Line integral of B along a closed loop equals μ₀ × current enclosed: ∮ B · dl = μ₀ I_enc. Powerful for fields with high symmetry.
-- NCERT Class 12 Physics, Ch. 4, p. 118Ampere's Law
8 MCQs3 revision cards9-step worked example
Source: NCERT Magnetic Effects of Current and MagnetismOfficial key: NTA-verifiedLast updated: 23 Sep 2026
Ampere's Law, explained for NEET
The word that decides most Ampere's-law questions is enclosed. The law reads ∮B·dl = μ₀I_enc, and I_enc counts only the current that pierces the surface bounded by your chosen loop. A wire sitting outside the loop still produces a magnetic field at every point on that loop — it simply contributes nothing to the integral around the closed path. Zero circulation is not zero field.
NCERT Class 12 Physics Chapter 4 states the law on page 118 as the magnetic analogue of Gauss's law: an integral statement relating a field's circulation to its source. Both share the same structure and the same limitation — the law is always true for steady currents, but it only yields B when you can find a loop on which B has constant magnitude and a fixed angle to dl. That requires symmetry. Without it, the equation holds and tells you nothing useful.
Two conditions of use travel with the law. First, the current must be steady; a time-varying electric flux requires the Maxwell–Ampere extension with a displacement-current term. Second, the path must be closed — an open arc has no defined circulation.
Signs matter. Currents through the loop in opposite directions subtract. Two antiparallel currents of equal magnitude threading the same loop give I_enc = 0, so ∮B·dl = 0 even though neither wire is off. The right-hand rule fixes which sense counts as positive: curl the fingers along the chosen direction of traversal, and the thumb points along positive current.
In NEET this appears as reasoning rather than arithmetic — which loop, which current, what the integral evaluates to. The applications to a straight wire and to a solenoid are separate topics in this unit.
Watch out: a loop enclosing no current gives a zero line integral, never a zero field.
Can you answer these Ampere's Law 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 Ampere's circuital law, ∮B·dl = μ₀I_enc, the quantity I_enc refers to which current?
Show answer and why every option is right or wrong
Answer: B. B is correct. As stated in NCERT Class 12 Physics Chapter 4, page 118, I_enc is the net current threading the surface bounded by the closed path — enclosure, not proximity or strength, is the criterion.
Why A is wrong: A is wrong because currents outside the loop contribute nothing to the circulation; only currents that pierce the bounded surface are counted.
Why C is wrong: C is wrong because distance from the loop plays no part in the law; a distant enclosed wire counts fully while a near external wire counts not at all.
Why D is wrong: D is wrong because field strength on the loop is irrelevant to I_enc; an external wire may dominate the field yet contribute zero to the integral.
MCQ 2Concept TrapPractice
A closed Amperian loop is drawn in a region where a long current-carrying wire lies entirely outside the loop. Which statement is correct?
Show answer and why every option is right or wrong
Answer: C. C is correct. The external wire produces a magnetic field at every point of the loop, but since I_enc = 0 the circulation vanishes — the positive and negative contributions along the path cancel exactly (NCERT Class 12 Physics Chapter 4, page 118).
Why A is wrong: A is wrong because it assumes a zero line integral implies a zero field; the wire's field is present on the loop regardless of where the loop is drawn.
Why B is wrong: B is wrong because it inverts the situation — the field is non-zero and the integral is zero, not the reverse.
Why D is wrong: D is wrong because I_enc = 0 forces the circulation to zero by the law, even though the field itself is non-zero.
MCQ 3Direct ApplicationPractice
Three straight wires pass perpendicularly through the plane of a closed Amperian loop. Two carry 4.0 A each in the same sense relative to the loop's traversal, and the third carries 6.0 A in the opposite sense. A fourth wire carrying 9.0 A lies outside the loop. The value of ∮B·dl is:
Show answer and why every option is right or wrong
Answer: A. A is correct. I_enc = 4.0 + 4.0 − 6.0 = 2.0 A; the 9.0 A wire is outside and is excluded. Applying ∮B·dl = μ₀I_enc gives μ₀ × 2.0 A (NCERT Class 12 Physics Chapter 4, page 118).
Why B is wrong: B is wrong because it sums every current in the diagram (4.0 + 4.0 + 6.0 + 9.0) — it both ignores the sign reversal and wrongly includes the external wire.
