Mini Physics
Part of Mini EducationSwitch subject
Search

Ampère’s Law

Key idea: Use Ampère’s law in integral form to compute magnetic fields for symmetric current distributions, with worked examples and exam tips.

  • GCE A-Level H3 Physics 2027
On this page

Learning objectives

  • show an understanding that ideal conductors form an equipotential volume, and that the electric field within an ideal conductor is zero
  • show an understanding that electric charge accumulates on the surfaces of a conductor, and that the electric field at the surface of a conductor is normal to the surface
  • recall and apply Gauss’s law 6 for electric and magnetic fields (knowledge of the differential form of Gauss’s law is not required), and
  • recall and apply Ampère’s law 7 relating the line integral of the magnetic field (in a vacuum) around a closed loop with the electric current enclosed by the loop to solve problems involving symmetric field configurations (knowledge of the differential form of Ampère’s law is not required) [Note further that candidates are not required to know Maxwell’s generalisation of Ampère’s law including the term related to the rate of change of electric flux, nor the Biot-Savart law.]
  • define the magnitude of the electric dipole moment as the product of the charge and the separation
  • show an understanding of and use the torque on an electric dipole and the potential energy of an electric dipole to solve related problems
  • define the magnitude of the magnetic dipole moment for a current loop as the product of the current and the area of the loop
  • show an understanding of and use the torque on a magnetic dipole and the potential energy of a magnetic dipole to solve related problems
  • solve problems involving symmetric charge distributions by relating the electric flux (in a vacuum) through a closed surface with the charge enclosed by that surface (ii) show an understanding that the magnetic flux through a closed surface is always zero, suggesting the non-existence of magnetic monopoles

Ampère’s law links the circulation of the magnetic field around a closed loop to the steady current enclosed. In H3, you use its magnetostatic integral form to calculate vector B for symmetric current distributions in vacuum.

1. Definitions (Must Know)

  • Line integral / circulation of vector B: ∮ vector B · d vector l where d vector l is a tangential vector element along the chosen loop.
  • Ampère’s law (integral form, in vacuum): ∮ vector B · d vector l = μ₀ I_encl
  • Enclosed current, I_encl: net current passing through any surface bounded by the loop (sign from right-hand rule orientation).
  • Vacuum permeability: μ₀ = 4π × 10⁻⁷ N A⁻².
  • Symbols used in this lesson: vector B (T), d vector l (m), I current (A), I_encl enclosed current (A), n turns per unit length (m⁻¹), μ₀ (N A⁻²), r (m).

2. Key Ideas (What Earns Marks)

  • In the H3 magnetostatic model, Ampère’s law applies to steady currents. It is computationally powerful when symmetry makes vector B · d vector l easy.
  • Choose an Amperian loop that matches the symmetry so that:
    • B is constant on the loop, and
    • vector B is parallel to d vector l where it contributes.
  • Sign matters: the direction you integrate is linked to the current direction by the right-hand rule.
  • H3 note: you are not required to include Maxwell’s correction (displacement current) or use Biot–Savart.

Common symmetry setups (quick table):

Current distributionBest Amperian loopWhat simplifies
Long straight wireCircle radius rB constant and tangent
Long solenoid (inside field)Rectangle loopOutside field approximated ~0, inside ~uniform
Coaxial cable (equal/opposite currents)Circle radius rI_encl depends on region
Circular Amperian loop around a long straight wire, with magnetic field tangent to the loop and current directed out of the page
For a long straight wire, cylindrical symmetry makes the field tangent to a centred circle and constant in magnitude along it.

3. Detailed Explanations

A. What Ampère’s law is telling you

The integral ∮ vector B · d vector l measures the “total tendency” of vector B to circulate around the loop. Ampère’s law says this circulation is set by the net current threading the loop.

B. The classic symmetric case: long straight wire

For a long straight current-carrying wire, symmetry tells you:

  • vector B is tangent to circles centred on the wire,
  • B depends only on the radius r.

Choose a circular Amperian loop of radius r: ∮ vector B · d vector l = B∮ dl = B(2π r) So: B(2π r) = μ₀ I ⇒ B = (μ₀ I)/(2π r)

C. When the loop encloses no net current

If I_encl = 0, Ampère’s law gives: ∮ vector B · d vector l = 0 This does not mean vector B = vector 0 everywhere on the loop; it means the signed circulation cancels.

4. Common Mistakes

  • Confusing Ampère’s law with Gauss’s law for magnetism (∮ vector B · d vector A = 0).
  • Choosing a loop where B is not constant and then incorrectly pulling B out of the integral.
  • Forgetting that d vector l is tangential: only the component of vector B along the loop contributes.
  • Treating ∮ vector B · d vector l = μ₀I as true for any time-varying situation without guidance (H3 scope keeps it simple and symmetric).

5. Exam Tips

  • Write a short “symmetry justification”: “By symmetry, vector B is tangent to … and has constant magnitude on …”
  • For straight-wire problems, the circular loop is almost always the intended choice.
  • Quote the result cleanly with units: B = (μ₀ I)/(2π r) (T)

6. Worked Examples

Modelled example 1

Field around a long straight wire

Core

Problem

A long straight wire carries current I = 5.0 A. Find the magnetic-field magnitude at r = 0.020 m from the wire.
Study the worked solution

  1. Choose the loop

    Method

    Use a circle of radius r centred on the wire.

    Reason

    Cylindrical symmetry makes B constant and tangent around this loop.

    Working

    ∮ vector B · d vector l = B(2π r)

  2. Apply Ampère's law

    Method

    B = μ₀I/(2π r).

    Reason

    The loop encloses the full current I.

