Priority 2 for the next ILTS. The wire, the coil, the arc and the solenoid — plus Ampere's law and the thick-wire graph that NEET has asked five times.
One big idea runs through this whole topic: a current makes a magnetic field, and the shape of the wire decides the shape of the field.
1. The starting point — Biot–Savart law
Chop the wire into a crumb so tiny that it counts as perfectly straight, even on a curve. Call that crumb I dl. Biot and Savart worked out how much magnetic field one crumb makes at a point P:
dB = (μ₀ / 4π) · (I dl sin θ) / r²
Read it as three simple statements:
More current → bigger field.
Further away → weaker field, and weakly fast — go twice as far and it drops to a quarter.
The sin θ measures how much the crumb and the direction to P disagree.
The blind spot. If P lies straight ahead of or straight behind the crumb, along the wire's own line, then θ = 0° and sin θ = 0, so the field there is zero. A torch is bright to the sides but dark along its own axis. This one fact solves several NEET questions instantly.
Watch the bar collapse to nothing twice per orbit — exactly when the point lies along the element's own line. That is the blind spot, and it is what makes straight sections pointing at the centre contribute zero.
The direction of dB is perpendicular to both the crumb and the line to P — so if you lay them flat on a table, the field sticks straight up out of the table. The value μ₀/4π = 10⁻⁷ T m/A is exact, and almost every numerical in this chapter starts by writing it down.
2. A long straight wire — the field goes in circles
Add up all the crumbs of a very long straight wire (or, faster, use Ampere's law) and you get:
B = μ₀ I / 2πr
The field lines are circles wrapped around the wire, like the rings that spread when you drop a stone in water. That is exactly what Oersted's iron filings showed in 1820 — now we can prove it.
This is what Oersted's iron filings showed in 1820 — and now the maths predicts it. Notice the inner circles march faster: the field is stronger close in.
Direction — the grip rule. Grasp the wire in your right hand with your thumb pointing along the current. Your curling fingers show the way the field goes round.
Two different right-hand rules — do not mix them up. Straight wire: thumb = current, curled fingers = field. Circular loop: curled fingers = current, thumb = field.
The fingers and thumb swap jobs. This is one of the most punished confusions in the chapter.
Only your distance from the wire matters — not which side you stand on, not how far along the wire you are. Note also that the field falls off as 1/r here, not 1/r².
3. Ampere's circuital law — the shortcut
Biot–Savart always works but is slow. Ampere's law is a shortcut that works beautifully when the situation is symmetrical:
∮ B · dl = μ₀ Ienclosed
In words: draw an imaginary loop, walk all the way around it, add up how much field points along your path at each step, and the total tells you how much current passes through the loop.
Picture it: standing outside a room and counting everyone who walks in through the door. You learn how many people are inside without ever going in.
The classic trap. If a wire lies outside your imaginary loop, then ∮B·dl = 0 for that loop — but the field B at points on the loop is not zero. The integral being zero does not mean the field is zero. NEET asks this repeatedly.
On the right the field is not zero anywhere on the loop — the walker feels it the whole way round. It simply adds up to nothing, because no current passes through. Zero integral does not mean zero field.
4. A thick wire — the graph NEET loves
Take a fat wire of radius a with current spread evenly across its whole cross-section.
Outside (r > a): your loop encloses the whole current, so you get the usual B = μ₀I/2πr, fading as 1/r.
Inside (r < a): your loop only catches part of the current. Since the current is spread evenly, the fraction caught is the area ratio:
Ienclosed = I · (πr² / πa²) = I r² / a² ⇒ B = μ₀ I r / 2πa²
which grows with r. Think of a crowd standing evenly across a field: rope off a small circle and you enclose only a few people; a bigger circle catches more.
Inside, the growing loop catches more and more current, so the field climbs. The moment it passes the surface it has caught all of it, and from then on only the distance matters — so the field falls. Peak exactly at r = a.
Make her able to DRAW this, not just recognise it. NEET has asked this exact graph in 2016, 2019, 2021, 2022 and 2026. One derivation covers five years of questions.
Two close relatives worth knowing: inside a hollow pipe carrying current, the field is zero (no current is enclosed); and for a coaxial cable, outside both conductors the equal and opposite currents enclose to zero.
5. A circular coil — the workhorse formula
Bend the wire into a circle. Every crumb sits the same distance from the centre and every crumb pushes the field the same way, so they all add up:
Bcentre = μ₀ N I / 2R
This is the single most-used formula in the whole chapter. Bigger current or more turns → stronger field. Bigger ring → weaker field.
Move off the centre, along the axis, and it becomes:
Baxis = μ₀ N I R² / 2(x² + R²)3/2
Put x = 0 and this collapses straight back to the centre formula — a useful check. Go very far away (x ≫ R) and it fades as 1/x³.
The reason it works is symmetry: each crumb's field points at a slant, but for every crumb on the top there is a matching one on the bottom whose sideways part points exactly the other way. All the sideways parts cancel in pairs, and only the along-the-axis parts survive.
That symmetry is the whole reason the messy sum collapses to B = μ₀NI / 2R at the centre.
6. Arcs and bent wires
A full circle sweeps 2π radians. An arc sweeps only θ radians, so it gives only that fraction of the field:
Barc = μ₀ I θ / 4πR
Shape
Angle
Field at the centre
Full circle
2π
μ₀I / 2R
Semicircle
π
μ₀I / 4R
Quarter circle
π/2
μ₀I / 8R
Arc of 60°
π/3
μ₀I / 12R
The arc gives exactly its fair share of a full loop. And the straight leads on either side point straight at the centre, so by the blind-spot rule they give nothing at all.
The two lines that solve almost every bent-wire question:
Straight sections that point directly at the centre contribute nothing (θ = 0, so sin θ = 0).
When a loop splits into two arcs, they behave like resistors in parallel — the longer arc has more resistance so it carries less current — and their fields point in opposite senses, so they partly cancel.
