AP Physics C E&M: Toroid Magnetic Field

Understanding the magnetic field of a toroid is a pivotal application of Ampere's Law that bridges fundamental electromagnetism with practical engineering design. Unlike an infinite solenoid, a toroid confines its magnetic field entirely within its core, making it an ideal model for transformers and inductors where minimal stray field is critical. Mastering this derivation solidifies your ability to manipulate Ampere's Law for symmetric current distributions and prepares you for both exam problems and real-world magnetic circuit analysis.

Ampere's Law: The Foundational Tool

Before tackling the toroid, you must be fluent with Ampere's Law. This law relates the integrated magnetic field around a closed loop to the total electric current passing through that loop. Its mathematical statement is: $$\oint \vec{B}\cdot d\vec{l}={\mu }_0{I}_{enc}$$. Here, the left side is the line integral of the magnetic field $\vec{B}$ along a closed Amperian loop, ${\mu }_0$ is the permeability of free space ($4\pi \times 10^{-7}\text{T}\cdot \text{m/A}$), and $I_{enc}$ is the net current enclosed by the loop.

The power of Ampere's Law lies in its simplicity for highly symmetric situations. You choose an Amperian loop where the magnetic field is either constant in magnitude and parallel to the path, or perpendicular to it. For the toroid, the symmetry is circular, guiding our loop choice. A common pitfall is applying the law to asymmetric situations; it is not a universal calculator but a clever shortcut for specific, symmetric geometries like solenoids, toroids, and long straight wires.

Geometry of a Toroidal Solenoid

A toroid is essentially a solenoid bent into a circle, forming a doughnut-shaped coil. Imagine taking a long, tightly wound helical coil (a solenoid) and connecting its ends. The core is often a circular ring, and wire is wrapped uniformly around it with $N$ total turns. The key geometric parameters are:

• $N$: The total number of turns of wire.
• $r$: The radial distance from the center of the toroid to a point inside the winding.
• $a$ and $b$: The inner and outer radii of the toroid's winding, respectively.

Crucially, the winding is so tight that each turn can be considered a closed loop concentric with the toroid's central axis. This geometry creates a magnetic field that is circular and concentric with the toroid's axis, with its magnitude depending on the radius $r$. There is no radial or longitudinal component inside the core. Visualizing this symmetry is the first step to correctly applying Ampere's Law.

Applying Ampere's Law to a Toroid

The goal is to find the magnetic field magnitude $B$ at a point inside the toroid core, at a radius $r$ from the center. We follow a systematic, step-by-step approach.

Step 1: Choose the Amperian Loop. Exploiting the symmetry, we select a circular Amperian loop of radius $r$ that lies inside the toroid core, concentric with the toroid's axis. This loop follows a magnetic field line.

Step 2: Evaluate the Line Integral $\oint \vec{B}\cdot d\vec{l}$. Along this circular path, the magnetic field $\vec{B}$ is everywhere tangential (parallel) to the path element $d\vec{l}$ and constant in magnitude (at that specific radius $r$). Therefore, the dot product simplifies to $Bdl$, and the integral becomes: $$\oint \vec{B}\cdot d\vec{l}=\oint Bdl=B\oint dl$$ The integral $\oint dl$ is simply the circumference of our circular loop, $2\pi r$. So, $$\oint \vec{B}\cdot d\vec{l}=B(2\pi r)$$

Step 3: Determine the Enclosed Current $I_{enc}$. This is the critical step. Our Amperian loop of radius $r$ threads through the center of the toroid, passing through each turn of the coil. If the current in the wire is $I$, and the loop encloses all $N$ turns of the toroid, then the total enclosed current is $N$ times $I$. Therefore, $I_{enc}=NI$.

Step 4: Solve for $B$ using Ampere's Law. We set the line integral equal to ${\mu }_0{I}_{enc}$: $$B(2\pi r)={\mu }_0(NI)$$ Solving for the magnetic field magnitude $B$ gives the central result: $$B=\frac{{\mu }_{0}NI}{2\pi r}$$

This equation reveals that inside a toroid, the magnetic field is not uniform; it varies inversely with the radial distance $r$ from the center. The field is strongest at the inner radius ($r=a$) and weakest at the outer radius ($r=b$).

