Force on a Current-Carrying Conductor

In short

A straight wire of length L carrying current I in a uniform magnetic field \vec B feels a force

\boxed{\;\vec F \;=\; I\,\vec L \times \vec B\;}, \qquad |\vec F| = BIL\sin\theta,

where \vec L points in the direction of conventional current and \theta is the angle between \vec L and \vec B. The direction is given by the right-hand rule: fingers from \vec L to \vec B, thumb points along \vec F.

Two parallel wires a distance d apart carrying currents I_1, I_2 exert a force per unit length on each other:

\frac{F}{L} \;=\; \frac{\mu_0 I_1 I_2}{2\pi d},

attractive when the currents flow the same way, repulsive when they flow opposite ways. The ampere used to be defined through this force — one ampere is the current that produces 2\times 10^{-7} N per metre between two infinitely long parallel wires one metre apart in vacuum.

A rectangular current loop of N turns, area A, placed in a uniform field \vec B, experiences a torque

\boxed{\;\tau \;=\; NIAB\sin\theta\;},

where \theta is the angle between the loop's normal and \vec B. This is the engine of every electric motor, every galvanometer, every moving-coil loudspeaker in the country.

Stand beside any Mumbai local train as it pulls out of Dadar station. What you hear as the rising whine is an electric motor driving wheels that each weigh half a tonne — the motor is lifting fifteen hundred people out of the platform. What physically pushes the train forward? A wire, carrying current, sitting in a magnetic field. The field pushes the wire sideways. Multiply that sideways push over thousands of turns of wire wound around an iron core and you get torque. Connect that torque to a gearbox and you get a train.

The same one-line idea — a current in a magnetic field feels a force — drives the pantograph pushing up against the 25 kV catenary above the tracks, the tiny motor in the ceiling fan over your bed, the voice coil vibrating the speaker in your earphones, the flashing needle in a school-lab galvanometer. Every one of these is an application of a single equation, \vec F = I\vec L \times \vec B. This article derives that equation, gives you the right-hand rule to predict its direction, computes the force between two parallel wires, arrives at the official definition of the ampere, and finally puts a current loop in a uniform field and extracts the torque formula that powers every motor on earth.

From "a charge in a field" to "a wire in a field"

You already know (or will meet in the Lorentz force article) that a point charge q moving with velocity \vec v in a magnetic field \vec B feels

\vec F_\text{charge} \;=\; q\,\vec v \times \vec B.

A wire carrying current is not a single charge — it is a river of charges. Inside a copper conductor, each free electron drifts with a small average velocity \vec v_d (the drift velocity, typically \sim 10^{-4} m/s for a household current — slower than an ant crawls). There are a lot of them: copper has about n = 8.5\times 10^{28} free electrons per cubic metre. The force on one electron is tiny, but the force on every electron in the wire adds up coherently because they all drift in the same direction. That is what gives you a measurable push on the wire.

Let the wire have cross-sectional area A and length L, with n free charge carriers per unit volume, each of charge q, all drifting at velocity \vec v_d. The wire sits in a uniform magnetic field \vec B.

Step 1. Count the carriers in the wire. The wire's volume is AL, so the number of carriers is

N_\text{carriers} \;=\; n A L.

Why: volume times number-density. If the wire were a bottle of water and the carriers were dissolved sugar molecules, this is how you would count them.

Step 2. Write the force on one carrier and sum over all of them. Each carrier feels q\vec v_d \times \vec B, and they all drift in the same direction, so the total force is

\vec F \;=\; N_\text{carriers}\;(q\,\vec v_d \times \vec B) \;=\; (nAL)\,q\,\vec v_d \times \vec B.

Why: the carriers drift coherently. Unlike their chaotic thermal velocities (which point in random directions and average to zero), the drift motion adds up. Only the drift contributes.

Step 3. Recognise that nAqv_d is the current I. The current density is J = nqv_d, and the current through the wire is I = JA = nAqv_d. Substituting:

\vec F \;=\; L\;(nAqv_d)\,\hat L \times \vec B \;=\; IL\,\hat L \times \vec B,

where \hat L is the unit vector along the direction of conventional current (opposite to electron drift for electrons, but the sign of q takes care of itself — see the pitfalls section).

