In short

The Biot-Savart law fed into specific wire shapes yields these closed-form results:

Long straight wire, perpendicular distance r:

B \;=\; \frac{\mu_0 I}{2\pi r}. \qquad \text{Field lines are circles around the wire.}

Axis of a circular loop of radius R, carrying current I, at distance x from the centre:

B(x) \;=\; \frac{\mu_0 I R^2}{2\,(R^2 + x^2)^{3/2}}. \qquad \text{Direction: along the axis by right-hand rule.}

Centre of the same loop (set x = 0):

B \;=\; \frac{\mu_0 I}{2R}.

Circular arc of radius R subtending an angle \theta (radians) at the centre — at that centre:

B \;=\; \frac{\mu_0 I}{4\pi R}\cdot \theta \;=\; \frac{\mu_0 I \theta}{4\pi R}.

Superposition. For any geometry made of straight segments and arcs, compute each piece's contribution at the point of interest and add them as vectors. Two antiparallel wires: fields add; two parallel wires: fields partially cancel; symmetric pieces with symmetric points often produce exact zero.

These five formulas cover almost every magnetostatics problem in the JEE Advanced and NEET syllabus. The derivations all follow the same pattern: write d\vec{B} from Biot-Savart, parametrise dl along the wire, integrate.

A single loop of copper wire sitting on a table, carrying a steady current of a few amperes, will create a magnetic field strong enough to deflect a compass needle placed at its centre. A dense coil of 1000 such loops, wound around a soft iron core — an MRI scanner magnet at AIIMS — produces a field of three teslas, 60,000 times Earth's. A rectangular coil driven by a small current — the voice coil of a loudspeaker — pushes and pulls a paper cone against a stationary magnet a million times a second to reproduce music. Every one of these is the same physics: the Biot-Savart law, applied to a specific shape of wire.

The Biot-Savart law tells you the magnetic field d\vec{B} produced by a tiny piece of wire d\vec{l} carrying current I:

d\vec{B} \;=\; \frac{\mu_0}{4\pi}\cdot\frac{I\,d\vec{l}\times\hat{r}}{r^2}.

The law is local — one piece of wire, the field it makes at one point. To find the field from a real wire, you integrate this contribution over the whole wire. That integration is usually the hard part. For the geometries below, the integrals collapse to closed form because of symmetry. For a messy shape (a bent wire loop in 3D, a spiral, a printed-circuit-board trace), you still set up the same integral and grind it numerically. The method never changes.

This article does four things. First, derive the field of an infinite straight wire — the cleanest integral in magnetostatics. Second, derive the field on the axis of a circular loop — the second cleanest, and the one that underlies every coil in an electric motor. Third, derive the field at the centre of a circular arc — which then unlocks every geometry made of arcs glued to straights. Fourth, apply superposition to compound geometries and see how symmetry simplifies problems like two parallel wires, a square loop, a half-infinite wire, and the Helmholtz coil pair.

The long straight wire — the cleanest magnetostatic integral

A long, straight wire carrying current I lies along the z-axis. You want the magnetic field at a point P at perpendicular distance r from the wire.

Geometry for the Biot-Savart integral of a long straight wireA vertical wire along the z-axis from bottom to top. A point P is at horizontal distance r. A small element dl at height z contributes dB at P, with the angle theta between r̂ and the wire. The triangle relates z, r, theta and the straight-line distance from the element to P. z I P dl r θ z r′ (dl → P) φ
Geometry for the Biot-Savart integral. An element $dl$ of the wire at height $z$ contributes $d\vec{B}$ at $P$, at perpendicular distance $r$. The angle $\theta$ between the wire and the line $dl \to P$ sets the cross-product magnitude; the straight-line distance $r' = \sqrt{r^2 + z^2}$ sets the inverse-square falloff.

Assumptions. The wire is infinitely long (or long enough that end-effects are negligible at P). Current I is steady. The medium is vacuum (\mu_0 is the free-space permeability).

Setting up the integral

Step 1. Pick a wire element d\vec{l} at height z, pointing in the +\hat{z} direction. Its length is dz.

d\vec{l} \;=\; dz\,\hat{z}.

Step 2. Write the vector from this element to P.

