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Pattern analysis · Sets 1 & 2 · 43 solved questions

Every question type this chapter asks

Thirteen recurring templates behind all the questions worked so far — the words that announce each one, the method it wants, and the variants an examiner can still spin from it.

Where the questions actually fell

43 solved · 7 pending
Read this as a study order, not a prediction. The counts come from the 43 questions worked through so far. Anything above four appearances deserves drilling first; everything on the list is fair game in an exam.
Charged particle: r, T, f, ratios7
Ampère’s law bodies6
Force on wires & loops4
Direction & cross products4
Re-bent wire (N² family)3
Rotating charge → current3
Axial field of a loop3
Superposition of two sources3
Instruments3
Arcs & composite loops2
Torque, moment, work2
Conceptual & statement-based3
Cyclotron1

The thirteen templates

Recognise → method → variants
T1

Radius, period or frequency of a charged particle

7 appearances · highest yield

Announced by

accelerated through a potential difference kinetic energy E specific charge enters normally / transverse field rotations per second ratio of masses

Method

  1. Read what is given: v, p, E or V.
  2. Pick the matching form of r — do not force one.
  3. For ratios, cancel everything shared before substituting.
r = mvqB = pqB = √2mEqB = 1B√2mV/q  ·  T = 2πmqB

Variants still possible

  • Ratio of radii given masses, or of charges given radii
  • Area enclosed (∝ E, ∝ 1/q²)
  • Two isotopes in a mass spectrometer
  • “Which quantity is independent of…” conceptual form
  • Pitch of a helix when v is not perpendicular

The standing trap

  • Writing r ∝ m when V is fixed. Same V gives a heavier particle a lower speed, so r ∝ √m.
  • Assuming T depends on speed. It never does.

Seen in: S1 Q2, Q12, Q14, Q15, Q25 · S2 Q13, Q23

T2

Ampère’s law bodies — wire, tube, solenoid, toroid

6 appearances

Announced by

uniformly distributed over its cross-section thin walled tube open space inside turns per cm / per metre average radius of the toroid

Method

  1. Locate the point: inside the metal, in a hollow, or outside.
  2. Ask only one question — how much current does my loop enclose?
  3. Apply B·2πr = μ₀Ienc.
solid inside: μ₀Ir2πa2 · outside: μ₀I2πr · solenoid: μ₀nI · toroid: μ₀NI2πr

Variants still possible

  • Coaxial cable — inner conductor plus outer return
  • Field at the end of a solenoid (half the value)
  • Graph of B versus r for a thick wire
  • Iron-cored solenoid: replace μ₀ by μ₀μr

The standing trap

  • “Thin walled” versus “uniformly distributed” — one gives zero inside, the other gives μ₀Ir/2πa².
  • Turns per centimetre quoted, metre needed.

Seen in: S1 Q4, Q20, Q21 · S2 Q2, Q10, Q11

T3

Force on a conductor, loop or beam

4 appearances

Announced by

loop placed near a long straight wire coplanar the loop will… two parallel wires beams of electrons / positrons

Method

  1. Delete the sides whose forces cancel by symmetry.
  2. Compute the near and far forces separately.
  3. Subtract — near side always wins, so a loop is always pulled in.
F = BIL sin θ · FL = μ₀I₁I₂2πd · beams: FBFE = v2c2

Variants still possible

  • Numerical net force on a square or rectangular loop
  • Triangular or circular loop near a wire
  • Loop carrying current in the reversed sense (repulsion)
  • Force per metre between three parallel wires
  • Current-carrying wire suspended in equilibrium against gravity

The standing trap

  • Adding the near and far forces instead of subtracting.
  • Treating a free charged beam like a neutral wire — beams repel, wires attract.