Step 4. Combine L and \hat L into the vector \vec L = L\hat L, a vector of length equal to the wire and pointing along the current.

\boxed{\;\vec F \;=\; I\,\vec L \times \vec B\;}

Why: this is the whole force, packaged neatly. I is what you read off the ammeter; \vec L is the geometry of the wire (length and direction); \vec B is the field at the wire. The cross product handles the perpendicular-to-both rule automatically. Notice: the expression does not involve the drift velocity explicitly. The slow-moving electrons have disappeared into the current I, and the force no longer cares that they are slow — it only cares how much charge is flowing per second.

Magnitude. If the wire makes an angle \theta with the field, the cross product gives

|\vec F| \;=\; BIL\sin\theta.

The force is maximum when the wire is perpendicular to the field (\theta = 90°, \sin\theta = 1) and zero when the wire is parallel to the field (\theta = 0°). A wire lying along the field lines feels nothing — the magnetic force only bites when the wire crosses field lines.

The right-hand rule — which way does it push?

The formula \vec F = I\vec L \times \vec B tells you the direction by the cross product. If you prefer the physical gesture, here is the right-hand rule in the form that actually helps at an exam desk:

Point your right hand's fingers along the current \vec L.
Curl them towards the field \vec B.
Your thumb points along \vec F.

Equivalently (the version physicists actually use in the lab): flatten your right hand, index finger along \vec L, middle finger along \vec B, and your thumb sticks out along \vec F at right angles to both.

The wire carries current to the right ($\vec L$ points right). The field $\vec B$ points out of the page (shown by the dots — the tip of an arrow coming at you). Fingers of your right hand along $\vec L$, curl them to $\vec B$ (rotate up out of the page): your thumb points out of the page. Force is out of the page.

A quick mnemonic worth carrying into the exam hall: FBI — Force (thumb), B field (first/index finger), I current (middle finger). The three are mutually perpendicular, right-hand orientation. If you mix up which finger is which, try it on a known case — current going east, field going north, force should be up. Adjust fingers until that works, then don't let go.

Deriving the formula end-to-end — a numerical check

A wire 0.5 m long carries 3 A through a uniform magnetic field of 0.4 T at right angles. The force is

|\vec F| \;=\; BIL\sin\theta \;=\; (0.4)(3)(0.5)\sin 90° \;=\; 0.6\;\text{N}.

That is about the weight of a 60 g object — a cricket ball cut in half, or a small lemon — felt continuously on a thin wire. A modest field and a modest current produce a force you can easily notice with your hand. Scale it up to the 750 A flowing through a Metro motor with the pole windings producing a 1.2 T field over several metres of effective conductor, and you are lifting trains.

What if the wire is not straight? Break it into tiny pieces d\vec L, each feeling d\vec F = I\,d\vec L \times \vec B, and integrate. For a curved wire in a uniform field, the integral simplifies to \vec F = I\,\vec L_\text{net} \times \vec B where \vec L_\text{net} is the straight-line vector from the start of the wire to the end — every curl and wiggle in between cancels when the field is uniform, because what matters is the net displacement. This is a useful trick: for a semicircular loop sitting in a uniform field, the force on the curved arc is the same as on a straight chord joining the endpoints.

Force between two parallel currents

Put two long, straight, parallel wires a distance d apart. The first wire carries current I_1; the second carries I_2. Each wire sits in the magnetic field produced by the other. That field, at distance d from a long straight wire, is (see the Biot–Savart article)

B_1 \;=\; \frac{\mu_0 I_1}{2\pi d},

and it wraps around wire 1 in circles. At the location of wire 2, wire 1's field is perpendicular to wire 2.

Step 1. Compute the force per unit length on wire 2.

The force on a length L of wire 2 is F = B_1 I_2 L (perpendicular current, \sin 90° = 1). Dividing by L:

\frac{F}{L} \;=\; B_1 I_2 \;=\; \frac{\mu_0 I_1 I_2}{2\pi d}.