\vec{r}' \;=\; r\,\hat{x} - z\,\hat{z}, \qquad r' \;=\; \sqrt{r^2 + z^2}, \qquad \hat{r}' \;=\; \frac{r\,\hat{x} - z\,\hat{z}}{\sqrt{r^2 + z^2}}.

Why: P sits at horizontal distance r from the wire and vertical height 0 relative to its foot; the element sits at height z. The vector from element to P is horizontal r minus vertical z.

Step 3. Compute the cross product.

d\vec{l}\times\hat{r}' \;=\; (dz\,\hat{z})\times\frac{r\,\hat{x} - z\,\hat{z}}{\sqrt{r^2+z^2}} \;=\; \frac{r\,dz}{\sqrt{r^2+z^2}}(\hat{z}\times\hat{x}) \;=\; \frac{r\,dz}{\sqrt{r^2+z^2}}\,\hat{y}.

Why: \hat{z}\times\hat{z} = 0 drops the second term. The remaining \hat{z}\times\hat{x} = \hat{y} from the right-hand-rule cyclic pattern (\hat{x}, \hat{y}, \hat{z}). The result points out of the plane of the page (toward you if P is to the right of an upward current).

Step 4. Plug into Biot-Savart.

d\vec{B} \;=\; \frac{\mu_0 I}{4\pi}\cdot\frac{1}{r'^2}\cdot d\vec{l}\times\hat{r}' \;=\; \frac{\mu_0 I}{4\pi}\cdot\frac{r\,dz}{(r^2 + z^2)^{3/2}}\,\hat{y}.

Why: 1/r'^2 = 1/(r^2+z^2) multiplies the cross-product magnitude r\,dz/\sqrt{r^2+z^2}, giving the combined (r^2+z^2)^{-3/2}. Every element contributes in the same \hat{y} direction at P, so no vector bookkeeping — just scalar integration.

Step 5. Integrate z from -\infty to +\infty.

B_y \;=\; \frac{\mu_0 I r}{4\pi}\int_{-\infty}^{\infty}\frac{dz}{(r^2 + z^2)^{3/2}}.

Step 6. Evaluate the standard integral. Use the trig substitution z = r\tan\phi, so dz = r\sec^2\phi\,d\phi and r^2 + z^2 = r^2\sec^2\phi.

\int \frac{dz}{(r^2+z^2)^{3/2}} \;=\; \int\frac{r\sec^2\phi\,d\phi}{r^3\sec^3\phi} \;=\; \frac{1}{r^2}\int\cos\phi\,d\phi \;=\; \frac{\sin\phi}{r^2}.

As z: -\infty \to +\infty, \phi: -\pi/2 \to +\pi/2, so \sin\phi: -1 \to +1. The definite integral is:

\int_{-\infty}^{\infty}\frac{dz}{(r^2 + z^2)^{3/2}} \;=\; \frac{2}{r^2}.

Why: the trig substitution turns the awkward (r^2+z^2)^{3/2} into r^3\sec^3\phi, which cancels with r\sec^2\phi from dz to give a clean \cos\phi\,d\phi/r^2. The bounds \pm\pi/2 reflect that z\to\pm\infty corresponds to looking along the wire.

Step 7. Plug back.

B \;=\; \frac{\mu_0 I r}{4\pi}\cdot\frac{2}{r^2} \;=\; \boxed{\;\frac{\mu_0 I}{2\pi r}\;}

Why: the r's in the numerator and denominator reduce to a single r in the denominator, and the prefactor \mu_0 I /(4\pi) multiplied by 2 gives \mu_0 I /(2\pi). The field falls as 1/r — not 1/r^2, as you might have guessed from the Biot-Savart law alone. The 1/r falloff comes from integrating a 1/r'^2 contribution along an infinite line; the "extra" dimension of integration eats one power.

Direction — the right-hand rule for a wire

The vector d\vec{B} at every element pointed in +\hat{y}, which is tangent to a horizontal circle around the wire at P. By the right-hand rule: point your right thumb along the current direction, and your fingers curl in the direction of the magnetic field lines. The field lines are circles concentric with the wire, in planes perpendicular to it.

