Formula
Field inside solenoid
B = μ₀ n I, where n is turns per unit length. Uniform inside, ~zero outside. Toroid: same.
-- NCERT Class 12 Physics, Ch. 4, p. 122Ampere's Law Solenoid
8 MCQs1 revision card9-step worked example
Source: NCERT Magnetic Effects of Current and MagnetismPYQ coverage: NEET 2020, 2022Official key: NTA-verifiedLast updated: 25 Sep 2026
Ampere's Law Solenoid, explained for NEET
The solenoid result is one line — B = μ₀nI — and the letter that costs marks is n, not B.
In that formula n is turns per metre, not the total number of turns. A solenoid with 500 turns wound over 25 cm has n = 500/0.25 = 2.0 × 10³ turns per metre. Substituting 500 directly inflates the field by a factor equal to the length in metres — here by 4. The examiner rarely hands you n; the stem gives N and a length, and the division is the question. Whenever the stem quotes a length, that length exists to be divided by.
NCERT Class 12 Physics Chapter 4, page 122 derives this by choosing a rectangular Amperian loop with one long side inside the solenoid, parallel to the axis, and the opposite side outside. Three of the four sides contribute nothing: the two short sides run perpendicular to B, and the outside long side sits where the field is taken as zero. So ∮B·dl collapses to B·L for the inside segment alone. The current threading that loop is nLI — n turns per metre, over length L, each carrying I. Equating gives BL = μ₀nLI, and L cancels: B = μ₀nI.
Read what the cancellation means. B does not depend on L, and it does not depend on the solenoid's radius or on where inside you stand. The interior field is uniform — that is the solenoid's whole point, and why it is the standard laboratory source of uniform B.
Two conditions carry that result: the solenoid must be long (length much greater than radius) and tightly wound. Both are what make "field outside ≈ 0" legitimate. Near the open ends the field weakens to roughly half its central value; the μ₀nI result is an interior, well-away-from-the-ends statement.
Watch-out: if a stem gives turns and a length, compute n before touching μ₀. Writing B = μ₀NI is the single most common way this one-step question is lost.
Can you answer these Ampere's Law Solenoid MCQs?
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
MCQ 1Easy RecallPractice
In the expression B = μ₀nI for the field inside a long solenoid, the symbol n denotes which quantity?
Show answer and why every option is right or wrong
Answer: B. B is correct: n is turns per unit length, with SI unit m⁻¹, as stated with the solenoid formula in NCERT Class 12 Physics Chapter 4, page 122.
Why A is wrong: A is wrong because total turns N is a pure number; using it in place of n is the standard solenoid slip, and it leaves the right-hand side with the wrong dimensions.
Why C is wrong: C is wrong because the number of winding layers is not a quantity in the formula; a tightly wound solenoid's field is set by turns per metre however those turns are stacked.
Why D is wrong: D is wrong because n is a linear density (turns per metre), not an areal density; the derivation divides turns by the length of the Amperian loop, not by cross-section.
MCQ 2Direct ApplicationPractice
A long solenoid has 800 turns wound uniformly over a length of 40.0 cm and carries a current of 2.00 A. Taking μ₀ = 4π × 10⁻⁷ T·m/A, the magnitude of the magnetic field at its centre is closest to:
Show answer and why every option is right or wrong
Answer: B. B is correct: n = 800/0.400 m = 2.00 × 10³ m⁻¹, so B = (4π × 10⁻⁷)(2.00 × 10³)(2.00) = 5.03 × 10⁻³ T, applying the solenoid formula given in NCERT Class 12 Physics Chapter 4, page 122.
Why A is wrong: A is wrong because it divides by length twice, or equivalently uses n = 800 m⁻¹; check that 800 turns spread over 0.400 m gives 2.00 × 10³ turns per metre, not 800.
Why C is wrong: C is wrong because it uses the length 40.0 cm without converting to metres: n = 800/40.0 = 20.0 turns per centimetre used as turns per metre, which is 100 times too small.
Why D is wrong: D is wrong because it multiplies the turns by the length instead of dividing, n = 800 × 0.400 = 320 m⁻¹, giving 8.04 × 10⁻⁴ T.
MCQ 3Concept TrapPractice
Two long, tightly wound solenoids carry the same current. Solenoid P has radius 1.0 cm; solenoid Q has radius 3.0 cm. Both have the same number of turns per metre. How do the magnetic fields at their respective interior centres compare?
Show answer and why every option is right or wrong
Answer: D. D is correct: B = μ₀nI contains no radius term, so equal n and equal I give equal interior fields regardless of cross-section — a consequence of the length L cancelling in the Amperian-loop derivation, NCERT Class 12 Physics Chapter 4, page 122.
Why A is wrong: A is wrong because it imports a direct dependence on radius that the solenoid result does not contain; the radius enters only through the 'long solenoid' condition (length ≫ radius), not through the field magnitude.
