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Electromagnetic Waves: Ampere's Law and Symmetry (Topic)

Electromagnetic Waves, Antennas, and RF: Ampère’s Law and Symmetry

EM11 used the Biot–Savart law to add magnetic-field contributions from many current elements. That source-integral method is systematic, but it can become mathematically expensive even when the current distribution has a simple symmetry.

Ampère’s law provides a second magnetostatic tool. Instead of summing contributions from every source element, it relates the circulation of the magnetic field around a closed path directly to the current enclosed by that path:

|∮-----------------|
|   B ⋅ dℓ = μ0Ienc.
--C-----------------
(1)

The central lesson of this article parallels EM07 on Gauss’s Law:

|--------------------------------------------------------------------------------|
|Amp  `ere’s law is simple to use for B only when the source has enough  symmetry.  |
---------------------------------------------------------------------------------
(2)

For steady currents, Ampère’s law is fully equivalent to the magnetostatic curl equation

|--------------|
|∇ × B  = μ0J. |
----------------
(3)

The time-varying case will later require Maxwell’s additional displacement-current term [1235].

1 Magnetic circulation

The integral

∮
   B  ⋅ dℓ
  C
(4)

is a line integral around a closed curve C.

For a small path element,

B  ⋅ dℓ = B dℓ cos𝜃,
(5)

where 𝜃 is the angle between the local magnetic field and the local direction of the path.

Three cases are immediately useful:

  • if B is tangent to the path in the same direction, then B d = B dℓ;
  • if B is perpendicular to the path, then the contribution is zero;
  • if B is constant along a path segment, it may be taken outside the integral over that segment.

These geometric simplifications are what make a carefully chosen Amperian path powerful.

2 Orientation and enclosed current

The direction in which the closed path is traversed determines the positive normal of any surface bounded by the path. Use the right-hand rule:

  1. curl the fingers of the right hand in the positive traversal direction around C;
  2. the thumb gives the positive surface-normal direction;
  3. currents piercing the surface in the thumb direction count as positive enclosed current.

Thus Ienc is an oriented quantity, not merely the sum of current magnitudes.

PIC

Figure. A circular Amperian loop around a long straight wire. The magnetic field is tangent to the circle and has constant magnitude at fixed radius.

3 Why symmetry matters

Ampère’s law always relates the closed-path circulation to enclosed steady current, but it does not automatically provide the local field magnitude.

To solve directly for B, one usually seeks a path for which symmetry guarantees one or more of the following:

  1. the direction of B is known;
  2. B is tangent to the path where it contributes;
  3. B has the same magnitude along a contributing path segment;
  4. other path segments make zero contribution because B d.

This is the magnetic counterpart of selecting a Gaussian surface that matches an electric-field symmetry.

4 Example 1: a known circular field

Suppose a magnetic field has constant magnitude

B  = 2.0 × 10−4T
(6)

and is everywhere tangent to a circular path of radius

s = 5.0 × 10−2 m.
(7)

Then

CB d = B Cdℓ (8)
= B(2πs). (9)

Therefore,

CB d = (2.0 × 104)2π(5.0 × 102) (10)
= 6.28 × 105 T m. (11)

The line integral measures magnetic circulation around the path.

5 The infinitely long straight wire

For an infinitely long straight wire carrying current I, cylindrical symmetry implies that magnetic-field lines are circles centered on the wire. At fixed distance s from the wire, the magnitude B is constant.

Choose a circular Amperian loop of radius s. Then

∮
   B ⋅ d ℓ = B(2πs ).
 C
(12)

The enclosed current is simply I, so Ampère’s law gives

B(2πs ) = μ0I.
(13)

Hence,

|------------|
|       μ0I- |
B (s) = 2πs .|
--------------
(14)

This is exactly the result derived by direct Biot–Savart integration in EM11, but symmetry makes the Ampère-law derivation much shorter.