Magnetic field lines around a straight currentCross-section perpendicular to a wire carrying current out of the page (dot in a circle at centre). Around it, four concentric dashed circles represent the magnetic field lines, with arrows showing counter-clockwise direction. I (out of page) field lines curl counter- clockwise (right-hand rule) B ∝ 1/r (falls with distance)
Cross-section through a straight wire carrying current out of the page. Magnetic field lines are concentric circles, pointing counter-clockwise by the right-hand rule. The density of circles falls as $1/r$ — closer to the wire, the field is stronger.

Finite wire — the small generalisation

For a wire of finite length, the integration limits in Step 5 are not \pm\infty but the two angles \phi_1, \phi_2 that the ends make with the perpendicular from P. The result is:

B_\text{finite} \;=\; \frac{\mu_0 I}{4\pi r}(\sin\phi_1 + \sin\phi_2),

where \phi_1 and \phi_2 are both measured as positive when the corresponding wire end is above/below the foot of the perpendicular (both contributing in the same direction). For an infinite wire, \phi_1 = \phi_2 = \pi/2, so \sin\phi_1 + \sin\phi_2 = 2, recovering the \mu_0 I/(2\pi r) result.

The finite-wire formula is worth remembering because it is how you build up arbitrary polygonal loops by summing straight segments.

Field on the axis of a circular loop

A circular loop of radius R lies in the xy-plane, centred at the origin, carrying current I (counter-clockwise as seen from +z). You want the magnetic field at the point P = (0, 0, x) on the axis.

Geometry for field on axis of a circular loopA circular loop seen in perspective. The axis passes through the centre; the point P sits on the axis at distance x. A wire element dl on the loop is at distance sqrt(R² + x²) from P; the Biot-Savart dB is perpendicular to dl and r̂. The component along the axis is dB cos α where α is the angle of dB from the axis. loop (radius R) axis P dl r′ = √(R² + x²) x dB dB cos α dB sin α α O R
A circular loop seen in perspective. The field at $P$ on the axis is contributed by every element $dl$ of the loop. Each $d\vec{B}$ is perpendicular to both $d\vec{l}$ and the line from $dl$ to $P$, with magnitude $dB = \mu_0 I\,dl/[4\pi(R^2 + x^2)]$. The components perpendicular to the axis cancel from opposite elements of the loop; the axial components $dB\cos\alpha$ add.

Setting up the integral

Step 1. Pick a loop element d\vec{l} somewhere on the loop. Its magnitude is dl = R\,d\phi, where \phi is the angle around the loop.

Step 2. The line from d\vec{l} to P has length r' = \sqrt{R^2 + x^2} — the same for every point on the loop, because every point on the loop is at distance R from the centre and P is at axial distance x. This constancy is the symmetry that makes the integral tractable.

Step 3. d\vec{l} is tangent to the loop (in the plane of the loop), and \hat{r}' points from d\vec{l} to P. The angle between them is 90° for every element — because the tangent at any point on the loop is perpendicular to the radial direction from the centre, and \hat{r}' from dl to P has both radial and axial components, but the axial component is perpendicular to the in-plane tangent d\vec{l}, and the in-plane component is radial (perpendicular to d\vec{l}). So |d\vec{l}\times\hat{r}'| = dl exactly.

Why: the tangent to the circle at any point is perpendicular to the radius at that point. The vector from the point to P lives in a vertical plane containing that radius and the axis, and this whole plane is perpendicular to d\vec{l}. Hence d\vec{l}\perp\hat{r}' and their cross product has magnitude dl\cdot 1.

Step 4. Magnitude of d\vec{B}:

dB \;=\; \frac{\mu_0 I}{4\pi}\cdot\frac{dl}{r'^2} \;=\; \frac{\mu_0 I}{4\pi}\cdot\frac{R\,d\phi}{R^2 + x^2}.

Step 5. The direction of d\vec{B} is perpendicular to both d\vec{l} and \hat{r}'. It lies in the vertical plane containing the axis and the current element, and it points away from the loop's interior axis line — making an angle \alpha with the axis itself, where

\cos\alpha \;=\; \frac{R}{\sqrt{R^2 + x^2}}.

Why: the geometry is a right triangle. The line from dl to P has axial leg x and radial leg R. d\vec{B} is perpendicular to this line in the vertical plane. d\vec{B} tilts so that its projection on the axis (dB\cos\alpha) and its projection perpendicular to the axis (dB\sin\alpha) relate by the same triangle. The cosine of the angle between dB and the axis is R/r' because the geometry swaps the roles of R and x when you rotate 90°.