Why C is wrong: C is wrong because it adds all three enclosed currents as positive (4.0 + 4.0 + 6.0), treating the opposing current as though direction did not matter.
Why D is wrong: D is wrong because it takes the correct enclosed net of 2.0 A but then adds the external 9.0 A, which contributes nothing to the circulation.
MCQ 4Concept TrapPractice
Ampere's circuital law is valid for any closed path around steady currents, yet it is useful for calculating B only in certain situations. The reason it can fail as a calculation tool is that:
Show answer and why every option is right or wrong
Answer: B. B is correct. The law remains true always for steady currents; extracting B requires a loop on which B has constant magnitude and a fixed angle to dl, which only symmetry supplies (NCERT Class 12 Physics Chapter 4, page 118).
Why A is wrong: A is wrong because the law's validity never depends on symmetry — only its usefulness as a solving tool does. It holds for every closed path around steady currents.
Why C is wrong: C is wrong because μ₀ is a constant of free space and has nothing to do with the geometry of the current distribution.
Why D is wrong: D is wrong because I_enc is perfectly well defined for any distribution; the difficulty lies in evaluating the line integral, not in counting the enclosed current.
MCQ 5Easy RecallPractice
Ampere's circuital law in the form ∮B·dl = μ₀I_enc requires which condition on the current?
Show answer and why every option is right or wrong
Answer: A. A is correct. The stated condition of use is a steady current; where a time-varying electric flux is present, the Maxwell–Ampere form with a displacement-current term is required (NCERT Class 12 Physics Chapter 4, page 118).
Why B is wrong: B is wrong because the law applies to any steady current distribution whatsoever, including loops, sheets and bundles of conductors.
Why C is wrong: C is wrong because the cross-sectional shape of the conductor is not a condition of the law at all.
Why D is wrong: D is wrong because it states the opposite of the requirement — an alternating current is precisely the case where the plain form is insufficient.
MCQ 6CalculationPractice
Two long parallel wires, each carrying current I in the same direction, pass perpendicularly through the plane of the page. A closed loop L₁ encloses both wires; a second closed loop L₂ encloses only one of them. The ratio of the circulation ∮B·dl around L₁ to that around L₂ is:
Show answer and why every option is right or wrong
Answer: D. D is correct. Circulation depends only on I_enc: for L₁, I_enc = 2I; for L₂, I_enc = I. The ratio is therefore 2 : 1 regardless of loop size or shape (NCERT Class 12 Physics Chapter 4, page 118).
Why A is wrong: A is wrong because it uses the field on the loop rather than the enclosed current; both wires do produce field on L₂, but only the enclosed one contributes to its circulation.
Why B is wrong: B is wrong because the circulation has no dependence on loop size — a larger or smaller L₂ around the same single wire gives the same value.
Why C is wrong: C is wrong because the law is linear in I_enc, not quadratic; doubling the enclosed current doubles the circulation.
MCQ 7CalculationPractice
A closed Amperian loop is traversed in a chosen sense. Two wires thread it: wire P carries 5.0 A and wire Q carries 5.0 A in the opposite direction through the loop. A student concludes that the magnetic field is zero at every point on the loop. The correct assessment is:
Show answer and why every option is right or wrong
Answer: C. C is correct. I_enc = 5.0 − 5.0 = 0, so the circulation vanishes; but each wire separately produces a field on the loop and these cancel only at special points, not everywhere (NCERT Class 12 Physics Chapter 4, page 118).
Why A is wrong: A is wrong because a vanishing line integral only means the positive and negative contributions around the path cancel — it places no constraint on B at any individual point.
Why B is wrong: B is wrong because it adds the antiparallel currents as magnitudes; opposite senses through the loop subtract, giving I_enc = 0, not 10 A.
Why D is wrong: D is wrong because equal and opposite currents in separated wires produce a non-zero net field almost everywhere; only the circulation, not the field, is forced to zero.
MCQ 8Direct ApplicationPractice
A closed Amperian loop is traversed in a given sense, and the enclosed current is found to be I_enc. If the direction of traversal around the same loop is reversed while the currents are unchanged, the value of ∮B·dl becomes:
Show answer and why every option is right or wrong
Answer: B. B is correct. The right-hand rule ties the positive current sense to the direction of traversal; reversing traversal flips dl everywhere and hence the sign of the integral (NCERT Class 12 Physics Chapter 4, page 118).
