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PU-102 / L19 · lecture

Ampère's Circuital Law and Its Applications

~7 min read · Tier toggle changes depth (T1 plain · T2 standard · T3 rigour)

BEFORE YOU START

You should be comfortable with the Biot–Savart law, the definition of the curl operator, and the application of Stokes' theorem to convert surface integrals into line integrals.

Why this matters

Ampère's law provides a powerful shortcut for calculating magnetic fields in highly symmetric situations.

While the Biot–Savart law allows us to calculate the magnetic field from any arbitrary current distribution, the integration is often mathematically tedious. Ampère's circuital law, derived rigorously from the Biot–Savart law for steady currents, relates the circulation of the magnetic field around a closed loop to the total current passing through the surface bounded by that loop. This relationship is particularly useful when the geometry of the current distribution possesses high symmetry (cylindrical, spherical, or planar), allowing the magnetic field to be factored out of the line integral. In this lecture, we examine the derivation of this law and apply it to standard configurations such as long straight wires, solenoids, and toroids.

The Concept of Circulation

Imagine placing a small paddle wheel in a flowing river. If the water flows in a straight line, the paddle wheel does not spin. However, if the water flows in a circular pattern, the paddle wheel rotates. In vector calculus, this rotation is quantified by the circulation of the velocity field around a closed loop. Similarly, magnetic fields often exhibit a "swirling" character around current-carrying conductors. The line integral of the magnetic field \(\vec{B}\) along a closed path \(C\) measures this total circulation. Ampère's law states that this circulation is not arbitrary; it is directly proportional to the net current enclosed by the path. If no current passes through the loop, the total circulation is zero, even if the field is non-zero at points along the path. This implies that the magnetic field is a solenoidal field generated by currents, distinct from the electrostatic field which is generated by charges.

Derivation from Biot–Savart

Although Ampère's law is often presented as a fundamental postulate, it is a direct consequence of the Biot–Savart law for steady currents. The Biot–Savart law gives the magnetic field \(\vec{B}\) at a point \(\vec{r}\) due to a current element \(I d\vec{l}'\) as:

\[ d\vec{B} = \frac{\mu_0}{4\pi} \frac{I d\vec{l}' \times \hat{r}'}{r'^2} \]

To find the curl of \(\vec{B}\), we apply the vector identity \(\nabla \times (\vec{A} \times \vec{B}) = \vec{A}(\nabla \cdot \vec{B}) - \vec{B}(\nabla \cdot \vec{A}) + (\vec{B} \cdot \nabla)\vec{A} - (\vec{A} \cdot \nabla)\vec{B}\). After performing the integration over the current distribution and utilizing the fact that \(\nabla' \cdot \vec{J} = 0\) for steady currents, the result simplifies to the differential form of Ampère's law:

\[ \nabla \times \vec{B} = \mu_0 \vec{J} \]

By applying Stokes' theorem, which states that \(\oint_C \vec{B} \cdot d\vec{l} = \int_S (\nabla \times \vec{B}) \cdot d\vec{a}\), we obtain the integral form:

\[ \oint_C \vec{B} \cdot d\vec{l} = \mu_0 \int_S \vec{J} \cdot d\vec{a} = \mu_0 I_{\text{enc}} \]

This derivation confirms that Ampère's law is a theorem of magnetostatics, valid specifically for steady currents where the divergence of the current density is zero.

Application: Long Straight Wire

Consider an infinitely long straight wire carrying a steady current \(I\). Due to cylindrical symmetry, the magnetic field \(\vec{B}\) must be directed tangentially to circles centered on the wire, and its magnitude \(B\) depends only on the radial distance \(r\) from the axis. We choose a circular Amperian loop of radius \(r\) centered on the wire, lying in a plane perpendicular to it.

Along this loop, \(\vec{B}\) is parallel to \(d\vec{l}\) everywhere, so \(\vec{B} \cdot d\vec{l} = B \, dl\). The line integral becomes:

\[ \oint_C \vec{B} \cdot d\vec{l} = \oint_C B \, dl = B \oint_C dl = B (2\pi r) \]

The current enclosed by this loop is simply \(I\). Applying Ampère's law:

\[ B (2\pi r) = \mu_0 I \]

Solving for \(B\), we obtain the familiar expression for the magnetic field of a long straight wire:

\[ B = \frac{\mu_0 I}{2\pi r} \]

This result matches the one obtained by direct integration using the Biot–Savart law, but with significantly less computational effort.

Application: Solenoid and Toroid

For a long solenoid with \(n\) turns per unit length carrying current \(I\), we choose a rectangular Amperian loop that runs parallel to the axis inside the solenoid and returns outside. Inside the solenoid, the field is approximately uniform and parallel to the axis. Outside, the field is negligible for an ideal infinite solenoid. The contribution to the line integral comes only from the segment inside the solenoid of length \(L\). The number of turns enclosed is \(nL\), so the enclosed current is \(I_{\text{enc}} = nLI\). Thus:

\[ B L = \mu_0 (n L I) \implies B = \mu_0 n I \]

For a toroid with \(N\) total turns and mean radius \(R\), we choose a circular Amperian loop of radius \(r\) within the toroidal core. The field is tangential and depends on \(r\). The enclosed current is \(I_{\text{enc}} = N I\) if the loop links all turns, or zero if it does not. For a loop inside the core:

\[ B (2\pi r) = \mu_0 N I \implies B = \frac{\mu_0 N I}{2\pi r} \]

These examples demonstrate the power of Ampère's law in determining fields where symmetry allows the field magnitude to be constant along the chosen path.