Step 6. Resolve into components. The axial component of d\vec{B} is dB\cos\alpha. The perpendicular component is dB\sin\alpha — but this perpendicular component from the element at angle \phi is cancelled by the contribution from the element at \phi + \pi (diametrically opposite). Hence, after integrating around the whole loop, only the axial components survive.

Step 7. Integrate only the axial component:

B_x \;=\; \oint dB\cos\alpha \;=\; \frac{\mu_0 I}{4\pi}\cdot\frac{R}{R^2 + x^2}\cdot\frac{R}{\sqrt{R^2+x^2}}\int_0^{2\pi} d\phi.
B_x \;=\; \frac{\mu_0 I R^2}{4\pi (R^2+x^2)^{3/2}}\cdot 2\pi \;=\; \boxed{\;\frac{\mu_0 I R^2}{2\,(R^2 + x^2)^{3/2}}\;}

Why: every factor except the azimuthal angle \phi is constant on the loop, so the integral \int d\phi over one turn contributes simply 2\pi. The R^2 in the numerator comes from combining R\,d\phi (the arclength element) with \cos\alpha = R/\sqrt{R^2+x^2}.

Special case: field at the centre

Set x = 0:

B_\text{centre} \;=\; \frac{\mu_0 I R^2}{2\,R^3} \;=\; \boxed{\;\frac{\mu_0 I}{2R}\;}

This is the formula you commit to memory. A 10\ \text{cm} loop carrying 1\ \text{A} produces a field at its centre of B = (4\pi\times 10^{-7})(1)/(2\times 0.1) = 6.3\ \mu\text{T} — about 1/8 of Earth's magnetic field. Wind the same loop 1000 times (every turn contributes, by superposition) and you get 6.3\ \text{mT}, a hundred times Earth's field.

Far from the loop — the dipole limit

When x \gg R, the axial field simplifies to:

B(x) \;\approx\; \frac{\mu_0 I R^2}{2 x^3} \;=\; \frac{\mu_0}{4\pi}\cdot\frac{2(I\pi R^2)}{x^3} \;=\; \frac{\mu_0}{4\pi}\cdot\frac{2m}{x^3},

where m = I\pi R^2 is the magnetic dipole moment of the loop (current times enclosed area). The 1/x^3 falloff is the characteristic signature of a dipole field — the same inverse-cube law that describes the field of a bar magnet at large distance. A current loop is a magnetic dipole; this is why the two look identical from far away.

Interactive: field along the axis

Drag the axial position x and watch B(x) fall off. The loop has R = 1\ \text{m} and I = 1\ \text{A} so that B is expressed in units of \mu_0 I/(2R) = \mu_0/2; read the vertical axis as multiples of this reference.

Interactive: axial magnetic field of a circular loop A curve of B(x) along the axis of a circular loop of unit radius, showing the cubic falloff at large |x| and the flat peak near the centre. axial distance x (in units of R) B(x) in units of μ₀I/(2R) −4 −2 2 4 1 0.5 B(x) = μ₀IR² / [2(R²+x²)^(3/2)] drag the red point along the x-axis
Drag the red point to sample $B(x)$. The curve peaks at the centre $x=0$ with $B = \mu_0 I/(2R)$. At $x = R$, $B$ is already about $0.35$ of its peak. At large $|x|$, $B$ falls as $1/x^3$ — the magnetic-dipole signature.

Helmholtz coil pair — the flattened peak

Two identical coils of radius R on the same axis, separated by a distance R, carrying current I in the same direction, produce a remarkably uniform field between them. This is a Helmholtz pair, used whenever you want a calibrated volume of uniform magnetic field — calibrating magnetometers at the Indian Institute of Geomagnetism in Mumbai, cancelling Earth's field in a MOKE microscopy lab at IIT Bombay, or studying magnetic-field effects on plants in a botany experiment.

By superposition, the field at the midpoint is the sum of each coil's field at axial distance R/2:

B_\text{mid} \;=\; 2\cdot\frac{\mu_0 I R^2}{2(R^2 + R^2/4)^{3/2}} \;=\; \frac{\mu_0 I R^2}{(5R^2/4)^{3/2}} \;=\; \left(\frac{4}{5}\right)^{3/2}\cdot\frac{\mu_0 I}{R}.

