Biot-Savart law

A compass needle deflects the instant a current flows in a nearby wire — Oersted's discovery in 1820 told us that moving charge makes a magnetic field. It took the French physicists Jean-Baptiste Biot and Félix Savart, working the very same year, to pin down exactly how much field and in which direction, for the smallest possible piece of current you can imagine: a current element $I\,d\vec l$ , a stub of wire almost too short to draw. Their law is the magnetic twin of Coulomb's law — same inverse-square fall-off with distance — except that a current element does not radiate its field symmetrically outward like a point charge. It fires the field sideways, out of the plane containing the wire and the point you are asking about, using a cross product instead of a plain direction vector. That one small difference — cross product instead of straight-line direction — is the whole plot of this chapter. Every magnetic field formula you will meet next, the loop's centre field $B=\mu_0I/2R$ , the straight wire's field $B=\mu_0I/2\pi r$ , even the solenoid's $B=\mu_0nI$ , is Biot-Savart's law summed (integrated) over every current element in that particular wire shape. Get the single-element law right — its $1/r^2$ , its $\sin\theta$ , its right-hand-rule direction — and the rest of the chapter is bookkeeping: choosing a wire shape, breaking it into elements, and adding up their contributions. This article builds that single-element law from the ground up, keeps faithful to the two original worked numbers from the first pass — a field of $1\ \mu\text{T}$ from a $1\ \text{cm}$ , $10\ \text{A}$ element viewed side-on, and exactly zero field viewed end-on — and stretches them into board-style and competitive-style problems, an explicit derivation walk-through, common exam traps, and a quick-fire quiz.

Key takeaways

Biot-Savart law: a current element $I\,d\vec l$ produces a field $d\vec B=\dfrac{\mu_0}{4\pi}\dfrac{I\,d\vec l\times\hat r}{r^2}$ at a point a distance $r$ away, with $\mu_0=4\pi\times10^{-7}\ \text{T}\,\cdot\,\text{m/A}$ .
Magnitude form: $dB=\dfrac{\mu_0}{4\pi}\dfrac{I\,dl\sin\theta}{r^2}$ , where $\theta$ is the angle between the element and the line to the point.
Direction rule: $d\vec B$ is perpendicular to the plane of $d\vec l$ and $\hat r$ , given by the right-hand (cross-product) rule — never along $\hat r$ itself.
Two limits worth memorising: the field is zero along the wire's own axis ( $\theta=0°$ ) and maximum broadside to it ( $\theta=90°$ ).
It is the seed law of the whole chapter: integrating it over a circular loop gives $B=\mu_0IR^2/2(R^2+x^2)^{3/2}$ and, at the centre, $B=\mu_0I/2R$ ; over an infinite straight wire it reproduces $B=\mu_0I/2\pi r$ , the same answer Ampere's circuital law gives instantly for symmetric cases.
Compare with Coulomb's law: both are inverse-square "source element" laws, but the electric one points straight along $\hat r$ while the magnetic one is perpendicular to both the current element and $\hat r$ .
Key idea: the Biot-Savart law gives the tiny magnetic field $dB=(\mu_0/4\pi)(I\,dl\sin\theta)/r^2$ made by one current element, perpendicular to both the element and the line to the point, and summing it over a whole conductor is how every other magnetic-field formula in this chapter is built.

The core insight

Every current element $I\,d\vec l$ produces a magnetic field $d\vec B=(\mu_0/4\pi)(I\,d\vec l\times\hat r)/r^2$ that falls off as $1/r^2$ , vanishes along the wire's own direction and peaks broadside to it, and this single elemental law, added up over a whole conductor, is the source of every magnetic field formula in the chapter.