7. The solenoid — a uniform field on demand
Wind the wire into a long spring shape. Between neighbouring turns the circling fields point opposite ways and rub each other out, so all that survives points neatly down the tube:
B = μ₀ n I
Three things worth noticing about this formula:
No R in it. The field does not depend on how fat the solenoid is. NEET has tested exactly this by giving a useless radius in the question.
No position in it. Same value everywhere inside — that is what "uniform" means, and it is why an MRI scanner is a tunnel.
n is turns per metre, not total turns. So n = N / L.
Outside a long solenoid the field is essentially zero. At the very ends it drops to half: Bend = μ₀nI / 2. And the field-line picture outside looks exactly like a bar magnet's — with an off switch.
No radius appears in B = μ₀nI — a fat solenoid and a thin one give the same field. And n is turns per metre, not total turns.
Coil versus solenoid — the commonest mix-up in this topic.
Circular coil
Solenoid
Formula
B = μ₀NI / 2R
B = μ₀nI
Meaning of the turns symbol
N = total turns
n = turns per metre
Does the radius matter?
Yes
No
Field shape
Strong only near the centre
Uniform right through the inside
8. Adding fields from more than one source
When two or more wires are present, work out each field separately with the right formula, then add them as arrows, not as numbers. Use the grip rule on each wire to find which way its field points at that spot.
Fields pointing the same way → add.
Fields pointing opposite ways → subtract.
Fields at right angles → use √(B₁² + B₂²).
For two parallel wires, remember the geometry: at a point between them, currents in the same direction give fields that oppose each other, and currents in opposite directions give fields that add. That feels backwards, so check it with your right hand every time.
Use the grip rule on each wire separately, then add the two arrows. Never trust memory on this one — it feels backwards, and it is the exact reversal that trips people up.
Part 2 — Formula sheet
Every field formula in sections 4.4 to 4.7, on one page.
Configuration
Field
Watch out for
Current element (Biot–Savart)
dB = (μ₀/4π) I dl sin θ / r²
Zero along the element's own line (θ = 0).
Constant
μ₀/4π = 10⁻⁷ T m/A
μ₀ = 4π × 10⁻⁷; and μ₀/2π = 2 × 10⁻⁷.
Long straight wire
B = μ₀I / 2πr
Falls as 1/r. Field lines are circles round the wire.
Semi-infinite wire (from one end)
B = μ₀I / 4πr
Exactly half the infinite-wire value.
Thick wire, outside (r > a)
B = μ₀I / 2πr
Whole current enclosed.
Thick wire, inside (r < a)
B = μ₀I r / 2πa²
B ∝ r. Zero on the axis, maximum at the surface.
Hollow pipe, inside
B = 0
No current enclosed by the loop.
Ampere's circuital law
∮ B·dl = μ₀ Ienclosed
Only enclosed current counts. Steady currents only.
Circular coil, at the centre
B = μ₀ N I / 2R
N = total turns. The most-used formula in the chapter.
Circular coil, on the axis
B = μ₀ N I R² / 2(x²+R²)3/2
Put x = 0 to recover the centre formula.
Coil, far along the axis
B ≈ μ₀ I R² / 2x³
Falls as 1/x³ — the dipole signature.
Arc of angle θ (radians)
B = μ₀ I θ / 4πR
θ must be in radians. 180° = π.
Semicircle
B = μ₀I / 4R
Half a full loop.
Quarter circle
B = μ₀I / 8R
A quarter of a full loop.
Long solenoid, inside
B = μ₀ n I
n = N/L, turns per metre. Radius is irrelevant.
Solenoid, at the end
B = μ₀ n I / 2
Exactly half the middle value.
Solenoid, outside
B ≈ 0
All the field is concentrated inside.
Speed of light link
ε₀ μ₀ = 1 / c²
Electricity and magnetism together give light.
Ten-second recognition drill. Before calculating anything, ask: is this a straight wire, a coil, an arc, or a solenoid? Getting that right is worth more marks than getting the arithmetic right, because the wrong formula guarantees a wrong answer no matter how neat the working.
If the question says…
Use
"long straight wire", "infinite conductor"
μ₀I / 2πr
"circular coil", "N turns", "at the centre"
μ₀NI / 2R
"semicircular", "arc", "bent wire"
μ₀Iθ / 4πR
"solenoid", "turns per unit length", "tightly wound long"
μ₀nI
"solid cylinder", "uniformly distributed", "inside the wire"
μ₀Ir / 2πa²
Part 3 — 50 questions with step-by-step solutions
Attempt each one on paper first, then open the solution. Every solution follows the same five steps: Given → Asked → Concept → Formula → Solution.
Q1Straight wire
A long straight wire carries a current of 5 A. The magnetic field at a point 10 cm from the wire is:
(a) 1 × 10⁻⁵ T
(b) 1 × 10⁻⁶ T
(c) 2 × 10⁻⁵ T
(d) 5 × 10⁻⁶ T
Show step-by-step solution
GivenI = 5 A, r = 10 cm = 0.1 m
AskedMagnetic field B
ConceptA long straight wire produces circular field lines whose strength falls off as 1/r.
FormulaB = μ₀I / 2πr = (2 × 10⁻⁷) I / r
SolutionUse the shortcut μ₀/2π = 2 × 10⁻⁷. Convert first: r = 10 cm = 0.1 m. B = 2 × 10⁻⁷ × 5 / 0.1 B = 10⁻⁶ / 0.1 = 1 × 10⁻⁵ T
Answer: 1 × 10⁻⁵ T
Q2Straight wire
A long straight conductor carries 10 A. The magnetic field at a distance of 2 m from it is:
(a) 1 × 10⁻⁶ T
(b) 2 × 10⁻⁶ T
(c) 1 × 10⁻⁵ T
(d) 5 × 10⁻⁷ T
Show step-by-step solution
GivenI = 10 A, r = 2 m
AskedMagnetic field B
ConceptSame formula — the numbers are already in SI units.
