Moving Charges and Magnetism

The shape of this set

Every one of the eleven was attempted — there are no blanks. The willingness to commit is complete; accuracy is the whole gap.

The eleven fall into three families. Five turn on a direction or a sign (Q7, Q11, Q19, and the two graph questions). Three turn on a formula being inverted or half-applied (Q8, Q15, Q19). Three turn on reading the exact geometry — solid against hollow, plateau against decay, inside against outside.

Two are worth singling out. Q11 is the answer to a near-identical question from an earlier paper, applied after the sign of B was changed — recognition used as a substitute for reading. And Q23 carries a genuine error in the paper's own answer key, flagged in full on that card.

Q 4Attempted · wrongForce between parallel currents

Three long straight and parallel wires carrying currents are arranged as shown. Wire A carries 30 A upward, wire B carries 10 A downward, and wire C carries 20 A upward. A and B are 3 cm apart; B and C are 10 cm apart. The force experienced by conductor B, of length 25 cm, is

Her answer2 × 10⁻⁴ N from left to right

Correct answer4 × 10⁻⁴ N from left to right

1 Given

Wire A: Iₐ = 30 A, directed upward.
Wire B: Iₒ = 10 A, directed downward.
Wire C: I_C = 20 A, directed upward.
Separations: A to B = 3 cm = 0.03 m; B to C = 10 cm = 0.10 m.
Length of B under consideration: L = 25 cm = 0.25 m.
Implicit: μ₀/2π = 2 × 10⁻⁷ T m A⁻¹.

2 Asked

The magnitude and direction of the net magnetic force on wire B.

3 Concept

Two parallel currents exert a force on each other of F/L = μ₀I₁I₂/2πd. The direction follows one rule and one rule only:

Parallel currents ATTRACT. Antiparallel currents REPEL.

B carries current downward while both A and C carry current upward, so B is antiparallel to both and is therefore repelled by both. A sits on B's left and pushes it right; C sits on B's right and pushes it left. The two forces oppose, so the answer is a difference, not a sum.

4 Method and baby steps

Force from A on B. F = (2 × 10⁻⁷)(30)(10)(0.25)/0.03.
Numerator: 2 × 10⁻⁷ × 300 × 0.25 = 1.5 × 10⁻⁵.
Divide by 0.03: Fₐ = 5 × 10⁻⁴ N, directed to the right (repulsion).
Force from C on B. F = (2 × 10⁻⁷)(20)(10)(0.25)/0.10.
Numerator: 2 × 10⁻⁷ × 200 × 0.25 = 1 × 10⁻⁵.
Divide by 0.10: F_C = 1 × 10⁻⁴ N, directed to the left (repulsion).
Combine. They point in opposite directions, so subtract:
F = 5 × 10⁻⁴ − 1 × 10⁻⁴ = 4 × 10⁻⁴ N.
Direction. The larger force is the one from A, which pushes B to the right. So the net force is from left to right.

5 Easy tricks and shortcuts

Collect the constants once and the arithmetic is trivial. Since F/L = 2 × 10⁻⁷ I₁I₂/d, work out I₁I₂/d for each pair first:

A–B: 300/0.03 = 10 000 · C–B: 200/0.10 = 2 000

The ratio is exactly 5 : 1, so the answer is 4 parts out of 5 — you can see “4” before touching a calculator. Then one multiplication by 2 × 10⁻⁷ × 0.25 fixes the size.

Direction check in one line: A is closer and carries more current, so A wins, and A repels B away from itself — to the right.

6 Solution

F = 4 × 10⁻⁴ N, directed from left to right

7 Diagram

8 Weakness area

The magnitude came out as 2 × 10⁻⁴ instead of 4 × 10⁻⁴ — exactly half. Since the two individual forces are 5 and 1 units, a result of 2 does not correspond to either the sum (6) or the difference (4). The most likely cause is a slip in one of the two divisions, most probably using 0.06 m or 0.05 m for a separation.

The habit to build: compute I₁I₂/d for each pair and write both numbers down before combining. Seeing 10 000 and 2 000 side by side makes the 5 : 1 ratio obvious and makes any arithmetic slip visible immediately.

The direction was correct, which is worth noting — the antiparallel-repels rule was applied properly to both wires.

Q 7Attempted · wrongSuperposing two perpendicular fields

Two infinitely long straight wires are arranged perpendicular to each other in mutually perpendicular planes. I₁ = 2 A along the y-axis and I₂ = 3 A along the negative z-axis, with AP = AB = 1 cm. The magnetic induction B⃗ at point P is

Her answer(3 × 10⁻⁵ T)ĵ + (−4 × 10⁻⁵ T)k̂

Correct answer(3 × 10⁻⁵ T)ĵ + (4 × 10⁻⁵ T)k̂

1 Given

I₁ = 2 A along +y, passing through point A.
I₂ = 3 A along −z (into the page), passing through point B.
AP = 1 cm = 0.01 m, so P is 0.01 m from wire 1.
AB = 1 cm, so PB = AP + AB = 2 cm = 0.02 m from wire 2.
P lies on the negative-x side of both wires.