Where it fits in the syllabus

In the CBSE Class 12 unit Moving Charges and Magnetism, the chapter opens with Oersted's qualitative discovery that current makes a magnetic field, and the Biot-Savart law is the very next step: the precise, quantitative rule for that field, stated and used exactly as printed in the syllabus, $d\vec B=(\mu_0/4\pi)(I\,d\vec l\times\hat r)/r^2$ . It is squarely inside the prescribed scope, and its main listed application — deriving the field on the axis of a circular current loop, $B=\mu_0IR^2/2(R^2+x^2)^{3/2}$ , with centre value $B=\mu_0I/2R$ — is the very next node in this chapter, so this article treats that derivation as the headline worked case. Ampere's circuital law, covered separately, gives the straight-wire and solenoid fields more quickly using symmetry, but it is Biot-Savart alone that must be used for a loop or any other non-symmetric geometry — that boundary is worth remembering for both boards and competitive exams.

The physical picture

Picture a wire carrying a steady current $I$ . Chop it, in your mind, into an enormous number of tiny straight segments, each of length $dl$ , so small that each looks like a straight arrow pointing along the local direction of current flow — call this vector $d\vec l$ . Biot and Savart's experimental result is that each such element, on its own, produces a small magnetic field $d\vec B$ at a field point $P$ a distance $r$ away, along the unit vector $\hat r$ pointing from the element to $P$ :

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

where: $\mu_0=4\pi\times10^{-7}\ \text{T}\,\cdot\,\text{m/A}$ is the permeability of free space (a fixed constant of nature, the magnetic counterpart of $\varepsilon_0$ in Coulomb's law), $I$ is the steady current in the wire (amperes), $d\vec l$ is the current-element vector (metres, pointing along the current), $\hat r$ is the unit vector from the element toward $P$ , and $r$ is the distance from the element to $P$ (metres). The physical reading: the field is strongest close to the element and dies away with the square of distance, exactly like an electric field from a point charge — but its direction comes from a cross product, not from $\hat r$ directly, so $d\vec B$ always points perpendicular to the little plane that contains the current element and the line joining it to $P$ . Taking magnitudes and letting $\theta$ be the angle between $d\vec l$ and $\hat r$ turns the vector law into a number you can actually compute:

dB = \dfrac{\mu_0}{4\pi}\dfrac{I\,dl\sin\theta}{r^2}

where: $\sin\theta$ carries all of the directional information that survived taking a magnitude — it is $1$ when $P$ lies broadside to the element ( $\theta=90°$ ) and $0$ when $P$ lies directly ahead of or behind the element, in line with the current itself ( $\theta=0°$ or $180°$ ). This single $\sin\theta$ factor explains a fact that surprises many students: a long straight wire produces zero field at a point lying on its own extended axis, however close that point is, because every element of the wire has $\theta=0$ for a point on that axis.

Figure

A short current element on the left with the current direction arrow, dashed lines to three field points at different angles, and the resulting field vectors drawn perpendicular to the plane at each point, largest at ninety degrees and shrinking to zero along the wire axis

Superposition: from one element to a real wire

No real wire is a single point-like element — the law as given is only the seed. To find the field of an actual finite or infinite conductor, you sum (in the limit, integrate) the $d\vec B$ contributions from every element along its length, keeping careful track of both magnitude and direction at each step, because $d\vec B$ from different elements can point in different directions and must be added as vectors, not scalars. This is exactly how the two named applications in this chapter are obtained: summing around a full circle of radius $R$ gives the loop's axial field $B=\mu_0IR^2/2(R^2+x^2)^{3/2}$ , and summing along an infinitely long straight wire (a separate, longer integration) reproduces $B=\mu_0I/2\pi r$ — the same result Ampere's circuital law delivers far more quickly for that symmetric case, though Ampere's law cannot be used for the loop's off-axis or general field.

Derivation ABuilding the current-element law and reading its two limits

Goal: starting from the way Biot and Savart's experiment is stated as a physical law — that the field element $d\vec B$ produced by a current element $I\,d\vec l$ at a field point a distance $r$ away depends on the current, the length of the element, the sine of the angle between the element and the line to the point, and falls as the inverse square of that distance — show precisely how these four experimental facts combine into the single vector formula $d\vec B=(\mu_0/4\pi)(I\,d\vec l\times\hat r)/r^2$ , and then extract the two special-angle limits ( $\theta=90°$ and $\theta=0°$ ) that most exam questions actually test, so that both the formula and its physical boundary behaviour are usable with confidence rather than memorised as an isolated symbol string.