The magnetic field at a distance r from a long straight wire is B. At a distance 2r from the same wire it becomes:
(a) B / 2
(b) B / 4
(c) 2B
(d) B
Show step-by-step solution
GivenField B at distance r
AskedField at distance 2r
ConceptFor a straight wire the field falls as 1/r — an inverse FIRST power, not inverse square. This is where marks are lost.
FormulaB ∝ 1 / r
SolutionB′ / B = r / 2r = 1/2 B′ = B / 2
Answer: B / 2
Q4Straight wire · superposition
Two long parallel wires 10 cm apart carry currents of 5 A and 10 A in the SAME direction. The magnetic field at the midpoint between them is:
(a) 2 × 10⁻⁵ T
(b) 6 × 10⁻⁵ T
(c) 4 × 10⁻⁵ T
(d) zero
Show step-by-step solution
GivenI₁ = 5 A, I₂ = 10 A, separation = 10 cm, same direction, point at the midpoint
AskedNet magnetic field at the midpoint
ConceptCompute each field separately, then use the grip rule. Between two wires carrying current the SAME way, the two fields point OPPOSITE ways, so they subtract.
FormulaB = (2 × 10⁻⁷) I / r; then add as vectors
SolutionEach wire is 5 cm = 0.05 m from the midpoint. B₁ = 2 × 10⁻⁷ × 5 / 0.05 = 2 × 10⁻⁵ T B₂ = 2 × 10⁻⁷ × 10 / 0.05 = 4 × 10⁻⁵ T Same direction currents ⇒ fields oppose at the midpoint ⇒ subtract. B = 4 × 10⁻⁵ − 2 × 10⁻⁵ = 2 × 10⁻⁵ T
Answer: 2 × 10⁻⁵ T
Q5Straight wire · superposition
The same two wires (5 A and 10 A, 10 cm apart) now carry currents in OPPOSITE directions. The field at the midpoint is:
(a) 2 × 10⁻⁵ T
(b) 4 × 10⁻⁵ T
(c) 6 × 10⁻⁵ T
(d) zero
Show step-by-step solution
GivenI₁ = 5 A, I₂ = 10 A, separation = 10 cm, opposite directions
AskedNet magnetic field at the midpoint
ConceptReverse one current and the two fields at the midpoint now point the SAME way, so they add.
FormulaB = (2 × 10⁻⁷) I / r; then add as vectors
SolutionB₁ = 2 × 10⁻⁵ T and B₂ = 4 × 10⁻⁵ T as before. Opposite currents ⇒ fields reinforce at the midpoint ⇒ add. B = 2 × 10⁻⁵ + 4 × 10⁻⁵ = 6 × 10⁻⁵ T
Answer: 6 × 10⁻⁵ T
Q6Straight wire · null point
Two long parallel wires 30 cm apart carry currents of 1 A and 4 A in the same direction. The magnetic field is zero at a point between them, at a distance from the 1 A wire of:
(a) 6 cm
(b) 10 cm
(c) 15 cm
(d) 24 cm
Show step-by-step solution
GivenI₁ = 1 A, I₂ = 4 A, separation d = 30 cm, same direction
AskedDistance x of the null point from the 1 A wire
ConceptBetween two same-direction wires the fields oppose, so somewhere they cancel exactly. Set the two magnitudes equal.
Formulaμ₀I₁ / 2πx = μ₀I₂ / 2π(d − x)
SolutionCancel μ₀/2π from both sides: I₁ / x = I₂ / (d − x) 1 / x = 4 / (30 − x) 30 − x = 4x ⇒ 30 = 5x ⇒ x = 6 cm Check: the null point is nearer the WEAKER current, as expected.
Answer: 6 cm
Q7Straight wire · concept
The magnetic field due to a long straight current-carrying wire, plotted against distance r from the wire, is:
(a) a straight line through the origin
(b) a rectangular hyperbola
(c) a parabola
(d) a horizontal straight line
Show step-by-step solution
GivenThin long straight wire
AskedShape of the B versus r graph
ConceptRecognise the mathematical form. B = constant/r is the equation of a rectangular hyperbola.
FormulaB = μ₀I / 2πr, so B ∝ 1/r
SolutionB is inversely proportional to r. A graph of y = k/x is a rectangular hyperbola. (For a THICK wire there is also a straight rising portion inside — do not confuse the two.)
Answer: a rectangular hyperbola
Q8Biot–Savart · blind spot
The magnetic field due to a small current element at a point lying along the axis of the element itself is:
(a) maximum
(b) half the maximum
(c) zero
(d) infinite
Show step-by-step solution
GivenPoint lying along the direction of the current element
AskedValue of dB there
ConceptThe sin θ factor in Biot–Savart vanishes when the point lies along the element's own line.
FormuladB = (μ₀/4π) I dl sin θ / r²
SolutionAlong the element's own line, θ = 0°. sin 0° = 0 So dB = 0 — the element has a blind spot straight ahead and straight behind.
Answer: zero
Q9Ampere's law
Ampere's circuital law states that the line integral of the magnetic field around a closed loop equals:
(a) μ₀ times the total current in the circuit
(b) μ₀ times the current enclosed by the loop
(c) μ₀ times the current outside the loop
(d) zero, always
Show step-by-step solution
GivenClosed Amperian loop with current passing through
AskedCorrect statement of the law
ConceptOnly current that actually passes through the surface bounded by the loop counts.
Formula∮ B·dl = μ₀ I_enclosed
SolutionCurrents that pass THROUGH the loop contribute. Currents lying outside the loop contribute nothing to the integral. Hence the answer is μ₀ × enclosed current.
Answer: μ₀ times the current enclosed by the loop
Q10Ampere's law · trap
A long straight wire lies completely OUTSIDE a closed Amperian loop. For that loop:
(a) both ∮B·dl and B are zero
(b) ∮B·dl is zero but B is not zero
(c) ∮B·dl is not zero but B is zero
(d) both are non-zero
Show step-by-step solution
GivenWire outside the Amperian loop
AskedValues of ∮B·dl and of B on the loop
ConceptThe integral counts only enclosed current, but the field itself exists everywhere around the wire.
