Formula
Field of long straight wire
B = μ₀ I / (2π r), at perpendicular distance r. Direction: right-hand rule (thumb along I, fingers curl with B).
-- NCERT Class 12 Physics, Ch. 4, p. 120Ampere's Law Straight Wire
8 MCQs1 revision card9-step worked example
Source: NCERT Magnetic Effects of Current and MagnetismPYQ coverage: NEET 2022Official key: NTA-verifiedLast updated: 25 Sep 2026
Ampere's Law Straight Wire, explained for NEET
The single most common way this topic is lost is a π. You reach for the field near a straight wire and write B = μ₀I/(2r), because the loop-centre formula is sitting in the same corner of your memory. That is a factor of π — a wrong answer that still looks dimensionally healthy, so nothing warns you.
Fix the two shapes apart by what r means in each. For a long straight wire, B = μ₀I/(2πr), and r is the perpendicular distance from the wire to the point you care about — a distance you choose. For a circular loop at its centre, B = μ₀I/(2R), and R is the loop's own radius — a property of the wire, not of the point. If the number in the question describes where you are standing, you need the π. If it describes how big the loop is, you do not.
NCERT Class 12 Physics Chapter 4 derives the straight-wire result on page 120 by wrapping a circular Amperian loop of radius r around the wire. B is constant in magnitude along that circle and everywhere parallel to it, so ∮B·dl collapses to B(2πr), and setting that equal to μ₀I gives the 2π. The π is not decoration; it is the circumference of the path you integrated along.
Two consequences worth holding. B falls as 1/r, not 1/r² — double your distance and the field halves. And the field lines are circles concentric with the wire, so B at a point is perpendicular to both the wire and the line joining wire to point; right-hand grip rule fixes the sense.
Watch-out: "distance from the wire" in a NEET stem always means perpendicular distance. If a stem gives a slant distance or places the point off the perpendicular foot, resolve first — plugging the slant length straight into 2πr is the second-biggest scorer of negative marks on this pattern.
Can you answer these Ampere's Law Straight Wire 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 = μ₀I/(2πr) for the magnetic field near a long straight current-carrying wire, the symbol r denotes which quantity?
Show answer and why every option is right or wrong
Answer: B. B is correct. In NCERT Class 12 Physics Chapter 4 (page 120), r is the radius of the circular Amperian loop drawn around the wire, which equals the perpendicular distance from the wire to the point where B is evaluated.
Why A is wrong: A is wrong because the wire's own thickness plays no part in the external field; the formula treats the wire as a line current, and it is valid only for r greater than the wire radius.
Why C is wrong: C is wrong because the derivation assumes an effectively infinite wire — no finite length enters the result at all. A student who substitutes a given wire length here has confused the geometry of the source with the geometry of the field point.
Why D is wrong: D is wrong because the field lines are concentric circles whose radii are the distances of the field points themselves, so 'radius of curvature at the surface' is a restatement of the wire radius and shares the error in option A.
MCQ 2Direct ApplicationPractice
A long straight wire carries a steady current of 5.00 A. Taking μ₀ = 4π × 10⁻⁷ T·m/A, the magnitude of the magnetic field at a perpendicular distance of 2.00 × 10⁻¹ m from the wire is
Show answer and why every option is right or wrong
Answer: A. A is correct. B = μ₀I/(2πr) = (4π × 10⁻⁷ × 5.00)/(2π × 2.00 × 10⁻¹); the π cancels, leaving (2 × 10⁻⁷ × 5.00)/(2.00 × 10⁻¹) = 5.00 × 10⁻⁶ T, matching the straight-wire formula of NCERT Class 12 Physics Chapter 4, page 120.
Why B is wrong: B is wrong because it is π times the correct value, the result of dropping the π from the denominator and computing μ₀I/(2r) — the long-wire versus circular-loop formula confusion this topic turns on.
Why C is wrong: C is wrong because it is 2π times the correct value, obtained by dropping the whole 2π from the denominator and computing μ₀I/r.
Why D is wrong: D is wrong because it is twice the correct value, from omitting the 2 in the denominator and computing μ₀I/(πr).