Seen in: S1 Q3, Q10, Q17 · S2 Q4

T4

Direction questions and the Lorentz split

4 appearances

Announced by

east to west / vertically upward released from rest projected towards north Bî with velocity vĵ at a short distance below it

Method

  1. If the charge starts at rest, that case gives E alone — magnetism cannot touch it.
  2. Subtract the electric part from the moving case to isolate the magnetic part.
  3. Fix direction by right hand, then check the sign of the charge.
F⃗ = q(E⃗ + v⃗ × B⃗) · î × ĵ = k̂ · ĵ × î = −k̂

Variants still possible

  • Velocity selector: qE = qvB, so v = E/B
  • Charge undeflected — find the required field
  • Which way does the particle curve, clockwise or anticlockwise
  • Field direction above versus below a wire

The standing trap

  • Reversing the cross-product order.
  • Forgetting an electron’s force is opposite to v × B.

Seen in: S1 Q8, Q22 · S2 Q8, Q14

T5

The re-bent wire — anything ∝ N²

3 appearances · nearly guaranteed

Announced by

it is then bent into same wire same material and same length smaller circular coil

Method

  1. Write the length constraint: N(2πr) = 2πR.
  2. Get r = R/N.
  3. Substitute into B = μ₀NI/2r — the N appears twice.
B = πμ₀N2IL ⇒ B ∝ N2

Variants still possible

  • Magnetic moment of the re-bent coil (M ∝ 1/N)
  • Wire bent into a square instead of a circle — compare fields
  • Same wire made into a solenoid
  • Given B and B′, find N

The standing trap

  • Answering nB. That is right only when a fresh, longer wire keeps the radius the same.

Seen in: S1 Q1, Q9 · S2 Q9

T6

Rotating charge converted into a current

3 appearances

Announced by

makes n rotations per second rotating with angular velocity ω charge uniformly distributed over its surface non-conducting disc / ring

Method

  1. Total charge first: λ·2πr for a ring, σ·πR² for a disc.
  2. Convert: I = Qf = Qω/2π.
  3. Ring → use the formula directly. Disc → integrate over rings.
I = Qω2π · Mring = ½qωR2 · Mdisc = ¼qωR2

Variants still possible

  • Field at the centre of a rotating disc (μ₀σωR/2)
  • Rotating charged sphere or spherical shell
  • Ratio M/L = q/2m for an orbiting particle
  • Bohr magneton / orbital magnetic moment of an electron

The standing trap

  • Treating a disc as one loop. The disc gives half the ring’s moment.
  • Missing that r cancels for a ring, so the answer holds no R at all.

Seen in: S1 Q6, Q13 · S2 Q3

T7

Field on the axis of a loop

3 appearances

Announced by

at a point on the axis at a distance x from the centre ratio of the induced fields variation of B as X varies

Method

  1. Never substitute numbers first — take the ratio of the two positions.
  2. μ₀, I and R² cancel, leaving pure geometry.
  3. Cube-root the ratio to strip the 3/2 power.
B = μ₀IR22(R2+x2)3/2 · BcentreBaxis = (R2+x2R2)3/2

Variants still possible

  • Find x where B falls to half or an eighth of the centre value
  • Helmholtz coils — separation equal to R for a uniform field
  • Inflection points at x = ±R/2
  • Far-field limit: B ≈ μ₀M/2πx³, matching a dipole

The standing trap

  • Confusing dB/dx = 0 (only at the centre) with d²B/dx² = 0 (at the inflections). Flat is not the same as straight.

Seen in: S1 Q11, Q23 · S2 Q15

T8

Two sources at one point — add, subtract or Pythagoras

3 appearances

Announced by

concentric coils at right angles to each other in opposite order / opposite sense perpendicular to the plane containing the wires

Method

  1. Compute each field on its own.
  2. Decide the angle between them — planes at 90° means fields at 90°.
  3. Combine by the right rule for that angle.
same plane: B₁ ± B₂ · at 90°: √B₁2+B₂2 · at θ: √B₁2+B₂2+2B₁B₂cos θ

Variants still possible

  • Find the current that makes the resultant zero
  • Angle of the resultant, not just its size
  • Coils at 60° or 120°
  • Wire plus loop at the same point

The standing trap

  • Adding perpendicular fields arithmetically. 3 + 4 = 7 is always planted as an option; 5 is the answer.