2 Asked

The resultant magnetic induction vector at P.

3 Concept

Each infinite wire contributes B = μ₀I/2πr, and the direction is given by the right-hand rule — equivalently by B̂ ∝ Î × r̂, where Î is the current direction and r̂ points from the wire to the field point.

The two contributions here turn out to be perpendicular to each other, so they cannot cancel or reinforce — they simply become the two components of the answer. All the difficulty is in the two directions.

4 Method and baby steps

Magnitudes first.
From wire 1: B₁ = (2 × 10⁻⁷)(2)/0.01 = 4 × 10⁻⁵ T.
From wire 2: B₂ = (2 × 10⁻⁷)(3)/0.02 = 3 × 10⁻⁵ T.
Direction from wire 1. Current is along +y, so Î = ĵ. P lies on the −x side, so r̂ = −î.
ĵ × (−î) = −(ĵ × î) = −(−k̂) = +k̂.
So B₁ = +4 × 10⁻⁵ k̂ T, out of the page.
Direction from wire 2. Current is along −z, so Î = −k̂. P lies on the −x side of B, so r̂ = −î.
(−k̂) × (−î) = +(k̂ × î) = +ĵ.
So B₂ = +3 × 10⁻⁵ ĵ T.
Add as vectors. The two lie along different axes, so they simply sit side by side:
B⃗ = 3 × 10⁻⁵ ĵ + 4 × 10⁻⁵ k̂ T.

5 Easy tricks and shortcuts

Keep the three cyclic cross products on hand and every direction becomes mechanical:
î × ĵ = k̂ · ĵ × k̂ = î · k̂ × î = ĵ
Reversing either factor flips the sign; reversing both leaves it unchanged. That second point is what saves you here — wire 2 has both a negative current direction and a negative position vector, so the two minus signs cancel and the answer is positive ĵ.

Magnitude shortcut: notice that doubling the current to 3 A while doubling the distance to 2 cm leaves B smaller, not equal — 3/2 against 2/1. A quick ratio check like this catches a mis-read distance.

6 Solution

B⃗ = 3 × 10⁻⁵ ĵ + 4 × 10⁻⁵ k̂ T (magnitude 5 × 10⁻⁵ T)

7 Diagram

8 Weakness area

Both magnitudes were right, and one of the two directions was right. The error is the sign of the k̂ component — it came out negative when it should be positive.

This is a double-negative situation: the position vector is −î and the current for wire 2 is −k̂. When two negatives appear in a cross product they cancel, and it is easy to carry one minus sign through and drop the other.

The habit to build: write Î and r̂ explicitly as signed unit vectors before taking the cross product, rather than reasoning about the geometry in words. Two lines of algebra cannot lose a sign; a mental picture can.

This is the same sign-reversal family flagged on the ILTS papers — Q112 (field at −3Q) and Q114 (field at 20 C) both turned on exactly this.

Q 8Attempted · wrongRadius at equal kinetic energy

A proton and an α-particle are projected with the same kinetic energy at right angles to a uniform magnetic field. Which one of the following statements will be true?

Her answerThe radius of the path of the α-particle will be greater than that of the proton

Correct answerThe α-particle and the proton will be bent in a circular path with the same radius

1 Given

Proton: mass m, charge q.
α-particle: mass 4m, charge 2q.
Both have the same kinetic energy K.
Both enter perpendicular to the same uniform field B.

2 Asked

How the radii of their circular paths compare.

3 Concept

The radius is r = mv/qB, so the radius measures momentum per unit charge. But the question gives energy, not speed, so momentum must be rewritten in terms of K:

K = ½mv² → p = mv = √(2mK)

which gives r = √(2mK)/qB. Now both particles differ in m and in q, and the whole question is whether those two differences cancel.

4 Method and baby steps

Write r in terms of K. From K = p²/2m we get p = √(2mK), so
r = √(2mK) / qB.
Proton. m and q are both 1 unit, so rₚ = √(2mK)/qB. Call this value r₀.
Alpha particle. Mass is 4m and charge is 2q, so
rα = √(2 × 4m × K)/(2q × B) = √(8mK)/(2qB).
Now √8 = 2√2, so this is 2√(2mK)/(2qB) = √(2mK)/qB.
Compare. Both expressions are identical, so rα = rₚ. The factor of 4 in the mass becomes a factor of 2 under the square root, and that is exactly cancelled by the factor of 2 in the charge.

5 Easy tricks and shortcuts

Reduce it to a single ratio and the cancellation is visible at once:

r ∝ √(m)/q when the kinetic energy is fixed.

Proton: √1 / 1 = 1. · Alpha: √4 / 2 = 2/2 = 1. Equal.

The companion result worth learning alongside it: if instead the two are accelerated through the same potential difference, then K = qV and r ∝ √(m/q), giving √1 : √2 — a ratio of 1 : √2.

So the same two particles give 1 : 1 at equal energy and 1 : √2 at equal voltage. Reading which one the question fixes is the entire decision.

