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NEET Physics · Chapter 4 gap pack 7 of 8

A Loop Tied in a Straight Wire

The (π + 1) and (π − 1) family — and the near-identical figure where the current divides instead, giving zero.

Part 1 — The concept, explained simply

A long wire with a loop tied in it, like a knot in a rope. Two familiar pieces sitting on top of each other — the only question is whether they add or fight.

1. What the figure actually shows

A long straight wire runs across the page. Somewhere in the middle it is twisted into a full circle of radius r, then continues straight on. The current flows all the way through: along the wire, round the loop, and out the other side.

Picture it: tie a loop in a skipping rope and lay it flat. The rope is still one continuous rope — it just goes round once on the way past.

2. Two pieces, both at distance r from the centre

The centre of the loop is the field point. Notice that the straight wire also passes at perpendicular distance r from it — because the wire is a tangent to the circle.

PieceField at the centreRewritten over 2πr
The full circular loopμ0i / 2rμ0iπ / 2πr
The long straight wireμ0i / 2πrμ0i / 2πr
The trick that makes the answer obvious. Put both over the same denominator 2πr. Then the loop counts as π and the wire counts as 1. The answer is always μ0i(π ± 1)/2πr — and the only decision left is the sign.

3. Deciding the sign

Use the grip rule on the straight wire and the loop rule on the circle, and see whether both fields point out of the page or in opposite senses.

O r same sense → ADD → (π + 1) opposite → SUBTRACT → (π − 1) the wire is a tangent, so it too is a distance r from O

Both answers have the same shape. Since π > 1, the loop always wins, so the result is never zero and never negative.

4. The two standard results

Same sense:   B = μ0i(π + 1) / 2πr Opposite sense:   B = μ0i(π − 1) / 2πr

5. The variant that catches people out

Sometimes the straight wire is split at the loop, so the current divides: part goes round one way, part the other. That is a completely different question — it becomes an arc-division problem where the two halves oppose each other. Read the figure carefully:

Two figures that look alike:
(a) Loop tied in the wire — the full current i flows round the whole circle AND continues along the wire. Answer: μ0i(π ± 1)/2πr.
(b) Wire splits into two arcs — the current divides between them and they circulate in opposite senses, so they partly or fully cancel. Answer: often zero.
Check whether the current divides or goes round once and carries on.

Part 2 — Formula sheet

ConfigurationField at the centreNote
Full loop aloneμ0i/2r = μ0iπ/2πrCounts as π in units of μ0i/2πr.
Infinite straight wire aloneμ0i/2πrCounts as 1.
Loop + wire, same senseμ0i(π+1)/2πrThe commonest answer.
Loop + wire, opposite senseμ0i(π−1)/2πrNever zero, since π > 1.
Semicircle + wire, same senseμ0i(π/2+1)/2πrSemicircle counts as π/2.
Loop + semi-infinite wireμ0i(π+½)/2πrSemi-infinite wire counts as ½.
Wire SPLIT into two arcsoften 0Different question — current divides, arcs oppose.
Split into unequal arcsuse I ∝ 1/resistanceLonger arc carries less current.
The standard-unit table — everything in units of μ0i/2πr
PieceCounts as
Full circleπ
Semicircleπ/2
Quarter circleπ/4
Infinite straight wire1
Semi-infinite straight wire½
Wire in line with the centre0
Add or subtract these plain numbers according to sense, then multiply by μ0i/2πr.

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.