Formula∮ B·dl = μ₀ I_enclosed
SolutionNo current is enclosed, so I_enclosed = 0 and therefore ∮B·dl = 0. But the wire still produces a field at every point in space, including on the loop. So B ≠ 0 even though the integral is zero.
Answer: ∮B·dl is zero but B is not zero
Q11Thick wire
For a long solid cylindrical conductor of radius a carrying a uniformly distributed current, the magnetic field INSIDE (r < a) varies as:
(a) 1 / r
(b) 1 / r²
(c) r
(d) r²
Show step-by-step solution
GivenSolid conductor of radius a, uniform current distribution, point at r < a
AskedDependence of B on r inside the conductor
ConceptOnly the current inside the Amperian loop counts, and that grows with the enclosed area.
FormulaB(2πr) = μ₀ I (πr²/πa²)
SolutionEnclosed current I_e = I r² / a² B × 2πr = μ₀ I r²/a² B = μ₀ I r / 2πa², so B ∝ r
Answer: r
Q12Thick wire
For the same solid conductor, the field OUTSIDE (r > a) varies as:
(a) r
(b) r²
(c) 1 / r
(d) 1 / r²
Show step-by-step solution
GivenPoint at r > a
AskedDependence of B on r outside
ConceptOutside, the loop encloses the entire current, so the wire behaves exactly like a thin one.
FormulaB(2πr) = μ₀ I
SolutionI_enclosed = I (the whole current). B = μ₀I / 2πr, so B ∝ 1/r
Answer: 1 / r
Q13Thick wire · graph
Which statement correctly describes the graph of B against r for a solid cylindrical conductor of radius a carrying a uniform current?
(a) B decreases as 1/r throughout
(b) B increases linearly up to r = a, then decreases as 1/r
(c) B is constant inside and decreases as 1/r outside
(d) B increases as r² up to r = a, then decreases as 1/r²
Show step-by-step solution
GivenSolid conductor of radius a, uniform current
AskedShape of the whole B versus r graph
ConceptCombine the inside and outside results. The field is zero on the axis and maximum at the surface.
FormulaInside: B ∝ r; Outside: B ∝ 1/r
SolutionInside (r < a): B = μ₀Ir/2πa² — a straight line rising from zero. At r = a: B is maximum, equal to μ₀I/2πa. Outside (r > a): B = μ₀I/2πr — a 1/r decay. So: linear rise to the surface, then 1/r fall.
Answer: B increases linearly up to r = a, then decreases as 1/r
Q14Thick wire · numeric
A solid wire of radius 2 cm carries a current of 10 A distributed uniformly. The magnetic field at a point 1 cm from the axis is:
(a) 2.5 × 10⁻⁵ T
(b) 5 × 10⁻⁵ T
(c) 1 × 10⁻⁴ T
(d) 1 × 10⁻⁵ T
Show step-by-step solution
Givena = 2 cm = 0.02 m, I = 10 A, r = 1 cm = 0.01 m (inside)
AskedMagnetic field B at r = 1 cm
ConceptThe point is inside the conductor, so use the inside formula — not the ordinary wire formula.
FormulaB = μ₀ I r / 2πa² = (2 × 10⁻⁷) I r / a²
Solutionr = 0.01 m is less than a = 0.02 m, so the point is INSIDE. B = 2 × 10⁻⁷ × 10 × 0.01 / (0.02)² = 2 × 10⁻⁷ × 0.1 / 4 × 10⁻⁴ = 2 × 10⁻⁷ × 250 = 5 × 10⁻⁵ T
Answer: 5 × 10⁻⁵ T
Q15Thick wire · numeric
For the same wire (radius 2 cm, current 10 A), the field at a point 4 cm from the axis is:
(a) 2.5 × 10⁻⁵ T
(b) 5 × 10⁻⁵ T
(c) 1 × 10⁻⁴ T
(d) zero
Show step-by-step solution
Givena = 0.02 m, I = 10 A, r = 4 cm = 0.04 m (outside)
AskedMagnetic field B at r = 4 cm
ConceptThis point is outside, so the whole current is enclosed. Compare your answer with the previous question — the result is striking.
FormulaB = μ₀ I / 2πr = (2 × 10⁻⁷) I / r
Solutionr = 0.04 m is greater than a = 0.02 m, so the point is OUTSIDE. B = 2 × 10⁻⁷ × 10 / 0.04 = 2 × 10⁻⁶ / 0.04 = 5 × 10⁻⁵ T Note: this equals the value at r = a/2 — a point inside and a point outside can have the SAME field strength.
Answer: 5 × 10⁻⁵ T
Q16Thick wire · concept
The magnetic field inside a hollow current-carrying cylindrical pipe (in the empty region inside) is:
(a) μ₀I / 2πr
(b) μ₀I / 2R
(c) zero
(d) μ₀I r / 2πR²
Show step-by-step solution
GivenHollow pipe carrying current along its walls, point inside the cavity
AskedField in the hollow region
ConceptDraw an Amperian loop inside the cavity: it encloses no current at all.
Formula∮ B·dl = μ₀ I_enclosed
SolutionAll the current flows along the walls, outside the loop. I_enclosed = 0 Therefore B × 2πr = 0, so B = 0 everywhere inside the cavity.
Answer: zero
Q17Ampere's law · numeric
An Amperian loop encloses two wires carrying 3 A and 2 A in opposite directions. The value of ∮B·dl for the loop is:
(a) μ₀ (1 A)
(b) μ₀ (5 A)
(c) μ₀ (6 A)
(d) zero
Show step-by-step solution
GivenEnclosed currents: 3 A one way, 2 A the other way
AskedValue of ∮B·dl
ConceptCurrents in opposite senses count with opposite signs, so take the algebraic sum.
