Solenoid and Toroid

The Solenoid — Construction

A solenoid is a long cylindrical coil produced by winding a single insulated wire in the shape of a tight helix. Two important parameters:

• $n$ = number of turns per unit length (units: turns·m⁻¹).
• $I$ = current through the wire.

We say a solenoid is ideal when:

1. Its length is much greater than its diameter (so the central region behaves as if the wire-wrappings were infinitely long).
2. The winding is tight (adjacent turns close together — almost a continuous current sheet on a cylinder).

For an ideal solenoid:

• The magnetic field inside is uniform, parallel to the axis, and strong. Its direction is given by the right-hand thumb rule: if you curl your fingers in the direction of current, the thumb points along the axis in the direction of $\vec B$.
• The magnetic field outside is essentially zero.

You can visualise the solenoid as a stack of many circular loops, each producing its own on-axis field; at any interior point, contributions from all loops add up to produce a strong uniform field, while at any exterior point they partly cancel.

[NEET Important] "Long solenoid" $\Rightarrow$ ideal solenoid approximation $\Rightarrow$ uniform $B$ inside, zero $B$ outside. This is a standard idealisation in JEE/NEET problems.

Applying Ampère's Law to the Solenoid

To find $B$ inside an ideal solenoid, we apply Ampère's circuital law to a rectangular Amperian loop straddling the wall of the solenoid.

Choosing the loop

Take a rectangle $abcd$ where:

• Side $ab$ of length $L$ lies inside the solenoid, parallel to the axis.
• Side $cd$ of length $L$ lies outside, parallel to the axis.
• Sides $bc$ and $da$ are short segments perpendicular to the axis, crossing the wall of the solenoid.

Evaluating $\oint \vec{B}\cdot d\vec{l}$

Split the integral into four parts:

• $ab$ (inside, along axis): $\vec{B}$ is along the axis (parallel to $d\vec{l}$ if you traverse $a\to b$ along $\vec{B}$). Contribution: $BL$.
• $cd$ (outside, opposite direction): $\vec{B}\approx 0$ outside an ideal solenoid. Contribution: $0$.
• $bc$ and $da$ (perpendicular crossings): inside, $\vec{B}\perp d\vec{l}$, so contribution is zero on the inside parts; outside, $\vec{B} = 0$. Contribution: $0$.

So $\oint \vec{B}\cdot d\vec{l} = BL$.

Enclosed current

The rectangle of length $L$ along the axis encloses $nL$ turns, each carrying $I$. So $I_{enc} = nLI$.

Putting it together

$$BL = \mu_0\,nLI \quad\Rightarrow\quad \boxed{\;B = \mu_0\,n\,I\;}$$

The field inside an ideal solenoid is uniform and depends only on $n$ and $I$ — not on the radius of the solenoid or on the position inside it. Outside: $B = 0$.

At the end of a long solenoid

By a careful argument (or by superposing the contributions of two semi-infinite solenoids), the field on the axis at one end of a long but finite solenoid is half the interior value:

$$B_{end} = \dfrac{\mu_0 n I}{2}$$

This is a 1-mark JEE recall — make sure you know which factor (½) applies.

[JEE Tip] A "long" solenoid is ideal only in its central region. The end correction matters whenever the question asks about the field at the mouth.

The Toroid — Solenoid Bent into a Donut

A toroid is what you get if you bend a long solenoid into a circle and join its ends — essentially a "donut-shaped solenoid". It has total $N$ turns around a central radius $r$ (the radius of the circle running through the centre of the tube).

The toroid has one big advantage over a straight solenoid: its field is entirely confined inside the tube, with no end effects. There's no "outside the ends" because the toroid has no ends.

Applying Ampère's law

Choose an Amperian loop that is a circle of radius $r$ concentric with the torus. By symmetry, $\vec{B}$ is tangent to this loop, constant in magnitude, and its direction follows the right-hand grip rule (curl your fingers along the current; the thumb points along $\vec B$).

$$\oint \vec{B}\cdot d\vec{l} = B\,(2\pi r)$$

Each of the $N$ turns of wire passes through this loop once, all carrying current $I$ in the same direction. So $I_{enc} = NI$.

$$B\,(2\pi r) = \mu_0\,N\,I \quad\Rightarrow\quad \boxed{\;B = \frac{\mu_0\,N\,I}{2\pi r}\;}$$

Connection to the solenoid formula

If we let $n = N/(2\pi r)$ — the number of turns per unit circumferential length of the toroid — then $B = \mu_0 n I$, exactly the same form as the solenoid.

Field outside the toroid

An Amperian loop drawn entirely outside the toroid encloses no net current. So

$$B_{outside} = 0$$

The toroid is the most "magnetically clean" device in elementary electromagnetism — no leakage field at all, in the ideal limit.

[JEE Tip] Toroid: $B$ inside depends on $r$ as $1/r$ — the field is not uniform across the toroid's cross-section (unlike the solenoid). For thin toroids, the variation is small; for fat ones, it matters.