Why: substitute the straight-wire field B_1. The force per metre is what matters, because wires are usually infinitely long (or long enough that the end effects don't matter).

Step 2. Find the direction using the right-hand rule twice.

Suppose both wires carry current in the same direction (say, both into the page). Wire 1's field at wire 2's location points sideways (say, to the left of wire 2). The force on wire 2 is I_2 \vec L \times \vec B_1: current into the page crossed with field to the left gives a force pointing toward wire 1. Symmetric argument for wire 1 feeling wire 2: the forces attract. The wires pull together.

If the currents are in opposite directions, reverse the sign of I_2 (or equivalently, flip the direction of wire 2's current); the force reverses — the wires repel.

\boxed{\;\frac{F}{L} \;=\; \frac{\mu_0 I_1 I_2}{2\pi d},\qquad\text{attractive if parallel, repulsive if antiparallel}\;}

Left: both currents into the page — the wires attract. Right: one into, one out of the page — the wires repel. The magnitude of the force per metre is $\mu_0 I_1 I_2 / (2\pi d)$ in both cases.

Why this is surprising the first time you see it

Two parallel wires are electrically neutral — neither carries net charge. By Coulomb's law there should be no force between them. Yet there is a force, and it is attractive when the currents are parallel — the opposite of the Coulomb intuition that "like repels like." The reason is that the force here is not electrostatic; it is the magnetic interaction between moving charges. The currents are the moving charges, and they bend each other's paths. The symmetry (parallel attract, antiparallel repel) is opposite to the electric case because the magnetic field circulates around the current rather than emanating from it.

In the Lorentz force article, you will see that this attraction is the special-relativistic partner of the Coulomb repulsion: if you transform into the frame where the charge carriers in one wire are at rest, the charge density of the other wire Lorentz-contracts asymmetrically and becomes slightly non-neutral, pulling on the first wire electrostatically. Magnetism is relativity's fingerprint on electric forces.

Explore the force — drag the current

Watching the formula F/L = \mu_0 I_1 I_2 / (2\pi d) is not the same as feeling how it scales. The interactive below fixes I_2 = 5 A and d = 2 cm and lets you drag I_1 from 0 to 20 A. The red curve is the force per metre; the dashed lines read out the current and the force.

Drag the red point along the horizontal axis to change $I_1$. The force per metre is linear in $I_1$ at fixed $I_2 = 5$ A and $d = 2$ cm. Two 20 A currents 2 cm apart — the kind of currents that flow in a washing-machine starter — produce about a millinewton per metre, barely perceptible. Now imagine the bundled conductors on a Delhi–Kolkata 765 kV transmission line at several thousand amperes: the forces become structural loads the towers must carry.

Definition of the ampere

Before 2019, the SI ampere was defined by exactly this force. The 1948 CGPM definition read:

One ampere is that constant current which, when maintained in two straight parallel conductors of infinite length, of negligible circular cross-section, and placed one metre apart in vacuum, would produce between these conductors a force equal to 2\times 10^{-7} newton per metre of length.

Plug into our formula: I_1 = I_2 = 1 A, d = 1 m,

\frac{F}{L} \;=\; \frac{\mu_0 \cdot 1 \cdot 1}{2\pi \cdot 1} \;=\; \frac{4\pi\times 10^{-7}}{2\pi} \;=\; 2\times 10^{-7}\;\text{N/m}.

The 2\times 10^{-7} is not a coincidence — the old definition chose the ampere's size so that \mu_0 came out to exactly 4\pi\times 10^{-7} T·m/A. Current and the magnetic constant were locked together by this definition.

Since the 2019 SI redefinition, the ampere is instead defined by fixing the elementary charge e = 1.602176634\times 10^{-19} C exactly, and \mu_0 is no longer exactly 4\pi\times 10^{-7} — it is measured. But for every practical purpose at school-level and JEE-level physics, \mu_0 = 4\pi\times 10^{-7} T·m/A is correct to 10 decimal places and the force-between-wires picture is the right physical intuition for the ampere.

Torque on a current loop — the heart of every motor

Put a rectangular wire loop in a uniform magnetic field. The sides of the loop feel forces; the forces can cancel or can combine to produce a torque. This is the principle on which every DC motor, every galvanometer, every loudspeaker works.