MCQ 3Concept TrapPractice
Two points P and Q lie in the same plane near a long straight current-carrying wire, at perpendicular distances 3.0 cm and 9.0 cm from it. The ratio of the field magnitudes B_P : B_Q is
Show answer and why every option is right or wrong
Answer: D. D is correct. For a long straight wire B ∝ 1/r, so B_P/B_Q = r_Q/r_P = 9.0/3.0 = 3. The nearer point has the larger field. This inverse-first-power dependence is the defining feature of the straight-wire result in NCERT Class 12 Physics Chapter 4, page 120.
Why A is wrong: A is wrong because it inverts the dependence, assigning the larger field to the farther point; B decreases with distance, so the point at 3.0 cm must carry the larger field.
Why B is wrong: B is wrong because it combines two errors: an inverse-square dependence, B ∝ 1/r², and an inverted ratio that gives the farther point the larger field.
Why C is wrong: C is wrong because it applies an inverse-square law, B ∝ 1/r², giving (9.0/3.0)² = 9. The straight-wire field falls as 1/r — the inverse square belongs to the Biot–Savart contribution of a single element, not to the integrated infinite-wire result.
MCQ 4Direct ApplicationPractice
The magnetic field at a perpendicular distance of 0.50 m from a long straight wire is found to be 8.0 × 10⁻⁶ T. Taking μ₀ = 4π × 10⁻⁷ T·m/A, the current in the wire is
Show answer and why every option is right or wrong
Answer: C. C is correct. Rearranging B = μ₀I/(2πr) gives I = 2πrB/μ₀ = (2π × 0.50 × 8.0 × 10⁻⁶)/(4π × 10⁻⁷). The π cancels: I = (0.50 × 8.0 × 10⁻⁶)/(2 × 10⁻⁷) = 20 A, using the relation of NCERT Class 12 Physics Chapter 4, page 120.
Why A is wrong: A is wrong because it is half the correct value, from writing I = πrB/μ₀ — the 2 in the numerator of the rearranged formula was dropped.
Why B is wrong: B is wrong because it is twice the correct value, from using I = 4πrB/μ₀, i.e. carrying the 4π of μ₀ into the numerator as well as leaving it in the denominator.
Why D is wrong: D is wrong because it is the result of inverting the distance, computing I = 2πB/(μ₀r) with r in the denominator instead of the numerator.
MCQ 5Easy RecallPractice
At a point near a long straight current-carrying wire, the direction of the magnetic field is
Show answer and why every option is right or wrong
Answer: C. C is correct. The field lines of a long straight wire are concentric circles in planes perpendicular to the wire, so B at any point is tangential to that circle — perpendicular both to the wire and to the radial line from wire to point. Its sense follows the right-hand grip rule (NCERT Class 12 Physics Chapter 4, page 120).
Why A is wrong: A is wrong because a field component along the current direction would violate the cross product in the Biot–Savart integrand, in which dl × r̂ is perpendicular to dl.
Why B is wrong: B is wrong because it describes an electric field pattern around a line charge, not a magnetic one. Magnetic field lines around a wire close on themselves; they have no radial component.
Why D is wrong: D is wrong for the same reason as option A: the direction of the current fixes the sense of circulation, not the direction of B itself.
MCQ 6Concept TrapPractice
A student is given a long straight wire carrying current I and is told the field is required at a point 4.0 cm away. The student writes B = μ₀I/(2 × 0.040) and obtains an answer. The numerical result is
Show answer and why every option is right or wrong
Answer: B. B is correct. The student has written the circular-loop centre formula μ₀I/(2R) in place of the straight-wire formula μ₀I/(2πr). Dropping π from the denominator makes the answer π times too large. In the Ampère derivation of NCERT Class 12 Physics Chapter 4, page 120, the 2πr is the circumference of the path integrated along and cannot be replaced by 2r.
Why A is wrong: A is wrong because μ₀ is a fixed constant, 4π × 10⁻⁷ T·m/A, and cannot absorb an extra π; the two expressions differ in value by a factor of about 3.14.