Formula∮ B·dl = μ₀ I_enclosed(net)
SolutionNet enclosed current = 3 − 2 = 1 A ∮B·dl = μ₀ × 1 = μ₀ (in SI units)
Answer: μ₀ (1 A)
Q18Circular coil
A circular coil of 100 turns and radius 10 cm carries a current of 1 A. The magnetic field at its centre is:
(a) 6.28 × 10⁻⁴ T
(b) 6.28 × 10⁻⁵ T
(c) 3.14 × 10⁻⁴ T
(d) 1.256 × 10⁻³ T
Show step-by-step solution
GivenN = 100, R = 10 cm = 0.1 m, I = 1 A
AskedMagnetic field at the centre
ConceptThe centre-of-coil formula. N here is the TOTAL number of turns.
If the radius of a current-carrying circular coil is doubled while the current stays the same, the field at its centre becomes:
(a) double
(b) half
(c) four times
(d) one quarter
Show step-by-step solution
GivenR → 2R, same I and N
AskedNew field at the centre
ConceptRead the dependence straight off the formula: B is inversely proportional to R.
FormulaB = μ₀NI / 2R ⇒ B ∝ 1/R
SolutionDoubling R halves B. B′ = B / 2
Answer: half
Q21Circular coil · shape
A wire carrying current I is bent into a single circular loop of radius R, giving a field B at its centre. The same wire is re-bent into n turns of smaller radius. The new field at the centre is:
(a) nB
(b) n²B
(c) B/n
(d) B/n²
Show step-by-step solution
GivenSame wire, 1 turn of radius R → n turns, same current
AskedNew field at the centre
ConceptThe wire length is fixed, so packing n turns shrinks the radius to R/n. Then apply the centre formula.
FormulaB = μ₀ N I / 2R
SolutionWire length: 2πR = n × 2πr, so r = R/n. B′ = μ₀ n I / 2(R/n) = μ₀ n² I / 2R B′ = n² B
Answer: n²B
Q22Circular coil · axial
The ratio of the magnetic field at the centre of a circular coil to that at an axial point a distance R from the centre is:
(a) √2 : 1
(b) 2 : 1
(c) 2√2 : 1
(d) 4 : 1
Show step-by-step solution
GivenAxial point at x = R
AskedRatio B_centre : B_axial
ConceptSubstitute x = R into the axial formula and compare with the centre formula.
At a point on the axis of a circular coil, very far from it (x ≫ R), the magnetic field varies as:
(a) 1 / x
(b) 1 / x²
(c) 1 / x³
(d) 1 / x⁴
Show step-by-step solution
GivenAxial point with x ≫ R
AskedDependence of B on x
ConceptDrop R² compared with x² in the denominator of the axial formula.
FormulaB = μ₀IR² / 2(x²+R²)^{3/2}
SolutionFor x ≫ R, (x² + R²)^{3/2} ≈ x³ B ≈ μ₀IR² / 2x³ So B ∝ 1/x³
Answer: 1 / x³
Q25Arc
The magnetic field at the centre of a semicircular wire of radius R carrying current I is:
(a) μ₀I / 2R
(b) μ₀I / 4R
(c) μ₀I / 8R
(d) μ₀I / πR
Show step-by-step solution
GivenSemicircular arc of radius R, current I
AskedField at the centre
ConceptA semicircle is exactly half a full loop, so it gives half the field.
FormulaB_arc = μ₀Iθ / 4πR, with θ = π
SolutionB = μ₀ I π / (4πR) B = μ₀I / 4R (Equivalently: half of the full-loop value μ₀I/2R.)
Answer: μ₀I / 4R
Q26Arc
The magnetic field at the centre of a quarter-circular arc of radius R carrying current I is:
(a) μ₀I / 2R
(b) μ₀I / 4R
(c) μ₀I / 8R
(d) μ₀I / 16R
Show step-by-step solution
GivenQuarter circle of radius R, current I
AskedField at the centre
ConceptA quarter circle sweeps π/2 radians — one quarter of a full loop.
FormulaB_arc = μ₀Iθ / 4πR, with θ = π/2
SolutionB = μ₀ I (π/2) / (4πR) B = μ₀I / 8R
Answer: μ₀I / 8R
Q27Arc
A wire is bent into an arc subtending 60° at its centre, radius R, carrying current I. The field at the centre is:
(a) μ₀I / 6R
(b) μ₀I / 12R
(c) μ₀I / 24R
(d) μ₀I / 3R
Show step-by-step solution
GivenArc angle 60°, radius R, current I
AskedField at the centre
ConceptConvert the angle to radians first — the arc formula needs radians, not degrees.
FormulaB = μ₀Iθ / 4πR
Solution60° = π/3 radians B = μ₀ I (π/3) / (4πR) B = μ₀I / 12R
Answer: μ₀I / 12R
Q28Arc · numeric
A straight wire carrying 12 A is bent into a semicircular arc of radius 2 cm, with straight segments extending outwards along the diameter on both sides. The field at the centre of the arc is:
(a) 0.9 × 10⁻⁴ T
(b) 1.9 × 10⁻⁴ T
(c) 3.8 × 10⁻⁴ T
(d) zero
Show step-by-step solution
GivenI = 12 A, R = 2 cm = 0.02 m, semicircle plus two straight segments along the diameter
AskedField at the centre
ConceptThe straight parts lie along the line to the centre, so they contribute NOTHING. Only the semicircle counts.
FormulaB = μ₀I / 4R
SolutionFor the straight segments, dl and r are parallel ⇒ sin θ = 0 ⇒ no contribution. B = (4π × 10⁻⁷ × 12) / (4 × 0.02) = (4π × 10⁻⁷ × 12) / 0.08 = 4π × 10⁻⁷ × 150 = 1.9 × 10⁻⁴ T
Answer: 1.9 × 10⁻⁴ T
Q29Arc · concept
In a bent-wire arrangement, straight segments that lie along the line joining them to the field point contribute:
(a) the maximum field
(b) half the field
(c) nothing to the field
(d) a field along the wire
Show step-by-step solution
GivenStraight segment pointing directly at the field point
AskedContribution to B
ConceptThis is the Biot–Savart blind spot again, and it is the single most useful shortcut for bent-wire questions.
