Biot-Savart law

Bring a compass near a current-carrying wire and the needle swings — Oersted's accidental discovery in 1820 that electricity and magnetism are linked. The Biot-Savart law turns that qualitative fact into a precise formula: it tells you exactly how much magnetic field a tiny piece of current-carrying wire produces at any point in space. It is the magnetic counterpart of Coulomb's law — instead of a point charge sourcing an electric field, a current element I dl⃗I\,d\vec l sources a magnetic field dB⃗d\vec B that falls off as 1/r21/r^2. Every magnetic field you will calculate in this unit — at the centre of a loop, on the axis of a loop, around a long straight wire — is built by summing (integrating) contributions of this basic law over the shape of the conductor. This article covers the vector law itself and the two loop results your ISC syllabus asks you to apply directly.

The core insight

A current element I dl⃗I\,d\vec l produces a magnetic field dB⃗=μ04πI dl⃗×r^r2d\vec B = \dfrac{\mu_0}{4\pi}\dfrac{I\,d\vec l\times\hat r}{r^2} at a point, and integrating this over a circular loop gives B=μ0I2RB=\dfrac{\mu_0 I}{2R} at its centre and B=μ0IR22(R2+x2)3/2B=\dfrac{\mu_0 I R^2}{2(R^2+x^2)^{3/2}} on its axis.

The core idea

Picture a wire carrying current II. Chop it into infinitesimal segments dl⃗d\vec l (a vector pointing along the current direction). Each segment behaves as an independent source of magnetic field, and the field it produces at a point P — located at position vector r⃗\vec r with unit vector r^\hat r from the segment — is:

dB⃗=μ04π I dl⃗×r^r2d\vec B = \frac{\mu_0}{4\pi}\,\frac{I\, d\vec l \times \hat r}{r^2}

where: dB⃗d\vec B is the field contribution (tesla, T), μ0=4π×10−7 T ⋅ m/A\mu_0 = 4\pi\times10^{-7}\ \text{T}\,\cdot\,\text{m/A} is the permeability of free space, II is the current (A), dl⃗d\vec l is the current-element length vector (m), r^\hat r is the unit vector from the element to P, and rr is the distance (m) between them. Four features are worth locking in before you use the law:

  • Because of the cross product, dB⃗d\vec B is perpendicular to the plane containing dl⃗d\vec l and r^\hat r; its direction follows the right-hand rule — curl your fingers from dl⃗d\vec l towards r^\hat r, and your thumb gives dB⃗d\vec B.
  • The magnitude form is dB=μ04πI dlsin⁡θr2dB = \dfrac{\mu_0}{4\pi}\dfrac{I\,dl\sin\theta}{r^2}, where θ\theta is the angle between dl⃗d\vec l and r^\hat r. This is the version you actually plug numbers into.
  • The field vanishes along the axis of the wire itself, since sin⁡θ=0\sin\theta = 0 when r^\hat r is parallel (or antiparallel) to dl⃗d\vec l — directly ahead of or behind a straight current element, there is no field.
  • Like Coulomb's law, the field obeys an inverse-square fall-off with distance, but the source is a moving-charge (current) element rather than a static charge, and the field direction depends on a cross product rather than pointing straight along r^\hat r.

Your syllabus does not require deriving the loop formulas from the integral — it asks you to know and apply two standard results, obtained by integrating dB⃗d\vec B around a full circular loop of radius RR carrying current II:

Bcentre=μ0I2RB_{\text{centre}} = \frac{\mu_0 I}{2R}
Baxis=μ0IR22(R2+x2)3/2B_{\text{axis}} = \frac{\mu_0 I R^2}{2(R^2+x^2)^{3/2}}

where: BcentreB_{\text{centre}} is the field at the loop's centre, BaxisB_{\text{axis}} is the field at distance xx from the centre along the axis perpendicular to the loop's plane, and RR, II have their usual meanings. Setting x=0x=0 in the axial formula collapses it to the centre formula — a quick self-check that you have remembered both correctly. The field direction along the axis, in either case, comes from the right-hand rule: curl the fingers along the current's circulation, and the thumb points along B⃗\vec B.