Let the loop have sides a (along one direction) and b (the other), current I flowing around it, and N turns wound in the same plane. The magnetic field \vec B is uniform and horizontal. The loop can rotate about a vertical axis through its centre. Let \theta be the angle between the normal to the loop (\vec n, determined by the right-hand rule on the current direction) and the field \vec B.

Looking down on a rectangular loop in a uniform horizontal field $\vec B$. The two sides of length $a$ (red) are perpendicular to $\vec B$ and feel equal-magnitude, opposite-direction forces $F = BIa$. These two forces form a couple — a net force of zero but a non-zero torque. The other two sides (of length $b$) lie along $\vec B$ and feel no force.

Step 1. Identify the forces on the four sides.

The two sides of length a are perpendicular to \vec B, each carrying current I. The force on each is

F_\text{side} \;=\; BIa.

Why: these two sides are perpendicular to \vec B, so \sin\theta = 1 in F = BIL\sin\theta. The current in these two sides flows in opposite directions (one down, one up, since the current goes around the loop), so the two forces point in opposite directions — equal magnitude, antiparallel.

The two sides of length b lie in the plane containing \vec B — they make some angle with \vec B, but crucially the component of their current perpendicular to \vec B is zero when the loop's normal and \vec B are coplanar (which we assume). So these sides contribute zero force in the generic rectangular-loop setup, or equivalently, they contribute forces that cancel axially. Either way, they do not produce the torque we are computing.

Step 2. Compute the torque produced by the two F_\text{side} forces.

The two forces of magnitude BIa act on lines separated perpendicularly by a distance b\sin\theta (where \theta is the angle between the loop normal and \vec B). A pair of equal-and-opposite forces separated by a perpendicular distance d_\perp is a couple, and its torque is simply

\tau \;=\; F \cdot d_\perp \;=\; (BIa)(b\sin\theta).

Why: a force couple produces a torque equal to the force times the perpendicular distance between the lines of action, regardless of the pivot you choose to measure it about. That is a property of couples — they give the same torque about any point.

Step 3. Recognise the area.

ab = A is the area of the loop. So

\tau \;=\; BIA\sin\theta.

Step 4. Account for N turns.

If the loop has N turns wound on top of each other, each turn feels the same torque, and they all add:

\boxed{\;\tau \;=\; NIAB\sin\theta\;}

Why: N turns mean N copies of the same loop, all carrying the same current. The torques stack. This is why motors have windings of hundreds of turns — each turn multiplies the torque without increasing the current.

Step 5. Write it as a cross product.

Define the magnetic moment of the loop as

\vec m \;=\; NI\vec A,

where \vec A = A\hat n points along the loop's normal (right-hand rule: fingers along current, thumb along \vec A). Then the torque is

\vec\tau \;=\; \vec m \times \vec B.

Why: |\vec m \times \vec B| = mB\sin\theta = (NIA)(B)\sin\theta, which matches what we derived. The cross-product form also gives the direction — the torque rotates the loop to align \vec m with \vec B, which is why a compass needle (a tiny current loop of circulating electrons) swings to align with Earth's field.

The loop is in stable equilibrium when \vec m \parallel \vec B (that is, \theta = 0: the loop's normal lies along the field, and the loop's plane is perpendicular to the field). It is in unstable equilibrium when \vec m points against \vec B (\theta = 180°). Everywhere else, the torque drives it toward alignment.

From torque to motor

A motor keeps spinning because the current in the loop reverses every time the loop passes the \theta = 0 stable point. The reversal is done by a commutator — two copper half-rings with brushes that swap the current's direction every half turn, so the torque never reverses sign. The loop is perpetually driven toward an equilibrium it will never reach, because the moment it approaches, the current flips and the equilibrium moves. A pendulum that never settles.

This is the inside of every DC motor on earth — from the micro-motor in a WhatsApp-era mobile vibration module, to the Westinghouse traction motor on a WAP-7 locomotive pulling the Rajdhani from Delhi to Howrah at 140 km/h.