The key property: the first and second derivatives of B along the axis both vanish at the midpoint, so the field is uniform to fourth order in the axial distance. That is why d = R is the magic separation.

Circular arc — the angle-scaling rule

A circular arc of radius R subtends an angle \theta at its centre O and carries current I. You want the field at O.

Circular arc of angle thetaA circular arc of radius R subtending angle theta at the centre O. Current I flows along the arc. The field at O is dB = (μ₀ I / 4π R²) dl, and each dl is perpendicular to the radius, so integrating over arclength R·theta gives B = μ₀ I θ / (4π R). O I θ dl = R dφ R dB out of page
A circular arc of radius $R$ and angle $\theta$ carrying current $I$. Every element $dl = R\,d\phi$ is perpendicular to the radius $\hat{r}$ from $dl$ to the centre $O$, so the cross product $d\vec{l}\times\hat{r}$ has magnitude $dl$. Every $d\vec{B}$ points in the same direction at $O$ (out of the page for counter-clockwise current). The total field is just the sum.

Derivation

Along the arc, dl = R\,d\phi, where \phi is the azimuthal angle. The vector from dl to O is the inward radial direction, perpendicular to d\vec{l}, so |d\vec{l}\times\hat{r}| = dl\cdot 1. The distance |\vec{r}| = R. Every element contributes in the same direction (by the right-hand rule, all out of the page for counter-clockwise current). So:

B \;=\; \int_0^\theta \frac{\mu_0 I}{4\pi}\cdot\frac{R\,d\phi}{R^2} \;=\; \frac{\mu_0 I}{4\pi R}\int_0^\theta d\phi \;=\; \boxed{\;\frac{\mu_0 I \theta}{4\pi R}\;}

Why: everything is constant on the arc — R, the right-angle between dl and \hat{r}, the direction of d\vec{B}. The only variable is the angle \phi, and the integral \int_0^\theta d\phi = \theta just counts arclength.

Sanity check: set \theta = 2\pi to recover a full loop. Result: B = \mu_0 I \cdot 2\pi/(4\pi R) = \mu_0 I/(2R) — matches the centre-of-loop formula. Set \theta = \pi (a semicircle): B = \mu_0 I/(4R), half the full-loop result. Set \theta = \pi/2 (quarter-circle): B = \mu_0 I/(8R).

Direction: by the right-hand rule. Curl your fingers in the direction of current flow along the arc; your thumb points in the direction of \vec{B} at the centre.

Superposition — building compound geometries

The Biot-Savart law is linear in d\vec{l}, so the field at a point due to a complicated wire is the sum of contributions from every piece. This means you can:

  1. Break any wire into simple pieces (straight segments, circular arcs).
  2. Compute each piece's contribution to the field at your point of interest.
  3. Add the contributions as vectors (magnitude and direction).

Example: the straight-arc-straight hairpin

Consider a current path that goes straight in, bends around a semicircular arc of radius R, and continues straight out on the other side — a hairpin shape. Field at the centre of the arc? Three pieces.

  1. Semicircle of radius R: field is \mu_0 I/(4R), direction out of the page (say).
  2. First straight segment (coming toward the arc): the field at the centre of the arc from a long straight wire passing through a point a distance R away is \mu_0 I/(2\pi R) — but only if the wire is infinite on both sides of the foot. Here the wire extends only on one side of the foot of the perpendicular. From the finite-wire formula with \phi_1 = \pi/2 (wire extends to infinity on one side) and \phi_2 = 0 (wire stops at the foot of the perpendicular), we get B = \mu_0 I/(4\pi R) — half the infinite result.
  3. Second straight segment: by symmetry, same magnitude, same direction.

Adding all three: B_\text{hairpin} = \mu_0 I/(4R) + 2\cdot\mu_0 I/(4\pi R) = (\mu_0 I/4R)(1 + 2/\pi).

When symmetry gives zero

If you stand at the geometric centre of a square current loop, the contributions from the four sides are equal and all point in the same direction (out of or into the page). The field is non-zero. But if you stand at the centre of a rectangular loop with different side lengths, the longer sides contribute more than the shorter sides — yet the direction is still the same for all four, so the magnitudes just add with different weights.

