In short

A magnetic field is a vector field \vec{B}(\vec{r}) that fills the space around moving charges. Its SI unit is the tesla (T): 1\ \text{T} = 1\ \text{N}/(\text{A}\cdot\text{m}). Earth's field at the surface is about 5\times 10^{-5} T; a refrigerator magnet is about 5\times 10^{-3} T; the field inside a 1.5-T AIIMS MRI scanner is about 30{,}000 times Earth's field.

The Biot-Savart law gives the magnetic field produced by a small current element. A wire carrying current I along a directed segment d\vec{l} produces, at a point displaced by \vec{r} = r\hat{r} from the element, an infinitesimal field

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

where \mu_0 = 4\pi\times 10^{-7}\ \text{T}\cdot\text{m}/\text{A} is the permeability of free space — a universal constant that sets the strength of the magnetic interaction in SI units.

Five features to remember.

  1. Inverse-square in r, like Coulomb's law.
  2. Cross product d\vec{l}\times\hat{r}: the field is perpendicular to both the current direction and the line joining the element to the field point.
  3. Vanishes along the line of current: if \hat{r}\parallel d\vec{l}, the cross product is zero. A point directly in front of the tip of a current element sees no field from that element.
  4. Right-hand thumb rule gives the direction: curl the fingers of your right hand from d\vec{l} toward \hat{r}; the thumb points along d\vec{B}.
  5. Superposition: the total field from an extended current is the vector integral \vec{B}(\vec{r}) = (\mu_0/4\pi)\int I\,d\vec{l}\times\hat{r}/r^2 over all current elements.

The Biot-Savart law is to magnetostatics what Coulomb's law is to electrostatics — the elemental rule from which every field (loop, solenoid, MRI magnet, planetary dynamo) can be computed by integration.

Hold a compass up to the back of a working laptop charger — the power brick with the fat cable running from the wall. The needle swings and settles, pointing in some direction that has nothing to do with north. Move the compass and the direction changes. Walk the compass slowly in a circle around the cable and the needle rotates with you, completing one full turn as you come back to where you started. The charger's cable is carrying a current — a few hundred milliamperes of DC-rectified mains — and that current is creating a magnetic field that steers your compass.

Now scale this up. An AIIMS MRI scanner threads its main coil with hundreds of amperes of direct current through thousands of turns of superconducting niobium-titanium wire, producing a 1.5-tesla field at the centre of the bore — strong enough to align proton spins in a patient's brain so they can be imaged. ISRO's satellite thruster control rooms run electromagnet-actuated valves, each valve a solenoid carrying a few tens of amperes. The transformers on the wooden poles outside every Indian colony are coils of wire handling hundreds of amperes at 11 kV, stepping it down to the 230 V that reaches your house. Every one of these devices is an application of the same rule: a current I flowing along a wire element d\vec{l} creates, at a point displaced by \vec{r} from the element, a magnetic field d\vec{B} given by the Biot-Savart law.

This chapter introduces the magnetic field \vec{B} as a vector field, explains what it means and how it is measured, states and unpacks the Biot-Savart law, and shows you how to use it. You will compute two fields from scratch — the on-axis field of a circular loop, and the field of a finite straight wire — and by the end you will be able to set up the integral for any current distribution you can draw.

What is a magnetic field?

An electric charge sitting at rest creates an electric field \vec{E} around it. A moving charge creates, in addition, a magnetic field \vec{B}. A current is just charge in motion through a wire, so every current-carrying wire surrounds itself with a magnetic field.

The magnetic field is defined operationally by the force it exerts on other moving charges. A test charge q moving with velocity \vec{v} through a region where the magnetic field is \vec{B} experiences a force

\vec{F} \;=\; q\,\vec{v}\times\vec{B}. \tag{1}

This is the Lorentz force law, and it is what the field does. You do not see \vec{B} directly; you see charges and compasses deflecting in response to it.

Two features of equation (1) carry through the whole subject.

First, \vec{B} is a vector field — it has a direction as well as a magnitude at each point in space. At your kitchen table, the Earth's \vec{B} points roughly horizontally toward magnetic north and slightly downward (the "angle of dip"); at the centre of a current loop, \vec{B} points along the loop's axis. The direction matters.