Why C is wrong: C is wrong because it has the direction of the error backwards: a smaller denominator gives a larger quotient, so the student's answer is too large, not too small.
Why D is wrong: D is wrong because the error is purely in the magnitude; the student's expression carries no directional information at all, so direction is not what went astray.
MCQ 7CalculationPractice
A long straight wire carries a current I. At a perpendicular distance r from the wire the field is B₀. The current is then increased to 3I and the field is measured at a new perpendicular distance 6r. The new field magnitude is
Show answer and why every option is right or wrong
Answer: C. C is correct. B = μ₀I/(2πr), so B ∝ I/r. Multiplying the current by 3 and the distance by 6 scales the field by 3/6 = 1/2, giving B₀/2. Both dependences come directly from the straight-wire relation of NCERT Class 12 Physics Chapter 4, page 120.
Why A is wrong: A is wrong because it inverts the whole dependence, treating B as proportional to r/I: 6/3 = 2. It would require the field to grow as you move away from the wire.
Why B is wrong: B is wrong because it divides by both factors, 1/(3 × 6) = 1/18, putting the current increase in the denominator along with the distance; B grows with current, so B ∝ I/r gives 3/6.
Why D is wrong: D is wrong because it multiplies the two scaling factors, 3 × 6, treating a larger distance as strengthening the field. The distance factor belongs in the denominator.
MCQ 8CalculationPractice
Two long straight parallel wires, separated by 0.20 m, carry currents of 6.0 A and 2.0 A in the same direction. Taking μ₀ = 4π × 10⁻⁷ T·m/A, the net magnetic field magnitude at the midpoint between them is
Show answer and why every option is right or wrong
Answer: D. D is correct. Each wire's field at the midpoint is found from B = μ₀I/(2πr) with r = 0.10 m: B₁ = 1.2 × 10⁻⁵ T and B₂ = 4.0 × 10⁻⁶ T. Because the currents run the same way, their circular field lines pass the midpoint in opposite senses, so the fields subtract: 1.2 × 10⁻⁵ − 4.0 × 10⁻⁶ = 8.0 × 10⁻⁶ T. The circular-field-line geometry is the one established on page 120 of NCERT Class 12 Physics Chapter 4.
Why A is wrong: A is wrong because it adds the two magnitudes, 1.2 × 10⁻⁵ + 4.0 × 10⁻⁶, which would be correct only for antiparallel currents. For parallel currents the senses of circulation oppose at a point between the wires.
Why B is wrong: B is wrong because it is the contribution of the 6.0 A wire alone; the 2.0 A wire's field has been ignored rather than subtracted.
Why C is wrong: C is wrong because it uses the full separation r = 0.20 m for both wires instead of the half-separation: (6.0 − 2.0) × 2 × 10⁻⁷/0.20 = 4.0 × 10⁻⁶ T. At the midpoint both wires are 0.10 m away.
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Ampere's Law Straight Wire: quick recall before you leave
How do you solve a Ampere's Law Straight Wire question? A worked example
Pattern: P.PHY.U13.BIOT_SAVART_LONG_WIRE — field from a long straight wire (in-scope; topic_codes include PHY.U13.AMPERES_LAW_STRAIGHT_WIRE).
- 1
Given
• Current, I = 3.50 A (3 significant figures)• Perpendicular distance, r = 7.00 × 10⁻² m (3 significant figures)• μ₀ = 4π × 10⁻⁷ T·m/A (defined constant)
- 2
Required
• Magnetic field magnitude B at the given point.
- 3
Concept
The wire is long and straight, and the point is specified by its perpendicular distance from the wire. That is the Ampère straight-wire geometry: a circular Amperian loop of radius r, coaxial with the wire, along which B has constant magnitude and lies everywhere tangential. This is the configuration treated on page 120 of NCERT Class 12 Physics Chapter 4. It is not a loop-centre problem — 7.00 × 10⁻² m describes where the field point sits, not the size of any loop of wire.
