Magnetic field of straight wire, circular loop, solenoid, toroid: Visual walkthrough

Step 1 — Chop the wire into tiny straight pieces

WHAT. We take a flat circular loop of wire, radius $a$ , carrying a current $I$ (charge flowing round the ring). We slice the wire into countless tiny segments. One such segment is a short straight arrow we call $d l$ .

WHY. A curved wire is hard to reason about all at once. But a piece so short that it looks straight is easy: the Biot–Savart law tells us the field of one straight current piece. If we can find the field of one piece, we just add up all the pieces. That "add up infinitely many tiny things" is exactly what the integral sign $\oint$ will mean.

PICTURE. Look at the green loop. The thick green arrow is one segment $d l$ — it points along the current (the way the charge is travelling at that spot). We want the field at the red point $P$ , which sits on the axis — the line through the loop's centre, perpendicular to the loop's plane — a distance $x$ from the centre.

Figure — Magnetic field of straight wire, circular loop, solenoid, toroid

Step 2 — Why Biot–Savart and not Ampère

WHAT. We recall the Biot–Savart law: the tiny field from one tiny piece is $d B = \frac{μ _{0}}{4 π} \frac{I d l \times r ^}{r ^{2}}$

WHY. Read the formula slowly, term by term, right where each symbol lives:

$d B$ — the tiny field contribution from this one segment.
$\frac{μ _{0}}{4 π}$ — a fixed constant of nature that sets the strength ( $μ_{0} = 4 π \times 1 0^{- 7} T\cdotm/A$ ).
$I d l$ — "how much current, times how long the piece is": a bigger, longer, stronger piece pushes a bigger field.
$\overset{r}{^}$ — a unit-length arrow pointing from the segment to the field point $P$ . It tells us which direction $P$ lies.
$r^{2}$ in the bottom — field weakens as the square of distance, just like gravity or Coulomb's law.
$d l \times \overset{r}{^}$ — the cross product. This is the key: it says $d B$ points perpendicular to both the wire piece and the line to $P$ . Magnetic field wraps around current, it does not point straight at $P$ .

Why not Ampere's Law? Ampère only helps when you can find a loop on which $B$ has the same value everywhere, so you can factor it out. On the axis of a coil there is no such loop — the field strength changes from point to point with no clean symmetry to exploit. So we are forced to add the pieces up by hand. That is what Biot–Savart is for.

PICTURE. The figure shows one segment, the vector $\overset{r}{^}$ reaching to $P$ , and the resulting $d B$ sticking out perpendicular to the plane containing $d l$ and $\overset{r}{^}$ . Notice $d B$ is tilted — it is not along the axis.

Step 3 — The distance $r$ is the same for every piece

WHAT. Every segment of the ring sits at the same slant-distance from $P$ : $r = a^{2} + x^{2}$

WHY. Picture a right triangle. One leg is $a$ (from the centre out to the wire), the other leg is $x$ (from the centre along the axis to $P$ ). The hypotenuse — the straight line from the wire to $P$ — is the slant distance $r$ . Pythagoras gives $r = a^{2} + x^{2}$ . Because the point $P$ is on the axis (dead centre), every piece of the ring is exactly this far away. So $r$ is a constant as we go round the loop — a huge simplification.

$a^{2}$ — the "outward" leg squared.
$x^{2}$ — the "along-axis" leg squared.
their sum's square root — the honest straight-line distance.

PICTURE. The blue right triangle: legs $a$ and $x$ , hypotenuse $r$ . Spin the segment around the ring and the triangle just rotates — $r$ never changes.

Step 4 — Each piece's field: the size, before the direction

WHAT. For our segment, the wire piece $d l$ is perpendicular to $\overset{r}{^}$ (the wire runs around the ring, while $\overset{r}{^}$ heads straight toward $P$ ). A cross product of two perpendicular directions has size equal to the product of the lengths, so $∣ d l \times \overset{r}{^} ∣ = d l \times 1 = d l$ . Hence the magnitude of the tiny field is $d B = \frac{μ _{0} I d l}{4 π ( a ^{2} + x ^{2} )}$

WHY. We substituted $r^{2} = a^{2} + x^{2}$ from Step 3 into the bottom, and used that the cross product's size is just $d l$ (because the angle between $d l$ and $\overset{r}{^}$ is $9 0^{\circ}$ , whose sine is $1$ ).

top $μ_{0} I d l$ — strength times current times piece-length.
bottom $4 π (a^{2} + x^{2})$ — the constant and the distance-squared.