Contrast this with standing midway between two antiparallel wires: one contributes \mu_0 I/(2\pi r) out of the page, the other contributes \mu_0 I/(2\pi r) into the page — the fields cancel exactly. Symmetry gives zero when contributions point in opposite directions.

The general rule: always ask, for each element, "which way does its d\vec{B} point at my point of interest?" Then sum vectors. Symmetry is not mystical — it is just the observation that some components are always equal and opposite.

Worked examples

Example 1: A hundred-turn loudspeaker voice coil

A loudspeaker voice coil is wound with N = 100 turns of fine copper wire around a cylindrical former of radius R = 1.5\ \text{cm}. The coil carries a steady current of I = 0.4\ \text{A}. Find the magnitude of the magnetic field at the centre of the coil, and compare it to Earth's field (B_\text{Earth} \approx 40\ \mu\text{T}).

Loudspeaker voice coil cross-sectionA coil of 100 turns wound tightly around a cylindrical former of radius 1.5 cm. A field B₀ indicated at the centre points along the coil axis. cylindrical former, radius R = 1.5 cm 100 turns B₀ axis I = 0.4 A
A loudspeaker voice coil with $100$ closely wound turns of radius $1.5\ \text{cm}$, carrying $0.4\ \text{A}$. The field at the centre is the sum of $100$ individual loop contributions.

Step 1. Write the field at the centre of a single loop.

B_1 \;=\; \frac{\mu_0 I}{2R}.

Why: this is the x=0 specialisation of the on-axis formula we derived. Every turn is a circular loop of radius R carrying current I.

Step 2. Apply superposition across all N turns.

B \;=\; N\cdot B_1 \;=\; \frac{N\mu_0 I}{2R}.

Why: the Biot-Savart law is linear in current. Each turn contributes the same field at the centre (all turns are concentric and coplanar to within a millimetre or so — thin coil). Their contributions add scalar-wise because they all point the same direction along the axis.

Step 3. Substitute numbers.

B \;=\; \frac{100\times (4\pi\times 10^{-7}\ \text{T m/A})\times 0.4\ \text{A}}{2\times 0.015\ \text{m}} \;=\; \frac{100\times 4\pi\times 0.4\times 10^{-7}}{0.03}\ \text{T}.

Compute the numerator: 100\times 4\pi\times 0.4 = 160\pi \approx 502.7. So:

B \;=\; \frac{502.7\times 10^{-7}}{0.03}\ \text{T} \;=\; 16.8\times 10^{-4}\ \text{T} \;=\; 1.68\ \text{mT}.

Why: group the numerical factors first, then divide by the small radius. The factor of 10^{-7} from \mu_0 combined with 1/0.03 (\sim 33) lands the answer in the milli-tesla range — reasonable for a hundred-turn coil.

Step 4. Compare to Earth's field.

\frac{B}{B_\text{Earth}} \;=\; \frac{1.68\times 10^{-3}}{40\times 10^{-6}} \;=\; 42.

The coil produces a field 42 times Earth's magnetic field.

Result: B = 1.68\ \text{mT}, about 42 times Earth's field.

What this shows: A modest coil with modest current produces a field tens of times Earth's — enough to drive a speaker cone against a permanent magnet's field via \vec{F} = I\vec{L}\times\vec{B}. The density of turns multiplies the effect linearly; this is why every electromagnet, relay, solenoid valve, and voice coil uses hundreds to thousands of turns. You cannot afford thousands of amperes — you can afford thousands of turns.

Example 2: Field at the centre of a square loop

A square loop of side 2a carries current I. Find the magnetic field at the centre.

Square loop with centre pointA square of side 2a with current I circulating counter-clockwise. The centre point C is at the middle. Dashed radial lines from C to the midpoints of each side show the perpendicular distance a. For one side, the angles phi₁ and phi₂ from the perpendicular to the ends of the side are both 45 degrees, giving sin phi₁ = sin phi₂ = 1/√2. I C a a side = 2a φ = 45°
A square loop of side $2a$. The centre $C$ is at perpendicular distance $a$ from the midpoint of every side. Each side, seen from $C$, is a finite wire of length $2a$ subtending half-angles $\phi_1 = \phi_2 = 45°$ on either side of the perpendicular.