Second, the SI unit of \vec{B} is fixed by equation (1). A charge of 1 C moving at 1 m/s through a 1 T field perpendicular to its motion feels a force of 1 N. So

1\ \text{T} \;=\; \frac{1\ \text{N}}{1\ \text{C}\cdot 1\ \text{m/s}} \;=\; \frac{1\ \text{N}}{1\ \text{A}\cdot 1\ \text{m}} \;=\; \frac{1\ \text{kg}}{1\ \text{A}\cdot 1\ \text{s}^2}.

The tesla is a large unit. Everyday magnetic fields are much smaller.

Magnetic field strengths on a logarithmic scaleA horizontal logarithmic axis spanning the range of magnetic fields humans encounter, from brain magnetic activity at 10 to the minus 13 tesla up to neutron star fields at 10 to the 10 tesla, with markers for Earth's field, fridge magnet, MRI, and laboratory high-field magnets. 10⁻¹⁴ 10⁻¹⁰ 10⁻⁶ 10⁻² 10² 10⁶ 10¹⁰ magnetic field strength (tesla, log scale) brain MEG 10⁻¹³ T Earth's field 5×10⁻⁵ T fridge magnet 5×10⁻³ T AIIMS MRI 1.5 T lab magnet 45 T neutron star 10¹⁰ T
Magnetic field strengths span 24 orders of magnitude. The Earth's field is 30{,}000 times weaker than an MRI scanner's; a neutron star's is $10^{13}$ times stronger than an MRI. The Biot-Savart law, written below, holds across this entire range.

A compass needle lines up with \vec{B} because the needle is itself a tiny magnet with a torque proportional to the field — so it rotates until its north pole points along \vec{B}. This is how you visualise the field: move a compass around a current-carrying wire and trace the directions it points. You will find the field lines form closed circles around the wire, as Hans Christian Ørsted first showed in 1820.

The Biot-Savart law

With Coulomb's law, an extended charge distribution creates an electric field you compute by summing contributions from infinitesimal charge elements:

\vec{E}(\vec{r}) \;=\; \frac{1}{4\pi\varepsilon_0}\int\frac{dq\,\hat{r}}{r^2}.

The magnetostatic analogue is the same strategy: every infinitesimal piece of a current-carrying wire contributes a tiny d\vec{B}, and you integrate over the whole wire. The rule that gives the contribution from one piece — discovered experimentally by Biot and Savart in 1820 — is the Biot-Savart law:

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

Read it slowly, term by term.

The magnitude of the infinitesimal field is

|d\vec{B}| \;=\; \frac{\mu_0}{4\pi}\,\frac{I\,|d\vec{l}|\,\sin\theta}{r^2}. \tag{3}

Two special cases are immediate.

The right-hand thumb rule for direction

The direction of d\vec{B} comes from the cross product d\vec{l}\times\hat{r}. Mechanical recipe: point your right hand's fingers along d\vec{l}, curl them toward \hat{r}, and your thumb points along d\vec{l}\times\hat{r}, which is the direction of d\vec{B}.

For a long vertical wire with current flowing upward, the field at a point east of the wire: d\vec{l} is upward (\hat{z}), \hat{r} is east (\hat{x}), so d\vec{l}\times\hat{r} = \hat{z}\times\hat{x} = \hat{y} — the field points north. Walk around the wire and the field rotates with you, forming closed circles centred on the wire. This is the \vec{B}-field pattern Ørsted saw around his current-carrying wire.

Biot-Savart geometryA current element I dl at the origin pointing along the y-axis. A field point P is at displacement vector r making angle theta with the current direction. The resulting dB points out of the page at P, shown as a circled dot, perpendicular to both I dl and r. x y (current direction) I dl r θ P dB (out of page) |dB| = (μ₀/4π) · I |dl| sin θ / r²
The Biot-Savart geometry. The red vector $I\,d\vec{l}$ is the current element; $\vec{r}$ is the displacement to the field point P, making angle $\theta$ with the current. The field $d\vec{B}$ at P is perpendicular to the plane of $d\vec{l}$ and $\vec{r}$ — here, out of the page, shown as a circled dot — with magnitude $(\mu_0/4\pi)\,I\,|d\vec{l}|\sin\theta/r^2$. Walk P around to the other side of the current line and $d\vec{B}$ flips to *into* the page — the field circles the wire.