FormuladB = (μ₀/4π) I dl sin θ / r²
SolutionFor such a segment, dl and r point along the same line ⇒ θ = 0°. sin 0° = 0, so dB = 0 for every element of that segment. The segment contributes nothing at all.
Answer: nothing to the field
Q30Circular coil · vectors
Two identical circular coils, each producing a field B at their common centre, are placed with their planes perpendicular to each other. The resultant field at the centre is:
(a) zero
(b) B
(c) √2 B
(d) 2B
Show step-by-step solution
GivenTwo identical coils, each giving B at the centre, planes perpendicular
AskedResultant field
ConceptThe field of each coil points along its own axis. Perpendicular planes mean perpendicular axes, so add as perpendicular vectors.
FormulaB_net = √(B₁² + B₂²)
SolutionThe two field vectors are at 90° to each other. B_net = √(B² + B²) = √(2B²) = √2 B
Answer: √2 B
Q31Circular coil · split loop
A circular loop is fed at two diametrically opposite points, so the current divides equally between the two semicircular halves. The magnetic 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 into two semicircles
AskedField at the centre
ConceptEach half gives a semicircle field, but the two currents travel in OPPOSITE senses around the loop, so their fields oppose.
FormulaB_semicircle = μ₀I′ / 4R, added as vectors
SolutionEach half carries I/2. Each semicircle would give μ₀(I/2)/4R = μ₀I/8R. One goes clockwise, the other anticlockwise ⇒ the fields point opposite ways. They are equal and opposite, so they cancel exactly: B = 0.
Answer: zero
Q32Circular coil · split loop
A circular loop of radius r is fed so that arc ABC carries twice the current of arc ADC, the two arcs being of the same length. If the total current entering is I₀, the field at the centre O is:
(a) μ₀I₀ / 6r
(b) μ₀I₀ / 12r
(c) μ₀I₀ / 4r
(d) zero
Show step-by-step solution
GivenTwo semicircular arcs, currents in the ratio 2 : 1, total current I₀
AskedField at the centre
ConceptSplit the current, compute each semicircle's field, then subtract because the two arcs carry current in opposite senses round the loop.
FormulaB_semicircle = μ₀I′ / 4r
SolutionCurrents divide as 2 : 1, so I₁ = 2I₀/3 and I₂ = I₀/3. B₁ = μ₀(2I₀/3)/4r = μ₀I₀/6r B₂ = μ₀(I₀/3)/4r = μ₀I₀/12r Opposite senses ⇒ subtract: B = μ₀I₀/6r − μ₀I₀/12r = μ₀I₀/12r
Answer: μ₀I₀ / 12r
Q33Solenoid
A long solenoid has 1000 turns per metre and carries a current of 5 A. The magnetic field inside it is:
(a) 6.28 × 10⁻³ T
(b) 6.28 × 10⁻⁴ T
(c) 1.256 × 10⁻² T
(d) 3.14 × 10⁻³ T
Show step-by-step solution
Givenn = 1000 turns/m, I = 5 A
AskedField inside the solenoid
ConceptThe turns per unit length is given directly, so substitute straight in.
A solenoid 50 cm long with 100 turns carries a current of 2.5 A. The field at its centre is:
(a) 6.28 × 10⁻⁴ T
(b) 3.14 × 10⁻⁴ T
(c) 6.28 × 10⁻³ T
(d) 1.256 × 10⁻³ T
Show step-by-step solution
GivenL = 50 cm = 0.5 m, N = 100 turns, I = 2.5 A
AskedField at the centre
ConceptThe TOTAL turns are given, so first convert to turns per metre. Forgetting this step is the commonest error in solenoid questions.
Formulan = N / L; B = μ₀ n I
Solutionn = 100 / 0.5 = 200 turns per metre B = 4π × 10⁻⁷ × 200 × 2.5 = 4π × 10⁻⁷ × 500 = 6.28 × 10⁻⁴ T
Answer: 6.28 × 10⁻⁴ T
Q35Solenoid
A long solenoid of radius 1 mm has 100 turns per mm and carries a current of 1 A. The magnetic field at its centre is:
(a) 12.56 × 10⁻² T
(b) 12.56 × 10⁻⁴ T
(c) 1.256 × 10⁻² T
(d) 6.28 × 10⁻² T
Show step-by-step solution
Givenradius = 1 mm, n = 100 turns per mm, I = 1 A
AskedField at the centre
ConceptThe radius is a deliberate distractor — it does not appear in the formula at all. Convert the turn density to per metre.
FormulaB = μ₀ n I
Solutionn = 100 per mm = 100 × 1000 = 10⁵ turns per metre B = 4π × 10⁻⁷ × 10⁵ × 1 = 4π × 10⁻² = 12.56 × 10⁻² T The radius plays no part whatsoever.
Answer: 12.56 × 10⁻² T
Q36Solenoid · concept
The magnetic field inside a long solenoid depends on:
(a) its radius only
(b) its length only
(c) the turns per unit length and the current
(d) the total number of turns only
Show step-by-step solution
GivenLong solenoid carrying steady current
AskedWhat B depends on
ConceptRead the formula: only n and I appear. There is no R and no position term.
FormulaB = μ₀ n I
SolutionThe formula contains only μ₀, n and I. The radius does not appear, so a fat and a thin solenoid with the same n and I give the same field. The position inside does not appear either — that is why the field is uniform.
Answer: the turns per unit length and the current
Q37Solenoid · concept
The magnetic field at one END of a long solenoid, compared with the field at its centre, is:
(a) the same
(b) half
(c) double
(d) zero
Show step-by-step solution
GivenLong solenoid, point at the end on the axis
AskedField at the end
ConceptAt the end, the 'very long on both sides' assumption fails — only half the winding lies to one side.