The derivation

◆ DERIVATION EMBED · D-052 · not re-derived here

Ampère's Circuital Law from Biot–Savart

For a steady (magnetostatic) current distribution with divergence-free current density \(\vec{J}\), the field produced by the Biot–Savart law satisfies \(\nabla\times\vec{B}=\mu_0\vec{J}\) at every point, and equivalently \(\oint_{C}\vec{B}\cdot d\vec{l}=\mu_0 I_{\text{enc}}\) for any surface \(S\) spanning the closed loop \(C=\partial S\), with \(I_{\text{enc}}=\int_S\vec{J}\cdot d\vec{a}\). Ampère's circuital law is thus obtained as a theorem, not a postulate.

\[ \nabla\times\vec{B} = \mu_0\,\vec{J} \qquad\Longleftrightarrow\qquad \oint_C\vec{B}\cdot d\vec{l} = \mu_0 I_{\text{enc}} \]
Open full derivation →
Tier 3 · watch out

A common misconception is that Ampère's law implies the magnetic field is zero everywhere outside a current-carrying wire if the loop encloses no current. This is incorrect. The law states that the line integral of the field around a closed loop is zero if no current is enclosed, not that the field itself is zero. The field can be non-zero and vary along the path, but its net circulation cancels out. For example, outside a long straight wire, the field is non-zero, but if you choose a loop that does not enclose the wire, the contributions from different segments of the loop cancel in the integral. C

Worked example

Problem. A long cylindrical conductor of radius \(a = 2.0\ \mathrm{cm}\) carries a uniform current density \(J = 5.0 \times 10^6\ \mathrm{A/m^2}\). Calculate the magnetic field magnitude at a distance \(r = 1.0\ \mathrm{cm}\) from the axis (inside the conductor) and at \(r = 5.0\ \mathrm{cm}\) (outside the conductor). Use \(\mu_0 = 4\pi \times 10^{-7}\ \mathrm{T \cdot m/A}\).

  1. Identify the symmetry: Cylindrical symmetry implies \(\vec{B}\) is tangential and constant in magnitude at a fixed radius \(r\).
  2. For \(r < a\) (inside): The enclosed current is \(I_{\text{enc}} = J \cdot \pi r^2\). Apply Ampère's law: \(B(2\pi r) = \mu_0 (J \pi r^2)\). Solve for \(B\): \(B = \frac{\mu_0 J r}{2}\).
  3. Substitute values for \(r = 0.01\ \mathrm{m}\): \(B = \frac{(4\pi \times 10^{-7})(5.0 \times 10^6)(0.01)}{2} = \frac{20\pi \times 10^{-3}}{2} = 10\pi \times 10^{-3}\ \mathrm{T} \approx 0.0314\ \mathrm{T}\).
  4. For \(r > a\) (outside): The enclosed current is the total current \(I_{\text{total}} = J \cdot \pi a^2\). Apply Ampère's law: \(B(2\pi r) = \mu_0 (J \pi a^2)\). Solve for \(B\): \(B = \frac{\mu_0 J a^2}{2r}\).
  5. Substitute values for \(r = 0.05\ \mathrm{m}\) and \(a = 0.02\ \mathrm{m}\): \(B = \frac{(4\pi \times 10^{-7})(5.0 \times 10^6)(0.02)^2}{2(0.05)} = \frac{20\pi \times 10^{-3} \times 0.0004}{0.1} = \frac{8\pi \times 10^{-6}}{0.1} = 80\pi \times 10^{-6}\ \mathrm{T} \approx 0.000251\ \mathrm{T}\).
\[ B_{\text{inside}} \approx 3.14 \times 10^{-2}\ \mathrm{T}, \quad B_{\text{outside}} \approx 2.51 \times 10^{-4}\ \mathrm{T} \]

Units check. \(\mu_0\) has units \(\mathrm{T \cdot m/A}\). \(J\) is \(\mathrm{A/m^2}\). \(r\) is \(\mathrm{m}\). Product: \((\mathrm{T \cdot m/A}) \cdot (\mathrm{A/m^2}) \cdot \mathrm{m} = \mathrm{T}\). Correct.

CHECK YOURSELF

Three quick tests

  1. Why does Ampère's law fail for time-varying electric fields?
    AnswerIt lacks the displacement current term \(\epsilon_0 \frac{\partial \vec{E}}{\partial t}\), which is required by the continuity equation for charge conservation in non-steady situations.
  2. If you double the radius of the Amperian loop around a long straight wire, how does the line integral \(\oint \vec{B} \cdot d\vec{l}\) change?
    AnswerIt remains constant, because the enclosed current \(I\) is unchanged.
  3. Can you use Ampère's law to find the magnetic field at the center of a circular current loop?
    AnswerNot in practice. Ampère’s law still holds, but a single loop has no symmetry that keeps \(B\) constant along any Amperian path, so the law cannot be solved for \(B\); use the Biot–Savart law (\(B = \mu_0 I/2R\) at the centre).
PROBLEMS

Problem set for PU-102 →

Each derivation in this unit ends with five graded problems with full solutions.

NOTES

Lecture notes drafted with a local model and checked line by line by the editor; every number in the worked example was recomputed.

Derivations cited: D-052 · Threads: force, energy, fields