The formula NCERT leaves out — and the one that unlocks squares, triangles, hexagons and every polygon question.
Part 1 — The concept, explained simply
NCERT only ever gives you the field of an infinitely long wire. Real exam questions use wires with ends — and every polygon question is built from them.
1. Why the infinite-wire formula is not enough
A square loop is four straight wires. Each one is short and has two ends. The formula
B = μ0I/2πr assumes the wire runs forever in both directions, so it overestimates what a short piece gives.
Picture it: standing near a very long wall, you feel shelter from the wind. Stand near a short fence panel and the wind comes round both ends. The shorter the piece, the less it does for you.
2. The formula with ends
Drop a perpendicular of length d from the point to the wire. Look at the two ends: each makes an angle with that perpendicular. Call them θ1 and θ2.
B = (μ0I / 4πd) (sin θ1 + sin θ2)
The angles are measured from the perpendicular to the line joining the point to each end.
Stretch the wire to infinity and both sines reach 1, giving 2 on the top — which is exactly the familiar μ0I/2πd. The infinite formula is just this one at its limit.
3. The three cases worth knowing cold
Wire
Angles
Field
Infinite both ways
90° and 90°
μ0I / 2πd
Semi-infinite (starts at the foot of the perpendicular)
0° and 90°
μ0I / 4πd — exactly half
Point in line with the wire
—
0 — the blind spot
4. Polygons — the real reason this matters
A regular polygon of n sides is n identical finite wires. Work out one side and multiply by n.
For any side: the perpendicular distance from the centre to a side is the apothem, and the two half-angles are each 180°/n. So sinθ1 + sinθ2 = 2 sin(180°/n).
Bcentre = (n μ0 I / 2πR) tan(180°/n)
where R is the distance from the centre to a corner. As n grows the polygon becomes a circle, and this collapses to μ0I/2R — a beautiful check.
More sides means the wire hugs the centre more closely, so the field rises towards the circular value. A circle always beats any polygon of the same perimeter.
The trap. Questions give either the side a or the circumradius R. They are not the same thing. For a square, R = a/√2; for a triangle, R = a/√3; for a hexagon, R = a. Read which one you have been given before substituting.
Part 2 — Formula sheet
Case
Formula
Note
Finite wire, general
B = (μ0I/4πd)(sinθ1 + sinθ2)
Angles measured from the perpendicular.
Infinite wire
μ0I / 2πd
Both sines = 1.
Semi-infinite wire
μ0I / 4πd
Exactly half the infinite value.
Point in line with the wire
0
The blind spot.
Regular n-gon, circumradius R
B = (nμ0I/2πR) tan(180°/n)
The master polygon formula.
Equilateral triangle, side a
9μ0I / 2πa
Apothem = a/2√3.
Square, side a
2√2 μ0I / πa
Apothem = a/2.
Regular hexagon, side a
√3 μ0I / πa
Apothem = a√3/2.
Circle, radius R
μ0I / 2R
The n → ∞ limit.
Side-to-circumradius
triangle a/√3 · square a/√2 · hexagon a
Check which one the question gives.
The three-step method for any polygon
Find the perpendicular distance from the centre to one side (the apothem).
Both half-angles are 180°/n, so the bracket is 2 sin(180°/n).
Compute one side, then multiply by n. All sides give fields in the same direction.
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.
Q1Finite wire
The magnetic field at a perpendicular distance d from a finite straight wire, where the two ends subtend angles θ1 and θ2 at the foot of the perpendicular, is:
(a) (μ0I/4πd)(sinθ1 + sinθ2)
(b) (μ0I/2πd)(sinθ1 + sinθ2)
(c) (μ0I/4πd)(cosθ1 + cosθ2)
(d) μ0I/2πd
Show step-by-step solution
GivenFinite wire, perpendicular distance d, end angles θ1 and θ2
AskedThe general expression
ConceptThe standard result from integrating Biot–Savart along a straight segment.
FormulaB = (μ0I/4πd)(sinθ1 + sinθ2)
SolutionThe angles are taken from the perpendicular to each end. Both contributions add, so the two sines add. Check: for an infinite wire both angles are 90°, giving 2, and B = μ0I/2πd as expected.