FormulaB_end = μ₀ n I / 2
SolutionAt the centre, turns extend far in both directions ⇒ B = μ₀nI. At the end, only one side contributes ⇒ B = μ₀nI / 2. So it is exactly half.
Answer: half
Q38Solenoid · ratio
A solenoid is stretched to twice its original length while the number of turns and the current stay the same. The field inside becomes:
(a) double
(b) half
(c) four times
(d) unchanged
Show step-by-step solution
GivenL → 2L, same N and I
AskedNew field inside
ConceptStretching spreads the same turns over a longer length, so n drops.
FormulaB = μ₀ n I with n = N/L
Solutionn′ = N / 2L = n / 2 B′ = μ₀ (n/2) I = B / 2
Answer: half
Q39Solenoid · concept
The magnetic field outside a very long current-carrying solenoid is:
(a) equal to that inside
(b) half that inside
(c) essentially zero
(d) double that inside
Show step-by-step solution
GivenPoint just outside a very long solenoid
AskedField outside
ConceptThe field spreads over a huge outside volume and becomes negligible — this is exactly what lets us drop one side of the Amperian rectangle.
FormulaAmpere's law applied to a rectangular loop
SolutionInside, the field is strong and uniform along the axis. Outside, it is spread over a vast region and is taken as zero. That assumption is what makes side cd of the Amperian rectangle contribute nothing.
Answer: essentially zero
Q40Solenoid · concept
A current-carrying solenoid behaves like:
(a) a single long straight wire
(b) a bar magnet
(c) an electric dipole
(d) a point charge
Show step-by-step solution
GivenLong solenoid carrying steady current
AskedEquivalent magnetic object
ConceptCompare the external field-line pattern — it is indistinguishable from a bar magnet's, but with an off switch.
FormulaRight-hand rule for a loop determines the poles
SolutionThe field lines emerge from one end and return to the other, exactly like a bar magnet. One face acts as the north pole, the other as the south. Switch the current off and the magnetism disappears — this is the electromagnet.
Answer: a bar magnet
Q41Solenoid · concept
Which Amperian loop is used to derive the field inside a long solenoid, and how many of its sides contribute?
(a) A circle; all of it contributes
(b) A rectangle straddling the winding; one side contributes
(c) A rectangle inside only; two sides contribute
(d) A square outside; none contributes
Show step-by-step solution
GivenDerivation of B = μ₀nI
AskedChoice of loop and contributing sides
ConceptThe rectangle is chosen so that three of its four sides give zero — that is the whole point of a clever Amperian loop.
Formula∮ B·dl = μ₀ I_enclosed
SolutionSide inside along the axis: B is parallel to the path ⇒ contributes B × h. The two sides crossing the wall: B is perpendicular to the path ⇒ zero. The side outside: B ≈ 0 ⇒ zero. So only one side contributes: Bh = μ₀(nh)I ⇒ B = μ₀nI.
Answer: A rectangle straddling the winding; one side contributes
Q42Biot–Savart
According to the Biot–Savart law, the magnetic field due to a current element is:
(a) along the current element
(b) along the line joining the element to the point
(c) perpendicular to the plane containing the element and that line
(d) always zero
Show step-by-step solution
GivenCurrent element I dl and displacement r to the point
AskedDirection of dB
ConceptThe cross product makes the field perpendicular to both vectors at once.
FormuladB ∝ I dl × r
SolutionA cross product is perpendicular to both of its inputs. So dB is perpendicular to both dl and r. Equivalently, it is perpendicular to the plane containing them.
Answer: perpendicular to the plane containing the element and that line
Q43Biot–Savart · numeric
A current element of length 1 cm carries a current of 10 A. The magnetic field at a point 0.5 m away, in a direction perpendicular to the element, is:
(a) 2 × 10⁻⁸ T
(b) 4 × 10⁻⁸ T
(c) 4 × 10⁻⁷ T
(d) 2 × 10⁻⁶ T
Show step-by-step solution
Givendl = 1 cm = 0.01 m, I = 10 A, r = 0.5 m, θ = 90°
AskedMagnetic field dB
ConceptDirect substitution into Biot–Savart, using μ₀/4π = 10⁻⁷ and sin 90° = 1.
Which of the following is a difference between the Biot–Savart law and Coulomb's law?
(a) Biot–Savart is not an inverse-square law
(b) The magnetic source is a vector (I dl) while the electric source is a scalar (q)
(c) Superposition does not apply to magnetic fields
(d) The magnetic field is along the displacement vector
Show step-by-step solution
GivenComparison of the two laws
AskedThe correct difference
ConceptBoth are inverse-square and both obey superposition. The real differences are the vector source, the perpendicular direction, and the angle dependence.
FormuladB ∝ I dl × r / r³; E ∝ q / r²
SolutionBoth ARE inverse-square laws, so the first option is wrong. Superposition applies to both, so the third is wrong. The magnetic field is PERPENDICULAR to r, not along it, so the fourth is wrong. The electric source q is a scalar; the magnetic source I dl carries a direction — this is the genuine difference.
Answer: The magnetic source is a vector (I dl) while the electric source is a scalar (q)
Q45Constants
The relation between ε₀, μ₀ and the speed of light c is:
(a) ε₀μ₀ = c
(b) ε₀μ₀ = 1/c
(c) ε₀μ₀ = c²
(d) ε₀μ₀ = 1/c²
Show step-by-step solution
GivenPermittivity ε₀, permeability μ₀, speed of light c
AskedThe correct relation
ConceptThe electric constant and the magnetic constant together produce the speed of light — the clue that led Maxwell to realise light is electromagnetic.
Assertion (A): The magnetic field inside a long solenoid does not depend on its radius. Reason (R): The field inside a long solenoid is uniform and directed along its axis.