Step 1. Analyse one side. The perpendicular from C to the side has length a; the side extends from -a to +a along the side's direction. The two ends of the side make angles with the perpendicular satisfying

\sin\phi_1 \;=\; \frac{a}{\sqrt{a^2 + a^2}} \;=\; \frac{1}{\sqrt 2}, \qquad \sin\phi_2 \;=\; \frac{1}{\sqrt 2}.

Why: from C, the line to each end of the side has length \sqrt{a^2 + a^2} = a\sqrt{2}. The component of this line along the side direction is a (half the side length). The sine of the angle to the perpendicular is (opposite over hypotenuse) = a/(a\sqrt 2) = 1/\sqrt 2.

Step 2. Apply the finite-wire formula for one side.

B_\text{side} \;=\; \frac{\mu_0 I}{4\pi a}(\sin\phi_1 + \sin\phi_2) \;=\; \frac{\mu_0 I}{4\pi a}\cdot\sqrt 2 \;=\; \frac{\mu_0 I\sqrt 2}{4\pi a}.

Step 3. Superpose all four sides. By the right-hand rule (curl fingers along the current direction around the loop, thumb gives \vec B), every side contributes a field at C pointing out of the page (for counter-clockwise current). They add scalarly:

B_\text{centre} \;=\; 4\cdot B_\text{side} \;=\; 4\cdot\frac{\mu_0 I\sqrt 2}{4\pi a} \;=\; \frac{\mu_0 I\sqrt 2}{\pi a}.

Why: all four contributions are in the same direction at C because the four sides carry currents that all circulate the same way around the loop. By symmetry their magnitudes are equal.

Step 4. Express in terms of the side length L = 2a if preferred:

B_\text{centre} \;=\; \frac{\mu_0 I\sqrt 2}{\pi\,(L/2)} \;=\; \frac{2\sqrt 2\,\mu_0 I}{\pi L}.

Step 5. Sanity check against a circular loop of the same enclosed area. A square of side 2a encloses area 4a^2. A circle of the same area has radius R = 2a/\sqrt\pi. The field at the centre of that circle is \mu_0 I/(2R) = \mu_0 I\sqrt\pi/(4a). Compare to our square-loop answer \mu_0 I\sqrt 2/(\pi a) \approx 0.45\,\mu_0 I/a; the circle gives \approx 0.44\,\mu_0 I/a. Almost equal — the field at the centre is not very sensitive to the shape as long as the enclosed area is fixed. That is a useful approximation for rough engineering estimates of a coil's central field.

Result: B_\text{centre} = \dfrac{\mu_0 I \sqrt 2}{\pi a} = \dfrac{2\sqrt 2\,\mu_0 I}{\pi L} for a square of side L = 2a.

What this shows: A square loop's central field is built up as the sum of four finite-wire contributions, each handled by the finite-wire Biot-Savart result. Symmetry guarantees all four contribute equally and in the same direction, turning a potentially long vector sum into a factor of 4. For a loop of any regular-polygon shape enclosing a given area, the field at the centre is within a few percent of the circular-loop value — a fact worth remembering when you need a quick estimate.

Common confusions

If your goal was to cover the JEE Advanced syllabus on magnetic-field geometries, you have everything you need. What follows is the level of rigour you would see at an IIT's EE department — vector forms, off-axis loop fields, dipole moments as a deeper concept, and the connection to the magnetic field of a rotating charge distribution.

The loop's dipole moment and the vector form of Biot-Savart

The magnetic dipole moment of a current loop is a vector:

\vec m \;=\; I\vec A,

where \vec A is the vector area of the loop — magnitude equal to the enclosed area, direction normal to the loop plane by the right-hand rule. For a circular loop of radius R, |\vec m| = I\pi R^2.

The on-axis field at distance x \gg R takes the clean form:

\vec B \;=\; \frac{\mu_0}{4\pi}\cdot\frac{2\vec m}{x^3} \quad (\text{on-axis, far field}).

The general far-field expression for the dipole — valid at any angle \theta from the axis — is:

\vec B(\vec r) \;=\; \frac{\mu_0}{4\pi r^3}\left[3(\vec m\cdot\hat r)\hat r - \vec m\right].

