UY1: Applications Of Ampere's Law

Use Ampere's law to obtain practical field models for solenoids, toroids, and current sheets with clear validity limits.

University Physics Year 1

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Learning objectives

Use vector and differential or integral calculus to express physical change and accumulation.
Use complex numbers, linear algebra, and differential equations to solve coupled physical models.
Choose an efficient mathematical method and validate the result physically.

This page gives the UY1 working model/result for Applications Of Ampere’s Law. You reuse it when you build fields/potentials by symmetry or superposition, and when you connect fields to forces, energy, and circuits.

Module path: Electromagnetism (UY1)
Practice: UY1 Electromagnetism Quiz
Full routing: UY1 Assessment Map
Math toolkit: Mathematics for Undergraduate Physics

1) At a glance

These are magnetostatic field models (steady currents, no time-varying electric-flux term). For time-varying situations, see Displacement Current and Faraday’s Law Of Induction & Lenz’s Law.
Long solenoid (long, tightly wound; away from the ends):

B_inside ≈ μ₀ nI, n = N/ℓ

Toroid (closely wound turns; field mostly confined to the core region):

B(r) = μ₀NI/(2π r) (inside the core region)

Infinite sheet current K (surface current density, A m⁻¹):

B = μ₀K/2 on each side (opposite directions)

Quick checks: solenoid field is approximately uniform inside; toroid field falls like 1/r; sheet-current field is approximately constant with distance.

Prerequisites: Ampere’s Law, Magnetic Field Of A Straight Current Carrying Conductor, Magnetic Field Of A Circular Current Loop
Next uses: Electromagnetic Induction Experiments, Self-Inductance & Inductors

2) Setup

Apply

∮ vector B · d vector ℓ = μ₀I_enc

with loops aligned to each geometry.

Assumptions used:

Long solenoid: length much larger than radius, dense winding.
Toroid: closely wound turns around ring.
Current sheet: effectively infinite extent.
Steady currents (magnetostatics) and negligible fringing/edge fields in the regions where the formulas are applied.

Mixing N (total turns) with n (turns per unit length): n = N/ℓ.
Using the solenoid formula right at the ends: the outside field is not exactly zero and the inside field is not perfectly uniform near edges.
For toroids: using one radius when the field point is at another radius. The model is explicitly B(r) ∝ 1/r.
For a sheet current: K is A m⁻¹ (current per unit width), not A m⁻².

3) Core derivation/explanation

Long solenoid

Take a rectangular Amperian loop partly inside, partly outside.

Outside field is approximately zero.
Inside field is approximately uniform and parallel to loop segment.
Enclosed current for length ℓ is (nℓ)I.

So:

Bℓ ≈ μ₀(nℓ)I ⇒ B ≈ μ₀nI.

Toroid

Use circular loop of radius r concentric with toroid.

vector B tangential and approximately constant on loop.
Enclosed current is NI if loop is in core region.

Thus:

B(2π r) = μ₀NI ⇒ B(r) = μ₀NI/(2π r).

Infinite current sheet

Using a rectangular loop straddling sheet with surface current density K:

2Bℓ = μ₀Kℓ ⇒ B = μ₀K/2

with opposite directions on opposite sides.

Checks (sanity)

Units: μ₀ n I and μ₀ N I/(2π r) both give tesla.
Solenoid: if n → 0 (no turns per length), B → 0.
Toroid: larger r gives smaller field; outside the windings the ideal model predicts very small field (stronger confinement than a straight wire).
Sheet: the magnitude does not depend on distance from the sheet (ideal infinite-extent model).

4) Worked example(s)

Example 1: solenoid

A long solenoid has turn density n = 1500 m⁻¹ and current I = 0.30 A.

B = μ₀nI = (4π × 10⁻⁷)(1500)(0.30) = 5.65 × 10⁻⁴ T.

Example 2: toroid

A toroid has N = 400, I = 0.50 A, and point of interest at r = 8.0 cm.

B = μ₀NI/(2π r) = ((4π × 10⁻⁷)(400)(0.50))/(2π(8.0 × 10⁻²)) = 5.0 × 10⁻⁴ T.

5) Practice set (with hints + answers)

A long solenoid has n = 2000 m⁻¹, I = 0.25 A. Find inside field.
For a toroid, if r doubles (same N,I), what happens to B?
A current sheet has K = 30 A m⁻¹. Find field magnitude on one side.

Hints

Use B ≈ μ₀nI for solenoid interior.
Toroid field scales as 1/r.
Use B = μ₀K/2 for sheet.

Answers

B = (4π × 10⁻⁷)(2000)(0.25) = 6.28 × 10⁻⁴ T.
It halves.
B = (4π × 10⁻⁷)(30)/2 = 1.88 × 10⁻⁵ T.

6) Summary + next steps

Ampere’s law gives fast, clean results when symmetry is strong.
Solenoids produce nearly uniform interior fields; toroids confine field around the core.
Always state the approximation limits (long, dense, or infinite geometry).

Next: Faraday’s Law Of Induction & Lenz’s Law Previous: Ampere’s Law Back To Electromagnetism