Seen in: S1 Q18, Q24 · S2 Q7

T9

Instruments — galvanometer, ammeter, voltmeter

3 appearances

Announced by

converted into an ammeter of range permissible current through its coil magnetic meridian radial field / concave poles needle deflects by

Method

  1. Ammeter → parallel shunt, equate the two branch voltages.
  2. Voltmeter → series resistance, use the full range voltage.
  3. Tangent galvanometer → the needle sits along the resultant of two ⟂ fields.
S = IgGI − Ig · R = VIg − G · tan θ = μ₀NI2RBH

Variants still possible

  • Fraction of total current through the coil
  • Effective resistance of the converted meter
  • Current and voltage sensitivity, and how to raise them
  • Why an ideal ammeter has zero resistance
  • Reduction factor of a tangent galvanometer

The standing trap

  • Swapping series and parallel. Ammeter goes in the circuit, so its resistance must be tiny.

Seen in: S2 Q1, Q12, Q24

T10

Arcs and composite loops

2 appearances

Announced by

semicircular portion of radius R the shape as shown in figure linear parts are very long central point O

Method

  1. Cut the shape into arcs and straight pieces.
  2. Delete every radial straight bit — it contributes nothing.
  3. Each arc uses its own radius and its own angle fraction.
Barc = θ2π·μ₀I2R = μ₀Iθ4πR · semi-infinite wire: μ₀I4πd

Variants still possible

  • Quarter circle plus two straight arms
  • Two arcs of the same radius, opposite senses (subtract)
  • Square or hexagonal loop — field at the centre
  • Answer demanded in î, ĵ, k̂ form

The standing trap

  • Using one radius for both arcs.
  • Reading the arc angle off a rough figure. Use the 360° closure and the printed options instead.

Seen in: S1 Q5, Q7

T11

Torque, magnetic moment and work

2 appearances

Announced by

work done in rotating it through from its equilibrium position placed in a magnetic field of magnetic moment of

Method

  1. Compute M = NIA first, with A in m².
  2. Read the starting angle. Equilibrium means θ₁ = 0.
  3. Use the energy difference, not the torque.
M = NIA · τ = MB sin θ · U = −MB cos θ · W = MB(cos θ₁ − cos θ₂)

Variants still possible

  • Maximum torque on a coil
  • Work for 0→90° (MB) or 90→180° (MB)
  • Stable versus unstable equilibrium
  • Period of small oscillations of a magnet or coil
  • Net torque on a loop near a straight wire (zero when coplanar)

The standing trap

  • Using MB instead of 2MB for a 180° turn.
  • Leaving the radius in centimetres inside A = πr².

Seen in: S2 Q3, Q5

T12

Conceptual, assertion–reason and statement sets

3 appearances · rising in new papers

Announced by

which of the following is false Assertion (A) … Reason (R) i) ii) iii) iv) is independent of the field around a wire has

Method

  1. Write the governing formula first, then read each statement against it.
  2. For assertion–reason, judge A and R separately before asking whether R explains A.
  3. Look for what is absent from the formula — that is usually the answer.
W = μ₀mI per lap · ∮B⃗·dl⃗ = μ₀I ≠ 0 · cylindrical symmetry

Variants still possible

  • Why a magnetic force does no work on a moving charge
  • Why Ampère’s law needs a symmetric loop
  • Biot–Savart versus Coulomb: similarities and differences
  • Why magnetic monopoles do not appear
  • Assertion–reason on radial fields, shunts, or the cyclotron

The standing trap

  • Assuming a closed path gives zero work. True for electrostatics, false for the field of a current.

