Arcs in one plane, straight wires in another. Frightening to look at, mechanical to solve once you split them up.
Part 1 — The concept, explained simply
Some questions bolt an arc in one plane onto straight wires in another. They look terrifying. They are actually three easy questions stapled together.
1. The method — break it into pieces
Never try to see the whole shape at once. Chop it into parts you already know, work out each one separately, then combine the answers as vectors.
Picture it: nobody builds a Lego model by staring at the finished picture. You follow the steps, one brick at a time, and the model appears.
Step 1: split → Step 2: solve each → Step 3: add as vectors
2. The pieces you will meet
Piece
Field at the centre
Direction
Full circle, radius R
μ0I / 2R
Along the axis of the circle
Semicircle
μ0I / 4R
Along the axis of the circle
Quarter circle
μ0I / 8R
Along the axis of the circle
Semi-infinite straight wire
μ0I / 4πR
Perpendicular to the plane containing wire and point
Straight wire pointing at the centre
0
— the blind spot
3. The step that makes it three-dimensional
Here is the part that trips people up. If the arc lies in one plane and the straight wires lie in another, their fields point in different directions — usually at right angles. So you cannot simply add the numbers.
Btotal = √(B1² + B2²) when they are perpendicular
Two arrows, perpendicular to each other. Adding their lengths would be wrong; you must combine them the way you combine any two perpendicular vectors.
4. The classic question, worked through
A wire carries current I. The straight portions are long and parallel to the x-axis; the semicircular portion of radius R lies in the y–z plane. Find B at the centre O.
Piece 1 — the semicircle. It lies in the y–z plane, so its axis is the x-axis. Magnitude μ0I/4R, direction along î.
Rewrite it as (μ0/4π)(πI/R) so both pieces share a common factor.
Piece 2 — the two straight wires. Each is semi-infinite as seen from O, at perpendicular distance R. Each gives μ0I/4πR, and both point the same way — along k̂. Together: (μ0/4π)(2I/R).
Combine. The two are perpendicular, so keep them as separate components: B = (μ0/4π)(I/R)(πî + 2k̂)
The habit that makes these easy. Pull out the common factor (μ0/4π)(I/R) early. Then the semicircle contributes π and each straight wire contributes 1 — and the answer almost writes itself. Every option in these questions is built in exactly that form.
Part 2 — Formula sheet
Piece
Contribution at the centre
In units of (μ0/4π)(I/R)
Full circle
μ0I/2R
2π
Semicircle
μ0I/4R
π
Quarter circle
μ0I/8R
π/2
Arc of angle θ (radians)
μ0Iθ/4πR
θ
Semi-infinite straight wire
μ0I/4πR
1
Infinite straight wire
μ0I/2πR
2
Wire in line with the centre
0
0
Two perpendicular contributions
B = √(B1²+B2²)
Keep as separate î, ĵ, k̂ components.
Two parallel contributions
B = B1 + B2
Add the numbers directly.
Two antiparallel contributions
B = |B1 − B2|
Subtract.
The right-hand column is the whole trick. Take out (μ0/4π)(I/R) and every piece becomes a plain number: 2π for a circle, π for a semicircle, 1 for a semi-infinite wire, 0 for an in-line wire. Then just collect them by 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.
Q1Classic
A wire carrying current I has long straight portions parallel to the x-axis and a semicircular portion of radius R in the y–z plane. The field at the centre O is:
(a) (μ0/4π)(I/R)(πî + 2k̂)
(b) (μ0/4π)(I/R)(πî − 2k̂)
(c) (μ0/4π)(I/R)(2î + πk̂)
(d) μ0I/4R only
Show step-by-step solution
GivenSemicircle radius R in the y–z plane; two long straight wires parallel to x
AskedField at O
ConceptSplit into the semicircle and the two straight wires; their fields are perpendicular, so keep them as separate components.