Why the cross product and the inverse-square?

The Biot-Savart law is an experimental law; you cannot derive it from anything more fundamental within magnetostatics. But two of its features follow from physical reasoning once you accept that a magnetic field comes from moving charge.

Why inverse-square. If the field of a point source fell off less steeply, the total flux through a sphere enclosing the source would grow as you moved the sphere away — violating source-conservation. Inverse-square is the unique falloff for which flux through any enclosing surface is constant, the natural rate for "influence spreading out" in three-dimensional space.

Why the cross product. A current element has a built-in direction (the direction of current flow). The resulting field cannot point along that direction: rotational symmetry around the current line would then force \vec{B} to be zero. It also cannot point along \hat{r}: any radial component would imply magnetic monopoles, which do not exist in nature. The only direction consistent with both constraints is perpendicular to both d\vec{l} and \hat{r} — exactly what the cross product delivers.

The experimental content is then the inverse-square law (2) and the numerical value of \mu_0/(4\pi). Everything else follows from symmetry.

Comparison with Coulomb's law

Side by side:

d\vec{E} \;=\; \frac{1}{4\pi\varepsilon_0}\,\frac{dq\,\hat{r}}{r^2}, \qquad d\vec{B} \;=\; \frac{\mu_0}{4\pi}\,\frac{I\,d\vec{l}\times\hat{r}}{r^2}.

Same 1/r^2, same universal-constant prefactor. Two differences:

  1. dq is a scalar (a charge); I\,d\vec{l} is a vector (a current element with a direction).
  2. The direction of d\vec{E} is along \hat{r} (radial); the direction of d\vec{B} is perpendicular to \hat{r} (tangential).

That one structural change — scalar source to vector source, radial response to perpendicular response — is the whole reason magnetism looks so different from electrostatics. Electric field lines begin on positive charges and end on negative ones; magnetic field lines form closed loops, never beginning or ending, because the only way to make them with d\vec{l}\times\hat{r} is to wrap them around the current.

Computing fields — superposition

A real wire is not a single current element; it is an extended distribution of them. To find the total \vec{B} at a point, integrate (2) over the whole current:

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

This is a vector integral. In practice you pick convenient axes, decompose d\vec{B} into components, exploit any symmetry (if one component cancels by symmetry, you only integrate the surviving component), and integrate. Two classic setups illustrate the method.

Explore the on-axis field of a current loop

Before deriving it in Example 1, visualise what happens. The interactive figure below lets you drag the loop's radius R while the axial distance x is fixed at 0.10 m, and read off the on-axis magnetic field B_\text{axis}(R). The current is fixed at 1 A. The formula (to be derived) is B = \mu_0 I R^2/[2(R^2+x^2)^{3/2}].

Interactive: on-axis field versus loop radius A plot of the axial magnetic field B at a fixed axial distance x = 0.10 m, as a function of the loop radius R, for a current of 1 ampere. The curve rises from zero, peaks near R = sqrt(2) x ≈ 0.14 m, and falls off for larger R. A draggable dot selects R and readouts show R and B. loop radius R (m) B on axis (T) 0 5 µT 10 µT 0 0.1 0.2 0.3 B_axis(R) drag the red dot along the axis
Drag the red dot to change the loop radius $R$, with the field point fixed at $x = 0.1$ m on the axis. The field peaks near $R = x\sqrt{2}\approx 0.14$ m (a standard calculus result): for very small $R$ the loop is too small to produce much field; for very large $R$ the distance $r = \sqrt{R^2+x^2}$ to any current element grows, suppressing the inverse-square contribution. $I = 1$ A.

Worked examples

Example 1: Magnetic field on the axis of a circular current loop

A circular wire loop of radius R carries steady current I. Find the magnetic field at a point P on the loop's axis, distance x from the centre.

