Gap 2 — The Toroid

Part 1 — The concept, explained simply

A toroid is a solenoid bent round into a doughnut so that its two ends join up. Cut from the rationalised NCERT — but still examined.

1. What a toroid is

Take a long solenoid, bend it into a ring, and join the ends. There is now no beginning and no end. The field lines, which used to leak out of the ends of a solenoid, are completely trapped inside the doughnut.

Picture it: a solenoid is a straight tunnel — wind escapes from both mouths. A toroid is a ring road — the wind just goes round and round forever, and none of it gets out.

Join the ends and the leaks disappear. That is the entire difference, and it is why a toroid has zero field both in the hole and outside.

2. Getting the formula

Draw a circular Amperian loop of radius r running round inside the doughnut. Walking round it, the field is along your path everywhere and has the same strength, so the sum is simply B × 2πr. The loop threads through all N turns, so it encloses NI:

B × 2πr = μ₀NI ⇒ B = μ₀NI / 2πr

Here r is the mean radius of the ring — the distance from the centre of the hole to the middle of the coil.

3. The three regions

Where	Enclosed current	Field
In the hole (r < inner radius)	none — the loop threads no turns	0
Inside the windings	NI	μ₀NI / 2πr
Outside the toroid	each turn goes in and out ⇒ net zero	0

4. Toroid versus solenoid

Write n = N/2πr, the number of turns per unit length along the ring. Then B = μ₀nI — identical to a solenoid. A toroid is a solenoid whose length is its own circumference.

The difference that matters in questions. In a solenoid the field is genuinely uniform across the inside. In a toroid it depends on r, so it is slightly stronger on the inner edge than on the outer edge. Only for a thin ring is it near enough uniform.

5. With an iron core

Wind the toroid on an iron ring and the field multiplies by the relative permeability:

B = μ₀μ_r N I / 2πr

Typical iron gives μ_r in the thousands, which is why transformers are built on toroidal cores.

Part 2 — Formula sheet

Quantity	Formula	Note
Field inside the windings	B = μ₀NI / 2πr	N = total turns, r = mean radius.
Same, per unit length	B = μ₀nI, n = N/2πr	Identical in form to a solenoid.
Field in the hole	0	The loop encloses no turns.
Field outside the toroid	0	Each turn enters and leaves ⇒ net enclosed current zero.
With a core	B = μ₀μ_rNI / 2πr	μ_r = relative permeability.
Relative permeability from B	μ_r = 2πrB / μ₀NI	Rearranged; a standard question.
Ratio of two toroids	B₁/B₂ = (N₁/r₁) ÷ (N₂/r₂)	Same current; only N/r matters.
Variation across the ring	B ∝ 1/r	Stronger at the inner edge.

The one-line comparison to memorise
Solenoid B = μ₀nI — no radius in it.
Toroid B = μ₀NI/2πr — radius is in it.
Both are zero outside; the toroid is also zero in its hole.

Part 3 — 20 questions with step-by-step solutions

Attempt each on paper first. Every solution follows the same five steps: Given → Asked → Concept → Formula → Solution.

Q1Basic

The magnetic field inside the windings of a toroid of N turns and mean radius r carrying current I is:

(a) μ₀NI/2r

(b) μ₀NI/2πr

(d) μ₀NI/4πr

Show step-by-step solution

GivenToroid, N turns, mean radius r, current I

AskedField inside

ConceptApply Ampere's law to a circular loop running inside the ring.

FormulaB × 2πr = μ₀NI

SolutionThe loop length is its circumference, 2πr.
It threads all N turns, so the enclosed current is NI.
B = μ₀NI/2πr.

Answer: μ₀NI/2πr

Q2Regions

The magnetic field in the empty hole at the centre of a toroid is:

(a) μ₀NI/2πr

(b) maximum

(d) half the value inside the windings

Show step-by-step solution

GivenPoint in the central hole of a toroid

AskedField there

ConceptAn Amperian loop drawn in the hole passes through no windings at all.

Formula∮B·dl = μ₀I_enclosed

SolutionA loop in the hole encloses none of the turns.
I_enclosed = 0, so B = 0.
The field exists only within the windings themselves.

Answer: zero

Q3Regions

The magnetic field outside a toroid is:

(a) μ₀NI/2πr

(b) zero

(d) stronger than inside

Show step-by-step solution

GivenPoint outside the toroid

AskedField there

ConceptA loop outside encloses each turn twice — once going in and once coming out.

Formula∮B·dl = μ₀I_enclosed

SolutionEvery turn crosses the enclosed surface twice, in opposite senses.
The net enclosed current is therefore zero.
B = 0 outside — the field is completely confined.

Answer: zero

Q4Ratio

Two toroids have 200 and 100 turns with mean radii 40 cm and 20 cm respectively, carrying the same current. The ratio of the fields inside them is:

(a) 1 : 1

(b) 2 : 1

(d) 1 : 2

Show step-by-step solution

GivenN₁=200, r₁=0.4 m; N₂=100, r₂=0.2 m; same I

AskedRatio B₁ : B₂

ConceptWith I fixed, the field depends only on the ratio N/r.