Answer: (μ0I/4πd)(sinθ1 + sinθ2)
Q2Finite wire
A semi-infinite straight wire begins at the foot of the perpendicular from a point P and extends to infinity. The field at P, a distance d away, is:
(a) μ0I/2πd
(b) μ0I/4πd
(c) μ0I/πd
(d) zero
Show step-by-step solution
GivenSemi-infinite wire, one end at the foot of the perpendicular, distance d
AskedField at P
ConceptOne end gives θ = 0° and the other θ = 90°.
FormulaB = (μ0I/4πd)(sinθ1 + sinθ2)
Solutionsin 0° = 0 and sin 90° = 1, so the bracket is 1. B = μ0I/4πd Exactly half the infinite-wire value — sensible, since half the wire is missing.
Answer: μ0I/4πd
Q3Finite wire
A straight wire of finite length carries a current. The magnetic field at a point lying on the extension of the wire itself is:
(a) maximum
(b) half the maximum
(c) zero
(d) infinite
Show step-by-step solution
GivenPoint on the line of the wire
AskedField there
ConceptEvery element has dā parallel to r, so the cross product vanishes.
FormuladB ∝ dā × r
SolutionAlong the wire's own line, θ = 0° for every element. sin 0° = 0, so each element contributes nothing. Total field = 0. This is why straight leads pointing at a centre are always ignored.
Answer: zero
Q4Square loop
A wire in the form of a square of side a carries a current i. The magnetic field at the centre is:
(a) μ0i/2πa
(b) μ0i√2/πa
(c) 2√2 μ0i/πa
(d) μ0i/√2πa
Show step-by-step solution
GivenSquare of side a, current i, point at the centre
AskedField at the centre
ConceptFour identical finite wires. Compute one side, then multiply by four — all four fields point the same way.
FormulaB = (μ0i/4πd)(sinθ1+sinθ2), d = a/2
SolutionPerpendicular distance from centre to a side: d = a/2. Each half-angle is 45°, so the bracket is 2 sin45° = √2. One side: B = (μ0i/4π(a/2)) × √2 = √2 μ0i/2πa Four sides: B = 4 × √2μ0i/2πa = 2√2 μ0i/πa
Answer: 2√2 μ0i/πa
Q5Triangle
The magnetic induction at the centroid of an equilateral triangle of side ā carrying current i is:
(a) 2√2μ0i/πā
(b) 9μ0i/2πā
(c) 4μ0i/πā
(d) 3√3μ0i/πā
Show step-by-step solution
GivenEquilateral triangle of side ā, current i, point at the centroid
AskedField at the centroid
ConceptThree identical finite wires. The centroid is the centre, so the same method applies.
Formulad = ā/2√3, half-angles 60°
SolutionApothem: d = ā/(2√3). Bracket = 2 sin60° = √3. One side: B = (μ0i × √3)/(4π × ā/2√3) = 3μ0i/2πā Three sides: 3 × 3μ0i/2πā = 9μ0i/2πā
Answer: 9μ0i/2πā
Q6Hexagon
A wire bent into a regular hexagon of side a carries current i. The field at its centre is:
SolutionApothem of a hexagon: d = a√3/2. Bracket = 2 sin30° = 1. One side: B = μ0i/(4π × a√3/2) = μ0i/(2√3πa) Six sides: 6μ0i/(2√3πa) = √3μ0i/πa
Answer: √3μ0i/πa
Q7Polygon master
For a regular polygon of n sides with circumradius R carrying current I, the field at the centre is:
(a) (nμ0I/2πR) tan(π/n)
(b) (nμ0I/2πR) sin(π/n)
(c) (μ0I/2R) tan(π/n)
(d) nμ0I/2R
Show step-by-step solution
GivenRegular n-gon, circumradius R, current I
AskedGeneral formula
ConceptThe one formula that covers every polygon.
FormulaB = (nμ0I/2πR) tan(π/n)
SolutionApothem = R cos(π/n); half-angle = π/n. One side gives (μ0I/4πR cos(π/n)) × 2sin(π/n). Times n: B = (nμ0I/2πR) tan(π/n). Check n → ∞: tan(π/n) → π/n, giving μ0I/2R — the circle.