Q1Standard
An infinite straight conductor carrying current i is bent to form a loop of radius r, with the current continuing along the wire. If both contributions at the centre are in the same sense, the field there is:
(a) μ0i/2r
(b) μ0i(π+1)/2πr
(c) μ0i(π−1)/2πr
(d) μ0i/2πr
Show step-by-step solution
GivenInfinite wire with a loop of radius r; full current i in both
AskedField at the centre
ConceptTwo pieces, both at distance r. Put them over the same denominator and add.
Formulaloop π units, wire 1 unit, in μ0i/2πr
SolutionLoop: μ0i/2r = μ0iπ/2πr ⇒ π units.
Straight wire: μ0i/2πr ⇒ 1 unit.
Same sense ⇒ add: (π+1) units.
B = μ0i(π+1)/2πr.
Answer: μ0i(π+1)/2πr
Q2Standard
For the same arrangement but with the loop's field opposing the wire's, the field at the centre is:
(a) μ0i(π+1)/2πr
(b) μ0i(π−1)/2πr
(c) zero
(d) μ0i/2πr
Show step-by-step solution
GivenLoop and wire contributions in opposite senses
AskedField at the centre
ConceptSame two magnitudes; subtract instead of adding.
Formula(π − 1) units
SolutionLoop ⇒ π units; wire ⇒ 1 unit.
Opposite senses ⇒ subtract.
B = μ0i(π−1)/2πr. Since π > 1 this is still positive.
Answer: μ0i(π−1)/2πr
Q3Units
In units of μ0i/2πr, a full circular loop of radius r contributes:
(a) 1
(b) π
(c) 2π
(d) ½
Show step-by-step solution
GivenFull loop of radius r
AskedContribution in standard units
ConceptRewrite μ0i/2r with 2πr in the denominator.
Formulaμ0i/2r = μ0iπ/2πr
Solutionμ0i/2r = μ0i × π/(2πr).
So it counts as π units.
Answer: π
Q4Units
In the same units, an infinite straight wire at distance r contributes:
(a) π
(b) 1
(c) 2
(d) ½
Show step-by-step solution
GivenInfinite wire at perpendicular distance r
AskedContribution in standard units
ConceptIt already has the standard form.
Formulaμ0i/2πr
SolutionThe wire gives exactly μ0i/2πr.
That is 1 unit by definition.
Answer: 1
Q5Semicircle
A long straight wire has a semicircular bend of radius r in it, the straight portions being tangential. If the two contributions are in the same sense, the field at the centre is:
(a) μ0i(π/2+1)/2πr
(b) μ0i(π+1)/2πr
(c) μ0i/4r
(d) μ0i(π/2−1)/2πr
Show step-by-step solution
GivenSemicircle radius r plus tangential straight portions, same sense
AskedField at the centre
ConceptSemicircle counts as π/2; the straight wire as 1.
Formula(π/2 + 1) units
SolutionSemicircle: μ0i/4r = μ0i(π/2)/2πr ⇒ π/2 units.
Straight wire ⇒ 1 unit.
Total = (π/2 + 1) units, so B = μ0i(π/2+1)/2πr.
Answer: μ0i(π/2+1)/2πr
Q6Split loop
A circular loop is fed at two diametrically opposite points, so the current divides equally between the two semicircular halves. The field at the centre is:
(a) μ0i/2r
(b) μ0i/4r
(c) μ0i/8r
(d) zero
Show step-by-step solution
GivenLoop fed at two opposite points; current divides equally
AskedField at the centre
ConceptThis is the SPLIT case, not the loop-in-a-wire case. The halves circulate in opposite senses.
Formulaeach half μ0(i/2)/4r
SolutionEach half carries i/2 and each is a semicircle: μ0i/8r.
One half goes clockwise, the other anticlockwise.
Equal and opposite ⇒ they cancel exactly. B = 0.
Answer: zero
Q7Split loop
A circular loop of radius r is fed so that one arc carries twice the current of the other, the arcs being of equal length. With total current I0, the field at the centre is:
(a) μ0I0/6r
(b) μ0I0/12r
(c) μ0I0/4r
(d) zero
Show step-by-step solution
GivenTwo semicircular arcs carrying currents in the ratio 2 : 1; total I0
AskedField at the centre
ConceptSplit the current, compute each semicircle, then subtract because the senses oppose.
Formulasemicircle: μ0I/4r
SolutionCurrents: 2I0/3 and I0/3.
B1 = μ0(2I0/3)/4r = μ0I0/6r.
B2 = μ0(I0/3)/4r = μ0I0/12r.
Opposite senses ⇒ subtract: μ0I0/12r.
Answer: μ0I0/12r
Q8Numerical
A long wire carrying 4 A has a loop of radius 5 cm formed in it, with both contributions in the same sense. The field at the centre is about:
(a) 5.2 × 10−5 T
(b) 6.6 × 10−5 T
(c) 1.6 × 10−5 T
(d) 3.4 × 10−5 T
Show step-by-step solution
Giveni = 4 A, r = 0.05 m, same sense
AskedField at the centre
ConceptUse the (π+1) result and substitute.
FormulaB = μ0i(π+1)/2πr = (2×10−7)i(π+1)/r
Solutionμ0/2π = 2 × 10−7.
B = 2×10−7 × 4 × (4.14)/0.05
= 8×10−7 × 4.14/0.05
= 6.6 × 10−5 T.
Answer: 6.6 × 10−5 T
Q9Recognition
In a figure, the current is seen to divide at the loop into two paths. This means the correct approach is:
(a) use the (π+1) formula
(b) treat it as two arcs carrying divided currents in opposite senses
(c) ignore the loop
(d) use Ampere's law
Show step-by-step solution
GivenCurrent divides at the loop
AskedCorrect approach
ConceptDividing current is the arc-division problem, not the loop-in-a-wire problem.
Formulaarcs oppose; use current division
SolutionIf the current divides, each arc carries only part of it.
The two arcs circulate in opposite senses round the centre.
So their fields subtract — often cancelling completely.
Answer: treat it as two arcs carrying divided currents in opposite senses
Q10Sign
Why can the loop-plus-wire arrangement never give zero field at the centre?
(a) because the currents are unequal
(b) because π > 1, so the loop always dominates
(c) because the wire contributes nothing
(d) because they are perpendicular
Show step-by-step solution
GivenLoop + wire, opposing senses
AskedReason the field is never zero
ConceptCompare the two coefficients in standard units.
Formulaloop π units vs wire 1 unit
SolutionThe loop counts as π ≈ 3.14 units.
The wire counts as 1 unit.
Even when opposing, the difference is (π−1) ≈ 2.14, never zero.
Answer: because π > 1, so the loop always dominates
Q11Semi-infinite
A semi-infinite straight wire ends at the edge of a circular loop of radius r, both carrying current i in the same sense. In standard units the total is:
(a) π + 1
(b) π + ½
(c) π − ½
(d) 2π + 1
Show step-by-step solution
GivenFull loop plus a semi-infinite wire, same sense
AskedTotal in standard units
ConceptA semi-infinite wire gives half the infinite value.
Formulaloop π; semi-infinite wire ½
SolutionLoop ⇒ π units.
Semi-infinite wire ⇒ ½ unit.
Total = (π + ½) units.
Answer: π + ½
Q12Quarter
In standard units of μ0i/2πr, a quarter circle contributes:
(a) π/2
(b) π/4
(c) π
(d) 1/4
Show step-by-step solution
GivenQuarter circle of radius r
AskedContribution in standard units
ConceptIt is a quarter of a full loop, and a full loop is π units.
Formulafull loop = π units
SolutionFull loop ⇒ π units.
A quarter of that is π/4 units.
(Check: μ0i/8r = μ0i(π/4)/2πr ✓)
Answer: π/4
Q13Numerical
A wire carrying 10 A has a loop of radius 10 cm, contributions opposing. The field at the centre is about:
(a) 4.3 × 10−5 T
(b) 8.3 × 10−5 T
(c) 6.3 × 10−5 T
(d) 2.0 × 10−5 T
Show step-by-step solution
Giveni = 10 A, r = 0.1 m, opposing senses
AskedField at the centre
ConceptUse the (π−1) result.
FormulaB = (2×10−7)i(π−1)/r
Solutionπ − 1 = 2.14.
B = 2×10−7 × 10 × 2.14/0.1
= 2×10−6 × 21.4
= 4.3 × 10−5 T.
Answer: 4.3 × 10−5 T
Q14Tangent
In the loop-in-a-wire figure, the perpendicular distance from the straight wire to the loop's centre is:
(a) 2r
(b) r
(c) r/2
(d) zero
Show step-by-step solution
GivenStraight wire tangential to a loop of radius r
AskedPerpendicular distance to the centre
ConceptThe straight wire touches the circle, so it is a tangent — and a tangent is exactly one radius from the centre.
Formulatangent distance = r
SolutionThe wire runs tangentially to the loop.
The perpendicular from a tangent to the centre is the radius.
So the distance is r, which is why both formulas share the same r.
Answer: r
Q15Concept
The reason both contributions can be written over the same denominator 2πr is that:
(a) the currents are the same
(b) both pieces are the same distance r from the centre