6 Example 2: field around a straight wire

Let

                          −2
I = 12 A,     s = 3.0 × 10   m.
(15)

Then

B = (4π-×-10−-7)(12-)
 2π(3.0 × 10−2) (16)
= 8.0 × 105 T. (17)

Thus,

|------------−-5-----------|
-B-=-8.0-×-10---T-=-80-μT.--
(18)

7 A cylindrical wire with uniform current density

Now suppose the wire has radius a and carries total current I distributed uniformly across its cross section.

The uniform current density is

      I
J =  --2.
     πa
(19)

For an Amperian circle of radius s < a, only the current inside radius s is enclosed:

Ienc = Jπs2 (20)
= Is2
--2
a. (21)

Ampère’s law gives

              s2-
B (2πs) = μ0I a2.
(22)

Therefore, inside the wire,

|------------------------|
|       -μ0Is            |
|B(s) = 2 πa2,     s < a.|
--------------------------
(23)

Outside the wire, s a, the entire current is enclosed, so

|------------------------|
|B (s) = μ0I-,    s ≥ a. |
---------2πs-------------|
(24)

Thus the field increases linearly with radius inside a uniformly current-filled wire and decreases as 1∕s outside.

PIC

Figure. A uniformly current-filled cylindrical wire. An inner Amperian loop encloses only the fraction of current lying inside its radius.

8 Example 3: field inside a uniform-current wire

A cylindrical wire has

a = 4.0 × 10−3 m,     I = 10 A.
(25)

Find the field at

s = 2.0 × 10−3 m.
(26)

Because s < a,

B = μ0Is-
2πa2 (27)
= (4π × 10 −7)(10)(2.0 × 10− 3)
------------------−3-2------
      2π (4.0 × 10   ) (28)
= 2.5 × 104 T. (29)

So

|------------------|
|B =  2.5 × 10 −4T. |
-------------------
(30)

9 The ideal long solenoid

A solenoid is a helical winding with many turns per unit length. Let

n =  N--
     L
(31)

be the number of turns per unit length.

For a sufficiently long ideal solenoid, symmetry and the long-solenoid approximation imply:

  • the magnetic field inside is approximately uniform and parallel to the solenoid axis;
  • the external field is small compared with the interior field away from the ends.

Choose a rectangular Amperian path with one long side of length inside the solenoid and parallel to the field, and the opposite long side outside where the idealized field is approximately zero. The short sides are perpendicular to B and contribute zero.

Therefore,

∮
   B ⋅ dℓ ≈ B ℓ.
 C
(32)

The number of windings piercing the surface is

nℓ,
(33)

so the enclosed current is

Ienc = (nℓ)I.
(34)

Ampère’s law gives

Bℓ = μ0n ℓI.
(35)

Hence,

|------------------------------------------|
|B  ≈ μ0nI     inside a long ideal solenoid.|
-------------------------------------------
(36)

PIC

Figure. A rectangular Amperian path for an ideal long solenoid. The interior segment is parallel to the nearly uniform magnetic field; the exterior contribution is idealized as negligible.

10 Example 4: ideal-solenoid field

Let

n = 1200 turns/m,      I = 0.80 A.
(37)

Then

B = μ0nI (38)
= (4π × 107)(1200)(0.80) (39)
= 1.21 × 103 T. (40)

Therefore,

|--------------|
-B-≈--1.21mT.--|
(41)

11 The ideal toroid

A toroid may be viewed as a solenoid bent into a closed ring. Suppose it has N turns carrying current I.

For an ideal toroid, symmetry implies that the magnetic field inside the winding region is approximately azimuthal and depends only on the distance s from the toroid’s central axis.

Choose a circular Amperian loop of radius s lying within the winding region. The field is tangent to the loop and has constant magnitude there, so

∮
   B ⋅ d ℓ = B(2πs ).
 C
(42)

The loop encloses N current crossings, so

Ienc = N I.
(43)

Thus,

B(2πs ) = μ0N I,
(44)

or

|--------------|
|       μ0N--I |
|B(s) =  2πs  .|
----------------
(45)

For the idealized toroid, the field is approximately zero in the central hole and outside the winding region because an appropriate Amperian loop there encloses zero net current.

