Solenoids and Toroids — Field Calculations

Solenoids and toroids are the bread and butter of magnetic field problems in Class 12 and JEE. Once we know how to apply Ampere’s law to these geometries, we get clean, memorable formulas. The catch: students often confuse the two, mix up “n” (turns per unit length) and “N” (total turns), and forget which formula has the $2\pi r$ in the denominator. Let’s sort all of that out.

A solenoid is a tightly wound helical coil — think of a long, thin spring of wire. A toroid is a solenoid bent into a doughnut shape so its two ends meet. Both produce strong, controllable magnetic fields when current flows, which is why they show up in everything from MRI machines to particle accelerators.

Key Terms & Definitions

Solenoid — A cylindrical coil with $N$ turns over length $L$ . Turns per unit length: $n = N/L$ .

Toroid — A circular ring with $N$ turns. Mean radius $r$ measured from the centre of the ring to the centre of the wound coil cross-section.

Ampere’s law — For any closed loop $C$ :

$\oint_C \vec{B} \cdot d\vec{l} = \mu_0 I_{\text{enc}}$

where $I_{\text{enc}}$ is the total current piercing any surface bounded by $C$ .

Permeability of free space — $\mu_0 = 4\pi \times 10^{-7}$ T·m/A.

Magnetic Field Inside an Ideal Solenoid

The classic result, derived using a rectangular Amperian loop straddling the solenoid wall:

$B = \mu_0 n I = \mu_0 \frac{N}{L} I$

Direction: along the axis. Use the right-hand rule on the current direction to find which way.

Outside the solenoid (and for an ideal infinitely long solenoid): $B = 0$ .

Why this formula? Inside the solenoid, the field is uniform along the axis. We construct a rectangular Amperian loop with one long side inside (length $L_{\text{loop}}$ ) and the other outside. Only the inside side contributes to the line integral. Total current through the loop is $n L_{\text{loop}} I$ . Solving gives $B = \mu_0 n I$ .

What about the ends?

At the very end of a finite solenoid (and on the axis), the field is half the inside value:

$B_{\text{end}} = \frac{\mu_0 n I}{2}$

This is asked surprisingly often in JEE Main as a 4-mark numerical.

Magnetic Field Inside a Toroid

For a toroid with $N$ total turns and mean radius $r$ :

$B = \frac{\mu_0 N I}{2\pi r}$

Direction: tangent to the circular axis of the toroid (azimuthal direction).

Outside the ring: $B = 0$ (no current pierces an external loop).

Why this formula? Take an Amperian loop that’s a circle of radius $r$ inside the toroid’s ring. By symmetry, $B$ is constant along this loop, so $\oint B\,dl = B \cdot 2\pi r$ . Total enclosed current: $NI$ (the loop encircles every turn). Solving gives the formula.

Connection to the Solenoid

If the toroid is “fat” (large $r$ , small cross-section), and we replace $2\pi r$ with the total length around the central axis, we get $B = \mu_0 (N/L) I = \mu_0 n I$ — the solenoid formula. So a toroid is just a solenoid bent into a closed loop.

Solved Examples — Easy to Hard

Easy (CBSE Level)

A solenoid has $1000$ turns spread over $50$ cm and carries a current of $2$ A. Find the magnetic field inside.

$n = 1000/0.5 = 2000$ turns/m.

$B = \mu_0 n I = 4\pi \times 10^{-7} \times 2000 \times 2 = 5.03 \times 10^{-3} \text{ T} \approx 5 \text{ mT}$

Medium (JEE Main)

A toroid of mean radius $10$ cm has $5000$ turns and carries $0.5$ A. Find the field inside.

$B = \frac{\mu_0 N I}{2\pi r} = \frac{4\pi \times 10^{-7} \times 5000 \times 0.5}{2\pi \times 0.1} = \frac{10^{-3}}{0.1} \times 5 \times 10^{-3} = 5 \times 10^{-3} \text{ T}$

Hard (JEE Advanced)

A solenoid of $n_1$ turns per metre is inserted coaxially inside a longer solenoid with $n_2$ turns per metre, both carrying the same current $I$ in the same direction. Find the field at the centre of the inner solenoid.

Inside the inner solenoid (and inside the outer): $B = \mu_0 (n_1 + n_2) I$ . The fields simply add because both are along the same axis. If currents flowed in opposite directions, subtract.