This is the length of the tilted arrow. Now we deal with its direction.

PICTURE. The right angle between the green wire arrow $d l$ and the yellow $\overset{r}{^}$ arrow is marked with a small square. The resulting $d B$ has length $d B$ and lies in the plane containing the axis and $\overset{r}{^}$ .

Step 5 — The great cancellation: only the axial part survives

WHAT. Split each tilted $d B$ into two pieces: one along the axis (call it $d B_{∥}$ ) and one perpendicular to the axis (call it $d B_{⊥}$ ). Now pair up the segment with the one directly opposite it across the ring. Their perpendicular parts point in exactly opposite directions and cancel to zero. Their axial parts point the same way and add.

WHY. This is the whole trick, and it is pure symmetry. For every piece pushing its $d B_{⊥}$ "up", the piece on the far side of the ring pushes an equal $d B_{⊥}$ "down". Sum over the whole ring: all sideways parts kill each other. Only the along-axis parts pile up. So we may throw away $d B_{⊥}$ entirely and keep only $d B_{∥} = d B cos α,$ where $α$ is the tilt angle of $d B$ away from the axis.

The angle $α$ is the same as the angle at $P$ in our triangle, and from the triangle $cos α = \frac{a}{a ^{2} + x ^{2}} .$

$cos α$ — "how much of the tilted arrow lies along the axis" ( $1$ = fully axial, $0$ = fully sideways).
$a$ on top — the outward leg; $a^{2} + x^{2}$ the hypotenuse. This ratio is opposite-over-hypotenuse from $P$ 's corner, which is precisely the fraction pointing axially.

PICTURE. Two opposite segments, their two tilted $d B$ arrows, and the dashed sideways components pointing in opposite directions (they annihilate). The solid axial components both point the same way along the axis.

Step 6 — Add up all the pieces (the integral)

WHAT. Every axial contribution has the same size (same $r$ , same $α$ everywhere on the ring), so summing is easy: $B = \oint d B cos α = \frac{μ _{0} I}{4 π} \frac{a}{( a ^{2} + x ^{2} ) ^{3/2}} \oint d l .$

WHY. We pulled the constant factors — everything except $d l$ — outside the sum, because they do not change as we walk around the loop. Then $\oint d l$ just means "add up all the tiny lengths around the ring" = the total circumference $2 π a$ .

The $\frac{1}{( a ^{2} + x ^{2} ) ^{3/2}}$ appeared because $d B$ gave one power $\frac{1}{a ^{2} + x ^{2}}$ and $cos α$ gave a factor $\frac{a}{a ^{2} + x ^{2}}$ — multiply: $\frac{a}{( a ^{2} + x ^{2} ) ^{3/2}}$ .

Substitute $\oint d l = 2 π a$ : $B = \frac{μ _{0} I}{4 π} \cdot \frac{a}{( a ^{2} + x ^{2} ) ^{3/2}} \cdot 2 π a = \frac{μ _{0} I a ^{2}}{2 ( a ^{2} + x ^{2} ) ^{3/2}}$

$μ_{0} I$ — strength × current.
$a^{2}$ on top — one $a$ from $cos α$ , one from the circumference.
$2$ in the bottom — the $4 π$ became $2$ after the $2 π$ from circumference cancelled.
$(a^{2} + x^{2})^{3/2}$ — the distance dependence: field dies off along the axis.

PICTURE. All axial arrows around the full ring, stacked head-to-tail into one big red axial arrow at $P$ — the total field.