Step 1. Biot and Savart's experimental observations, taken together, say that the magnitude of the field element scales as $dB\propto I$ , $dB\propto dl$ , $dB\propto\sin\theta$ , and $dB\propto 1/r^2$ . Multiplying these four proportionalities into a single relation and inserting a constant of proportionality $k$ gives $dB=k\,\dfrac{I\,dl\sin\theta}{r^2}$ .

Step 2. In SI units the constant is written $k=\mu_0/4\pi$ , where $\mu_0=4\pi\times10^{-7}\ \text{T}\,\cdot\,\text{m/A}$ is the permeability of free space, chosen (historically) so that the ampere comes out with a convenient, exactly defined size in the force-between-wires experiment covered later in this chapter. Substituting gives the working magnitude formula, $dB=\dfrac{\mu_0}{4\pi}\dfrac{I\,dl\sin\theta}{r^2}$ .

Step 3. To restore the direction, recall that $\sin\theta$ is precisely the magnitude of the cross product of a unit vector along $d\vec l$ with the unit vector $\hat r$ , and that a cross product automatically points perpendicular to both vectors — exactly the direction the right-hand rule assigns to $d\vec B$ in the laboratory. Writing $d\vec l$ for the full vector element (magnitude $dl$ , direction along the current) recovers the vector law, $d\vec B=\dfrac{\mu_0}{4\pi}\dfrac{I\,d\vec l\times\hat r}{r^2}$ , with the order $d\vec l\times\hat r$ fixed by the experiment — reversing the order flips the field's direction.

Step 4. Read the broadside limit. Set $\theta=90°$ , so $\sin\theta=1$ and the formula collapses to its largest possible value for a given $I$ , $dl$ , $r$ : $dB_{\max}=\dfrac{\mu_0}{4\pi}\dfrac{I\,dl}{r^2}$ . Physically, a point directly to the side of a short current element feels the strongest possible field from that element.

Step 5. Read the axial limit. Set $\theta=0°$ (or $180°$ ), so $\sin\theta=0$ and $dB=0$ identically, regardless of how large $I$ , $dl$ are or how small $r$ is. This is why a long straight wire, built from countless collinear elements, contributes exactly zero field to any point lying on its own extended line — a fact examiners like to test because it looks counter-intuitive next to Coulomb's law, where a point charge's field is never zero anywhere in space.

Worked examples

Board level - field beside and in line with a short element

A short straight current element of length $dl=1\ \text{cm}$ carries a current $I=10\ \text{A}$ . Find the magnetic field it produces (a) at a point $P$ a distance $r=0.1\ \text{m}$ away, where the line joining the element to $P$ makes an angle of $90°$ with the element, and (b) at a point $Q$ the same distance away but lying along the direction of the current element itself.

Step 1. Convert to SI and list the knowns: $dl=1\ \text{cm}=0.01\ \text{m}$ , $I=10\ \text{A}$ , $r=0.1\ \text{m}$ . For point $P$ , $\theta=90°$ so $\sin\theta=1$ ; for point $Q$ , $\theta=0°$ so $\sin\theta=0$ .

Step 2. Write the magnitude formula and substitute $\mu_0/4\pi=10^{-7}\ \text{T}\,\cdot\,\text{m/A}$ (the standard shortcut value used throughout this chapter): $dB=10^{-7}\times\dfrac{I\,dl\sin\theta}{r^2}$ .

Step 3. Compute for $P$ : $dB_P=10^{-7}\times\dfrac{(10)(0.01)(1)}{(0.1)^2}=10^{-7}\times\dfrac{0.1}{0.01}=10^{-7}\times10=1\times10^{-6}\ \text{T}$ .