Exam-Specific Tips

JEE Main: 1 question almost every year, usually a direct numerical on $B = \mu_0 n I$ . Watch for trick questions where they give total turns and length separately — compute $n$ first.

JEE Advanced: Often combines solenoid + Faraday’s law to compute induced emf in a coil placed inside. The 2023 paper asked about a solenoid carrying time-varying current and induced electric field outside.

CBSE Boards: Derivations of both $B = \mu_0 n I$ (using rectangular Amperian loop) and the toroid formula are 3-mark questions. Practice drawing the correct Amperian loop diagram.

Common Mistakes to Avoid

Mistake 1 — Using $N$ instead of $n$ for solenoids. The formula $B = \mu_0 n I$ uses turns per unit length, not total turns. If the question gives $N$ and $L$ , compute $n = N/L$ first.

Mistake 2 — Using mean radius vs inner/outer radius for toroids. The formula uses the mean radius $r$ — the centreline of the doughnut, not the inner or outer edge. Inner radius alone gives a wrong answer.

Mistake 3 — Treating the field at the end of a solenoid as full strength. It’s half: $B_{\text{end}} = \mu_0 n I / 2$ . CBSE asks this in MCQs to catch this error.

Mistake 4 — Forgetting that the ideal solenoid has zero field outside. Real solenoids have fringing fields at the ends, but for problems treating “ideal” solenoids, outside field is zero.

Mistake 5 — Confusing the directions of $\vec{B}$ in solenoid vs toroid. Solenoid: along the axis. Toroid: tangential (azimuthal). They’re perpendicular!

Practice Questions

A solenoid is $0.5$ m long with $2000$ turns. What current is needed to produce $0.01$ T inside?
A toroid of $1000$ turns and mean radius $0.05$ m carries $1$ A. Find the field inside.
A solenoid of $n$ turns/m carries current $I$ . A small coil of radius $r$ and $N$ turns is placed at its centre with axis aligned. Find the flux through the small coil.
Two solenoids of equal length and turns are placed coaxially. Currents $I_1$ and $I_2$ flow in opposite directions. Find the field at the common centre.
A toroid has different inner and outer radii $r_1$ and $r_2$ . Should you use $r_1$ , $r_2$ , or $(r_1 + r_2)/2$ in the formula? Why?
A long solenoid of length $1$ m has $500$ turns and carries $4$ A. A circular loop of radius $2$ cm is placed inside, perpendicular to the axis. Find the flux through the loop.
The current in a solenoid changes from $0$ to $5$ A in $0.1$ s. If $n = 10000$ turns/m and area $= 4$ cm², find the induced emf in a single turn.
Show that the magnetic field at the centre of a finite solenoid of length $L$ and radius $R$ at one end approaches $\mu_0 n I / 2$ as $L \gg R$ .

Q1: $I = B/(\mu_0 n) = 0.01/(4\pi \times 10^{-7} \times 4000) \approx 2$ A.

Q2: $B = (4\pi \times 10^{-7} \times 1000 \times 1)/(2\pi \times 0.05) = 4 \times 10^{-3}$ T = 4 mT.

Q5: Use the mean radius $(r_1 + r_2)/2$ . The Amperian loop sits along the central circle of the toroid.

FAQs

Q: Why is the field outside an ideal solenoid zero?

By Ampere’s law applied to a loop entirely outside the solenoid, the enclosed current is zero. Combined with the assumption that field strength decays at infinity, this forces $B = 0$ outside.

Q: How do real MRI machines use solenoids?

An MRI uses a superconducting solenoid producing $\sim 1.5–3$ T inside. To get such large fields, you need huge $nI$ — achieved by very high $n$ (densely wound) and large $I$ (kiloamperes through superconducting wire with zero resistance).

Q: Can I treat a solenoid as a stack of circular loops?

Yes — the field of a solenoid is mathematically the sum of fields from each turn, treated as a circular loop. Integrating gives $B = \mu_0 n I$ on the axis, away from the ends.

Q: What’s the difference between B-field and H-field in solenoids?

In vacuum or air, $B = \mu_0 H$ , so they’re proportional. In a material with permeability $\mu$ , $B = \mu H$ . JEE problems mostly stay with $B$ in vacuum, so this distinction rarely matters at the entrance level.

Q: Why are toroids better than solenoids for some applications?

Toroids confine the field entirely within the ring — no leakage. This makes them ideal for transformers and inductors where stray fields would interfere with neighbouring circuits.