---
title: "State Biot-Savart's law"
subject: "Electromagnetism"
chapter: 3
key: "semester_3/electromagnetism:biot-savart-law-state-explain"
source: https://www.bctengineer.com/q/semester_3/electromagnetism/biot-savart-law-state-explain
site: BCT Engineering
---

# State Biot-Savart's law

Electromagnetism · Chapter 3: Magnetic field

- Asked in: 2067 Mangsir, 4 marks; 2067 Shrawan, 2 marks (asked 2 times)
- Video: [Biot Savart Law for Magnetic Field Intensity | Magnetism | Electromagnetics Theory](https://www.youtube.com/watch?v=CiGmxrzDAgA) — Engineering Funda
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/electromagnetism/biot-savart-law-state-explain
- Whole chapter: https://www.bctengineer.com/md/semester_3/electromagnetism/chapter-3

## Question

State and explain Biot-Savart's law.

## Short answer

Biot–Savart law: dH = I dL × a_R / (4πR²). It gives the magnetic field intensity produced by a differential current element I dL at a distance R. The direction follows the right-hand rule (dL × a_R). For a closed loop, integrate around the circuit.

## Answer

**Statement.** The magnetic field intensity dH produced at a point P by a differential current element I dL is directly proportional to the product I dL and the sine of the angle between the element and the line joining it to P, and inversely proportional to the square of the distance R between them.

```text
dH = I dL × a_R / (4πR²)       (A/m)
dH = I dL sinθ / (4πR²)        (magnitude)
```
Here R is the distance from the element to P, a_R is the unit vector from the element to P, and θ is the angle between dL and a_R.

```text
        I dL
   →→→→→●→→→→→
         \  θ
          \ R
           \
            ● P     dH is perpendicular to the plane of dL and R
                    (into the page in the figure; right-hand rule)
```

#### Explanation
- The direction of dH is that of dL × a_R, given by the right-hand rule: point the right thumb along the current, and the fingers curl in the direction of H.
- For a complete circuit, H = ∮ I dL × a_R/(4πR²).
- For surface or volume currents: H = ∫ K × a_R dS/(4πR²) or ∫ J × a_R dv/(4πR²).
- The law is the magnetic counterpart of Coulomb's law, since the field falls as 1/R². An isolated current element cannot exist, but the law is valid for closed circuits.

**Example.** For an infinite straight wire, integrating the law gives H = I/(2πρ) a_φ.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. dH from a current element varies as:
   A. I dL sinθ / R
   B. I dL sinθ / (4πR²)
   C. I dL cosθ / R²
   D. I dL / R³
   Answer: B. dH = I dL sinθ/(4πR²), proportional to I dL and sinθ, inversely to R². (Key point: Statement)

2. The direction of dH is:
   A. Along dL
   B. Along a_R
   C. Perpendicular to the plane of dL and R (dL × a_R)
   D. Along the wire
   Answer: C. dH ∝ dL × a_R, found by the right-hand rule. (Key point: Direction)

3. Integrating over an infinite straight wire gives:
   A. I/(4πρ)
   B. I/(2πρ) a_φ
   C. I/(2πρ²)
   D. I/ρ
   Answer: B. The Biot-Savart law reproduces H = I/(2πρ) a_φ. (Key point: Application)

## Related in Electromagnetism

- [H of infinite filament by Biot-Savart](https://www.bctengineer.com/question.md?key=semester_3%2Felectromagnetism%3Abiot-savart-infinite-filament-derive) — Electromagnetism, chapter 3

## Related in other subjects

- [Biot-Savart vs Ampere's law, coil axis](https://www.bctengineer.com/question.md?key=semester_1%2Fengineering_physics%3Abiot-ampere-compare-coil-axis) — Engineering Physics, chapter 8