Why the Field is Zero Outside the Toroid

A defining feature of an ideal toroid is that the magnetic field is completely confined to the space within the windings (the core). To understand why, consider Ampere's Law applied to two external loops:

1. Loop outside the entire toroid (radius $r>b$): This loop encloses the entire toroid. However, the net current passing through it is zero. For every turn of wire carrying current into the page on the outer arc, there is a corresponding turn on the inner arc carrying current out of the page. They cancel, so $I_{enc}=0$. Ampere's Law then gives $B(2\pi r)=0$, so $B=0$.
2. Loop inside the "hole" of the toroid (radius $r<a$): This loop inside the central empty space encloses no current at all ($I_{enc}=0$). Again, Ampere's Law dictates that $B=0$.

This perfect confinement is an idealization. In a real toroid with finite-sized, spaced windings, there will be a small stray external field, but for all practical AP Physics and introductory engineering purposes, the external field is considered negligible.

Comparison with an Ideal Solenoid

It is highly instructive to compare the toroid with its linear counterpart, the ideal solenoid. For an infinite solenoid, the magnetic field inside is uniform and given by $B={\mu }_{0}nI$, where $n=N/L$ is the number of turns per unit length. The field outside an infinite solenoid is zero.

The toroid formula $B=\frac{{\mu }_{0}NI}{2\pi r}$ can be rewritten to highlight this connection. Notice that $N/(2\pi r)$ is the number of turns per unit length along the circular path of radius $r$. If we define $n(r)=N/(2\pi r)$ as the local turns per unit length, the formula becomes $B={\mu }_{0}n(r)I$. This is analogous to the solenoid formula, but with $n$ now depending on radius. Conceptually, a small segment of a toroid looks like a segment of a solenoid. The key difference is geometry: the solenoid's uniformity arises from its (idealized) infinite length, while the toroid's field varies due to its curved, closed shape.

Common Pitfalls

1. Misapplying the Enclosed Current for a Partial Loop. A frequent exam trap is an Amperian loop that does not encircle all $N$ turns. For example, if a loop is drawn at a radius that places it outside the winding core, $I_{enc}$ is zero, not $NI$. Always ask: "Does my loop pierce through the center of the coil, or does it loop around outside it?" The answer determines $I_{enc}$.

2. Treating the Field as Uniform. After deriving $B={\mu }_{0}NI/(2\pi r)$, a common mistake is to treat $B$ as a constant. Remember, $r$ is a variable. If a problem asks for the field at the center of the toroid's cross-section (at an average radius $r_{avg}=(a+b)/2$), you must use that specific $r$ value in the formula. Confusing the radius $r$ in the formula with the inner ($a$) or outer ($b$) radius is a specific error to avoid.

3. Forgetting the "Ideal" Conditions. The derived formula and the conclusion of zero external field depend on ideal conditions: tightly wound, circular, symmetric turns. In a problem, if the winding is sparse or non-uniform, Ampere's Law in its simple form may not be directly applicable. Always check the problem statement for clues that the toroid is "closely wound" or "ideal."

Summary

• The magnetic field inside an ideal toroid is confined to the core and is given by $B=\frac{{\mu }_{0}NI}{2\pi r}$, where $r$ is the radial distance from the toroid's center. The field strength is inversely proportional to $r$.
• The derivation is a direct, symmetry-based application of Ampere's Law, using a circular path inside the toroid that encloses all $N$ turns of current-carrying wire.
• The magnetic field outside an ideal toroid (both in the central hole and beyond the outer radius) is zero because any Amperian loop drawn in these regions encloses zero net current.
• The toroid formula is the curved analogue of the solenoid formula $B={\mu }_{0}nI$, with the turns-per-unit-length $n$ becoming a function of radius: $n(r)=N/(2\pi r)$.
• On exams, carefully identify the correct enclosed current $I_{enc}$ for your chosen Amperian loop and remember that $B$ is not uniform across the toroid's cross-section.