Why A is wrong: A is wrong because the two traversals are alternative single evaluations, not simultaneous ones — nothing cancels within a single circulation.
Why C is wrong: C is wrong because it treats the sign of I_enc as absolute; the magnitude of the enclosed current is fixed, but its sign is set by the chosen sense of traversal.
Why D is wrong: D is wrong because reversing direction does not halve the path; every element dl is reversed, and all of them still contribute.
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Ampere's Law: quick recall before you leave
How do you solve a Ampere's Law question? A worked example
- 1
Given
A closed Amperian loop is drawn in a plane. Four long straight wires pass perpendicularly through that plane:• Wire 1: I₁ = 8.0 A, threading the loop in the positive sense (right-hand rule relative to the chosen traversal)• Wire 2: I₂ = 3.0 A, threading the loop in the positive sense• Wire 3: I₃ = 5.0 A, threading the loop in the negative sense• Wire 4: I₄ = 7.0 A, lying entirely outside the loop
μ₀ = 4π × 10⁻⁷ T·m/A (defined constant in this problem) - 2
Required
The value of ∮B·dl around the loop.
- 3
Concept
Ampere's circuital law relates the circulation of B around a closed path to the net current threading the surface bounded by that path. Two decisions do the whole job: which wires are enclosed, and what sign each enclosed current carries. Wire 4 is outside — it produces a field at every point of the loop but contributes nothing to the circulation. Wire 3 threads the loop opposite to wires 1 and 2, so it enters the sum with a minus sign.
- 4
Formula
∮B·dl = μ₀ I_enc, where I_enc is the signed sum of currents threading the bounded surface.
- 5
Substitution
I_enc = I₁ + I₂ − I₃ = 8.0 A + 3.0 A − 5.0 A
(Wire 4 is excluded: it is not enclosed.)
∮B·dl = (4π × 10⁻⁷ T·m/A) × I_enc - 6
Calculation
I_enc = 8.0 + 3.0 − 5.0 = 6.0 A
∮B·dl = 4π × 10⁻⁷ × 6.0 = 24π × 10⁻⁷ = 7.5398... × 10⁻⁶ T·m
The factor 4π in μ₀ is exact by definition here, as is π itself — mathematical constants and defined values do not contribute to the significant-figure count. The three given currents each carry two significant figures, and their sum 6.0 A is likewise quoted to two, so the answer is reported to two significant figures. - 7
Final answer
∮B·dl = 7.5 × 10⁻⁶ T·m (equivalently 24π × 10⁻⁷ T·m)
- 8
Common trap
The instinct is to add every current that appears in the figure — 8.0 + 3.0 + 5.0 + 7.0 = 23 A — because all four wires are "there." Two separate errors hide in that sum. First, wire 4 is outside the loop: its field is real on the path, but its circulation contribution is exactly zero. Second, wire 3 threads the loop in the opposite sense, so it subtracts rather than adds. Read the geometry before reaching for arithmetic: enclosed or not, and if enclosed, with which sign.
A related slip is concluding that because ∮B·dl would be zero for a loop with I_enc = 0, the field on that loop is zero too. It is not. The circulation being zero says the contributions cancel around the path; it says nothing about B at any individual point. - 9
Similar NEET-style question
A closed loop in a plane is threaded by two long wires carrying 12 A and 4.0 A in opposite senses through the loop, while a third wire carrying 20 A passes outside it. If the direction of traversal around the loop is then reversed, the value of ∮B·dl is:
(A) μ₀ × 36 A (B) μ₀ × 8.0 A (C) −μ₀ × 8.0 A (D) zero
What to remember before solving Ampere's Law questions
Which Ampere's Law formulas do you need for NEET?
1 formula — click to collapse
Ampere's circuital law
Line integral of B around closed loop = mu0 x current enclosed.
| Symbol | Quantity | SI Unit |
|---|---|---|
| I_enc | current enclosed | A |
| mu0 | permeability | T*m/A |
Valid when
- Steady current (Maxwell-Ampere needed if displacement current present)
- Closed Amperian loop
More in Magnetic Effects of Current and Magnetism: 3 exam traps and mistakes · 10 formulas · 5 question patterns from its other lessons.
Ampere's Law questions from past NEET papers
No question in our NEET 2020–2025 set targets this topic directly.
All 17 past-paper questions from Magnetic Effects of Current and Magnetism →
Sources
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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