Worked examples

Example 1: Force on a horizontal wire in Earth's field

A 2-m long straight horizontal copper wire carries a steady current of 15 A in the east-to-west direction in Bangalore, where the horizontal component of Earth's magnetic field is B_H = 40\,\mu\text{T} pointing geographic north. Find the magnitude and direction of the magnetic force on the wire.

Plan view: the wire runs east-west; current flows east to west; Earth's horizontal field points north. The cross product $\vec L \times \vec B$ points out of the page (upward, out of the ground).

Step 1. Identify the geometry. \vec L points west (along the current), magnitude 2 m. \vec B points north, magnitude 40\times 10^{-6} T. The angle between them is 90°, so \sin\theta = 1.

Why: east-to-west current and north-pointing field are perpendicular by construction. This is the maximum-force orientation.

Step 2. Compute the magnitude.

|\vec F| \;=\; BIL\sin 90° \;=\; (40\times 10^{-6})(15)(2)(1) \;=\; 1.2\times 10^{-3}\;\text{N} \;=\; 1.2\;\text{mN}.

Why: plug numbers straight in. The answer is about the weight of a grain of rice — Earth's field is very weak, so the force is gentle unless the current is huge.

Step 3. Find the direction by the right-hand rule.

Fingers along \vec L (west), curl to \vec B (north): thumb points up, out of the ground.

Why: west cross north = up, by the standard right-hand orientation of the compass rose (N-E-S-W reading anticlockwise from above). Alternatively: west × north = (−east) × north = −(east × north) = −(down) = up.

Result. The wire feels a force of magnitude 1.2 mN directed vertically upward.

What this shows. Earth's magnetic field exerts a real, measurable force on every current-carrying wire in the country. For normal household wiring (low currents, a few amperes) the force is milligram-weight-sized and irrelevant. For the 765 kV transmission line near Nagpur carrying 1000+ A, the vertical force per km of a 1 km span is a couple of newtons — still small, but it is there, and engineers have to know about it.

Example 2: Torque on a galvanometer coil

The moving coil of a school-lab galvanometer is a rectangular coil with N = 150 turns, dimensions 2\text{ cm} \times 3\text{ cm}, placed in a radial magnetic field of B = 0.20 T (the field is engineered to always be parallel to the coil's plane so that \theta is effectively 90° — this is the radial-field trick that makes the torque proportional to current, independent of angle). A current of 5 mA flows through the coil. Find the torque on the coil.

A galvanometer's permanent magnets and central soft-iron cylinder shape the field so that it always crosses the coil's sides perpendicularly. The coil feels a constant-magnitude torque $\tau = NIAB$ for any orientation.

Step 1. Compute the coil area.

A \;=\; (0.02)(0.03) \;=\; 6\times 10^{-4}\;\text{m}^2.

Step 2. Apply the torque formula with \sin\theta = 1.

\tau \;=\; NIAB \;=\; (150)(5\times 10^{-3})(6\times 10^{-4})(0.20).

Why: the radial field is engineered so that \sin\theta = 1 at every angle the coil rotates through. That is the entire point of the curved pole pieces and the central soft-iron cylinder — they make the galvanometer's scale linear in current.

Step 3. Multiply out.

\tau \;=\; 150 \times 0.005 \times 0.0006 \times 0.20 \;=\; 9\times 10^{-5}\;\text{N·m}.

Why: arithmetic. 150 × 0.005 = 0.75; 0.75 × 0.0006 = 4.5×10⁻⁴; × 0.20 = 9×10⁻⁵.

Step 4. Compare with the restoring torque of the spring. A galvanometer's hairspring has a torsion constant k such that in equilibrium \tau_\text{mag} = k\phi, where \phi is the angle of deflection. For a typical lab galvanometer, k \sim 2\times 10^{-7} N·m/rad, giving a deflection

\phi \;=\; \frac{\tau}{k} \;=\; \frac{9\times 10^{-5}}{2\times 10^{-7}} \;=\; 450\;\text{rad}.