Why: this is the magnetostatic dipole field, identical in form to the electrostatic dipole field with \vec E replaced by \vec B and the electric dipole moment \vec p by \vec m. The 3(\vec m\cdot\hat r)\hat r - \vec m structure captures the fact that along the axis (\hat r \parallel \vec m), the factor is 3\vec m - \vec m = 2\vec m; perpendicular to the axis (\hat r \perp \vec m), it is -\vec m — the field points opposite to the moment, with half the axial magnitude.

Every atom with a nonzero orbital angular momentum or spin has such a dipole moment, and the far-field expression above is how nuclear magnetic resonance (NMR) at BARC, the Raman Research Institute, or AIIMS images tissue: the spin dipoles radiate according to this equation when the Larmor precession is excited by an RF pulse.

Off-axis field of a circular loop — the elliptic-integral formula

The on-axis result is a special case. For a point off the axis, the integral does not collapse to elementary functions; it produces complete elliptic integrals. Let (r_\perp, z) be cylindrical coordinates of the field point, with z along the loop axis and r_\perp the distance from the axis. The axial and radial components of \vec B are:

B_z \;=\; \frac{\mu_0 I}{2\pi}\cdot\frac{1}{\sqrt{(R+r_\perp)^2 + z^2}}\cdot\left[K(k) + \frac{R^2 - r_\perp^2 - z^2}{(R-r_\perp)^2 + z^2}\cdot E(k)\right],
B_{r_\perp} \;=\; \frac{\mu_0 I}{2\pi r_\perp}\cdot\frac{z}{\sqrt{(R+r_\perp)^2 + z^2}}\cdot\left[-K(k) + \frac{R^2 + r_\perp^2 + z^2}{(R-r_\perp)^2 + z^2}\cdot E(k)\right],

where K(k) and E(k) are the complete elliptic integrals of the first and second kind, and the modulus is

k^2 \;=\; \frac{4R r_\perp}{(R + r_\perp)^2 + z^2}.

This is what the inside of an MRI scanner coil sees at any general point. The on-axis limit (r_\perp \to 0) recovers our \mu_0 I R^2/[2(R^2+z^2)^{3/2}] formula after using the limiting behaviour K(0) = E(0) = \pi/2.

Rotating charged ring as a current loop

A ring of radius R carrying uniform charge Q spinning at angular velocity \omega about its axis is equivalent to a current loop. The current is the charge per period:

I_\text{eff} \;=\; \frac{Q}{T} \;=\; \frac{Q\omega}{2\pi}.

The field at the centre is:

B \;=\; \frac{\mu_0 I_\text{eff}}{2R} \;=\; \frac{\mu_0 Q\omega}{4\pi R}.

This is the elementary version of a classical atom (before quantum mechanics): electrons orbiting a nucleus as current loops, each contributing a magnetic moment \vec m = I\pi R^2 \hat{n}. This picture is wrong in detail — electrons don't follow classical orbits — but the dimensional scaling and the existence of orbital magnetic moments is a real phenomenon, captured in the Bohr magneton \mu_B = e\hbar/(2m_e).

The SST-1 tokamak at IPR Gandhinagar

The Steady State Superconducting Tokamak (SST-1) at the Institute for Plasma Research in Gandhinagar uses magnetic fields to confine a plasma at millions of degrees for fusion research. The toroidal field coils generate a field along the torus, while poloidal field coils (circular loops stacked along the torus's short direction) shape the plasma. Every one of these coils is, at root, a Biot-Savart integration problem. The engineering is in controlling 10^4 amperes through superconducting windings at cryogenic temperatures — but the underlying physics is what this article derived.

A numerical warning

When you evaluate B = \mu_0 I/(2R) for R = 0.15\ \text{m} and I = 1\ \text{A}, you get B = 4.2\ \mu\text{T} — almost exactly 1/10 of Earth's field. This is useful to remember: a single loop of radius \sim 10\ \text{cm} carrying \sim 1\ \text{A} produces a field comparable to Earth's. This is why Oersted's 1820 experiment worked — a compass needle pivoting in Earth's field will be noticeably deflected by the field of a nearby current-carrying wire, and this observation launched the whole of electromagnetism.

Where this leads next