\cdot(\text{arm})$ .
Work to rotate from $\theta_1$ to $\theta_2$ : $W = mB(\cos\theta_1 - \cos\theta_2)$ — from the energy $U = -mB\cos\theta$ of Current Loops as Magnetic Dipoles. Aligned → anti-aligned costs $2mB$ .
Shape never matters: any planar coil with the same $m = NIA$ feels the same torque — circles, squares, blobs.
KEY IDEA: every coil-torque problem is the same ledger — write $mB\sin\theta$ with $\theta$ from the normal, write the opposing torque about the same axis, and balance or integrate.

The core insight

In a uniform field a current coil experiences the torque $\tau = NIAB\sin\theta$ about its own centre ( $\theta$ from the normal); every exam variant — plane-angle, hinged, spring-held, work-to-rotate — is this one torque balanced against, or integrated against, something mechanical.

Where this lives today

The torque-balance ledger runs the modern optics lab. A galvo scanner — the device steering laser light in lidar units, barcode readers, laser shows, and confocal microscopes — is a small coil (or magnet-and-coil pair) on a torsion suspension: drive current in, magnetic torque twists the mirror against the spring, and the equilibrium angle tracks the current exactly as in Competitive 2, thousands of times per second. MEMS micromirror chips shrink the same balance to silicon hinges micrometres wide. And the archetype's laboratory grandparent still matters: torsion-balance magnetometers held oscillating magnets in known fields for a century of Earth-field surveys — an inheritance this track picks up again in Earth's Magnetism: Dip, Declination and B_H and the dip-and-declination problem gym coming later in E5.

The setup

One canonical picture powers the whole gym. A planar coil — the exam favourite is a rectangle of width $a$ and height $b$ , but the algebra never uses the shape — carries current $I$ through $N$ turns inside a uniform field $\vec B$ . The coil's identity card is its magnetic moment $\vec m = NI\vec A$ , a single arrow along the normal to the plane (right-hand curl: fingers with the current, thumb gives $\vec m$ — the drill for this lives in Direction Triage — Hands, Thumbs & Cross Products). The field wants $\vec m$ parallel to $\vec B$ and twists with $\tau = mB\sin\theta$ until something says no.

Figure

The canonical coil-torque setup — one moment arrow, one angle, one twist toward alignment

Why zero net force but nonzero torque? In a uniform field the force on each current segment is matched by an equal and opposite force on the segment across the coil — the pushes cancel as a total but act along different lines, leaving a pure couple. That cancellation argument, and the face-by-face construction of the couple, is Current Loops as Magnetic Dipoles: Torque & Moment's derivation; here we take its result and put it to work. The one geometric fact to internalise before any problem: $\theta$ lives between $\vec m$ and $\vec B$ — between the normal and the field. Every other angle in the question is a costume.

The master equations

The torque itself, quoted from Current Loops as Magnetic Dipoles with the moment $m = NIA$ packaging the coil:

\tau = NIAB\sin\theta = mB\sin\theta \qquad (1)

where: $N$ — turns; $A$ — area of any planar shape; $\theta$ — the angle between the normal $\vec m$ and $\vec B$ . Read it as a dial: $\theta = 0$ (plane perpendicular to field, moment aligned) gives zero torque and stable rest; $\theta = 90°$ (plane parallel to the field — the counter-intuitive one) gives the maximum $\tau_{\max} = NIAB$ ; $\theta = 180°$ gives zero again but balanced on a knife's edge. A galvanometer manufacturer's whole trick, as The Moving-Coil Galvanometer tells it, is to bend the field radial so the coil always sits at the $\sin\theta = 1$ sweet spot — the engineered escape from this dial. When the coil must hold an angle, equation (1) enters a ledger against a mechanical torque about the same axis:

mB\sin\theta = \tau_{\text{mech}} \qquad (2)

where: $\tau_{\text{mech}}$ — spring force times its arm ( $kx \cdot r$ ), or weight times its horizontal arm ( $Mg \cdot r_\perp$ ), or whatever the examiner has bolted on. The discipline is mechanical, straight from Torque and Rigid-Body Equilibrium in mechanics: pick the rotation axis once, express every torque about that same axis, set the sum to zero. The magnetic side is always one term, equation (1) in some trigonometric costume fixed by the geometry. And when the coil must move between angles, torque problems become energy problems. The orientation energy $U = -mB\cos\theta$ (statement and stability reading in Current Loops as Magnetic Dipoles) prices every angle, so rotating from $\theta_1$ to $\theta_2$ costs