Geometry for a circular loop field on axisA circular loop seen edge-on as an ellipse centred at the origin. Axial point P is at distance x from the centre along the axis. A current element dl at the top of the loop is shown as a dot out of the page, with displacement vector r pointing to P making angle alpha with the axis, where sin(alpha) = R over r. Vector dB at P is perpendicular to r; its axial component dB sin(alpha) survives the integration while its radial component dB cos(alpha) cancels by symmetry. axis loop, radius R O P x I dl r α dB dB sin α (axial) dB cos α (radial, cancels)
The loop of radius $R$ seen edge-on; P sits on the axis distance $x$ from O. A current element $I\,d\vec{l}$ at the top of the loop points out of the page. $\vec{r}$ goes from the element to P, making angle $\alpha$ with the axis where $\cos\alpha = x/r$ and $\sin\alpha = R/r$. $d\vec{B}$ is perpendicular to $\vec{r}$ (since $d\vec{l}\perp\vec{r}$ for every loop element), rotated $90°$ from $\vec{r}$ in the same plane. Its axial component is $dB\sin\alpha$ (surviving the integration); its radial component $dB\cos\alpha$ cancels with the diametrically opposite element.

Step 1. Identify \theta for every loop element.

Every current element on the loop is tangent to the loop (lying in the loop's plane). The line from the element to P lies in a plane containing both the element's radius vector and the axis. The element itself is perpendicular to this plane, so the angle \theta between d\vec{l} and \hat{r} is exactly 90° for every element. Therefore \sin\theta = 1, and

|d\vec{B}| \;=\; \frac{\mu_0}{4\pi}\,\frac{I\,|d\vec{l}|}{r^2}, \quad r = \sqrt{R^2 + x^2}. \tag{5}

Why: the fact that \sin\theta = 1 for every point on the loop is what makes the loop a clean example — all elements contribute with full strength 1/r^2, and r is the same for every element (each is the same distance \sqrt{R^2+x^2} from P). This symmetry is what makes the integration tractable.

Step 2. Decompose d\vec{B} into axial and radial components.

By the right-hand rule, d\vec{B} is perpendicular to \vec{r} (and perpendicular to the current element, which is out of the page). So d\vec{B} lies in the plane of \vec{r} and the axis, rotated 90° from \vec{r}. Define \alpha as the angle \vec{r} makes with the axis; then \cos\alpha = x/r and \sin\alpha = R/r.

A vector rotated 90° from \vec{r} in that plane has an axial projection of magnitude |d\vec{B}|\sin\alpha and a radial projection of magnitude |d\vec{B}|\cos\alpha.

Why: if \vec{r} points at angle \alpha above the axis, then a vector perpendicular to \vec{r} (in the same plane) points at angle (90°-\alpha) above the axis. Its axial projection is \cos(90°-\alpha) = \sin\alpha. The figure shows this decomposition: axial dashed line has length dB\sin\alpha; radial dashed line has length dB\cos\alpha.

Step 3. Use symmetry — the radial component cancels.

For every current element at angular position \phi on the loop, there is an element at \phi + 180°. Their radial d\vec{B} contributions point in opposite directions (one toward +y, one toward -y) and cancel in pairs. Only the axial component survives.

\vec{B} \;=\; \hat{x}\int dB\,\sin\alpha \;=\; \hat{x}\int\frac{\mu_0}{4\pi}\,\frac{I\,|d\vec{l}|}{r^2}\cdot\frac{R}{r} \;=\; \frac{\mu_0\,I\,R}{4\pi\,r^3}\,\hat{x}\int d\ell.

Why: cancelling the radial component by symmetry is a trick that returns again and again in magnetostatics — spotting a symmetry that kills an unwanted component reduces a 3-D vector integral to a 1-D scalar integral. And r = \sqrt{R^2+x^2} is constant around the loop (every element is the same distance from P), so r^3 comes outside the integral.

Step 4. Integrate around the loop.

The integral \int d\ell is the loop's circumference, 2\pi R:

\vec{B}_\text{axis} \;=\; \frac{\mu_0\,I\,R}{4\pi\,r^3}\cdot 2\pi R\,\hat{x} \;=\; \frac{\mu_0\,I\,R^2}{2\,r^3}\,\hat{x}.