Step 4. Compute for $Q$ : $dB_Q=10^{-7}\times\dfrac{(10)(0.01)(0)}{(0.1)^2}=0$ .

Answer: $dB_P=1\times10^{-6}\ \text{T}=1\ \mu\text{T}$ , directed out of the plane containing the element and $P$ (by the right-hand rule); $dB_Q=0\ \text{T}$ exactly, because $Q$ lies on the element's own axis.

Competitive level - net field from two perpendicular current elements

Two short current elements sit at the origin, each carrying $I=10\ \text{A}$ and length $dl=0.01\ \text{m}$ : element 1 points along the $x$ -axis, element 2 points along the $y$ -axis. Both contribute to the field at a point $P$ on the $z$ -axis, a distance $r=0.1\ \text{m}$ from the origin, directly above both elements. Find the magnitude of the total field at $P$ .

Step 1. Note the geometry: since $P$ is on the $z$ -axis while both elements lie in the $xy$ -plane through the origin, the line from either element to $P$ makes a $90°$ angle with that element's own direction (a point straight "up" off a horizontal element is always broadside to it), so $\theta=90°$ and $\sin\theta=1$ for both.

Step 2. Compute each element's contribution using the same magnitude formula as the board example: $dB_1=dB_2=10^{-7}\times\dfrac{(10)(0.01)(1)}{(0.1)^2}=1\times10^{-6}\ \text{T}$ , reusing the identical arithmetic as the $1\ \mu\text{T}$ result above, since $I$ , $dl$ , $r$ , and $\theta$ are unchanged.

Step 3. Find the directions by the right-hand rule ( $d\vec l\times\hat r$ ): $d\vec B_1$ (from the $x$ -directed element) points along $-\hat y$ , while $d\vec B_2$ (from the $y$ -directed element) points along $+\hat x$ — the two field contributions are perpendicular to each other, not parallel, so they cannot simply be added as numbers.

Step 4. Combine as perpendicular vectors using Pythagoras, exactly as with perpendicular electric field components: $dB_{\text{net}}=\sqrt{dB_1^2+dB_2^2}=\sqrt{(1\times10^{-6})^2+(1\times10^{-6})^2}=\sqrt{2}\times10^{-6}\ \text{T}\approx1.41\times10^{-6}\ \text{T}$ .

Answer: the net field at $P$ has magnitude $\approx1.41\ \mu\text{T}$ , directed along $(\hat x-\hat y)/\sqrt2$ in the $xy$ -plane sense of the two perpendicular contributions — not simply double the single-element answer, because the two fields add as vectors, not scalars. Timing note: this is a 3-mark-style question dressed up with vector combination — the arithmetic for each element is identical to the board example (reuse it instantly, do not recompute from scratch), and the only new work is recognising the two field vectors are perpendicular and combining them with Pythagoras; budget under 3 minutes once the single-element result is memorised.

Common traps

Treating the Biot-Savart law like Coulomb's law and forgetting the $\sin\theta$ and cross-product nature — the field is not simply along $\hat r$ , and it vanishes exactly along the direction of current flow, not just weakens.
Trying to use $d\vec B$ from one element as though it were the total field of a finite or infinite wire — the total field always requires summing (integrating) contributions from every element along the conductor, as in the loop and straight-wire derivations later in the chapter.
Confusing $\mu_0/4\pi$ (used directly in the Biot-Savart formula, value $10^{-7}\ \text{T}\,\cdot\,\text{m/A}$ ) with $\mu_0$ itself ( $4\pi\times10^{-7}\ \text{T}\,\cdot\,\text{m/A}$ ) — mixing these up gives an answer off by a factor of $4\pi\approx12.6$ .
Getting the field direction backwards by computing $\hat r\times d\vec l$ instead of $d\vec l\times\hat r$ — the order in a cross product fixes the sign, and swapping it flips the field to point the wrong way.
Adding two or more field contributions as plain numbers when they point in different directions — as in the competitive example above, perpendicular (or otherwise non-parallel) field vectors must be combined by vector addition, not simple arithmetic sum.