That is absurd — the needle cannot rotate 450 radians. What actually happens is that this is an over-sensitive galvanometer; real instruments use a much stiffer spring (or a larger k) so that 5 mA deflects the needle by, say, \pi/3 rad (about 60°) full-scale.

Why: the numerical absurdity is the point — it shows that 5 mA through 150 turns in a 0.2 T field is a large signal for a sensitive galvanometer, which is why sensitive ones use currents in microamperes. The physics — \tau \propto I — is correct; the geometry of a real instrument is just tuned to keep the deflection on-scale.

Result. The torque on the coil is \tau = 9\times 10^{-5} N·m (90\,\muN·m).

What this shows. The galvanometer converts current into torque; the spring converts torque into deflection. The combination converts current into needle position — an analog measurement of a dynamic electrical quantity by a mechanical position you can read with your eye. The whole thing is an application of \tau = NIAB\sin\theta with the radial-field trick that sets \sin\theta = 1.

Example 3: Two transmission-line conductors

A 765 kV Indian transmission line runs two bundled conductors 40 cm apart, each carrying 900 A in the same direction. Find the force per metre between them, and say whether they attract or repel.

Two 900-A conductors 40 cm apart, both carrying current into the page, attract each other with about 0.4 N per metre of length.

Step 1. Apply F/L = \mu_0 I_1 I_2 / (2\pi d).

\frac{F}{L} \;=\; \frac{(4\pi\times 10^{-7})(900)(900)}{2\pi (0.40)}.

Step 2. Simplify the \pi's.

\frac{F}{L} \;=\; \frac{4\pi\times 10^{-7} \times 810000}{2\pi \times 0.40} \;=\; \frac{2\times 10^{-7} \times 810000}{0.40}.

Why: 4\pi / (2\pi) = 2, so the \pi's cancel cleanly. This is why we write \mu_0 = 4\pi\times 10^{-7} — the factor of 2\pi in the denominator of the wire-force formula eats two \pi's and leaves a clean 2\times 10^{-7}.

Step 3. Arithmetic.

\frac{F}{L} \;=\; \frac{0.162}{0.40} \;=\; 0.405\;\text{N/m}.

Step 4. Direction. Both currents flow in the same direction, so the force is attractive — the conductors pull towards each other.

Result. The force per metre is about 0.4 N/m, attractive.

What this shows. Over the 400 m span between two transmission towers, that is roughly 160 N of attractive force between the two conductors of a bundle — enough that if the spacers (the metal struts that keep the conductors apart) failed, the cables would slap together under short-circuit currents ten times higher, producing kilonewton forces that can rip out insulators. This is why bundle spacers are engineered as carefully as they are, and why a transmission-line failure in a storm is so violent — the cables are strung under enormous magnetic tension the moment short-circuit currents flow.

Common confusions

"The wire has no net charge, so there can be no force." False. The wire is electrically neutral but the carriers inside it are moving, and the magnetic force acts on moving charge. The total charge is zero; the current is not zero; and magnetic force responds to current, not to net charge.
"The force depends on the speed of the electrons." No. The drift speed is slow (~10^{-4} m/s), but the force involves nAqv_d, which equals the current I. The force on the wire is determined by I, not by v_d separately. You can scale v_d down and n up while keeping I constant, and the force stays the same.
"Parallel currents repel because like charges repel." No. Static like charges repel (Coulomb's law). Parallel currents attract. The physical mechanism is magnetic, not electric, and the sign is opposite. Memorise this one: parallel → attract, antiparallel → repel. This is the magnetic counterpoint to the electrostatic intuition.
"The torque on a loop is zero when the plane of the loop is perpendicular to \vec B." This is a common sign mistake. The torque is NIAB\sin\theta where \theta is the angle between the loop's normal and \vec B. When the loop's plane is perpendicular to \vec B, the normal is parallel to \vec B, \theta = 0, and the torque is zero. When the loop's plane is parallel to \vec B (normal perpendicular to \vec B), \theta = 90° and the torque is maximum. Draw a picture and check.
"The formula F = BIL\sin\theta is only for straight wires." For curved wires in a uniform field, you can replace \vec L with the straight-line vector from the start of the wire to the end — all the intermediate shape cancels. So the formula extends cleanly as long as the field is uniform. For non-uniform fields, integrate: \vec F = I\int d\vec L \times \vec B.
"The magnetic force does work on the wire." Strictly, the magnetic force on each individual charge does no work, because it is always perpendicular to the charge's velocity. But the wire does move, and energy does flow through the motor shaft. What is going on? The magnetic force redirects the charges' momentum sideways; the resulting sideways push on the lattice of copper ions is where the mechanical work comes from. The electric field in the battery (or generator) is the ultimate source of the energy. This is one of the subtlest points in all of electromagnetism — for the full story, see the Lorentz-force article.

