---
title: "Ampere's law, field of straight wire"
subject: "Engineering Physics"
chapter: 8
key: "semester_1/engineering_physics:ampere-law-straight-wire-field"
source: https://www.bctengineer.com/q/semester_1/engineering_physics/ampere-law-straight-wire-field
site: BCT Engineering
---

# Ampere's law, field of straight wire

Engineering Physics · Chapter 8: Electromagnetism

- Asked in: 2074 Chaitra, 5 marks; 2073 Shrawan, 5 marks; 2071 Shrawan, 5 marks; 2067 Asar, 5 marks; 2067 Asoj, 5 marks (asked 5 times)
- Video: [PHYS 102 | Ampere's Law 2 - Ampere's Law for a Straight Wire](https://www.youtube.com/watch?v=HVJyIZmxw4k) — Professor Hafner
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/engineering_physics/ampere-law-straight-wire-field
- Whole chapter: https://www.bctengineer.com/md/semester_1/engineering_physics/chapter-8

## Question

State Ampere's law. Using it, calculate the magnetic field inside, outside (and on the surface) of a long straight current carrying conductor of circular cross-section, and plot the magnetic field versus distance from the center of the conductor.

## Short answer

Ampere's law: ∮B·dl = μ₀ I_enc. For a long straight wire of radius R, B = μ₀ I r/(2πR²) inside (r<R) and B = μ₀ I/(2πr) outside (r>R), peaking at μ₀ I/(2πR) on the surface.

## Answer

**Ampere's law.** The line integral of the magnetic field B around any closed loop equals μ0 times the net current enclosed by the loop:

∮ B·dl = μ0 I_enc

#### Long straight conductor of radius R carrying current I (uniform current density)
Current density J = I/(πR²). By symmetry the field lines are circles about the axis, with B tangential and constant on a circular loop of radius r. Use circular Amperian loops.

```text
      Amperian loop (radius r)
          .--------.
        /   .----.   \
       |   | wire |   |  -> current I out of page
        \   '----'   /
          '--------'
```

#### 1. Outside (r > R)
Enclosed current = I.

B (2πr) = μ0 I

**B = μ0 I / (2πr)**

#### 2. On the surface (r = R)

**B = μ0 I / (2πR)**  (maximum value)

#### 3. Inside (r < R)
Enclosed current = J (πr²) = I r²/R².

B (2πr) = μ0 I r²/R²

**B = μ0 I r / (2πR²)**

The field inside grows linearly with r; outside it falls as 1/r.

#### Graph of B versus r
```text
   B
   |        . B max = μ0I/(2πR)
   |      /  `.
   |    /      ` .   B ~ 1/r
   |  /             ` . .
   | /  B ~ r
   |/_____________________ r
   0            R
```

B is zero at the axis, rises linearly to the maximum at r = R, then decays as 1/r.

## Check yourself

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

1. Ampere's law states ∮B·dl equals
   A. ε0 times enclosed charge
   B. μ0 times enclosed current
   C. μ0 times total current in the circuit
   D. B times area
   Answer: B. The line integral of B around a closed loop is μ0 I_enc. (Key point: Ampere's law)

2. Outside the wire (r > R), B is
   A. μ0 I r/(2πR^2)
   B. μ0 I/(2πR)
   C. μ0 I/(2πr)
   D. Zero
   Answer: C. The loop encloses all I, so B(2πr) = μ0 I. (Key point: Outside field)

3. Inside the wire (r < R), the enclosed current is
   A. I
   B. I r^2/R^2
   C. I r/R
   D. Zero
   Answer: B. Uniform J gives I_enc = J πr^2 = I r^2/R^2, so B = μ0 I r/(2πR^2). (Key point: Inside field)

4. The graph of B vs r
   A. Rises linearly to a maximum at r = R, then falls as 1/r
   B. Falls from the axis
   C. Is constant inside, 1/r outside
   D. Rises as r^2 everywhere
   Answer: A. B is zero at the axis, peaks at the surface μ0 I/(2πR), then decays as 1/r. (Key point: Graph shape)

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