Formulasemicircle μ0I/4R; each straight wire μ0I/4πR
SolutionSemicircle in the y–z plane ⇒ its field is along î: magnitude μ0I/4R = (μ0/4π)(πI/R). Each straight wire is semi-infinite from O, distance R: (μ0/4π)(I/R), both along k̂. Two of them: (μ0/4π)(2I/R) along k̂. B = (μ0/4π)(I/R)(πî + 2k̂).
Answer: (μ0/4π)(I/R)(πî + 2k̂)
Q2Semicircle + leads
A long straight wire is bent into a semicircular arc of radius R at its middle, with the two straight portions lying along the diameter. The field at the centre is:
(a) μ0I/2R
(b) μ0I/4R
(c) μ0I/8R
(d) μ0I/2πR
Show step-by-step solution
GivenSemicircle radius R with straight leads along the diameter
AskedField at the centre
ConceptThe straight leads point directly at the centre, so they are blind.
Formulasemicircle μ0I/4R; in-line wires give 0
SolutionFor the straight leads, dℓ and r are parallel ⇒ contribution zero. Only the semicircle counts. B = μ0I/4R.
Answer: μ0I/4R
Q3Quarter + leads
A wire consists of two long straight semi-infinite portions along the x and y axes, joined by a quarter circle of radius R in the x–y plane. The field at the centre of the quarter circle is:
(a) (μ0I/4πR)(π/2)
(b) (μ0I/4πR)(π/2 + 2)
(c) (μ0I/4πR)(π/2 + 1)
(d) μ0I/8R only
Show step-by-step solution
GivenQuarter circle radius R plus two semi-infinite straight arms along the axes
AskedField at the centre
ConceptHere the straight arms are NOT in line with the centre, so each contributes. All three pieces lie in the same plane, so they simply add.
Formulaquarter π/2; each semi-infinite arm 1
SolutionIn units of (μ0I/4πR): quarter circle gives π/2. Each semi-infinite arm gives 1, and there are two of them ⇒ 2. All three fields are perpendicular to the plane and in the same sense, so they add. B = (μ0I/4πR)(π/2 + 2).
Answer: (μ0I/4πR)(π/2 + 2)
Q4Perpendicular
Two contributions to the field at a point are 3 × 10−5 T and 4 × 10−5 T, at right angles to each other. The resultant is:
In units of (μ0I/4πR), the contribution of a full circular loop at its centre is:
(a) π
(b) 2π
(c) 1
(d) 2
Show step-by-step solution
GivenFull circle of radius R
AskedContribution in the standard unit
ConceptRewrite μ0I/2R with 4π in the denominator.
Formulaμ0I/2R = (μ0I/4πR) × 2π
Solutionμ0I/2R = (μ0I/4πR) × (4π/2) = (μ0I/4πR) × 2π. So a full circle counts as 2π units.
Answer: 2π
Q6Mixed planes
An arc lies in the x–y plane and a straight wire lies along the z-axis. Their fields at a common point are:
(a) parallel, so they add
(b) antiparallel, so they subtract
(c) perpendicular, so combine by Pythagoras
(d) both zero
Show step-by-step solution
GivenArc in x–y plane, wire along z
AskedHow the fields combine
ConceptAn arc's field is along its own axis; a wire's field circles around it. Different planes give different directions.
FormulaCombine as vectors
SolutionAn arc in the x–y plane produces a field along the z-axis. A wire along z produces a field circling in the x–y plane. These directions are perpendicular, so use √(B1²+B2²).
Answer: perpendicular, so combine by Pythagoras
Q7Three-quarter
A wire is bent into three quarters of a circle of radius R (the remaining quarter is missing). The field at the centre due to the arc is:
(a) 3μ0I/8R
(b) μ0I/8R
(c) μ0I/2R
(d) 3μ0I/4R
Show step-by-step solution
GivenArc covering 270° = 3π/2 radians, radius R
AskedField at the centre
ConceptAn arc gives its fair share of the full-loop value.