If you came here to learn \vec F = I\vec L \times \vec B, the parallel-wire force, and the torque formula, and you can apply them — you have everything you need. What follows is the deeper structure: how this force relates to the general Lorentz force, the derivation for arbitrary geometry, the work-energy question, and the relativistic reading of parallel-wire attraction.

The Ampère-force law as a consequence of the Lorentz force

Everything in this article is a consequence of the microscopic Lorentz force \vec F = q\vec v \times \vec B applied to a current — no new physics. The steps from the Lorentz force to \vec F = I\vec L \times \vec B (our Step 1–4 above) are the whole derivation. Nothing else is assumed. In particular:

The "ampere-on-meter" definition is the Lorentz force between moving charges, with the \mu_0 constant fitting in so that the numbers come out to 2\times 10^{-7} N/m.
The torque on a loop is the sum of Lorentz forces on every infinitesimal segment.
The right-hand rule is the cross-product right-hand rule.

This is why electromagnetism has the structure it does: the Lorentz force is the one microscopic law, and every macroscopic magnetic effect (motor torque, transformer force, pinch effect in plasmas) follows from it by summing over charges.

Arbitrary wire in a non-uniform field

For a wire of arbitrary shape in a non-uniform field, the force is

\vec F \;=\; I\int_C d\vec L \times \vec B(\vec r),

where the integral runs along the wire's curve C and \vec B(\vec r) is the field at each point. This is a line integral of a vector.

For a closed loop in a uniform field, the net force is zero:

\vec F_\text{net} \;=\; I\oint d\vec L \times \vec B_0 \;=\; I\vec B_0 \times \oint d\vec L,

but \oint d\vec L = 0 for any closed curve (you return to your starting point). So the force vanishes. What is left — as we saw — is the torque, which does not vanish.

For a closed loop in a non-uniform field, the net force is not zero; it depends on the gradient of \vec B. This is what lets a magnet pick up iron: the magnetised iron acts like a loop whose \vec m aligns with the external field, and the gradient of \vec B pulls the loop toward regions of stronger field. In vector form,

\vec F_\text{on dipole} \;=\; \nabla(\vec m \cdot \vec B).

For \vec m \parallel \vec B, this is m\nabla B — the dipole is pulled toward stronger field.

Does the magnetic force do work?

Strictly, no. The Lorentz force on a point charge is always perpendicular to its velocity, so \vec F\cdot\vec v = 0 and the work-rate is zero, at every instant, for every charge. And yet, when you run current through a motor in a magnetic field, the motor turns, raising a load, doing positive work.

The resolution sits in the decomposition of the electrons' velocity inside a wire. An electron in a wire has two velocity components: (a) the drift velocity \vec v_d along the wire, and (b) the mechanical velocity \vec u of the wire itself as it moves through the field. The total velocity is \vec v = \vec v_d + \vec u. The magnetic force q(\vec v_d + \vec u)\times \vec B has a component along \vec v_d (in the direction -\vec u \times \vec B \cdot \hat v_d) that opposes the drift — that is a back-EMF, and it is what the battery does work against. The battery's electric field pushes the electrons along despite this back-EMF, and the power \vec E\cdot \vec J delivered by the battery is exactly the power that emerges from the motor shaft. The magnetic force is an intermediary — it redirects energy from the electrical circuit to mechanical motion without doing net work itself.

This bookkeeping is subtle, and most textbooks fudge it. The honest statement is: the magnetic field is a catalyst; the work is done by the electric field in the battery, transmitted through the magnetic interaction, and delivered as torque to the shaft.