FormulaB ∝ N/r

SolutionToroid 1: N/r = 200/0.4 = 500.
Toroid 2: N/r = 100/0.2 = 500.
Equal, so the ratio is 1 : 1.

Answer: 1 : 1

Q5Ratio

Two toroids have 400 and 200 turns with mean radii 30 cm and 60 cm, carrying equal currents. The ratio of their fields is:

(a) 2 : 1

(b) 1 : 4

(d) 4 : 1

Show step-by-step solution

GivenN₁=400, r₁=0.3; N₂=200, r₂=0.6

AskedRatio of fields

ConceptAgain only N/r matters.

FormulaB ∝ N/r

SolutionToroid 1: 400/0.3 = 1333.
Toroid 2: 200/0.6 = 333.
Ratio = 1333/333 = 4, so 4 : 1.

Answer: 4 : 1

Q6Numerical

A toroid of 500 turns and mean radius 25 cm carries a current of 2 A. The field inside is about:

(a) 4 × 10⁻⁴ T

(b) 8 × 10⁻⁴ T

(d) 1.6 × 10⁻³ T

Show step-by-step solution

GivenN = 500, r = 0.25 m, I = 2 A

AskedField inside

ConceptDirect substitution.

FormulaB = μ₀NI/2πr = (2×10⁻⁷)NI/r

SolutionUse μ₀/2π = 2 × 10⁻⁷.
B = 2×10⁻⁷ × 500 × 2 / 0.25
= 2×10⁻⁴/0.25 = 8 × 10⁻⁴ T

Answer: 8 × 10⁻⁴ T

Q7Core

The current in the windings of a toroid is 2 A. There are 400 turns and the mean radius is 40 cm. If the field inside is 1 T, the relative permeability of the core is:

(a) 1000

(b) 1500

(d) 2500

Show step-by-step solution

GivenI = 2 A, N = 400, r = 0.4 m, B = 1 T

AskedRelative permeability μ_r

ConceptFind what the field would be with no core, then see how many times bigger the actual field is.

Formulaμ_r = 2πrB / μ₀NI

SolutionWithout a core: B₀ = 2×10⁻⁷ × 400 × 2 / 0.4 = 4 × 10⁻⁴ T.
μ_r = B / B₀ = 1 / (4 × 10⁻⁴)
= 2500.

Answer: 2500

Q8Comparison

Compared with a solenoid, the formula for the field inside a toroid:

(a) contains the radius, whereas the solenoid's does not

(b) contains no radius, like the solenoid's

(d) has no turns term

Show step-by-step solution

GivenB = μ₀nI vs B = μ₀NI/2πr

AskedThe key difference

ConceptCompare the two expressions term by term.

Formulasolenoid μ₀nI; toroid μ₀NI/2πr

SolutionThe solenoid formula has no radius at all.
The toroid formula has r in the denominator.
So the toroid field varies across the ring; the solenoid's does not.

Answer: contains the radius, whereas the solenoid's does not

Q9Variation

Across the cross-section of a thick toroid, the magnetic field is:

(a) uniform

(b) stronger near the inner edge

(d) zero

Show step-by-step solution

GivenThick toroid, points at different r within the windings

AskedHow B varies

ConceptB ∝ 1/r, so smaller r gives a bigger field.

FormulaB = μ₀NI/2πr

SolutionThe inner edge has the smallest r.
Since B ∝ 1/r, that is where B is greatest.
Only for a thin ring is the variation negligible.

Answer: stronger near the inner edge

Q10Concept

A toroid can be regarded as:

(a) a straight wire bent into a circle

(b) a solenoid bent into a ring with its ends joined

(d) two parallel wires

Show step-by-step solution

GivenStructure of a toroid

AskedCorrect description

ConceptIts geometry and its formula both follow from the solenoid.

Formulan = N/2πr gives B = μ₀nI

SolutionA solenoid is a coil wound along a straight axis.
Bend that axis into a circle and join the ends — that is a toroid.
Its 'length' becomes its own circumference, 2πr.

Answer: a solenoid bent into a ring with its ends joined

Q11Numerical

A toroid has 1000 turns, mean radius 20 cm, and carries 5 A. The field inside is:

(a) 5 × 10⁻³ T

(b) 2.5 × 10⁻³ T

(d) 5 × 10⁻⁴ T

Show step-by-step solution

GivenN = 1000, r = 0.2 m, I = 5 A

AskedField inside

ConceptStraight substitution.

FormulaB = (2×10⁻⁷)NI/r

SolutionB = 2×10⁻⁷ × 1000 × 5 / 0.2
= 10⁻³/0.2
= 5 × 10⁻³ T

Answer: 5 × 10⁻³ T

Q12Turns density

For a toroid, the number of turns per unit length is given by:

(a) N/2πr

(b) N/r

(d) N/πr²

Show step-by-step solution

GivenToroid of N turns and mean radius r

AskedTurns per unit length n

ConceptThe 'length' of a toroid is the circumference of its mean circle.