    Working

    B(2π r) = μ₀I

  3. Substitute

    Method

    B = 5.0 × 10⁻⁵ T.

    Reason

    Use SI units before evaluating.

    Working

    B = ((4π × 10⁻⁷)(5.0))/2π(0.020) = 5.0 × 10⁻⁵ T

Guided practice 2

Loop encloses zero net current

About 4 min

Problem

An oriented Amperian loop encloses currents + 3 A and -3 A in opposite directions. Find ∮ vector B · d vector l.

Try this before viewing the solution

Hints

Hint 1: find net current first
Add the currents algebraically using the signs fixed by the loop orientation.
Hint 2: apply the law
Substitute I_encl into ∮ vector B · d vector l = μ₀I_encl.
View solution step by step

  1. Sum enclosed current

    Method

    I_encl = 0.

    Reason

    The equal currents pass through the bounded surface in opposite signed directions.

    Working

    I_encl = 3 + (-3) = 0 A

  2. Find circulation

    Method

    ∮ vector B · d vector l = 0.

    Reason

    Ampère’s law depends on net enclosed current.

    Working

    ∮ vector B · d vector l = μ₀(0) = 0

Common misconception 3

Concept check: what does “zero circulation” mean?

Find and correct the mistake

Learner claim

A learner says that ∮ vector B · d vector l = 0 proves B = 0 everywhere on the loop. Explain what is wrong with the claim.

Try this before viewing the solution

What zero circulation establishes

View solution step by step

  1. Identify the measured quantity

    Method

    The integral sums the tangential component of vector B around the loop.

    Reason

    Each contribution is the signed dot product vector B · d vector l.

    Working

    ∮ B_∥ dl

  2. Interpret a zero sum

    Method

    Positive and negative contributions may cancel even if B is locally non-zero.

    Reason

    A zero integral constrains the total, not every integrand value.

    Working

    non-zero terms can have signed sum 0

  3. Correct the claim

    Method

    Only additional symmetry could turn zero circulation into B = 0 for a particular loop.

    Reason

    For example, constant tangential B would give B∮ dl = 0.

    Working

    ∮ vector B · d vector l = 0not ⇒ vector B = vector 0

Examiner practice 4

Field inside a long solenoid (standard result)

3 marks

Examination question

A long solenoid has n = 800 turns per metre and carries current I = 0.50 A. Assuming its external field is negligible, find the magnetic-field magnitude inside. [3 marks]

Try this before viewing the solution

View solution step by step

  1. State the long-solenoid result

    1 mark

    Method

    B = μ₀nI.

    Reason

    A suitable rectangular Amperian loop has one contributing length inside; the enclosed current is the number of turns crossed times I.

    Working

    Bl = μ₀(nl)I ⇒ B = μ₀nI

  2. Substitute

    1 mark

    Method

    Insert the turn density and current in SI units.

    Reason

    n is already in m⁻¹.

    Working

    B = (4π × 10⁻⁷)(800)(0.50)

  3. Report the field

    1 mark

    Method

    B = 5.0 × 10⁻⁴ T.

    Reason

    The value is given to two significant figures with the magnetic-field unit.

    Working

    B = 5.0 × 10⁻⁴ T

Challenge 5

Coaxial cable: why the external field can be zero

Minimal support

Independent transfer

An ideal coaxial cable has an inner conductor carrying + I and an outer conductor carrying -I. For a circular Amperian loop of radius r outside the outer conductor, determine B and justify why zero circulation is sufficient here.

Try this before viewing the solution

Hints

Hint 1: combine the currents
Find the signed net current enclosed by a loop outside both conductors.
View solution step by step

  1. Find enclosed current

    Method

    I_encl = 0.

    Reason

    The equal inner and outer currents have opposite directions.

    Working

    I_encl = +I + (-I) = 0

  2. Apply Ampère's law

    Method

    The circulation is zero.

    Reason

    It equals μ₀I_encl.

    Working

    ∮ vector B · d vector l = 0

  3. Use cylindrical symmetry

    Method

    B is tangent to the circle and has constant magnitude.

    Reason

    The ideal concentric current distribution is cylindrically symmetric.

    Working

    ∮ vector B · d vector l = B(2π r)

  4. Conclude

    Method

    B = 0 outside the ideal cable.

    Reason

    Here symmetry prevents non-zero tangential contributions from cancelling around the circle.

    Working

    B(2π r) = 0 ⇒ B = 0

7. Mind Stretchers

Mind stretcher 1: Choosing a better loopExtension

You choose a rectangular loop around a long straight wire and get stuck because B is not constant along the loop.

What loop choice makes the integral easy, and why?

Answer

A circular loop centred on the wire. Symmetry makes B constant along the circle and tangent to the path, so ∮ vector B · d vector l = B(2π r).

Mind stretcher 2: “I_encl = 0” doesn’t always mean “B = 0”Extension

In some geometries, you can have ∮ vector B · d vector l = 0 but still have a non-zero B at points on the loop.

What does this tell you about interpreting the integral equation?

Answer

It tells you Ampère’s law constrains the net circulation around the loop. Zero circulation means the signed contributions cancel; it does not guarantee the field is zero at every point unless symmetry lets you argue B is constant and tangent along the path.

8. Optional/Enrichment

A. Syllabus boundary

The H3 syllabus uses Ampère’s law for symmetric steady-current configurations. Maxwell’s displacement-current term and the Biot–Savart law are explicitly outside the required scope.

For a university-level walkthrough:

Next step

Continue to Electric Dipoles to move from field symmetry to the torque and energy of an oriented source.

Continue with the next resource in this course.

Course and syllabus information
Course
GCE A-Level H3 Physics
Edition
GCE A-Level H3 Physics 2027