Parallel-wire attraction as a relativistic effect

Take two parallel wires with current flowing in the same direction. In the lab frame, the wires are neutral (equal densities of positive ions and negative electrons). The electrons drift in the +x direction at v_d; the ions sit still.

Transform into the frame of the electrons in wire 1. Now the wire-1 electrons are stationary, but the wire-1 ions are moving in the −x direction at v_d. The wire is still neutral in this frame — charge density is Lorentz-invariant only in a specific sense, and for a current-carrying wire, the two species transform differently.

Specifically: the electrons of wire 2 are moving in the +x direction at v_d in the lab frame; in the electron-1 frame, they are at rest (if the electrons all drift at the same speed). The ions of wire 2 are at rest in the lab; in the electron-1 frame, they are moving at −v_d. The length-contraction factors for the two species differ — the moving species has its linear number density slightly increased by the Lorentz factor — and this makes wire 2 appear slightly positively charged in the electron-1 frame.

A slightly positively charged wire 2 exerts an electrostatic attraction on the negatively charged electrons of wire 1 (and repels wire 1's ions, but more weakly because the effect is weighted by Lorentz factors). Tally the forces and what looked like a magnetic attraction in the lab frame becomes an electrostatic attraction in the electron-1 frame.

The two descriptions agree exactly. The magnetic force between parallel currents is the relativistic transformation of the electrostatic force between charges in a different frame. Magnetism is relativity applied to electrostatics at small velocities — and the "small velocity" is the electron drift velocity, which is about 10^{-13} times the speed of light, and yet the magnetic effect is huge because the total number of electrons involved is astronomical.

This is the deepest thing to know about the force between parallel currents: it is not a separate fundamental interaction. It is Coulomb's law, seen from a moving frame.

The DRDO railgun and other extreme applications

The DRDO (Defence Research & Development Organisation) electromagnetic railgun programme at Hyderabad accelerates small projectiles using the same \vec F = I\vec L \times \vec B force, scaled up enormously. A rail gun consists of two parallel conducting rails shorted at one end by a conducting armature (the "projectile"). A current of 10^6 A or more flows down one rail, through the armature, and back up the other rail. The armature is essentially a current-carrying rod sitting in the magnetic field of the two rails, and the \vec L \times \vec B force pushes it along the rails at ten times the speed of a rifle bullet.

At these currents, the force per metre between the two rails is

\frac{F}{L} \;=\; \frac{\mu_0 I^2}{2\pi d} \sim \frac{(4\pi\times 10^{-7})(10^6)^2}{2\pi (0.1)} \;=\; 2000\;\text{N/m}.

Two kilonewtons per metre of repulsive force between the rails. This is why rail guns are hard to build — the force that accelerates the projectile also tries to blow apart the gun. The engineering problem is to contain the recoil on the rails themselves, not the projectile.

Indian Railways electrification runs off the same physics, gentler. The 25 kV catenary wire overhead runs horizontally; the locomotive's pantograph pushes upward against it with a contact force of 70–120 N; the pantograph carries several hundred amperes; and the \vec L \times \vec B interaction between the contact current and the catenary's return-current path in the rails produces a small sideways force that pushes the catenary slightly — a nuisance the engineering team has to design around when laying new track. Every kilometre of electrified line in India is a slow lesson in the force-between-parallel-currents.

Where this leads next

Force on a Moving Charge — the Lorentz Force — the microscopic law \vec F = q\vec v\times\vec B from which everything in this article descends.
Motion of Charged Particles in Magnetic Fields — what the Lorentz force does to free charges: circular motion, helices, the cyclotron frequency.
Force Between Parallel Currents and Moving Charges — the velocity selector, the Hall effect, and the fine print of parallel-current attraction.
Magnetic Dipole Moment — \vec m = NI\vec A as a fundamental object, the bar-magnet analogy, and potential energy U = -\vec m\cdot\vec B.
Ampère's Circuital Law — the law that generates the fields for which we computed forces.
Electromagnetic Induction: Faraday's Law — the reverse process: a moving wire in a field generates a current.