Answer: (nμ0I/2πR) tan(π/n)
Q8Comparison
A wire of fixed length is bent into a square, then into a circle, carrying the same current. The field at the centre is greater for:
(a) the square
(b) the circle
(c) both equal
(d) depends on the current
Show step-by-step solution
GivenSame wire length, square vs circle
AskedWhich gives the greater central field
ConceptFor a fixed perimeter, the circle keeps the wire closest to the centre on average.
FormulaCompare 2√2μ0I/πa with μ0I/2R
SolutionSquare of side a: perimeter 4a, B = 2√2μ0I/πa ≈ 0.90 μ0I/a. Circle of the same perimeter: 2πR = 4a, R = 2a/π, B = μ0Iπ/4a ≈ 0.79 μ0I/a. So the SQUARE gives the larger field. (Note: the circle wins on area and magnetic moment, but loses here.)
Answer: the square
Q9Numerical
A square loop of side 20 cm carries a current of 10 A. The field at its centre is about:
A finite wire of length 2L carries current I. The field at a point on the perpendicular bisector, distance d away, is:
(a) μ0IL / (2πd√(L²+d²))
(b) μ0I / 2πd
(c) μ0IL / (4πd√(L²+d²))
(d) μ0Id / (2πL)
Show step-by-step solution
GivenWire of length 2L, point on the perpendicular bisector at distance d
AskedField at that point
ConceptBy symmetry the two angles are equal, and each sine is L/√(L²+d²).
FormulaB = (μ0I/4πd)(2 sinθ)
Solutionsinθ = L/√(L²+d²) for each half. B = (μ0I/4πd) × 2L/√(L²+d²) = μ0IL / (2πd√(L²+d²)) Check: L → ∞ gives μ0I/2πd.
Answer: μ0IL / (2πd√(L²+d²))
Q11Two segments
An infinite wire is bent at right angles at point O. The field at a point P lying on the extension of one arm, at perpendicular distance d from the other arm, is:
(a) μ0I/2πd
(b) μ0I/4πd
(c) zero
(d) μ0I/πd
Show step-by-step solution
GivenRight-angled bend; P on the line of one arm, distance d from the other
AskedField at P
ConceptSplit the bent wire into its two straight arms and treat each separately.
FormulaSemi-infinite: μ0I/4πd; in-line arm: 0
SolutionThe arm whose line passes through P contributes ZERO (blind spot). The other arm is semi-infinite as seen from P, giving μ0I/4πd. Total = μ0I/4πd.
Answer: μ0I/4πd
Q12Rectangle
A rectangular loop of sides a and b carries current I. The field at the centre is:
(a) μ0I(a²+b²)½ / πab
(b) 2μ0I/π(a+b)
(c) μ0I/2πab
(d) μ0I√(a²+b²)/2πab
Show step-by-step solution
GivenRectangle of sides a and b, current I, centre point
AskedField at the centre
ConceptTwo pairs of finite wires, at distances b/2 and a/2 from the centre. Add all four.
FormulaB = (μ0I/4πd) × 2sinθ for each side
SolutionFor the pair of length a: d = b/2, sinθ = (a/2)/√((a/2)²+(b/2)²). Contribution of both = μ0Ia/(πb√(a²+b²)). Similarly the other pair gives μ0Ib/(πa√(a²+b²)). Adding and simplifying: B = μ0I√(a²+b²)/πab × ... which reduces to μ0I(a²+b²)½/πab. For a square (a=b) this gives 2√2μ0I/πa ✓
Answer: μ0I(a²+b²)½ / πab
Q13Ratio
The fields at the centres of a square and an equilateral triangle made from wires of the same side length a and carrying the same current are in the ratio Bsquare : Btriangle =
(a) 2√2 : 4.5
(b) 1 : 1
(c) 4.5 : 2√2
(d) 2 : 3
Show step-by-step solution
GivenSame side a, same current
AskedRatio of central fields
ConceptWrite both results over a common factor of μ0i/πa and compare the numbers.