W_{\text{ext}} = U_2 - U_1 = mB\,(\cos\theta_1 - \cos\theta_2) \qquad (3)

where: $W_{\text{ext}}$ — work done by the external agent rotating slowly (the field does $-W_{\text{ext}}$ ). Two landmarks worth memorising: aligned to perpendicular ( $0° \to 90°$ ) costs $mB$ ; aligned to anti-aligned ( $0° \to 180°$ ) costs $2mB$ , the price of a full flip. Equation (3) is the sole equation in this gym that is not a balance — it is a bill.

Derivation AWork to rotate a coil from the energy ledger

We want equation (3), the work an external agent must supply to rotate a coil quasi-statically from angle $\theta_1$ to $\theta_2$ in a uniform field. The strategy is pure energy bookkeeping: Current Loops as Magnetic Dipoles prices each orientation at $U = -mB\cos\theta$ , and for slow rotation (no kinetic energy appears) the agent pays exactly the difference in stored energy. This route needs no integral of torque — the energy function has already done that integral once and for all. Watch the sign convention: $W_{\text{ext}}$ is what you pay; the field's own work is its negative.

Step 1. Quote the price list: each orientation of the moment in the field carries potential energy $U(\theta) = -mB\cos\theta$ — lowest ( $-mB$ ) when aligned, highest ( $+mB$ ) when anti-aligned.

Step 2. Rotate slowly from $\theta_1$ to $\theta_2$ . Slow means the coil starts and ends at rest: no kinetic energy term enters the ledger.

Step 3. Energy conservation for the agent-plus-field system: the agent's work equals the rise in stored energy,

W_{\text{ext}} = U(\theta_2) - U(\theta_1) = (-mB\cos\theta_2) - (-mB\cos\theta_1) = mB(\cos\theta_1 - \cos\theta_2)

Why "slow" matters: rotate fast and part of your work becomes rotational kinetic energy, and equation (3) becomes an inequality — the exam phrase "slowly" or "quasi-statically" is the licence for the clean formula.

Step 4. The work done by the magnetic field is the mirror entry: $W_{\text{field}} = -W_{\text{ext}} = mB(\cos\theta_2 - \cos\theta_1)$ — positive when the coil swings toward alignment, as a restoring twist should. Verified: landmarks. $\theta_1 = 0, \theta_2 = 90°$ : $W_{\text{ext}} = mB(1 - 0) = mB$ ✓. Full flip $0 \to 180°$ : $mB(1-(-1)) = 2mB$ ✓. Round trip $\theta_1 = \theta_2$ : zero — the energy is a function of position only ✓. Numbers from Board 2 below: $m = 0.628$ A·m², $B = 0.20$ T, $0 \to 60°$ : $0.628 \times 0.20 \times (1 - 0.5) = 0.063$ J ✓.

Figure

The two equilibrium costumes — a hinged coil balancing gravity and a spring-held coil at maximum torque

The four costumes examiners rotate

variant	what is given	the working equation
free torque	$N, I, A, B$ and an angle	$\tau = NIAB\sin\theta$ — first decode whether the angle is from the normal or the plane
plane-angle trap	"plane makes angle $\alpha$ with $\vec B$ "	$\theta = 90° - \alpha$ , so $\tau = NIAB\cos\alpha$
hinged against gravity	hinged coil, mass $M$ , deflection $\theta$	$mB \times (\text{trig of } \theta) = Mg \cdot r_\perp$ about the hinge
spring-held	spring $k$ at arm $r$ from the axis	$mB\sin\theta = kx\,r$ ; current reversal flips the magnetic side
work ladder	rotate from $\theta_1$ to $\theta_2$	$W_{\text{ext}} = mB(\cos\theta_1 - \cos\theta_2)$

