Ampère's Law — Solenoids and Toroids
Ampère's Law — Solenoids and Toroids
Ampère's law is one of the four Maxwell equations — the foundation of all electromagnetic theory. It relates the magnetic field circulating around a closed loop to the total current passing through that loop. In its integral form, the line integral of B dot dL around a closed path equals mu-zero times the enclosed current. This powerful law allows us to calculate the magnetic field in situations with high symmetry, particularly inside solenoids and toroids.
A solenoid is a coil of wire wound in a helix. When current flows through it, the individual circular field loops from each turn add up inside the coil to produce a nearly uniform magnetic field. Outside the solenoid, the fields from adjacent turns mostly cancel, making the external field negligible. The field inside an ideal solenoid is B equals mu-zero times n times I, where n is the number of turns per unit length and I is the current. This field is uniform — the same everywhere inside the solenoid, regardless of position. This makes solenoids invaluable for creating controlled, uniform magnetic fields in laboratories, MRI machines, and particle accelerators.
A toroid is a solenoid bent into a doughnut shape. All the magnetic field lines are confined inside the core, with no external field at all. The field inside a toroid is B equals mu-zero N I divided by 2 pi r, where N is the total number of turns and r is the radial distance from the center. Toroids are used in transformers, inductors, and magnetic confinement fusion devices like tokamaks.
The mathematics in this lesson extends the concepts of line integrals and symmetry arguments. You will learn to choose Amperian loops, calculate enclosed currents, and derive the fields inside solenoids and toroids. These are the same analytical techniques used by engineers designing electromagnets, MRI scanners, and fusion reactors.
$$\oint \vec{B} \cdot d\vec{\ell} = \mu_0 I_{\text{enc}}$$
The line integral of $\vec{B}$ around any closed loop equals $\mu_0$ times the total current enclosed by the loop.
An ideal solenoid with $n$ turns per meter carrying current $I$:
$$B = \mu_0 n I$$
The field is uniform inside and essentially zero outside.
A toroid with $N$ total turns carrying current $I$, at radius $r$ from the center:
$$B = \frac{\mu_0 N I}{2\pi r}$$
The field exists only inside the toroid — it is zero outside.
A solenoid is 30 cm long with 600 turns and carries 4 A. Find the magnetic field inside.
Step 1: Find the turns per meter: $n = 600/0.30 = 2000$ turns/m.
Step 2: Apply the solenoid formula:
$$B = \mu_0 n I = (4\pi \times 10^{-7})(2000)(4) = 1.005 \times 10^{-2} \text{ T} \approx 10.1 \text{ mT}$$
An MRI machine requires a 1.5 T field. If the solenoid is 1.5 m long, how many turns are needed with a current of 300 A?
Step 1: Solve for $n$: $n = B/(\mu_0 I) = 1.5/(4\pi \times 10^{-7} \times 300)$
Step 2: $n = 1.5/(3.77 \times 10^{-4}) \approx 3{,}979$ turns/m
Step 3: Total turns: $N = n \times L = 3979 \times 1.5 \approx 5{,}969$ turns
A toroid has 500 turns, inner radius 8 cm, outer radius 12 cm, and carries 6 A. Find the field at the mean radius.
Step 1: Mean radius: $r = (0.08 + 0.12)/2 = 0.10$ m
Step 2: Apply toroid formula:
$$B = \frac{\mu_0 N I}{2\pi r} = \frac{(4\pi \times 10^{-7})(500)(6)}{2\pi (0.10)}$$
Step 3: $B = \frac{1.2\pi \times 10^{-3}}{0.2\pi} = 6.0 \times 10^{-3}$ T $= 6.0$ mT
Practice Problems
Show Answer Key
1. $B = (4\pi\times10^{-7})(1000)(5) = 6.28$ mT
2. $I = B/(\mu_0 n) = 0.02/(4\pi\times10^{-7}\times4000) = 3.98$ A
3. $n = 2000/0.50 = 4000$ turns/m. $B = (4\pi\times10^{-7})(4000)(3) = 15.1$ mT
4. $B = \mu_0(800)(5)/(2\pi\times0.15) = 5.33$ mT
5. Ampère's law with a loop outside encloses zero net current (equal currents in both directions), so $B_{\text{outside}} = 0$.
6. $n = B/(\mu_0 I) = 0.1/(4\pi\times10^{-7}\times10) \approx 7{,}958$ turns/m