- 4
Formula
B = μ₀I/(2πr)
- 5
Substitution
B = (4π × 10⁻⁷ × 3.50) / (2π × 7.00 × 10⁻²)
- 6
Calculation
The π in the numerator and denominator cancel:
B = (4 × 10⁻⁷ × 3.50) / (2 × 7.00 × 10⁻²)
B = (1.40 × 10⁻⁶) / (1.400 × 10⁻¹)
B = 1.00 × 10⁻⁵ T
Note on exact quantities: the 4 and the 2 in the formula, the π, and the defined constant μ₀ are exact — none of them limits the significant-figure count. The precision is set by the two measured quantities, I = 3.50 A and r = 7.00 × 10⁻² m, each carrying 3 significant figures. - 7
Final answer
B = 1.00 × 10⁻⁵ T, directed tangentially to the circle of radius 7.00 × 10⁻² m centred on the wire, with the sense given by the right-hand grip rule.
Written as 1.00 × 10⁻⁵ T rather than 0.00001 T so that the three significant figures are unambiguous. - 8
Common trap
Writing B = μ₀I/(2r) here gives 3.14 × 10⁻⁵ T — π times too large. The pull comes from the circular-loop centre formula μ₀I/(2R), which has no π. Before substituting, ask what the given length is: here 7.00 × 10⁻² m is a distance from the wire to a point you chose, so it is r and the π stays. Had the problem said "a circular loop of radius 7.00 × 10⁻² m, field at its centre," the same number would be R and the π would go.
- 9
Similar NEET-style question
A long straight wire carries a current of 12.0 A. At what perpendicular distance from the wire is the magnetic field magnitude 4.00 × 10⁻⁵ T? (Take μ₀ = 4π × 10⁻⁷ T·m/A.) Answer: 6.00 × 10⁻² m.
What to remember before solving Ampere's Law Straight Wire questions
Which Ampere's Law Straight Wire formulas do you need for NEET?
1 formula — click to collapse
B near long straight wire
Field at perpendicular distance r from infinite straight wire carrying I.
| Symbol | Quantity | SI Unit |
|---|---|---|
| B | field | T |
| I | current | A |
| r | perp distance | m |
Valid when
- Wire >> r (effectively infinite)
- r > wire radius
Where do students lose marks on Ampere's Law Straight Wire?
These are the exact patterns that cause wrong answers in NEET. Each trap includes when it triggers and how to avoid it.
2 items — click to collapse
Category: Similar Terms
Student uses μ_0 I/(2πr) (straight wire) when asked about circular loop centre (μ_0 I/(2R)) — or vice versa.
When it triggers
Magnetic field magnitude question with specific geometry (straight wire, loop, solenoid).
How to avoid
Straight wire: B = μ_0 I/(2πr) at perpendicular distance r. Circular loop centre: B = μ_0 I/(2R). Solenoid: B = μ_0 nI. Match formula to geometry.
Root cause: formula misuse
Correction
Straight wire (long, infinite): B = μ_0 I/(2πr). Circular loop, centre: B = μ_0 I/(2R). The 'π' is in straight-wire; missing in loop.
More in Magnetic Effects of Current and Magnetism: 1 exam trap or mistake · 10 formulas · 4 question patterns from its other lessons.
Ampere's Law Straight Wire questions from past NEET papers
1 question from NEET 2022. Answers verified against NTA official keys. — click to collapse
NEET 2022
1A linearly decreasing function of distance upto the boundary of the wire and then a linearly increasing one for the outside region.
2Uniform and remains constant for both the regions.
3A linearly increasing function of distance upto the boundary of the wire and then linearly decreasing for the outside region.
4A linearly increasing function of distance r upto the boundary of the wire and then decreasing one with 1 r dependence for the outside region.
NTA Answer: Option 4(final)
All 17 past-paper questions from Magnetic Effects of Current and Magnetism →
How does NEET ask about Ampere's Law Straight Wire?
1 recurring pattern from past papers — click to collapse
Field from long straight wire: B = mu_0 I/(2 pi r).
Direct ApplicationEasy
Common distractors
forgets pi in denominator
Confuses with circular loop formula
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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