(c) the field is uniform
(d) μ0 is constant
Show step-by-step solution
GivenLoop-in-a-wire arrangement
AskedWhy the common denominator works
ConceptThe loop's radius and the wire's perpendicular distance are both r.
Formulaloop μ0i/2r; wire μ0i/2πr
SolutionThe loop has radius r.
The tangential wire is also a distance r from the centre.
With the same r in both, the two expressions share a denominator and combine neatly.
Answer: both pieces are the same distance r from the centre
Q16Mixed
A long wire has a SEMICIRCULAR bend of radius r, with the straight portions along the diameter (not tangential). The field at the centre is:
(a) μ0i(π/2+1)/2πr
(b) μ0i/4r
(c) μ0i(π+1)/2πr
(d) zero
Show step-by-step solution
GivenSemicircle with straight leads ALONG the diameter
AskedField at the centre
ConceptLeads along the diameter point straight at the centre, so they are blind — unlike the tangential case.
Formulain-line wire contributes 0
SolutionThe straight leads lie on the line through the centre ⇒ contribution zero.
Only the semicircle counts.
B = μ0i/4r.
Answer: μ0i/4r
Q17Comparison
Two figures both show a loop of radius r in a wire carrying i. In figure A the straight portions are tangential; in figure B they lie along a diameter. The fields at the centre differ because:
(a) the currents differ
(b) in B the straight parts are blind, in A they are not
(c) the radii differ
(d) μ0 differs
Show step-by-step solution
GivenTangential leads vs diametric leads
AskedReason for the difference
ConceptThe blind-spot rule depends on whether the wire's line passes through the field point.
FormuladB ∝ sinθ
SolutionTangential leads do not pass through the centre ⇒ they contribute.
Diametric leads lie on a line through the centre ⇒ sinθ = 0 ⇒ nothing.
So the same-looking figures give different answers.
Answer: in B the straight parts are blind, in A they are not
Q18Numerical
A loop of radius 2 cm carrying 6 A has a tangential wire carrying the same current in the same sense. The total field at the centre is about:
(a) 2.5 × 10−4 T
(b) 1.9 × 10−4 T
(c) 6.0 × 10−5 T
(d) 1.2 × 10−4 T
Show step-by-step solution
Givenr = 0.02 m, i = 6 A, same sense
AskedTotal field at the centre
ConceptApply the (π+1) result.
FormulaB = (2×10−7)i(π+1)/r
SolutionB = 2×10−7 × 6 × 4.14 / 0.02
= 1.2×10−6 × 4.14/0.02
= 2.5 × 10−4 T.
Answer: 2.5 × 10−4 T
Q19Method
The fastest reliable way to handle any loop-plus-wire figure is to:
(a) integrate Biot–Savart from scratch
(b) express every piece in units of μ0i/2πr, then add or subtract the plain numbers
(c) use Ampere's law
(d) assume the wire contributes nothing
Show step-by-step solution
GivenAny composite loop-and-wire figure
AskedBest method
ConceptThe standard-unit table turns the whole problem into simple arithmetic.
Formulafull loop π, semicircle π/2, wire 1, semi-infinite ½, in-line 0
SolutionConvert every piece to standard units.
Decide the sense of each and attach a sign.
Add the numbers, then multiply by μ0i/2πr once at the end.
Answer: express every piece in units of μ0i/2πr, then add or subtract the plain numbers
Q20Summary
Which of these arrangements gives ZERO field at the centre?
(a) loop plus tangential wire, same sense
(b) loop plus tangential wire, opposite senses
(c) loop fed at two opposite points so the current divides equally
(d) semicircle with diametric leads
Show step-by-step solution
GivenFour arrangements
AskedWhich gives zero
ConceptOnly the equally divided loop produces two exactly opposite, exactly equal contributions.
Formulasplit loop halves cancel
SolutionLoop + wire, same sense ⇒ (π+1), non-zero.
Loop + wire, opposite ⇒ (π−1), still non-zero.
Semicircle with diametric leads ⇒ μ0i/4r, non-zero.
Equally divided loop ⇒ two equal semicircles in opposite senses ⇒ exactly zero.
Answer: loop fed at two opposite points so the current divides equally