Seen in: S1 Q11, Q19 · S2 Q10, Q12, Q25

T13

Cyclotron

1 appearance · a standing one-marker

Announced by

cyclotron dees resonance condition used to accelerate

Method

  1. Magnetic field bends inside the dees; the electric field in the gap does all the accelerating.
  2. Frequency is fixed because T does not depend on speed or radius.
  3. Maximum energy is set by the dee radius.
f = qB2πm · Kmax = q2B2R22m · vmax = qBRm

Variants still possible

  • Numerical: find f, Kmax or the number of revolutions
  • Why electrons are unsuitable (mass and relativity, not sign)
  • Why neutrons cannot be accelerated at all
  • Effect of doubling B or the dee radius on the output energy

The standing trap

  • Saying “only positive particles”. The sign is irrelevant; zero charge is what disqualifies a neutron.

Seen in: S1 Q16

In the syllabus, not yet asked

The likely gaps
Fair game in any paper from this chapter
TopicFormula to have readyTypical question
Helical path / pitchpitch = v cos θ · T = 2πmv cos θ / qBCharge enters at an angle other than 90°
Velocity selectorv = E/BWhich speed passes undeflected through crossed fields
Finite straight wireB = μ₀I(sin θ₁ + sin θ₂)/4πdField at a point off the end of a short wire
Field at the centre of a polygonsquare: 2√2 μ₀I/πLSame wire bent into a square versus a circle
Moment-to-angular-momentum ratioM/L = q/2mGyromagnetic ratio of an orbiting electron
Galvanometer sensitivitycurrent: NAB/k · voltage: NAB/kRHow to increase sensitivity, and its cost
Helmholtz coilsseparation = R for uniformityWhy two coils give a flat field between them
Coaxial cableoutside both conductors: B = 0Field in each of the three regions
Two loops or a wire plus a loopsuperpositionFind the current that makes the net field zero
Mass spectrometerr = √(2mV/q)/BSeparating two isotopes

Trigger phrase → what to do

Read the question for these words first
If the question saysTypeImmediately
“it is then bent into n turns”T5r = R/n, expect a square in the answer
“accelerated through the same potential difference”T1r ∝ √m, expect a square root
“uniformly distributed over its cross-section”T2Ienc = I·r²/a²
“thin walled tube” / “open space inside”T2Answer is zero — stop calculating
“at right angles to each other”T8Pythagoras, never a plain sum
“in opposite order”T8Subtract the two fields
“coplanar with a long straight conductor”T3Near minus far; the loop is pulled in
“released from rest”T4That case isolates E; magnetism does nothing
“rotations per second” / “angular velocity ω”T6Convert to a current before anything else
“on the axis at a distance”T7Take a ratio; cube-root it
“permissible current through its coil”T9Shunt in parallel
“from its equilibrium position”T11θ₁ = 0, so a 180° turn costs 2MB
“specific charge”T1Keep q/m as one block; don’t split it
“turns per cm”T2Convert to per metre, or use a ratio
“which of the following is false”T12Write the formula, then test each line against it

Tactics that worked across all 43

Marks earned without full calculation
  1. Take the ratio before you take the calculator. Q2, Q7, Q9, Q15 and Q23 across the two sets all collapse when the two cases are divided — μ₀, I and even unit conversions vanish.
  2. Use the options as a measuring instrument. When a figure is ambiguous, compute the two or three plausible readings and see which one appears. That is how the 240°/120° split was fixed in Set 1 Q7.
  3. Check dimensions to delete options. A magnetic field formula without μ₀, a radius with charge under a root, an answer printed in m/s when m/s² was asked — all struck off on sight.
  4. Look for what the formula does not contain. The solenoid ignores its radius, the spinning ring ignores its radius, the loop force ignores L, the time period ignores speed. Every one of those is a question in itself.
  5. Spot the zero before you compute. Radial segments, hollow interiors, toroid holes, stationary charges, uniform-field translation. When a question hands you full numerical data but asks about such a region, the data is bait.
  6. Notice engineered numbers. A radius of 2π cm, a current of 0.9 A beside c = 3 × 10⁸, a 216:125 ratio, a 3–4–5 triangle. Clean numbers signal the intended route; if your working turns ugly, you have taken a wrong turn.