FormulaB = μ0Iθ/4πR
Solutionθ = 3π/2 radians. B = μ0I(3π/2)/4πR = 3μ0I/8R. (Equivalently three quarters of μ0I/2R.)
Answer: 3μ0I/8R
Q8Method
The first step in any composite-geometry field question is to:
(a) apply Ampere's law
(b) split the shape into standard pieces
(c) find the magnetic moment
(d) assume the field is uniform
Show step-by-step solution
GivenComposite wire shape
AskedCorrect first step
ConceptNever analyse the whole shape at once; each piece has a known result.
FormulaSuperposition
SolutionIdentify the arcs, the straight segments and any in-line pieces. Write down each contribution separately from the standard list. Then combine them according to direction.
Answer: split the shape into standard pieces
Q9Two semicircles
Two semicircular arcs of radii R and 2R share the same centre and are joined to form a closed loop carrying current I. The field at the centre is:
(a) μ0I/4R + μ0I/8R
(b) 3μ0I/8R
(c) μ0I/8R
(d) μ0I/2R
Show step-by-step solution
GivenSemicircles of radii R and 2R, concentric, same current, same sense
AskedField at the centre
ConceptEach semicircle contributes μ0I/4r with its own radius. Same sense means they add.
FormulaB = μ0I/4r for each
SolutionInner semicircle: μ0I/4R. Outer semicircle: μ0I/4(2R) = μ0I/8R. Same rotational sense ⇒ add: μ0I/4R + μ0I/8R = 3μ0I/8R.
Answer: 3μ0I/8R
Q10Opposite senses
The same two concentric semicircles are joined so that the currents circulate in OPPOSITE senses. The field at the centre is:
(a) 3μ0I/8R
(b) μ0I/8R
(c) μ0I/4R
(d) zero
Show step-by-step solution
GivenSemicircles radii R and 2R, opposite senses
AskedField at the centre
ConceptOpposite senses mean the two fields point opposite ways, so subtract.
FormulaB = |B1 − B2|
Solutionμ0I/4R and μ0I/8R, pointing opposite ways. B = μ0I/4R − μ0I/8R = μ0I/8R.
Answer: μ0I/8R
Q11Numerical
A semicircular arc of radius 5 cm carries 4 A, with straight leads along the diameter. The field at the centre is:
(a) 1.26 × 10−5 T
(b) 2.51 × 10−5 T
(c) 5.03 × 10−5 T
(d) 6.28 × 10−6 T
Show step-by-step solution
GivenR = 0.05 m, I = 4 A, semicircle with in-line leads
AskedField at the centre
ConceptLeads are blind, so only the semicircle counts.
In composite figures, a straight segment contributes nothing when:
(a) it is very short
(b) its line passes through the field point
(c) it is perpendicular to the arc
(d) the current is small
Show step-by-step solution
GivenStraight segment in a composite shape
AskedWhen it contributes zero
ConceptThe Biot–Savart blind spot: sinθ = 0 along the segment's own line.
FormuladB ∝ sinθ
SolutionIf the point lies on the segment's line, then θ = 0° for every element. sin 0° = 0, so every element gives nothing. This is why diameter leads on a semicircle are always ignored.
Answer: its line passes through the field point
Q13Full circle + wire
An infinite straight wire has a circular loop of radius r formed in it, with the current continuing along the wire. If both contributions at the centre are in the same sense, the field is:
(a) μ0i(π+1)/2πr
(b) μ0i(π−1)/2πr
(c) μ0i/2r
(d) μ0i/2πr
Show step-by-step solution
GivenInfinite wire with a circular loop of radius r formed in it
AskedField at the centre
ConceptTwo pieces: the full loop and the infinite straight wire, both at distance r from the centre.
Formulaloop μ0i/2r; wire μ0i/2πr
SolutionLoop: μ0i/2r = μ0iπ/2πr. Straight wire: μ0i/2πr. Same sense ⇒ add: μ0i(π+1)/2πr.
