The Biot-Savart Law — Stating the Source Equation

We have, up to now, looked at how magnetic fields push on moving charges and current-carrying wires. Now we flip the question: given a current, what magnetic field does it produce? The fundamental answer was supplied in 1820 by two French physicists, Jean-Baptiste Biot and Félix Savart.

Illustration of Biot-Savart geometry

Consider a steady current II flowing through a wire. Take a tiny element dl⃗d\vec l of the wire. Let r⃗\vec r be the position vector drawn from the element to a field point P, of magnitude rr and unit vector r^\hat r. Then the magnetic field contributed by this element at P is:

dB⃗=μ04π I dl⃗×r^r2\boxed{d\vec B = \frac{\mu_0}{4\pi}\,\frac{I\,d\vec l\times\hat r}{r^2}}

Here μ0=4π×10−7 \mu_0 = 4\pi\times10^{-7}\,T·m/A is the permeability of free space.

Analogy with Coulomb's Law

Coulomb's law Biot-Savart law
dE⃗=14πε0dq r^r2d\vec E = \frac{1}{4\pi\varepsilon_0}\frac{dq\,\hat r}{r^2} dB⃗=μ04πI dl⃗×r^r2d\vec B = \frac{\mu_0}{4\pi}\frac{I\,d\vec l\times\hat r}{r^2}
Source: charge element dqdq Source: current element I dl⃗I\,d\vec l
Field along r^\hat r (radial) Field perpendicular to plane of dl⃗d\vec l and r^\hat r
∝1/r2\propto 1/r^2 ∝1/r2\propto 1/r^2

Direction, Magnitude and Units of dB⃗d\vec B

Direction — perpendicular to the dl⃗d\vec l–r^\hat r plane

Because dB⃗d\vec B involves dl⃗×r^d\vec l\times\hat r, it is perpendicular to both dl⃗d\vec l and r^\hat r. Use the right-hand rule:

Point the fingers of your right hand along dl⃗d\vec l; curl them toward r^\hat r. Your thumb gives the direction of dB⃗d\vec B.

Magnitude

Writing ∣dl⃗×r^∣=dlsin⁡θ|d\vec l\times\hat r| = dl\sin\theta where θ\theta is the angle between dl⃗d\vec l and r^\hat r:

dB=μ04πI dl sin⁡θr2dB = \frac{\mu_0}{4\pi}\frac{I\,dl\,\sin\theta}{r^2}

  • θ=0∘\theta = 0^\circ or 180∘180^\circ (on the wire's axis): dB=0dB=0.
  • θ=90∘\theta = 90^\circ (point perpendicular to dl⃗d\vec l): dBdB is maximum.
  • Doubling II doubles dBdB; doubling rr quarters dBdB.

Units

μ0\mu_0 has units T·m/A. Thus μ0I/(4πr2)\mu_0 I/(4\pi r^2) has units T/m, and multiplying by dldl (m) gives dBdB in tesla.

The total field at P is obtained by integration over the entire circuit:

B⃗(P)=μ0I4π∮dl⃗×r^r2 .\vec B(P)=\frac{\mu_0 I}{4\pi}\oint\frac{d\vec l\times\hat r}{r^2}\,.

When all current elements contribute dB⃗d\vec B in the same direction—for example, points on the axis of a circular loop or any external point around a straight wire—the Biot-Savart integral reduces to a simple scalar sum.

Application 1 — Magnetic Field Due to an Infinite Straight Wire

illustration of the magnetic field around a long straight current-carrying wire

Place an infinitely long straight wire along the yy-axis, carrying current II upward. We want B⃗\vec B at a point P at perpendicular distance dd from the wire.

  • Choose a current element at position yy: dl⃗=dy y^d\vec l = dy\,\hat y.
  • The vector from this element to P has length r=d2+y2r=\sqrt{d^2+y^2} and sin⁡θ=d/r\sin\theta = d/r.
  • Every dB⃗d\vec B points into or out of the page at P (same sense everywhere), so we integrate magnitudes:

dB=μ0I4πdy sin⁡θr2=μ0I4πd dy(d2+y2)3/2.dB = \frac{\mu_0 I}{4\pi}\frac{dy\,\sin\theta}{r^2} = \frac{\mu_0 I}{4\pi}\frac{d\,dy}{(d^2+y^2)^{3/2}}.
Integrate yy from −∞-\infty to +∞+\infty:

B=μ0I d4π∫−∞∞dy(d2+y2)3/2=μ0I2πd.B = \frac{\mu_0 I\,d}{4\pi}\int_{-\infty}^{\infty}\frac{dy}{(d^2+y^2)^{3/2}} = \frac{\mu_0 I}{2\pi d}.