Limiting cases

case	result	meaning
$\theta = 0$ (plane ⊥ $B$ )	$\tau = 0$ , $U = -mB$	stable equilibrium — the coil's parking spot
$\theta = 90°$ (plane ∥ $B$ )	$\tau = mB$ (max), $U = 0$	strongest twist exactly where the flux through the coil is zero
$\theta = 180°$	$\tau = 0$ , $U = +mB$	unstable — any nudge and the coil flips all the way
doubled turns, same wire length	$N \to 2N$ , $A \to A/\sqrt{?}$	not simply $2\tau$ : fixed wire length means smaller loops — read the stem before scaling
any planar shape, same $NIA$	same $\tau$	the moment is the whole identity: circle, square, blob — the field cannot tell

Common trapThe angle in the torque formula is the angle the plane makes with the field

It is not — and this single confusion outsells every other error in the archetype. $\theta$ in $\tau = NIAB\sin\theta$ is measured from the normal (the moment arrow $\vec m$ ) to $\vec B$ . When a stem says "the plane of the coil makes $30°$ with the field", the normal makes $60°$ , and the torque carries $\sin 60° = 0.866$ , not $\sin 30° = 0.5$ — the wrong reading is always planted among the options. Fastest safe habit: convert every angle to the normal's angle before touching the formula, and remember the two anchors — plane parallel to field means maximum torque, plane perpendicular means zero. If your intuition objects that "parallel should mean aligned and calm", it is picturing the plane; the moment is then square-on to the field and maximally twisted.

Worked examples

Board level 1 — the plane-angle torque (CBSE/ISC pattern, 3 marks)

A square coil of side $10$ cm has $20$ turns and carries $12$ A. Its plane makes an angle of $30°$ with a uniform field $B = 0.80$ T. Find the magnitude of the torque on the coil.

Step 1. Decode the angle. The plane is at $\alpha = 30°$ to $\vec B$ , so the normal is at $\theta = 90° - 30° = 60°$ .

Step 2. Assemble the moment pieces: $A = (0.10)^2 = 1.0\times10^{-2}$ m², so $NIA B = 20 \times 12 \times 1.0\times10^{-2} \times 0.80 = 1.92$ N·m — this is $\tau_{\max}$ , the value at $\sin\theta = 1$ .

Step 3. Apply the true angle:

\tau = NIAB\sin 60° = 1.92 \times 0.866 = \boxed{1.66\ \text{N}\,\cdot\,\text{m}}

Step 4. Sanity: the planted distractor $1.92 \times \sin 30° = 0.96$ N·m is what the plane-angle misreading produces — it will be among the options. Yours is the larger value, as it must be: a plane at only $30°$ to the field is close to the maximum-torque orientation, not the calm one.

Board level 2 — the work ladder (CBSE/ISC pattern, 3 marks)

A circular coil of $50$ turns and radius $4.0$ cm carries $2.5$ A. It sits in a uniform field $B = 0.20$ T with its moment aligned with the field. Find (a) the work needed to rotate it slowly to $60°$ from the field, and (b) the torque holding it there.

Step 1. The moment: $m = NIA = 50 \times 2.5 \times \pi(0.04)^2 = 50 \times 2.5 \times 5.03\times10^{-3} = 0.628$ A·m².

Step 2. Work, equation (3), from $\theta_1 = 0$ to $\theta_2 = 60°$ :

W_{\text{ext}} = mB(\cos 0° - \cos 60°) = 0.628 \times 0.20 \times (1 - 0.5) = \boxed{6.3\times10^{-2}\ \text{J}}

Step 3. Holding torque at $60°$ — the external agent must supply exactly what the field exerts:

\tau = mB\sin 60° = 0.628 \times 0.20 \times 0.866 = \boxed{0.109\ \text{N}\,\cdot\,\text{m}}

Step 4. Sanity: a full flip would cost $2mB = 0.251$ J; our quarter-ish rotation costing $0.063$ J sits sensibly inside that ceiling.