Answer: μ0i(π+1)/2πr
Q14Opposite
For the same arrangement but with the two contributions in opposite senses, 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 opposing
AskedField at the centre
ConceptSame two magnitudes, but now they subtract.
FormulaB = |Bloop − Bwire|
Solutionμ0iπ/2πr − μ0i/2πr = μ0i(π−1)/2πr. Note π > 1, so the loop always wins and the result is positive.
Answer: μ0i(π−1)/2πr
Q15Direction
A semicircular arc lies in the y–z plane. The field it produces at its centre points along:
(a) the y-axis
(b) the z-axis
(c) the x-axis
(d) in the y–z plane
Show step-by-step solution
GivenArc in the y–z plane
AskedDirection of its field at the centre
ConceptAn arc's field at its centre lies along the axis of the circle, which is perpendicular to the arc's plane.
FormulaRight-hand rule for a loop
SolutionThe plane of the arc is y–z. The axis perpendicular to that plane is the x-axis. So the field points along ±x, i.e. along î.
Answer: the x-axis
Q16Numerical
In a composite figure the arc contributes 6 × 10−6 T along î and the straight wires contribute 8 × 10−6 T along k̂. The magnitude of the total field is:
(a) 14 × 10−6 T
(b) 10 × 10−6 T
(c) 2 × 10−6 T
(d) 7 × 10−6 T
Show step-by-step solution
GivenB1 = 6×10−6 î, B2 = 8×10−6 k̂
AskedMagnitude of the total
ConceptThe components are along perpendicular axes, so use Pythagoras.
In units of (μ0I/4πR), an arc subtending 60° at the centre contributes:
(a) π/6
(b) π/3
(c) π/2
(d) 1/3
Show step-by-step solution
GivenArc of 60°, radius R
AskedContribution in standard units
ConceptThe contribution in these units is simply the angle in radians.
FormulaB = (μ0I/4πR) × θ
Solution60° = π/3 radians. In units of (μ0I/4πR), the contribution is just θ. So it is π/3.
Answer: π/3
Q18Two wires
Two semi-infinite straight wires both end at the same point and both lie at perpendicular distance R from the field point, with their fields in the same sense. Together they contribute:
(a) μ0I/4πR
(b) μ0I/2πR
(c) μ0I/πR
(d) zero
Show step-by-step solution
GivenTwo semi-infinite wires, each at distance R, same sense
AskedTotal contribution
ConceptEach gives the semi-infinite value; same sense means simple addition.
Formulaeach μ0I/4πR
SolutionEach contributes μ0I/4πR. Same sense ⇒ add. Total = μ0I/2πR — which equals one infinite wire, as it should.
Answer: μ0I/2πR
Q19Concept
Why can the contributions from an arc and a straight wire in different planes NOT simply be added as numbers?
(a) because the currents differ
(b) because their fields point in different directions
(c) because arcs have no field
(d) because μ0 differs
Show step-by-step solution
GivenComposite shape spanning two planes
AskedReason for vector addition
ConceptMagnetic field is a vector, so only parallel contributions add arithmetically.
FormulaVector superposition
SolutionThe arc's field lies along its own axis. The straight wire's field circles around the wire. In different planes these directions differ, so the contributions must be combined as vectors.
Answer: because their fields point in different directions
Q20Strategy
A composite figure has an arc, two in-line straight leads and one perpendicular semi-infinite wire. The number of pieces that actually contribute is:
(a) all four
(b) three
(c) two
(d) one
Show step-by-step solution
GivenArc + two in-line leads + one perpendicular semi-infinite wire
AskedHow many contribute
ConceptIn-line leads are blind; everything else counts.
Formulain-line contribution = 0
SolutionThe two in-line leads point at the centre ⇒ zero each. The arc contributes. The perpendicular semi-infinite wire contributes. So two pieces contribute.