Figure
A circular current loop with a current element shown producing a field vector perpendicular to the plane containing the element and the radius vector to a point, and the axial field arrow along the central axis
Biot-Savart law — a current element sources the field, the loop sums it dl r, r-hat dB (out of loop plane) centre O P, distance x on axis B(centre) = mu0 I / 2R B(axis) = mu0 I R squared / 2(R squared + x squared) to the three-halves setting x = 0 in B(axis) reproduces B(centre)
Derivation AField at the centre of a circular current loop

The goal is to build the centre-of-loop formula B=μ0I/(2R)B = \mu_0 I/(2R) from the Biot-Savart law by exploiting the loop's symmetry, showing where the 2R2R and the absence of any trigonometric factor come from, rather than treating the result as handed down from nowhere — this is the reasoning your syllabus expects you to recognise even though you will not be asked to reproduce the full integral in an exam.

Step 1. Set up the geometry. Take a current element dl⃗d\vec l anywhere on a circular loop of radius RR carrying current II, and look at the field it produces at the centre O. The position vector from the element to O has magnitude r=Rr = R (every point on the loop is a distance RR from the centre), and it points radially inward, exactly perpendicular to dl⃗d\vec l, which lies tangent to the circle. So the angle θ\theta between dl⃗d\vec l and r^\hat r is 90°90° for every single element on the loop — this is the special feature that makes the centre calculation clean.

Step 2. Write the magnitude of one element's contribution. Using dB=μ04πI dlsin⁡θr2dB = \dfrac{\mu_0}{4\pi}\dfrac{I\,dl\sin\theta}{r^2} with θ=90°\theta = 90° (so sin⁡θ=1\sin\theta=1) and r=Rr=R: dB=μ04πI dlR2dB = \dfrac{\mu_0}{4\pi}\dfrac{I\,dl}{R^2}.

Step 3. Check the direction is common to all elements. By the right-hand rule, every element's dB⃗d\vec B at the centre points the same way — straight along the loop's axis, perpendicular to the loop's plane. Because all contributions point in the same direction, they add as scalars rather than needing vector resolution; this is why the centre calculation avoids any messy component-wise integration.

Step 4. Sum over the whole loop. Since μ0\mu_0, II and RR are the same for every element, they come out of the sum, leaving B=μ0I4πR2∑dlB = \dfrac{\mu_0 I}{4\pi R^2}\sum dl. The sum of all the dldl around the loop is simply its circumference, 2πR2\pi R. Substituting: B=μ0I4πR2(2πR)=μ0I2RB = \dfrac{\mu_0 I}{4\pi R^2}(2\pi R) = \dfrac{\mu_0 I}{2R}, which is exactly the prescribed centre-of-loop formula.

Worked example

A circular coil of radius 0.05 m0.05\ \text{m} carries a current of 4 A4\ \text{A}. Find the magnetic field (a) at the centre of the coil, and (b) at a point on its axis, 0.04 m0.04\ \text{m} from the centre, for a second coil of radius 0.03 m0.03\ \text{m} carrying 2 A2\ \text{A}.

Step 1. Centre field of the first coil. Use B=μ0I2RB = \dfrac{\mu_0 I}{2R} with I=4 AI=4\ \text{A}, R=0.05 mR=0.05\ \text{m}: B=(4π×10−7)(4)2(0.05)B = \dfrac{(4\pi\times10^{-7})(4)}{2(0.05)}. The numerator is 16π×10−7=5.03×10−6 T ⋅ m16\pi\times10^{-7} = 5.03\times10^{-6}\ \text{T}\,\cdot\,\text{m}, and dividing by 2R=0.1 m2R=0.1\ \text{m} gives B=5.03×10−60.1=5.03×10−5 TB = \dfrac{5.03\times10^{-6}}{0.1} = 5.03\times10^{-5}\ \text{T}, directed along the axis (right-hand rule).

Step 2. Axial field of the second coil — set up the numbers. Use B=μ0IR22(R2+x2)3/2B = \dfrac{\mu_0 I R^2}{2(R^2+x^2)^{3/2}} with I=2 AI=2\ \text{A}, R=0.03 mR=0.03\ \text{m}, x=0.04 mx=0.04\ \text{m}. First, R2+x2=0.0009+0.0016=0.0025 m2R^2+x^2 = 0.0009+0.0016 = 0.0025\ \text{m}^2, and since 0.0025=0.05\sqrt{0.0025}=0.05, (R2+x2)3/2=(0.05)3=1.25×10−4 m3(R^2+x^2)^{3/2} = (0.05)^3 = 1.25\times10^{-4}\ \text{m}^3.