Step 7 — Edge case: the field right at the centre

WHAT. Put $x = 0$ (the point $P$ sits at the centre of the loop): $B = \frac{μ _{0} I a ^{2}}{2 ( a ^{2} + 0 ) ^{3/2}} = \frac{μ _{0} I a ^{2}}{2 a ^{3}} = \frac{μ _{0} I}{2 a}$

WHY. At the centre there is no tilt at all — every piece's $d B$ already points straight along the axis ( $α = 0$ , $cos α = 1$ ). The formula must reflect this, and it does: setting $x = 0$ collapses $(a^{2} + x^{2})^{3/2}$ to $a^{3}$ , leaving the clean $μ_{0} I / (2 a)$ . Smaller loops (small $a$ ) give stronger centre fields — the current is closer.

PICTURE. At the centre, all the tiny arrows fan out but each already lies flat along the axis; no cancellation is even needed here.

Step 8 — Limiting case: very far away

WHAT. When $P$ is far along the axis, $x ≫ a$ , so $a^{2}$ is negligible next to $x^{2}$ : $B \approx \frac{μ _{0} I a ^{2}}{2 x ^{3}} \propto \frac{1}{x ^{3}} .$

WHY. Drop the tiny $a^{2}$ inside the bracket: $(a^{2} + x^{2})^{3/2} \approx (x^{2})^{3/2} = x^{3}$ . The field now dies as $1/ x^{3}$ — the hallmark of a magnetic dipole. From far away a current loop is indistinguishable from a tiny bar magnet. The quantity $I a^{2}$ (up to a factor $π$ ) is the loop's magnetic moment — its "dipole strength".

PICTURE. A curve of $B$ versus $x$ : tall and finite at the centre, sagging, and trailing off like $1/ x^{3}$ far out — with the exact curve and its far-field approximation drawn together.

The one-picture summary

The whole derivation on one canvas: chop the loop (Step 1) → one tilted $d B$ per piece via Biot–Savart (Step 2) → same slant distance $r$ for all (Step 3) → sideways parts cancel, axial parts survive (Step 5) → sum the axial parts over the circumference $2 π a$ (Step 6) → the boxed answer, with its centre and far-field faces (Steps 7–8).

Recall Feynman retelling — say it to a friend

We wanted the magnetic "swirl" strength at a point sitting on the line poking through the middle of a current ring. We couldn't use the easy shortcut (Ampère) because there's no nice symmetric path here, so we did it the honest way: cut the ring into a million little current-arrows. Each little arrow makes a little magnetic arrow, and — this is the neat part — because the point is dead-centre on the axis, every little current-arrow is the same distance away. Each little magnetic arrow is tilted, so we split it into an "along-the-axis" part and a "sideways" part. Pair each arrow with the one directly across the ring: their sideways parts point opposite and wipe each other out, while their along-axis parts team up. So we only ever needed to add up the axis parts — and since they're all identical, adding them up is just "one piece times the circumference $2 π a$ ." Do the bookkeeping and out drops $μ_{0} I a^{2} / [2 (a^{2} + x^{2})^{3/2}]$ . Stand at the centre and it becomes $μ_{0} I /2 a$ ; stand far away and the ring looks like a tiny bar magnet fading as $1/ x^{3}$ .

Recall

Why must we use Biot–Savart, not Ampère, for the loop's axis? ::: There is no closed path on which $B$ is constant, so $B$ cannot be pulled out of Ampère's integral — we must add contributions directly. On the axis, why is $r = a^{2} + x^{2}$ the same for every wire segment? ::: Because $P$ is on the axis: every segment sits one leg $a$ out and one leg $x$ along, forming congruent right triangles. Which components of $d B$ cancel and why? ::: The perpendicular-to-axis parts; the diametrically opposite segment supplies an equal and opposite perpendicular part. What does $\oint d l$ evaluate to and why can constants leave the integral? ::: $2 π a$ (the circumference); the other factors are identical for every segment, so they are constant round the loop. Field at the centre of the loop ::: $B = \frac{μ _{0} I}{2 a}$ Far-field fall-off of an axial loop field ::: $B \propto 1/ x^{3}$ (dipole behaviour)

Related tools and where this leads: Biot-Savart Law · Ampere's Law · Bar magnet and magnetic dipole · Magnetic force on a current-carrying conductor · Magnetic flux · Faraday's law of induction.