Substituting r^3 = (R^2+x^2)^{3/2}:

\boxed{\;\vec{B}_\text{axis} \;=\; \frac{\mu_0\,I\,R^2}{2\,(R^2 + x^2)^{3/2}}\,\hat{x}.\;} \tag{6}

Why: equation (6) is the on-axis field of a circular loop — one of the most useful results in magnetostatics. It underlies the field inside an MRI coil, the on-axis field of a Helmholtz pair, and the magnetic-dipole moment idea m = IA. The direction is given by the right-hand curl rule: curl the fingers of your right hand around the loop in the direction of the current, and the thumb points along \vec{B}.

Step 5. Two limiting cases — sanity checks.

At the centre (x = 0): r = R, so \vec{B}_\text{centre} = \mu_0 I/(2R). The field is maximum at the centre.

Far from the loop (x\gg R): (R^2+x^2)^{3/2}\approx x^3, so \vec{B}_\text{axis}\approx \mu_0 I R^2/(2x^3) = \mu_0 m/(2\pi x^3) with m = I\pi R^2 — the characteristic magnetic-dipole 1/x^3 falloff.

Result: \vec{B}_\text{axis}(x) = \mu_0\,I\,R^2/[2(R^2+x^2)^{3/2}], along the axis in the right-hand-rule direction.

What this shows: One loop, one current, three-dimensional geometry — and you still end up with a compact closed-form expression. Biot-Savart plus one symmetry argument plus one elementary integral gives the complete answer.

Example 2: Magnetic field of a finite straight wire

A straight wire segment carries steady current I. At a field point P a perpendicular distance a from the wire, the ends of the segment are seen at angles \phi_1 and \phi_2 from the perpendicular (measured at P, with the convention that both angles are positive when the ends lie on opposite sides of the foot of the perpendicular). Show that the field at P is

B \;=\; \frac{\mu_0 I}{4\pi a}(\sin\phi_1 + \sin\phi_2).
Geometry for the field of a finite straight wireA horizontal wire segment along the y-axis. Point P is at perpendicular distance a to the right. The foot of the perpendicular from P to the wire sits at the origin. Segment extends from y = -L1 below to y = +L2 above. The line from the upper end of the wire to P makes angle phi2 with the perpendicular PO; the line from the lower end makes phi1 on the opposite side. A current element dl at position y, displacement vector r pointing to P, and dB out of the page at P are all shown. wire, current I ↑ O a (perpendicular distance) P φ₂ φ₁ dl at y r φ dB out of page
The wire runs along the $y$-axis with current $I$ flowing upward. P is the field point at perpendicular distance $a$ to the right. Let $y$ be the position of a current element $d\vec{l}$ along the wire, and let $\phi$ be the angle at P between the perpendicular and the line to the element. The wire extends from $y = -L_1$ (at angle $\phi_1$ below) to $y = +L_2$ (at angle $\phi_2$ above).

Step 1. Set up coordinates.

Let the wire lie along the y-axis, with O the foot of the perpendicular from P. A current element d\vec{l} at height y (i.e. d\vec{l} = \hat{y}\,dy, current flowing in the +\hat{y} direction) is at displacement

\vec{r} \;=\; a\,\hat{x} - y\,\hat{y}, \qquad r = \sqrt{a^2 + y^2}.

Why: \vec{r} points from the element at (0, y) to the field point at (a, 0), which is (a-0, 0-y) = (a, -y).

Step 2. Cross product d\vec{l}\times\hat{r}.

d\vec{l} \;=\; \hat{y}\,dy, \qquad \hat{r} \;=\; \frac{a\,\hat{x} - y\,\hat{y}}{r}.
d\vec{l}\times\hat{r} \;=\; \frac{dy}{r}(\hat{y}\times(a\,\hat{x} - y\,\hat{y})) \;=\; \frac{dy}{r}(a\,\hat{y}\times\hat{x} - y\,\hat{y}\times\hat{y}) \;=\; \frac{dy}{r}(-a\,\hat{z}).