Step 3. Compute the numerator and divide. μ0IR2=(4π×10−7)(2)(0.0009)=2.26×10−9 T ⋅ m3\mu_0 I R^2 = (4\pi\times10^{-7})(2)(0.0009) = 2.26\times10^{-9}\ \text{T}\,\cdot\,\text{m}^3. Then B=2.26×10−92×1.25×10−4=2.26×10−92.5×10−4=9.05×10−6 TB = \dfrac{2.26\times10^{-9}}{2\times1.25\times10^{-4}} = \dfrac{2.26\times10^{-9}}{2.5\times10^{-4}} = 9.05\times10^{-6}\ \text{T}.

Final answer: the centre field of the first coil is 5.03×10−5 T\mathbf{5.03\times10^{-5}\ T}, and the axial field of the second coil at x=0.04 mx=0.04\ \text{m} is 9.05×10−6 T\mathbf{9.05\times10^{-6}\ T} — noticeably smaller because that point lies off the loop's plane, illustrating how the axial formula reduces towards zero as xx grows for fixed RR.

Competitive stretch

Same second coil (R=0.03 mR=0.03\ \text{m}, I=2 AI=2\ \text{A}): what current would the first coil (R=0.05 mR=0.05\ \text{m}) need to match that axial field of 9.05×10−6 T9.05\times10^{-6}\ \text{T} at its own centre? Set μ0I′2(0.05)=9.05×10−6\dfrac{\mu_0 I'}{2(0.05)} = 9.05\times10^{-6}, so I′=9.05×10−6×0.14π×10−7=9.05×10−71.257×10−6≈0.72 AI' = \dfrac{9.05\times10^{-6}\times0.1}{4\pi\times10^{-7}} = \dfrac{9.05\times10^{-7}}{1.257\times10^{-6}} \approx 0.72\ \text{A} — a much smaller current suffices at the centre, since axial points are always weaker than the centre for a given loop.

Common traps

  • Dropping the sin⁡θ\sin\theta factor and writing dB=μ0I dl/(4πr2)dB = \mu_0 I\,dl/(4\pi r^2) — this is only correct when θ=90°\theta = 90°, as it happens to be for the centre-of-loop case, but not in general.
  • Confusing the Biot-Savart law (a current-element law that must be summed/integrated for any finite wire or loop) with Ampere's circuital law (a shortcut that works directly, without summing, only in situations with high symmetry).
  • Mixing up the centre-of-loop and axial formulas, or forgetting that the axial formula must reduce to the centre formula when x=0x=0 — a fast way to catch a memory slip.
  • Reporting a magnitude without direction. The right-hand rule fixes the direction of B⃗\vec B, and reversing the current direction reverses the field — an answer with no stated direction is incomplete.

Ask the tutor: Want to see the full integration that builds the axial-field formula step by step, or work through why Ampere's law gives the same field faster for a straight wire — ask the AI tutor to walk through it.

Why does the Biot-Savart law use a cross product instead of just multiplying like Coulomb's law?

Because the source is a moving charge (a current element), not a stationary one, and a moving charge's magnetic field depends on the direction of motion relative to the point where you measure it, not just the distance. The cross product dl⃗×r^d\vec l \times \hat r automatically makes dB⃗d\vec B perpendicular to both the current direction and the line to the point, and makes it vanish when the point lies directly ahead of or behind the element (θ=0°\theta=0° or 180°180°). Coulomb's law has no such directional dependence because a static point charge's field only cares about distance, not orientation.

Do I need to memorise the integration that gives the loop formulas?

No — for ISC Class 12, you are expected to state the vector Biot-Savart law correctly, apply the right-hand rule for direction, and use the two loop results (B=μ0I/2RB=\mu_0 I/2R at the centre and B=μ0IR2/2(R2+x2)3/2B=\mu_0 I R^2/2(R^2+x^2)^{3/2} on the axis) directly in numerical problems. The reasoning behind the centre-field result — why every element contributes at θ=90°\theta=90° and how the sum of dldl becomes the circumference 2πR2\pi R — is worth understanding conceptually, since it explains why the formula looks the way it does, but reproducing the full axial integral is not required.

How is the Biot-Savart law connected to Ampere's circuital law, which comes next in this unit?

Both describe the same underlying magnetic field produced by currents, but they are used differently. The Biot-Savart law works element by element and needs you to sum (integrate) contributions from every part of the current distribution — it works for any shape, symmetric or not. Ampere's circuital law instead relates the field circulating around a closed loop to the current enclosed, and it gives you the answer in one line, but only in situations with enough symmetry (like an infinite straight wire or a long solenoid) to pull BB out of the integral. Think of Ampere's law as the fast shortcut that Biot-Savart's brute-force summation would eventually also produce.

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