Why B is wrong: B is wrong because it imports an inverse-radius dependence borrowed from a different geometry; nothing in B = μ₀nI falls off with radius.
Why C is wrong: C is wrong because it assumes a 1/radius² dependence; the interior field of a long solenoid is uniform in the cross-section and independent of it.
MCQ 4Direct ApplicationPractice
A long solenoid is to produce an interior field of 6.28 × 10⁻³ T while carrying a current of 2.50 A. Taking μ₀ = 4π × 10⁻⁷ T·m/A, how many turns per metre must it have?
Show answer and why every option is right or wrong
Answer: A. A is correct: rearranging the solenoid formula from NCERT Class 12 Physics Chapter 4, page 122 gives n = B/(μ₀I) = (6.28 × 10⁻³)/((4π × 10⁻⁷)(2.50)) = 2.00 × 10³ m⁻¹.
Why B is wrong: B is wrong because it divides by 2I rather than I, importing a spurious factor of 2 that belongs to a different geometry, not to B = μ₀nI.
Why C is wrong: C is wrong because it omits the current entirely, computing B/μ₀ = 5.00 × 10³; rearranging B = μ₀nI for n puts I in the denominator.
Why D is wrong: D is wrong because it divides by I² = 6.25 instead of by I, as if the current appeared twice in B = μ₀nI.
MCQ 5Easy RecallPractice
In the standard Amperian-loop derivation of the solenoid field, a rectangular loop is chosen with one long side inside the solenoid parallel to the axis. Which single assumption makes the opposite long side contribute zero to ∮B·dl?
Show answer and why every option is right or wrong
Answer: C. C is correct: the outside long side contributes nothing because the exterior field of a long solenoid is taken as zero, which is the assumption the derivation in NCERT Class 12 Physics Chapter 4, page 122 rests on.
Why A is wrong: A is wrong because steadiness is what licenses Ampere's law in its μ₀I_enc form at all; it does not by itself kill the outside segment's contribution.
Why B is wrong: B is wrong because the perpendicularity of the short sides is a real part of the derivation, but it eliminates those two sides, not the outer long side asked about here.
Why D is wrong: D is wrong because tight winding justifies treating the enclosed current as nLI with no gaps; the zero contribution of the outer side comes from the exterior field vanishing.
MCQ 6CalculationPractice
A long solenoid is wound with 1.20 × 10³ turns over a length of 60.0 cm and carries current I. The winding is then removed and re-wound, using the same wire and the same current, as 1.20 × 10³ turns over a length of 30.0 cm. By what factor does the interior field change?
Show answer and why every option is right or wrong
Answer: C. C is correct: n = N/L rises from 1.20 × 10³/0.600 = 2.00 × 10³ m⁻¹ to 1.20 × 10³/0.300 = 4.00 × 10³ m⁻¹, and since B = μ₀nI with I fixed, the field doubles — the formula is from NCERT Class 12 Physics Chapter 4, page 122.
Why A is wrong: A is wrong because it treats the field as depending on total turns N, which is unchanged here; B depends on N/L, and L has halved.
Why B is wrong: B is wrong because it inverts the dependence: shortening the solenoid packs the same turns more densely, raising n and therefore raising B.
Why D is wrong: D is wrong because B is linear in n, not quadratic; halving L doubles n and so doubles B.
MCQ 7Concept TrapPractice
A student measures the field along the axis of a real long solenoid and finds that near an open end it is about half the value measured deep inside. Which statement best accounts for this?
Show answer and why every option is right or wrong
Answer: B. B is correct: μ₀nI is an interior, far-from-the-ends result resting on the long-solenoid conditions stated with the formula in NCERT Class 12 Physics Chapter 4, page 122; near an end the field lines fan out and the idealisation fails.
Why A is wrong: A is wrong because the current is the same in every turn of a series winding; the end-effect is geometric, not a change in the current.
Why C is wrong: C is wrong because the end-weakening is a longitudinal effect along the axis; it says nothing about radial dependence, and the interior field remains radius-independent.
Why D is wrong: D is wrong because substituting total turns N for n is dimensionally wrong everywhere, at the ends as much as at the centre; it is not an end-correction.
MCQ 8CalculationPractice
Solenoid X has 600 turns over 30.0 cm and carries 3.00 A. Solenoid Y has 400 turns over 50.0 cm. What current must Y carry for its interior field to equal X's?
Show answer and why every option is right or wrong
Answer: C. C is correct: n_X = 600/0.300 = 2.00 × 10³ m⁻¹ and n_Y = 400/0.500 = 8.00 × 10² m⁻¹; equal fields require n_X I_X = n_Y I_Y, so I_Y = (2.00 × 10³)(3.00)/(8.00 × 10²) = 7.50 A, using the solenoid formula from NCERT Class 12 Physics Chapter 4, page 122.
Why A is wrong: A is wrong because it inverts the ratio, giving Y the smaller current when Y's sparser winding (lower n) demands a larger current for the same field.