So d\vec{l}\times\hat{r} points in the -\hat{z} direction (into the page, if you hold the page so \hat{x} is right and \hat{y} is up). For a field point on the other side of the wire, the field would point the other way. Every element contributes a d\vec{B} in the same direction (-\hat{z}), so the integration is scalar.

Why: \hat{y}\times\hat{x} = -\hat{z} by the standard right-hand rule for Cartesian unit vectors. This confirms that all contributions from the wire add constructively — the field at P is the integral of magnitudes, with no component cancellation.

Step 3. Magnitude of d\vec{B}.

|d\vec{B}| \;=\; \frac{\mu_0}{4\pi}\,\frac{I\,a\,dy}{r^3} \;=\; \frac{\mu_0 I a}{4\pi}\cdot\frac{dy}{(a^2 + y^2)^{3/2}}. \tag{7}

Why: from step 2, |d\vec{l}\times\hat{r}| = a\,dy/r. Dividing by r^2 gives a\,dy/r^3 = a\,dy/(a^2+y^2)^{3/2}.

Step 4. Integrate from y = -L_1 to y = +L_2.

B \;=\; \frac{\mu_0 I a}{4\pi}\int_{-L_1}^{L_2}\frac{dy}{(a^2+y^2)^{3/2}}.

To evaluate, change variables: let y = a\tan\phi, so dy = a\sec^2\phi\,d\phi and (a^2+y^2)^{3/2} = a^3\sec^3\phi. Then

\frac{dy}{(a^2+y^2)^{3/2}} \;=\; \frac{a\sec^2\phi\,d\phi}{a^3\sec^3\phi} \;=\; \frac{\cos\phi\,d\phi}{a^2}.

When y = -L_1, \tan\phi = -L_1/a, so \phi = -\phi_1. When y = +L_2, \tan\phi = L_2/a, so \phi = +\phi_2. The integral becomes

B \;=\; \frac{\mu_0 I a}{4\pi}\cdot\frac{1}{a^2}\int_{-\phi_1}^{\phi_2}\cos\phi\,d\phi \;=\; \frac{\mu_0 I}{4\pi a}\,[\sin\phi]_{-\phi_1}^{\phi_2} \;=\; \frac{\mu_0 I}{4\pi a}(\sin\phi_2 - \sin(-\phi_1)).

Why: the substitution y = a\tan\phi is the standard trick for integrals of (a^2+y^2)^{-3/2} — it turns the algebraic expression into a trigonometric one whose antiderivative is \sin\phi.

Step 5. Simplify.

Using \sin(-\phi_1) = -\sin\phi_1:

\boxed{\;B \;=\; \frac{\mu_0 I}{4\pi a}(\sin\phi_1 + \sin\phi_2).\;} \tag{8}

Step 6. Two limiting cases.

Infinite wire (L_1, L_2 \to \infty): \phi_1 = \phi_2 = 90°, \sin\phi_1 = \sin\phi_2 = 1, so

B_\text{infinite} \;=\; \frac{\mu_0 I}{4\pi a}\cdot 2 \;=\; \frac{\mu_0 I}{2\pi a}. \tag{9}

This is the classic field of an infinite straight wire, which you will see again in the next chapter and in Ampère's law.

Semi-infinite wire starting at O (e.g. wire extending from y = 0 to +\infty, with P on the perpendicular bisector end): \phi_1 = 0, \phi_2 = 90°, so B = \mu_0 I/(4\pi a) — exactly half the infinite-wire field.

Result: B = (\mu_0 I)/(4\pi a)\,(\sin\phi_1 + \sin\phi_2).

What this shows: The finite-wire formula contains both the infinite wire and every partial-wire case as special values of \phi_1, \phi_2. In any geometry where you can identify the endpoints' angles as seen from the field point, you get the field without re-doing the integral. This is why the formula is so useful — it is the one-shot result for every polygonal-wire problem in JEE Advanced.

Common confusions

If you came here to understand the Biot-Savart law and compute simple fields, you have what you need. What follows is for JEE Advanced readers who want the vector-calculus version, the connection to Ampère's law, the numerical value of \mu_0 after 2019, and the special role of the Lorentz gauge.

