Academy

Biot-Savart Law

Magnetic field contributed by a current element Idl at position vector r from the element. The cross product encodes both magnitude (sinθ) and direction. Integrate over a complete circuit for total B.
Class 12
Derivation

Statement

The magnetic field dBd\vec{B} at a field point P due to a current element IdlI\,d\vec{l}:

dB=μ04πIdl×r^r2\boxed{d\vec{B} = \frac{\mu_0}{4\pi} \frac{I\,d\vec{l} \times \hat{r}}{r^2}}

r\vec{r} is the position vector from the element to P; r^=r/r\hat{r} = \vec{r}/r.

Magnitude

dB=μ04πIdlsinθr2dB = \frac{\mu_0}{4\pi} \frac{I\,dl\,\sin\theta}{r^2}

θ\theta is the angle between dld\vec{l} and r^\hat{r}.

  • dB=0dB = 0 when θ=0\theta = 0 or π\pi: points on the line of the element feel no field.
  • dBdB is maximum when θ=90°\theta = 90°.

Direction

dBd\vec{B} is perpendicular to the plane containing dld\vec{l} and r^\hat{r}, given by the right-hand rule on dl×r^d\vec{l} \times \hat{r}.

Comparison with Coulomb's Law

Coulomb (electric)Biot-Savart (magnetic)
Sourcepoint charge qqcurrent element IdlI\,dl
FielddE1/r2dE \propto 1/r^2dB1/r2dB \propto 1/r^2
Directionalong r^\hat{r}perpendicular to dld\vec{l} and r^\hat{r}

Both are inverse-square laws. The cross product is the essential structural difference.

Superposition

For a complete circuit:

B=μ0I4πdl×r^r2\vec{B} = \frac{\mu_0 I}{4\pi} \oint \frac{d\vec{l} \times \hat{r}}{r^2}
Note
A current element $I\,d\vec{l}$ cannot exist in isolation — Biot-Savart is always applied by integrating over a complete closed circuit.