PIC

Figure. Circular Amperian path through an ideal toroid. Inside the winding region, B is tangent to the path and has constant magnitude at fixed radius.

12 Example 5: ideal-toroid field

Let

N  = 400,     I = 0.50 A,     s = 0.10m.
(46)

Then

B = μ0N--I
 2πs (47)
= (4π × 10 −7)(400)(0.50 )
-------2π-(0.10-)------- (48)
= 4.0 × 104 T. (49)

So

|------------------|
|B =  4.0 × 10 −4T. |
-------------------
(50)

13 When Ampère’s law is true but not directly useful

Consider two separated current-carrying wires or a finite bent Conductor with little symmetry. Ampère’s law still gives the circulation around any chosen closed path, but the local field magnitude may vary around the path and may not remain tangent to it.

Then one cannot replace

∮
   B  ⋅ dℓ
  C
(51)

with a simple product such as

B (2 πs).
(52)

The difficulty is not a failure of Ampère’s law. The difficulty is that symmetry is insufficient to remove B from the line integral.

In such cases, Biot–Savart integration, superposition, numerical field methods, or later Maxwell-equation techniques may be more useful.

14 Example 6: deciding whether a circular path is useful

Suppose a current distribution contains a single infinitely long straight wire, but the proposed circular path is not centered on the wire.

Ampère’s law still gives

∮
   B ⋅ dℓ = μ I  .
 C           0 enc
(53)

However, the distance from the wire to points on the off-center circle changes around the path. Therefore B is not constant on the path, and the field is not everywhere tangent to that circle.

So the off-center circle is a poor path for directly solving for B.

The correct symmetry-matched path is a circle centered on the wire.

15 From integral Ampère’s law to the differential form

Stokes’ theorem relates the circulation of a vector field around a closed curve to the flux of its curl through any surface bounded by that curve:

|∮----------∫----------------|
|                            |
|   B ⋅ dℓ =   (∇ × B ) ⋅ dA.|
--C-----------S---------------
(54)

Ampère’s law states

∮
   B ⋅ dℓ = μ0Ienc.
 C
(55)

Using the current-density relation

      ∫
I   =    J ⋅ dA,
 enc    S
(56)

we obtain

∫                    ∫
   (∇ × B ) ⋅ dA = μ0   J ⋅ dA.
  S                    S
(57)

Therefore,

∫
   (∇ × B  − μ0J ) ⋅ dA = 0.
  S
(58)

For arbitrary surfaces in magnetostatics, the local relation is

|--------------|
|∇ × B  = μ0J. |
----------------
(59)

This equation connects directly back to the curl operator introduced in EM03.

16 Example 7: verify the differential form

Consider the field

B  = − μ0J0-yˆx + μ0J0-xˆy.
         2         2
(60)

The curl has only a z component:

(∇× B)z = ∂By-
 ∂x ∂Bx--
∂y (61)
= μ0J0-
  2 (       )
  − μ0J0-
     2 (62)
= μ0J0. (63)

Thus,

-----------------
|∇ × B  = μ  J ˆz.|
------------0-0--|
(64)

This corresponds to the current density

|--------|
J-=--J0ˆz.-
(65)

17 The magnetostatic limitation

The form

∮
   B ⋅ d ℓ = μ I
 C           0 enc
(66)

is the magnetostatic form of Ampère’s law. It assumes steady-current conditions.

For genuinely time-varying electromagnetic fields, this expression is incomplete. Maxwell discovered that a changing electric flux contributes to magnetic-field circulation even where no conduction current passes through the chosen surface.