Why B is wrong: B is wrong because it compares total turns (600 against 400) and ignores the lengths; the comparison must be made on n = N/L, not on N.
Why D is wrong: D is wrong because it assumes equal currents give equal fields; that holds only when the turns-per-metre also match, and here they differ by a factor of 2.50.
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Ampere's Law Solenoid: quick recall before you leave
How do you solve a Ampere's Law Solenoid question? A worked example
Pattern: long solenoid, plug in turns per metre and current (the topic's own PYQ pattern; years observed 2020, 2022).
- 1
Given
Total number of turns, N = 750 (exact, a count)
Length of solenoid, L = 25.0 cm = 0.250 m
Current, I = 1.60 A
μ₀ = 4π × 10⁻⁷ T·m/A (exact by definition in the convention used here)
The solenoid is long and tightly wound; the point is at the interior centre. - 2
Required
The magnitude of the magnetic field B at the centre of the solenoid.
- 3
Concept
Inside a long, tightly wound solenoid the field is uniform and parallel to the axis. Ampere's law applied to a rectangular loop with one long side inside and one outside gives a field that depends only on the turn density and the current — not on the solenoid's length, radius, or the position of the interior point.
- 4
Formula
B = μ₀nI, with n = N/L.
- 5
Substitution
First convert the length and form n:
n = N/L = 750/0.250 m = 3.00 × 10³ m⁻¹
Then
B = (4π × 10⁻⁷ T·m/A)(3.00 × 10³ m⁻¹)(1.60 A) - 6
Calculation
(4π × 10⁻⁷)(3.00 × 10³) = 3.770 × 10⁻³
(3.770 × 10⁻³)(1.60) = 6.032 × 10⁻³
Note on exact quantities: N = 750 is a count of turns, and μ₀ = 4π × 10⁻⁷ T·m/A is an exact defined constant in this convention. Neither limits significant figures. The measured inputs are L = 0.250 m and I = 1.60 A, each to three significant figures, so the answer is reported to three. - 7
Final answer
B = 6.03 × 10⁻³ T, directed along the solenoid's axis.
- 8
Common trap
Substituting the total turn count 750 in place of n. That gives B = (4π × 10⁻⁷)(750)(1.60) = 1.51 × 10⁻³ T — exactly one quarter of the correct value, because the length 0.250 m was never divided out. The tell is the stem quoting a length at all: in a one-step solenoid question the length exists to be divided into N. A second, cheaper version of the same error is dividing by 25.0 instead of 0.250, giving an answer 100 times too small.
- 9
Similar NEET-style question
A long solenoid of length 80.0 cm is wound with 2.40 × 10³ turns and carries 0.500 A. Find the interior field. *(n = 3.00 × 10³ m⁻¹; B = 1.88 × 10⁻³ T.)*
What to remember before solving Ampere's Law Solenoid questions
Which Ampere's Law Solenoid formulas do you need for NEET?
1 formula — click to collapse
B inside long solenoid
Uniform field inside long solenoid; n = turns per unit length. Approximately zero outside.
| Symbol | Quantity | SI Unit |
|---|---|---|
| n | turns per metre | 1/m |
| I | current | A |
Valid when
- Long solenoid (length >> radius)
- Tightly wound
Where do students lose marks on Ampere's Law Solenoid?
These are the exact patterns that cause wrong answers in NEET. Each trap includes when it triggers and how to avoid it.
1 item — click to collapse
Root cause: formula misuse
Correction
B = μ_0 n I where n = N/L (turns per metre). If N=100 over 50 cm, then n=200/m and B = μ_0 × 200 × I.
More in Magnetic Effects of Current and Magnetism: 2 exam traps and mistakes · 10 formulas · 4 question patterns from its other lessons.
Ampere's Law Solenoid questions from past NEET papers
2 questions from NEET 2020, 2022. Answers verified against NTA official keys. — click to collapse
NEET 2022
16.28 × 10⁻⁴ T
26.28 × 10⁻² T
312.56 × 10⁻² T
412.56 × 10⁻⁴ T
NTA Answer: Option 3(final)
NEET 2020
13.14×10⁻⁵ T
26.28×10⁻⁴ T
33.14×10⁻⁴ T
46.28×10⁻⁵ T
NTA Answer: Option 2(final)
All 17 past-paper questions from Magnetic Effects of Current and Magnetism →
How does NEET ask about Ampere's Law Solenoid?
1 recurring pattern from past papers — click to collapse
Long solenoid: B = mu_0 n I. Plug in turns/m and current.
Direct ApplicationEasy
Common distractors
uses total N not n
Forgets to divide by length
Sources
NCERT refs: Class 12 Physics Chapter 4, p.122
Page numbers are the ones printed in the current NCERT textbook (2023 rationalised edition), unless marked pre-2023. The books are free at ncert.nic.in.
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