Competitive level 1 — the hinged coil against gravity (JEE-Main/NEET pattern)

A rectangular coil of $50$ turns, width $a = 10$ cm (the hinged side, horizontal) and height $b = 20$ cm, mass $M = 0.40$ kg, hangs vertically from a frictionless hinge along its top edge. A uniform vertical field $B = 0.20$ T points upward, and the coil carries $I = 2.0$ A in the sense that makes the field lift it. Find the angle from the vertical at which it hangs in equilibrium. Take $g = 10$ m/s².

Step 1. Set the ledger about the hinge. When the plane tips by $\theta$ from the vertical, the moment $\vec m$ (normal to the plane) sits at $90° - \theta$ from the vertical field, so the magnetic torque about the hinge is $\tau_B = mB\sin(90° - \theta) = NIabB\cos\theta$ .

Step 2. Gravity pulls at the centre, $b/2$ down the plane from the hinge; its horizontal arm is $(b/2)\sin\theta$ , so $\tau_g = Mg(b/2)\sin\theta$ , opposing the lift.

Step 3. Balance and watch $b$ cancel:

NIabB\cos\theta = Mg\frac{b}{2}\sin\theta \;\;\Rightarrow\;\; \tan\theta = \frac{2NIaB}{Mg}

Step 4. Numbers: $\tan\theta = \dfrac{2 \times 50 \times 2.0 \times 0.10 \times 0.20}{0.40 \times 10} = \dfrac{4.0}{4.0} = 1$ , so $\boxed{\theta = 45°}$ .

Step 5. Sanity: both torques at $45°$ evaluate to $0.283$ N·m — the ledger closes. And the height $b$ dropping out is the question's hidden teaching: doubling the coil's height doubles both the area (more magnetic torque) and the weight's arm in exact proportion. Timing note: three minutes if you write both torques about the hinge immediately; ten if you try to track the four side-forces separately. The moment $\vec m$ exists precisely so you never touch individual sides in a uniform field.

Competitive level 2 — spring balance and the current reversal (JEE-Advanced flavour)

A square coil of side $10$ cm, $100$ turns, carries $1.0$ A. It is mounted on a frictionless vertical axis through its centre, with a uniform horizontal field $B = 0.50$ T lying in the plane of the coil. A horizontal spring of constant $k = 500$ N/m, attached at the midpoint of one vertical side (arm $= 5.0$ cm) and acting perpendicular to the plane, holds it in equilibrium. Find (a) the spring's extension, (b) the net torque on the coil at the instant the current is reversed, and (c) the final state.

Step 1. Magnetic torque in this orientation: the plane contains $\vec B$ , so $\theta = 90°$ and $\tau_B = NIAB = 100 \times 1.0 \times (0.10)^2 \times 0.50 = 0.50$ N·m — the maximum the field can exert on this coil.

Step 2. Spring ledger about the axis: $kx \times r = \tau_B$ , so

x = \frac{\tau_B}{k\,r} = \frac{0.50}{500 \times 0.050} = 0.020\ \text{m} = \boxed{2.0\ \text{cm}}

Step 3. Reverse the current. The moment $\vec m$ flips $180°$ , so the magnetic torque reverses direction at the same magnitude $0.50$ N·m. The spring, still stretched $2.0$ cm at that first instant, keeps pulling the same way as before — which is now the same sense as the new magnetic torque:

\tau_{\text{net}} = 0.50 + 0.50 = \boxed{1.0\ \text{N}\,\cdot\,\text{m}}

Step 4. Final state: the coil swings through the spring's natural length and settles mirrored — spring compressed $2.0$ cm (a coil spring pushes as well as pulls), magnetic torque again balanced at $0.50$ N·m. The snap direction is always toward the reversed moment's new alignment.