Result:

B=μ0I2πd\boxed{B = \frac{\mu_0 I}{2\pi d}}

Geometry of field lines: concentric circles around the wire; sense by right-hand thumb rule.
Note the 1/d1/d fall-off arises from integrating an inverse-square element over an infinite line.

Application 2 — Magnetic Field Due to a Finite Straight Wire

For a straight segment of current II, drop a perpendicular from P to the wire at foot F; let d=PFd=PF. Draw lines from P to the segment ends, making angles α1\alpha_1, α2\alpha_2 from the perpendicular. A direct integration gives:

B=μ0I4πd(sin⁡α1+sin⁡α2)\boxed{B = \frac{\mu_0 I}{4\pi d}\bigl(\sin\alpha_1 + \sin\alpha_2\bigr)}

Direction: perpendicular to the plane of the wire and P, by the right-hand rule.

Special cases:

  • Infinite wire: α1=α2=90∘⇒B=μ0I/(2πd)\alpha_1=\alpha_2=90^\circ\Rightarrow B=\mu_0I/(2\pi d).
  • Semi-infinite wire: one angle 0∘0^\circ, the other 90∘90^\circ, gives B=μ0I/(4πd)B=\mu_0I/(4\pi d).

When solving finite-wire problems, sketch the geometry first. Mark angles from the perpendicular, using negative signs if P lies off one end (so that sin⁡(−α)=−sin⁡α\sin(-\alpha)=-\sin\alpha).

Memory Capsule

Biot-Savart law (memorise): dB⃗=μ04πI dl⃗×r^r2,μ0=4π×10−7 T⋅m/Ad\vec B=\frac{\mu_0}{4\pi}\frac{I\,d\vec l\times\hat r}{r^2},\qquad \mu_0=4\pi\times10^{-7}\,\mathrm{T·m/A}

Property Details
Direction Perpendicular to dl⃗d\vec l and r^\hat r (right-hand rule)
Magnitude dB=(μ0/4π)I dlsin⁡θ/r2dB=(\mu_0/4\pi)I\,dl\sin\theta/r^2
Maximum θ=90∘\theta=90^\circ
Zero θ=0∘,180∘\theta=0^\circ,180^\circ

Key results to remember:

  • Infinite straight wire, perp. dist. dd: B=μ0I/(2πd)B=\mu_0I/(2\pi d).
  • Finite straight wire, perp. dist. dd: B=(μ0I/4πd)(sin⁡α1+sin⁡α2)B=(\mu_0I/4\pi d)(\sin\alpha_1+\sin\alpha_2).

Numerical anchor: μ0/(4π)=10−7 \mu_0/(4\pi)=10^{-7}\,T·m/A.

Solved Examples

Example 1: Field at a point near an infinite wire

A long straight wire carries current 5 5\,A. Find the magnetic field at a point 10 10\,cm from the wire.

Solution. Use B=μ0I/(2πd)B=\mu_0I/(2\pi d) with I=5 I=5\,A, d=0.10 d=0.10\,m.
Since μ0/(2π)=2×10−7 \mu_0/(2\pi)=2\times10^{-7}\,T·m/A,

B=2×10−7×50.10=1×10−5 T=10 μT.B=\frac{2\times10^{-7}\times5}{0.10}=1\times10^{-5}\,\mathrm{T}=10\,\mu\text{T}.

Example 2: Doubling the current, doubling the distance

A long straight wire produces a field of 4×10−5 4\times10^{-5}\,T at 5 5\,cm. What is the field at 10 10\,cm if the current is doubled?

Solution. B∝I/dB\propto I/d.
Bnew=4×10−5×22=4×10−5 T.B_{\text{new}}=4\times10^{-5}\times\frac{2}{2}=4\times10^{-5}\,\mathrm{T}.