You should be able to…

State the Biot-Savart law in both vector form, $d\vec B=(\mu_0/4\pi)(I\,d\vec l\times\hat r)/r^2$ , and magnitude form, $dB=(\mu_0/4\pi)(I\,dl\sin\theta)/r^2$ , and explain what each symbol represents.
Compute the magnitude of $dB$ from a given current element, current, distance and angle, using the shortcut $\mu_0/4\pi=10^{-7}\ \text{T}\,\cdot\,\text{m/A}$ .
Explain, using the $\sin\theta$ factor, why the field from a straight current element is exactly zero along its own axis and maximum broadside to it.
Find the direction of $d\vec B$ using the right-hand rule applied to $d\vec l\times\hat r$ , and recognise when this order must not be swapped.
Combine field contributions from two or more current elements as vectors (not scalars) when they point in different directions, including the perpendicular-components case.
State how the Biot-Savart law is the starting point for deriving the circular-loop field $B=\mu_0IR^2/2(R^2+x^2)^{3/2}$ , and how it relates to the Ampere's-law shortcuts for a straight wire and solenoid.

Ask the tutor: Want to see the full step-by-step integration that takes this elemental law all the way to the circular loop's centre field $B=\mu_0I/2R$ ? Ask the AI tutor to walk through that derivation next.

Why is there a 4 pi in the formula at all — where does it come from?

It is not an arbitrary decoration; it comes from how $\mu_0$ was historically defined so that the ampere (the SI base unit of current) has a clean, exactly specified size in the force-between-two-wires experiment you will meet later in this chapter. Writing the constant as $\mu_0/4\pi$ rather than folding the $4\pi$ into a redefined $\mu_0$ keeps the $4\pi$ visible in "spherical" formulas (like this one, which effectively spreads the effect of the source over a sphere of area $4\pi r^2$ ) and absent from formulas built for straight-line symmetry, like $B=\mu_0I/2\pi r$ for a long wire. You do not need to derive this convention — just remember that $\mu_0/4\pi=10^{-7}\ \text{T}\,\cdot\,\text{m/A}$ exactly, and use that shortcut number in every Biot-Savart magnitude calculation.

How is this different from what I already know about Coulomb's law?

They share the same skeleton — both are inverse-square laws describing the field produced by an infinitesimally small "source element," a point charge $dq$ on one side, a current element $I\,d\vec l$ on the other — but they differ in the crucial detail of direction. Coulomb's law points the field straight along the line from source to field point: $d\vec E\propto \hat r/r^2$ . The Biot-Savart law instead uses a cross product, $d\vec B\propto (d\vec l\times\hat r)/r^2$ , which is perpendicular to that line and to the current element itself. That single change is why magnetic fields circle around wires instead of pointing radially outward from them, and it is the reason the right-hand rule, not simple radial symmetry, is the tool you reach for in this chapter.

Do I actually need to integrate the Biot-Savart law myself in the board exam?

For the exact prescribed application — the magnetic field on the axis of a circular current loop — yes, the derivation of $B=\mu_0IR^2/2(R^2+x^2)^{3/2}$ starting from the elemental $d\vec B$ is explicitly listed and can appear as a 3-to-5-mark derivation question, so it is worth practising fully (that derivation is the natural next node after this one). Outside that one named case, you are expected to state and apply the Biot-Savart law itself, not re-derive the straight-wire or solenoid results from scratch — those are obtained far more efficiently from Ampere's circuital law, which is examined separately.

Is the toroid, or deriving Ampere's law from Biot-Savart, something I should learn here?

No — both are explicitly outside the current CBSE 2025-26 prescription for this chapter. The toroid's field derivation and a formal proof of Ampere's circuital law starting from the Biot-Savart law are advanced extensions found in some older textbooks and reference sites, but they are not listed content for your board exam, so do not spend revision time deriving them; focus instead on the loop-axis derivation and the direct applications named in the syllabus (straight wire, solenoid — both via Ampere's law, taken as a separate quotable tool).