Formulan = N / (2πr)

SolutionTotal length along the ring = 2πr.
n = total turns / length = N/2πr.
Substituting gives B = μ₀nI, the solenoid form.

Answer: N/2πr

Q13Scaling

If the number of turns of a toroid is doubled and its mean radius is also doubled, the field inside:

(a) doubles

(b) halves

(d) becomes four times

Show step-by-step solution

GivenN → 2N, r → 2r, same current

AskedNew field

ConceptTrack N/r through the formula.

FormulaB ∝ N/r

SolutionN doubles ⇒ B doubles.
r doubles ⇒ B halves.
Net effect: unchanged.

Answer: stays the same

Q14Ampere

In applying Ampere's law to a toroid, the Amperian loop chosen is:

(a) a rectangle straddling the winding

(b) a circle running inside the ring

(d) a sphere enclosing the toroid

Show step-by-step solution

GivenDerivation of the toroid field

AskedChoice of loop

ConceptChoose the shape along which B is constant and tangential — here a circle following the ring.

Formula∮B·dl = B(2πr)

SolutionInside the windings the field runs in circles round the ring.
A circular loop of radius r follows the field exactly.
So B is constant and tangential all the way, giving B × 2πr.

Answer: a circle running inside the ring

Q15Application

Toroidal cores are preferred in transformers mainly because:

(a) they are cheaper

(b) the field is confined inside, so there is little leakage

(d) they need no current

Show step-by-step solution

GivenPractical use of toroids

AskedReason for the preference

ConceptThe field outside a toroid is zero, so almost no flux escapes.

FormulaB_outside = 0

SolutionA solenoid leaks flux from both ends.
A toroid has no ends, so essentially all the flux stays in the core.
That means less energy loss and less interference with nearby circuits.

Answer: the field is confined inside, so there is little leakage

Q16Numerical

A toroid of mean radius 10 cm carries 2 A and produces a field of 4 × 10⁻⁴ T inside (no core). The number of turns is:

(a) 50

(b) 100

(d) 400

Show step-by-step solution

Givenr = 0.1 m, I = 2 A, B = 4 × 10⁻⁴ T

AskedNumber of turns N

ConceptRearrange the toroid formula for N.

FormulaN = Br / (2×10⁻⁷ × I)

SolutionN = (4×10⁻⁴ × 0.1) / (2×10⁻⁷ × 2)
= 4×10⁻⁵ / 4×10⁻⁷
= 100 turns.

Answer: 100

Q17Concept

Which of the following is true for a toroid?

(a) B is zero inside the hole and outside, non-zero within the windings

(b) B is uniform everywhere inside the hole

(d) B is independent of the current

Show step-by-step solution

GivenField distribution of a toroid

AskedThe correct statement

ConceptCheck each of the three regions with Ampere's law.

Formula∮B·dl = μ₀I_enclosed

SolutionHole: no turns enclosed ⇒ B = 0.
Windings: NI enclosed ⇒ B = μ₀NI/2πr.
Outside: net enclosed current zero ⇒ B = 0.

Answer: B is zero inside the hole and outside, non-zero within the windings

Q18Core

A toroid with an iron core of relative permeability 800 has 200 turns, mean radius 10 cm and carries 1 A. The field inside is about:

(a) 0.32 T

(b) 0.16 T

(d) 0.08 T

Show step-by-step solution

Givenμ_r = 800, N = 200, r = 0.1 m, I = 1 A

AskedField inside

ConceptCompute the vacuum value first, then multiply by μ_r.

FormulaB = μ_r × (2×10⁻⁷)NI/r

SolutionVacuum value: 2×10⁻⁷ × 200 × 1 / 0.1 = 4 × 10⁻⁴ T.
With core: 800 × 4 × 10⁻⁴
= 0.32 T.

Answer: 0.32 T

Q19Comparison

A solenoid and a toroid have the same turns per unit length and carry the same current. Their internal fields are:

(a) equal

(b) the toroid's is larger

(d) the toroid's is zero

Show step-by-step solution

GivenSame n and I for both

AskedComparison of fields

ConceptWrite the toroid formula in terms of n and compare.

FormulaB = μ₀nI for both

SolutionToroid: B = μ₀NI/2πr, and n = N/2πr.
So B = μ₀nI — the same expression as for a solenoid.
With the same n and I, the fields are equal.

Answer: equal

Q20Trap

A question gives a toroid's TOTAL number of turns and its mean radius. A student uses B = μ₀nI with n = total turns. The error is:

(a) none

(b) n must be turns per unit length, N/2πr

(d) μ₀ should be μ₀/4π

Show step-by-step solution

GivenStudent substitutes total turns for n

AskedIdentify the mistake

ConceptThe same trap as the solenoid: n means turns per METRE.

Formulan = N/2πr

SolutionN is the total number of turns.
n is turns per unit length, so n = N/2πr.
Using N in place of n inflates the answer by a factor of 2πr.

Answer: n must be turns per unit length, N/2πr