Step 5. Sanity: the momentary doubling is the signature of every reversal problem — the restorer becomes a conspirator for one instant. If an option says "net torque zero just after reversal", it has forgotten the spring is still stretched. Timing note: four minutes. Parts (a) and (c) are symmetric twins; the only thinking is part (b)'s "who points which way now" — sketch the two torque arrows before computing anything.

Exam traps

The plane-angle swap. Angle from the plane $\alpha$ means $\sin\theta \to \cos\alpha$ . Both wrong readings ( $\sin\alpha$ kept, or angles mixed mid-problem) sit among the options every single year. Convert to the normal's angle on line one.

Dropping $N$ . The moment is $NIA$ , not $IA$ — a hundred-turn coil is a hundred times the torque. Symmetric slip: using $N$ twice, once inside a quoted $m$ and once again in the formula.

Area units. Sides arrive in centimetres; $A$ must be in m². A $10$ cm square is $10^{-2}$ m², and the $10^{-4}$ -per-cm² conversion silently misplaced is two orders of magnitude gone.

Work signs. Equation (3) is the external agent's bill for slow rotation; the field's work is its negative. "Work done by the field in rotating from aligned to $90°$ " is $-mB$ , not $+mB$ . And a full flip costs $2mB$ , never zero — $U$ returns to the same magnitude but opposite sign.

Equilibrium without stability. $\tau = 0$ at both $\theta = 0$ and $\theta = 180°$ ; only the first is stable. If the question asks "will it return when nudged", answer from $U$ : minimum at alignment, maximum anti-aligned.

Reversal problems solved as fresh equilibria. The instant after a reversal, the mechanical partner (spring, gravity) is still in its old configuration — torques momentarily add. Compute the instant first, the new equilibrium second.

Constants. Pure torque problems need no $g$ ; hinged and hanging variants do — CBSE/NEET run $g = 9.8$ m/s² unless told otherwise, JEE stems usually hand you $g = 10$ m/s² (Competitive 1 above says so explicitly). State whichever you use.

Origins · 1 minJoseph Henry and how this was discovered

Joseph Henry

1797–1878 · Albany Academy and the Smithsonian Institution

Joseph Henry taught at a boys' academy in Albany, New York, wound insulated wire in his spare hours, and in 1831 built what he cheerfully called a "philosophical toy": a wire rocker that see-sawed endlessly between two battery cups, tipped back and forth by the torque of a magnetic field on its current. It was the first machine to turn electromagnetic torque into sustained motion — the ancestor of every motor, and precisely this archetype's physics welded into brass: a current-carrying conductor in a field, twisting against gravity, over and over. Henry went on to build the strongest electromagnets of his era, to co-discover induction, and to run the new Smithsonian Institution; the SI unit of inductance carries his name. But the rocking engine came first — equilibrium problems, escaping equilibrium once a second.

History

1820

— Ørsted's needle and Ampère's force experiments establish that currents both make and feel magnetic fields.

1821

— Faraday builds his electromagnetic rotator: a current-carrying wire circles a magnet, continuous motion from magnetic force.

1831

— Henry's rocking-beam engine: a coil see-saws between torque and gravity — this archetype as working machinery.

1882

— d'Arsonval tames the same torque with a spiral spring and a radial field: the measuring instrument descends from the toy.

2019

— micro-mirror scanners in lidar and projectors twist on the same ledger, kilohertz-fast, micrometres wide.

Going further

Texts. NCERT Physics Class 12, Ch. 4 (Moving Charges and Magnetism — torque on a current loop, worked example with plane angles) · HC Verma, Concepts of Physics Vol. 2, Ch. 34 exercises (hinged loops, work to rotate) · Irodov, Problems in General Physics §3.2 (spring-held and hinged coil stretches).

Provenance. All numbers recomputed for this article: $\tau = 1.66$ N·m with distractor $0.96$ (Board 1); $m = 0.628$ A·m², $W = 6.3\times10^{-2}$ J, $\tau = 0.109$ N·m, flip ceiling $0.251$ J (Board 2); $\tan\theta = 1$ , both torques $0.283$ N·m (Competitive 1); $x = 2.0$ cm, momentary $1.0$ N·m (Competitive 2).