The vector-potential form

You can rewrite Biot-Savart as the curl of a magnetic vector potential \vec{A}:

\vec{B}(\vec{r}) \;=\; \nabla\times\vec{A}(\vec{r}), \qquad \vec{A}(\vec{r}) \;=\; \frac{\mu_0}{4\pi}\int\frac{I\,d\vec{l}'}{|\vec{r}-\vec{r}'|}.

The vector potential \vec{A} is the magnetic analogue of the electric scalar potential V. It is determined up to a gauge choice (you can add any gradient \nabla\chi and the physics is unchanged). In the Lorenz gauge \nabla\cdot\vec{A} = -(1/c^2)\partial V/\partial t, the potentials (V, \vec{A}) form a 4-vector that transforms cleanly under special relativity — magnetic and electric fields are two components of one electromagnetic 4-tensor. That is the insight that unified them in the nineteenth century and put them on the footing Maxwell left behind.

The vector-potential approach is how Biot-Savart is implemented in numerical electromagnetics codes (including the MRI coil design simulations at AIIMS and the particle-accelerator magnets at BARC): you integrate for \vec{A}, then take the curl to get \vec{B}. It is almost always less work than integrating \vec{B} directly, because \vec{A} has no cross-product in its definition.

The differential form — what you cannot see in a single-element view

Taking the curl of the Biot-Savart \vec{B} gives

\nabla\times\vec{B} \;=\; \mu_0\,\vec{J}

where \vec{J} is the volume current density (A/m²). Integrating this over any surface and using Stokes' theorem produces Ampère's circuital law:

\oint\vec{B}\cdot d\vec{\ell} \;=\; \mu_0\,I_\text{enclosed},

which is the subject of Chapter 183. So Ampère's law is a consequence of Biot-Savart, not an independent law — the two are equivalent in magnetostatics. Ampère is just the symmetry-exploitation shortcut for problems where a closed-loop integral of \vec{B} is easier than a volume integral of \vec{J}. Problems without symmetry still demand Biot-Savart.

Also from the curl form:

\nabla\cdot\vec{B} \;=\; 0

— there are no magnetic monopoles. This is Gauss's law for magnetism, and it is structurally necessary: if Biot-Savart is written as a curl, its divergence is automatically zero (because \nabla\cdot(\nabla\times\vec{A}) = 0 always).

The 2019 SI redefinition — why \mu_0 is no longer exactly 4\pi\times 10^{-7}

Until 2019, the ampere was defined by setting \mu_0 = 4\pi\times 10^{-7} T·m/A exactly, and the tesla was derived from this. In the revised SI (effective 20 May 2019), the ampere is defined by fixing the elementary charge e at exactly 1.602{,}176{,}634\times 10^{-19} C, and \mu_0 becomes a measured quantity — though measured so precisely that to more than nine significant figures it still equals 4\pi\times 10^{-7} T·m/A. For every calculation in this chapter (and in JEE Advanced), you use \mu_0 = 4\pi\times 10^{-7} without hesitation; the post-2019 change matters only to metrology laboratories measuring fundamental constants at NPL (National Physical Laboratory) in Delhi.

Relativistic consistency — magnetism as a relativistic effect

A subtle consequence of the Biot-Savart law: what one observer sees as a pure electric field, another (moving) observer sees as a mixture of electric and magnetic fields. Specifically, if you have a static line of positive charges, an observer at rest sees only an electric field. An observer moving parallel to the line sees the charges moving (a current!) and therefore sees a magnetic field as well. The two observers agree on the physics (the force on a test charge); they disagree on whether that force is "electric" or "magnetic." Magnetism is, in this precise sense, the relativistic aspect of electricity — you do not need v \approx c to see this, because the magnetic field is proportional to v/c^2 times the electric field, and even slow drift velocities in a wire produce measurable magnetic fields because of the enormous density of charges.

This is why \mu_0\varepsilon_0 = 1/c^2: the constant \mu_0 is exactly what is needed to make magnetic forces consistent with Coulomb's law under Lorentz boosts. If you accept Coulomb's law, the Biot-Savart law is forced on you by special relativity — there is nothing genuinely new in magnetism beyond electricity-plus-relativity.

Where this leads next