The corrected law, introduced later in this series, has the structure

|∮-----------------------------|
|                        d-ΦE- |
|   B ⋅ d ℓ = μ0Ienc + μ0𝜖0 dt .|
--C----------------------------
(67)

For EM12, the important point is only the boundary of validity:

|--------------------------------------------------------|
|the simple Amp  `ere law  used here is a steady-current law. |
----------------------------------------------------------
(68)

The displacement-current term will become essential when the series turns to electromagnetic waves and antennas.

18 Common mistakes

  • Mistake: assuming Ampère’s law gives B directly for every current geometry. The law gives circulation; symmetry is what may turn the line integral into a simple algebraic expression.
  • Mistake: using total current when an Amperian loop encloses only part of a distributed current.
  • Mistake: forgetting the orientation sign of enclosed current.
  • Mistake: assuming a circular Amperian path is useful merely because it is a circle. It must be centered on the symmetry axis for the straight-wire problem.
  • Mistake: treating the ideal solenoid and toroid formulas as exact for every finite winding geometry.
  • Mistake: applying the magnetostatic Ampère law without modification to time-varying antenna currents.

19 What EM12 adds to the series

EM11 introduced the magnetostatic source integral

        μ I ∫  d ℓ′ × (r − r′)
B (r) = -0--   --------′-3--.
         4π  C    |r − r |
(69)

EM12 adds a complementary circulation law:

|∮-----------------|
|                  |
|   B ⋅ dℓ = μ0Ienc.|
--C-----------------
(70)

With sufficient symmetry, that law gives the straight-wire, solenoid, and toroid fields with very little integration.

Using Stokes’ theorem, the same magnetostatic physics can be written locally as

|--------------|
-∇-×-B--=-μ0J.--
(71)

The next stage of the series will move from magnetostatics toward time-varying fields and induction, where electric and magnetic fields begin to generate one another dynamically.

References

[1]   David J. Griffiths, Introduction to Electrodynamics, 4th ed., Cambridge University Press, 2017.

[2]   Edward M. Purcell and David J. Morin, Electricity and Magnetism, 3rd ed., Cambridge University Press, 2013.

[3]   Samuel J. Ling, Jeff Sanny, and William Moebs, University Physics, Volume 2, OpenStax, 2016, sections on Ampère’s law, solenoids, and toroids.

[4]   Richard P. Feynman, Robert B. Leighton, and Matthew Sands, The Feynman Lectures on Physics, Volume II, Addison-Wesley, 1964, chapters on magnetic fields, circulation, and Maxwell’s equations.

[5]   Massachusetts Institute of Technology, 8.02 Physics II: Electricity and Magnetism, MIT OpenCourseWare, materials on Ampère’s law, solenoids, toroids, and magnetic-field symmetry.


"Electromagnetic Waves: Ampere's Law and Symmetry" is owned by bloftin.
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Other names:  EM12
Keywords:  Ampere's law, magnetic field, magnetic circulation, Amperian loop, symmetry, straight wire, current density, solenoid, toroid, Stokes theorem, curl, magnetostatics, Maxwell correction, antenna current

Attachments:
Electromagnetic Waves: Ampere's Law and Symmetry - Exercises and Complete Worked Solutions (Example) by bloftin

Cross-references: formulas, algebraic, waves, boundary, conduction, EM03, operator, relation, flux, vector field, Stokes theorem, Conductor, section, field, magnitudes, curl, Gauss's Law, EM07, magnetic field, magnetostatic, EM11
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This is version 1 of Electromagnetic Waves: Ampere's Law and Symmetry, born on 2026-09-18.
Object id is 1231, canonical name is ElectromagneticWavesAmperesLawAndSymmetry.
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Classification:
Physics Classification41.20.Gz (Magnetostatics; magnetic shielding, magnetic induction, boundary-value problems)
 03.50.De (Classical electromagnetism, Maxwell equations )
 41.20.-q (Applied classical electromagnetism)
 41.20.Jb (Electromagnetic wave propagation; radiowave propagation )
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