Prerequisites. Current Loops as Magnetic Dipoles: Torque & Moment · Force on a Current-Carrying Conductor. Related. The Moving-Coil Galvanometer (the spring-balanced special case) · Torque and Rigid-Body Equilibrium (mechanics) · Forces & Torques on Magnetic Dipoles (UG deep dive). Coming up. The Rod on Rails and its energy audits in E6 · dip-and-declination torsion problems later in E5.

Why is there a torque but no net force on the coil?

Because the field is uniform. Each segment of current feels $F = BIL$ , but the segment carrying the same current in the opposite direction across the coil feels the exact opposite force — the vector sum over the whole loop is zero. The forces cancel as a total yet act along different lines, and a pair of equal, opposite, offset forces is a couple: pure twist, no shove. In a non-uniform field the cancellation fails and the coil is also pulled bodily — that force story lives at the UG level, not in this gym.

Why is the torque maximum when the plane is parallel to the field? That feels backwards.

Your intuition is tracking the plane; the physics tracks the normal. Plane parallel to $\vec B$ means the moment arrow $\vec m$ is square-on to the field — $90°$ of misalignment, the most the field can object to. Equivalently, in that orientation the side-forces on the coil have the longest lever arms about the rotation axis. When the plane is perpendicular to the field ( $\vec m$ aligned), the same side-forces still exist but pull radially outward — they stretch the coil and turn nothing.

Who actually does the work when a coil rotates — me or the field?

Both ledgers exist and exams sample each. Rotating against the field's preference (toward anti-alignment), you pay $W_{\text{ext}} = mB(\cos\theta_1 - \cos\theta_2) > 0$ and the field does the negative of that. Released to swing back, the field's torque does positive work, converting orientation energy to kinetic. What the magnetic force on free charges never does is work on a moving charge directly — that no-work theorem belongs to Force on a Moving Charge — but torque on a current-carrying rigid body transferring energy through the wire's structure is a different, legitimate channel.

Does the shape of the coil really never matter?

For torque in a uniform field, never: the derivation in Current Loops as Magnetic Dipoles shows any planar loop can be tiled into rectangles whose internal edges cancel, so only $m = NIA$ survives. A circle, a square and an amoeba with equal $NIA$ are indistinguishable to the field. Shape re-enters only when the field varies across the coil, or when you ask for forces within the loop (the stretching tension), or in induction problems later in E6 where the flux geometry counts.

In spring problems, why does the torque double at the instant of current reversal?

Because two independent agents each keep their torque. Reversing $I$ flips the magnetic torque instantly — electrons do not care about the spring. The spring's torque depends on its stretch, which cannot change instantly. So for one instant old-spring-torque and new-magnetic-torque point the same way: $0.50 + 0.50 = 1.0$ N·m in Competitive 2. The coil then accelerates, the stretch changes, and the system hunts for the mirrored equilibrium. Any answer built on "equilibrium, therefore zero" at that instant has confused a snapshot with a steady state.

Why does the coil height cancel in the hinged-coil problem?

Follow $b$ through the ledger: the magnetic torque is $NI(ab)B\cos\theta$ — one power of $b$ through the area — while gravity's torque is $Mg(b/2)\sin\theta$ — one power of $b$ through the arm. Same power both sides, cancel. Physically: making the coil taller buys more area for the field to push on and, in exactly the same proportion, a longer lever for gravity to resist with. Examiners love planting an unused $b$ value to reward students who notice; if your final formula still contains $b$ , re-derive.

Why does a galvanometer not need any of this angle bookkeeping?

Because its designers deleted the angle. As The Moving-Coil Galvanometer explains, the concave pole pieces and iron core make the field radial: whatever the coil's rotation, the field in the gap lies in the coil's plane, pinning the geometry at the $\sin\theta = 1$ orientation permanently. The torque is then $NIAB$ at every deflection — linear in $I$ — and the only ledger left is against the hairspring, $NIAB = k\varphi$ . This gym is the general case; the galvanometer is the special case money can buy.