Example 3: Field due to a current element

A current element I dl⃗=2×10−4 x^I\,d\vec l=2\times10^{-4}\,\hat x A·m is at the origin. Find dB⃗d\vec B at (0,0.5,0)(0,0.5,0) m.

Solution. r⃗=0.5 y^\vec r=0.5\,\hat y, so r^=y^\hat r=\hat y, r=0.5r=0.5 m.
dB⃗=μ04πI dl⃗×r^r2=(10−7)2×10−4(x^×y^)0.25=8×10−11 z^ T.d\vec B=\frac{\mu_0}{4\pi}\frac{I\,d\vec l\times\hat r}{r^2}=(10^{-7})\frac{2\times10^{-4}(\hat x\times\hat y)}{0.25}=8\times10^{-11}\,\hat z\,\mathrm{T}.

Example 4: Two parallel wires, field at midpoint

Two long parallel wires carry I1=6 I_1=6\,A and I2=4 I_2=4\,A in the same direction, separated by 20 20\,cm. Find BB at the midpoint.

Solution. Each wire at distance 0.10 0.10\,m:
B1=μ0I12π(0.10)=1.2×10−5 T,B2=0.8×10−5 T.B_1=\frac{\mu_0I_1}{2\pi(0.10)}=1.2\times10^{-5}\,\mathrm{T},\quad B_2=0.8\times10^{-5}\,\mathrm{T}.
The fields oppose, so Bnet=∣B1−B2∣=4×10−6 B_{\rm net}=|B_1-B_2|=4\times10^{-6}\,T, toward the weaker wire.

Example 5: Field due to a finite wire — symmetric case

A wire of length 2L=20 2L=20\,cm carries I=10 I=10\,A. Find BB at a point 10 10\,cm from its centre.

Solution. L=d=0.10L=d=0.10 m, so sin⁡α=L/L2+d2=1/2\sin\alpha=L/\sqrt{L^2+d^2}=1/\sqrt2.
B=μ0I4πd(2sin⁡α)=(10−7)100.1022=10−52 T≈14.1 μT.B=\frac{\mu_0I}{4\pi d}(2\sin\alpha)=(10^{-7})\frac{10}{0.10}\frac{2}{\sqrt2}=10^{-5}\sqrt2\,\mathrm{T}\approx14.1\,\mu\mathrm T.

Example 6: Semi-infinite wire

A wire extends from the origin to +∞+\infty on the xx-axis, carrying I=8 I=8\,A. Find BB at (0,5cm,0)(0,5\text{cm},0).

Solution. Semi-infinite: one angle 0∘0^\circ, one 90∘90^\circ.
B=μ0I4πd=(10−7)80.05=1.6×10−5 T.B=\frac{\mu_0I}{4\pi d}=(10^{-7})\frac{8}{0.05}=1.6\times10^{-5}\,\mathrm{T}.

Example 7: Anti-parallel currents

Two long parallel wires, 4 4\,cm apart, carry 5 5\,A in opposite directions. Find BB at a point 1 1\,cm from one wire and 5 5\,cm from the other.

Solution. B1=1.0×10−4 B_1=1.0\times10^{-4}\,T at 0.01 m, B2=2.0×10−5 B_2=2.0\times10^{-5}\,T at 0.05 m.
Outside the gap, they subtract: Bnet=1.0×10−4−2.0×10−5=8.0×10−5 B_{\rm net}=1.0\times10^{-4}-2.0\times10^{-5}=8.0\times10^{-5}\,T.

Example 8: Recovering Coulomb-style 1/r21/r^2

A short element I dl=1 I\,dl=1\,mm at I=1I=1 A produces dBdB at r=0.10 r=0.10\,m. Compare with the electric field of q=1 q=1\,nC at the same rr.

Solution.
Magnetic: dB=10−710−30.01=10−8 T.dB=10^{-7}\frac{10^{-3}}{0.01}=10^{-8}\,\mathrm{T}.
Electric: E=9×10910−90.01=900 V/m.E=9\times10^9\frac{10^{-9}}{0.01}=900\,\mathrm{V/m}.
Both fall as 1/r21/r^2, but magnetic fields from small currents are much weaker.