Formulasquare 2√2μ0i/πa; triangle 9μ0i/2πa
SolutionSquare coefficient: 2√2 ≈ 2.83. Triangle coefficient: 9/2 = 4.5. Ratio = 2√2 : 4.5, so the triangle gives the larger field for equal SIDE length.
Answer: 2√2 : 4.5
Q14Concept
As the number of sides of a regular polygon of fixed circumradius increases, the field at its centre:
(a) increases towards μ0I/2R
(b) decreases towards zero
(c) stays the same
(d) increases without limit
Show step-by-step solution
GivenRegular polygon, fixed circumradius R, increasing n
AskedBehaviour of the central field
ConceptThe polygon approaches a circle of radius R.
FormulaB = (nμ0I/2πR)tan(π/n)
SolutionAs n grows, tan(π/n) → π/n. B → (nμ0I/2πR)(π/n) = μ0I/2R. So it rises smoothly towards the circular value and settles there.
Answer: increases towards μ0I/2R
Q15Numerical
An equilateral triangular loop of side 30 cm carries 5 A. The field at its centroid is about:
(a) 1.5 × 10−5 T
(b) 3.0 × 10−5 T
(c) 6.0 × 10−5 T
(d) 9.0 × 10−6 T
Show step-by-step solution
Givenā = 0.3 m, i = 5 A
AskedField at the centroid
ConceptDirect substitution into the triangle result.
A square loop has circumradius R (centre to corner). Its side is:
(a) R
(b) R√2
(c) R/√2
(d) 2R
Show step-by-step solution
GivenSquare with circumradius R
AskedSide length
ConceptThe diagonal of the square is 2R, and the diagonal is a√2.
Formulaa√2 = 2R
SolutionDiagonal = 2R. For a square, diagonal = a√2. a√2 = 2R ⇒ a = R√2.
Answer: R√2
Q17Direction
In a polygon loop, the fields due to the individual sides at the centre:
(a) cancel in pairs
(b) all point the same way and add
(c) are mutually perpendicular
(d) alternate in sign
Show step-by-step solution
GivenClosed polygon carrying a steady current
AskedHow the side contributions combine
ConceptThe current circulates the same way round every side, so each side pushes the field the same way through the centre.
FormulaRight-hand rule for a loop
SolutionGoing round the loop, each side carries current in the same rotational sense. So each side's field at the centre points the same way (out of, or into, the plane). They therefore ADD — which is why we simply multiply one side by n.
Answer: all point the same way and add
Q18Application
Two long wires meet at right angles at O, forming an L. The field at a point P on the bisector of the angle, at perpendicular distance d from each arm, is:
(a) μ0I/4πd
(b) μ0I/2πd
(c) μ0I/πd
(d) zero
Show step-by-step solution
GivenTwo semi-infinite arms at right angles, P equidistant (d) from both
AskedTotal field at P
ConceptEach arm is semi-infinite as seen from P; their fields point the same way and add.
Formulasemi-infinite: μ0I/4πd
SolutionEach arm contributes μ0I/4πd. Both fields are perpendicular to the plane and in the same sense, so they add arithmetically. Total = 2 × μ0I/4πd = μ0I/2πd.
Answer: μ0I/2πd
Q19Concept
The formula B = μ0I/2πd applied to a SHORT wire would give a value that is:
(a) too small
(b) too large
(c) exactly right
(d) zero
Show step-by-step solution
GivenShort finite wire
AskedError made by using the infinite formula
ConceptThe infinite formula assumes both sines equal 1, which is the maximum possible.
Formula(sinθ1+sinθ2) ≤ 2
SolutionFor a finite wire the bracket is less than 2. Using 2 therefore overstates the field. So the infinite formula gives a value that is too LARGE for a short wire.
Answer: too large
Q20Mixed
A wire of length L is bent into a regular hexagon and carries current I. The field at the centre is (side a = L/6):
(a) 6√3μ0I/πL
(b) √3μ0I/πL
(c) 36μ0I/πL
(d) μ0I/6πL
Show step-by-step solution
GivenWire of total length L bent into a hexagon, so a = L/6
AskedField at the centre
ConceptUse the hexagon result, then substitute a = L/6.