(a) Both A and R are true, and R is the correct explanation of A
(b) Both A and R are true, but R is not the correct explanation of A
(c) A is true but R is false
(d) A is false but R is true
Show step-by-step solution
GivenLong solenoid, B = μ₀nI
AskedTruth of A and R, and whether R explains A
ConceptBoth statements are correct facts about a solenoid, but check carefully whether the second one actually causes the first.
FormulaB = μ₀ n I
SolutionA is TRUE: the formula B = μ₀nI contains no radius term. R is TRUE: the field inside is indeed uniform and axial. But uniformity is not the REASON the radius drops out — that comes from the Amperian loop calculation, where the enclosed current depends on n and h only. So both are true, but R is not the correct explanation.
Answer: Both A and R are true, but R is not the correct explanation of A
Q47Assertion–Reason
Assertion (A): The magnetic field at the centre of a solid current-carrying wire is zero. Reason (R): An Amperian loop of vanishing radius drawn at the axis encloses no current.
(a) Both A and R are true, and R is the correct explanation of A
(b) Both A and R are true, but R is not the correct explanation of A
(c) A is true but R is false
(d) A is false but R is true
Show step-by-step solution
GivenSolid conductor with uniform current, point on the axis
AskedTruth of A and R
ConceptUse the inside-the-wire result B = μ₀Ir/2πa² and set r = 0.
FormulaB = μ₀ I r / 2πa²
SolutionA is TRUE: putting r = 0 gives B = 0 on the axis. R is TRUE: a loop of zero radius encloses zero current. And that is exactly WHY the field is zero there — so R correctly explains A.
Answer: Both A and R are true, and R is the correct explanation of A
Q48Match the following
Match each configuration with the field it produces:
(a) Long straight wire (b) Centre of a circular coil of N turns (c) Inside a long solenoid (d) Centre of a semicircular arc
(i) μ₀nI (ii) μ₀I/2πr (iii) μ₀I/4R (iv) μ₀NI/2R
(a) a–ii, b–iv, c–i, d–iii
(b) a–ii, b–i, c–iv, d–iii
(c) a–iii, b–iv, c–i, d–ii
(d) a–ii, b–iv, c–iii, d–i
Show step-by-step solution
GivenFour standard configurations
AskedCorrect matching
ConceptThis is the recognition drill in question form — the single most valuable skill in this topic.
FormulaStandard field formulas
Solution(a) Long straight wire ⇒ μ₀I/2πr ⇒ (ii) (b) Coil of N turns at the centre ⇒ μ₀NI/2R ⇒ (iv) (c) Inside a solenoid ⇒ μ₀nI ⇒ (i) (d) Semicircular arc ⇒ μ₀I/4R ⇒ (iii)
Answer: a–ii, b–iv, c–i, d–iii
Q49Match the following
Match each situation with how the field varies with distance:
(a) Outside a long straight wire (b) Inside a solid current-carrying wire (c) Far along the axis of a small loop (d) Inside a hollow current-carrying pipe
(i) B ∝ r (ii) B = 0 (iii) B ∝ 1/r (iv) B ∝ 1/r³
(a) a–iii, b–i, c–iv, d–ii
(b) a–i, b–iii, c–iv, d–ii
(c) a–iii, b–ii, c–iv, d–i
(d) a–iv, b–i, c–iii, d–ii
Show step-by-step solution
GivenFour different regions
AskedCorrect matching of the distance dependence
ConceptThese four dependences are the ones NEET mixes up in graph questions.
FormulaStandard results from Ampere's law and Biot–Savart
Solution(a) Outside a straight wire: B = μ₀I/2πr ⇒ B ∝ 1/r ⇒ (iii) (b) Inside a solid wire: B = μ₀Ir/2πa² ⇒ B ∝ r ⇒ (i) (c) Far along a loop's axis: B ∝ 1/x³ ⇒ (iv) (d) Inside a hollow pipe: no enclosed current ⇒ B = 0 ⇒ (ii)
Answer: a–iii, b–i, c–iv, d–ii
Q50Recognition
A question describes 'a tightly wound long coil of 500 turns over a length of 25 cm carrying 4 A'. Which formula applies to the field at its centre?
(a) B = μ₀NI/2R
(b) B = μ₀nI
(c) B = μ₀I/2πr
(d) B = μ₀Iθ/4πR
Show step-by-step solution
Given500 turns over 25 cm, current 4 A, tightly wound and long
AskedThe correct formula to use
ConceptThe words 'long', 'tightly wound' and a turns-per-length description signal a solenoid, not a coil. Getting this right is worth more than the arithmetic.
FormulaB = μ₀ n I with n = N/L
Solution'Long, tightly wound coil' with a LENGTH given = solenoid. So use B = μ₀nI, not the coil formula. n = 500 / 0.25 = 2000 turns per metre B = 4π × 10⁻⁷ × 2000 × 4 = 1.005 × 10⁻² T
Answer: B = μ₀nI
Before the next ILTS — how to use this pack
Read Part 1 once, slowly. Every trap in Part 3 is explained there first.
Copy the formula sheet out by hand. Then cover it and rewrite it from memory until it comes out complete.
Q1–Q10 (straight wire) in one sitting. These are the fastest marks here.
Q11–Q20 (Ampere's law and the thick wire) next. Derive the inside-and-outside result once yourself, then sketch the graph from memory — that single derivation covers five past papers.
Q21–Q34 (coils and arcs) the following day. Say out loud whether each question is a full loop, a semicircle or an arc before writing anything.
Q35–Q43 (solenoid) — quick and mechanical. The only real risk is the turns-per-metre conversion.
Q44–Q50 are Biot–Savart theory, assertion–reason and matching. Save them for final revision.
The four mistakes that cost the most marks in this topic
Using the coil formula for a solenoid, or the other way round. Check whether the question gives a total number of turns with a radius (coil) or turns over a length (solenoid).
Treating the straight-wire field as inverse-square. It is 1/r, not 1/r².
Forgetting that straight sections pointing at the centre contribute nothing.
Not converting cm to m, and — in solenoid questions — not converting turns per mm to turns per metre.