Solenoid vs Toroid — A Comparison & Applications

Feature	Solenoid	Toroid
Shape	Long straight cylinder	Donut (closed ring)
Field formula	$B = \mu_0 n I$	$B = \mu_0 N I/(2\pi r)$
Field uniform?	Yes (central region)	No — varies as $1/r$ across cross-section
Field outside	Approx. zero (ideal limit)	Exactly zero (no end effects)
End correction	Yes, $B_{end} = B_{center}/2$	None
Best for	Strong uniform fields	Confined magnetic fields

Practical applications

• Solenoid valves: small solenoid pulls an iron plunger when energised — used in washing machines, automatic doors, hydraulic systems, and car starters.
• Electromagnets: a solenoid wrapped around an iron core can lift cars in scrapyards (core multiplies $B$ by hundreds).
• MRI machines: superconducting solenoids generate fields of several tesla over a region big enough for a human body.
• Loudspeakers: a solenoid attached to the speaker cone — current oscillations push it, producing sound.
• Toroidal transformers: power-supply transformers with toroidal cores have lower leakage flux and are quieter than open-core transformers.
• Tokamak fusion reactors: enormous toroidal magnets confine plasma at $100$ million K.

Memory Capsule

Solenoid — the must-remember formulas

Location	Field
Inside (central)	$B = \mu_0 n I$
Outside (ideal)	$B = 0$
At one end (on axis)	$B = \mu_0 n I/2$

$n$ = turns·m⁻¹. Uniform interior field; direction by right-hand rule.

Toroid — the must-remember formula

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

$N$ = total turns. Field varies as $1/r$. Outside (and in the central hole): $B = 0$.

Key conceptual points

• Both follow Ampère's law: solenoid uses a rectangular loop; toroid uses a circular loop.
• A toroid is essentially a solenoid bent into a circle.
• A solenoid has end-effects; a toroid does not.

One-line takeaway

Solenoid: stores a uniform $B$ inside; toroid: keeps the field entirely confined.

Exam Tips

• For solenoids, choose a rectangular Amperian loop with one long side inside (parallel to $\vec B$) and the opposite side outside (where $B\approx0$). Perpendicular sides contribute zero.
• For toroids, choose a circular Amperian loop at the mean radius $r$, concentric with the torus. The symmetry makes $B$ tangent and constant.
• Always apply the right-hand rule (thumb along $\vec B$, fingers in direction of current) to fix the field direction.

Solved Examples

Example 1: Field inside a solenoid

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

Solution. First compute $n$:

$$n = \dfrac{N}{L} = \dfrac{1000}{0.50} = 2000\;\text{turns/m}$$

Then apply $B = \mu_0 n I$:

$$B = (4\pi\times 10^{-7})(2000)(2) = 1.6\pi\times 10^{-3}\;\text{T} \approx 5.03\times 10^{-3}\;\text{T}$$

$$B \approx 5.0\;\text{mT}$$

Example 2: Field at the end of a solenoid

For the same solenoid as Example 1 ($n=2000$ turns·m$^{-1}$, $I=2$ A), find the magnetic field at one end (on the axis).

Solution. The end-correction formula gives $$B_{end}=\frac{B_{inside}}{2} = \frac{\mu_0nI}{2} = \frac{5.0\;\text{mT}}{2} = 2.5\text{ mT}\,.$$ Alternatively, use $B_{end}=\mu_0nI/2$ directly.

Example 3: Designing a solenoid for a given $B$

A solenoid is to produce a field of $0.02$ T inside it, using a current of $5$ A. How many turns per metre are needed?

Solution. From $B = \mu_0 n I$:

$$n = \dfrac{B}{\mu_0 I} = \dfrac{0.02}{(4\pi\times 10^{-7})(5)} = \dfrac{0.02}{6.28\times 10^{-6}} \approx 3183\;\text{turns/m}$$

That's about $3.2$ turns per millimetre — achievable with fine wire wound tightly.

Example 4: Toroid — basic numerical

A toroid has 2000 turns and a mean radius of 0.20 m. It carries a current of 4 A. Find the magnetic field along the central circle inside the toroid.

Solution. $$B=\frac{\mu_0NI}{2\pi r} = \frac{(4\pi\times10^{-7})(2000)(4)}{2\pi(0.20)} = 8.0\times10^{-3}\;\text{T} = 8.0\;\text{mT}\,.$$

Example 5: Solenoid vs toroid with same $n$

A solenoid has $n=1000$ turns·m$^{-1}$, $I=3$ A. A toroid is made by joining the ends of the same length of coil into a donut of mean radius 0.25 m.

(a) $B_{sol}=\mu_0nI=3.77\times10^{-3}$ T.

(b) For the toroid, $N=n\cdot2\pi r=1000\times2\pi(0.25)=500\pi\approx1571$ turns, $$B_{tor}=\frac{\mu_0NI}{2\pi r}=3.77\times10^{-3}\;\text{T}$$ Same as the solenoid.

Example 6: Effect of doubling current and halving turns

A solenoid currently produces $B=0.01$ T. The current is doubled while the turn density is halved. What is the new field?

Solution. $$B'=\mu_0\,(n/2)\,(2I)=\mu_0nI=0.01\;\text{T}\,.$$

Example 7: Toroid field across its cross-section

A toroid has $N=1500$, $I=6$ A, inner radius $0.15$ m, outer radius $0.25$ m. Find $B$ at $r_1$ and $r_2$.

$$B_1=\frac{(4\pi\times10^{-7})(1500)(6)}{2\pi(0.15)}=12.0\;\text{mT},\quad B_2=7.2\;\text{mT}.$$

Example 8: Solenoid in an MRI-style application

A superconducting solenoid has $n=2000$ turns·m$^{-1}$ and must produce 1.5 T.

(a) $$I=\frac{B}{\mu_0n}\approx597\;\text{A}.$$

(b) In copper this current would generate enormous heat. MRI solenoids use superconducting wire cooled by liquid helium to carry hundreds of amperes with zero resistance.