6 Solution

rα = rₚ — both follow circular paths of the same radius

7 Diagram

8 Weakness area

The answer chosen says the alpha particle's radius is larger, which is what you get by looking at the mass alone and forgetting that its charge is also doubled.

The structure of the formula is what matters here: r = √(2mK)/qB has mass under a square root but charge outside it. So quadrupling the mass only doubles the numerator, while doubling the charge doubles the denominator — and they cancel exactly. Any reasoning that considers only one of the two will go wrong.

The habit to build: when two particles are compared, write the quantity as a single ratio in m and q before substituting anything. Here r ∝ √(m)/q takes ten seconds and removes all ambiguity.

Note that this exact comparison also appears as Q31 on the practice paper just built — with the other condition (equal potential difference) and therefore the other answer. Worth learning as a matched pair.

Q 9Attempted · wrongHollow cylinder: B against distance

The magnetic field due to a hollow cylinder of radius R carrying a current is represented by which graph?

Her answerA straight rise from the origin to a peak at R, then a falling curve

Correct answerZero from the axis out to R, then a sudden jump at R followed by a 1/d fall

1 Given

A hollow cylinder of radius R.
Current flows along its length, distributed over the surface.
B is plotted against distance d from the axis.

2 Asked

Which graph correctly shows B against d.

3 Concept

Everything follows from Ampere's law, ∮B·dl = μ₀I_enclosed, applied to a circular loop of radius d centred on the axis.

The only question to ask is: how much current does my loop enclose?

For a hollow cylinder all the current sits on the surface at radius R. A loop drawn inside encloses nothing at all, so B = 0 there. The moment the loop passes R it suddenly encloses the whole current, and B jumps.

4 Method and baby steps

Region 1: d < R (inside). The Amperian loop encloses no current, so B(2πd) = μ₀(0) and B = 0 everywhere inside.
At d = R. The enclosed current jumps discontinuously from 0 to I, so B jumps to μ₀I/2πR — a vertical step in the graph.
Region 2: d > R (outside). The whole current I is enclosed, so B(2πd) = μ₀I and B = μ₀I/2πd, falling as 1/d.
Shape. Flat at zero, a vertical step at R, then a hyperbolic decay. That is the second graph.

5 Easy tricks and shortcuts

Hold the two cylinder cases as a contrasting pair — they are the only two that ever appear:

SOLID: enclosed current grows as d², so B rises linearly from the origin → a RAMP up to R, then 1/d.
HOLLOW: enclosed current is zero then suddenly all of it → a STEP at R, then 1/d.

Ramp for solid, step for hollow. Both peak at R and both decay as 1/d outside — only the inside differs.

A one-second check on any such graph: look at what happens at d = 0. If the curve leaves the origin, the conductor is solid. If it runs along the axis at zero, it is hollow.

6 Solution

B = 0 for d < R; B = μ₀I/2πd for d > R — the graph with a step at R

7 Diagram

8 Weakness area

The graph chosen — a linear rise to a peak at R — is the correct graph for a SOLID conductor. So the shape was recalled accurately; it was simply attached to the wrong geometry.

That distinction is not cosmetic. Inside a solid conductor the enclosed current grows in proportion to the area, so B climbs steadily. Inside a hollow one there is nothing enclosed at all, so B is flatly zero — which is also why a metal tube shields its interior.

The habit to build: before looking at any option, ask “solid or hollow?” and say the answer out loud — ramp or step. The word decides the graph.

This same pair was tested as Q112 and Q128 on the ILTS paper, and the two were also swapped there. It is now the third appearance, which makes it worth a dedicated line in the notes.

Q 11Attempted · wrongIon drift in crossed E and B

An ionised gas contains both positive and negative ions, initially at rest. If it is subjected simultaneously to an electric field along the +x direction and a magnetic field along the +z direction, then

Her answerall ions effectively deflect towards the +y direction

Correct answerall ions effectively deflect towards the −y direction

1 Given

Ions initially at rest — so any velocity they acquire comes from E.
Electric field E⃗ along +x.
Magnetic field B⃗ along +z (out of the page).
Both positive and negative ions are present.

2 Asked

The direction in which each type of ion deflects.

3 Concept

Nothing moves until the electric field starts it, so the sequence matters:

Step 1. E accelerates each ion — positive ions along +x, negative ions along −x.
Step 2. Once moving, each feels F⃗ = q(v⃗ × B⃗).

For the negative ion two things are reversed: its charge and its velocity. Two sign reversals in the same product cancel, so the magnetic force comes out in the same direction for both. Every ion drifts the same way, whatever its sign.

4 Method and baby steps

Positive ion. E pushes it along +x, so v⃗ = vî.
F⃗ = q(vî) × (Bk̂) = qvB(î × k̂) = qvB(−ĵ) = −y direction.
Negative ion. E pushes it along −x, so v⃗ = −vî, and its charge is −q.
v⃗ × B⃗ = (−vî) × (Bk̂) = −vB(î × k̂) = −vB(−ĵ) = +vBĵ.
Now multiply by the charge: F⃗ = (−q)(+vBĵ) = −qvBĵ = −y direction again.
Both ions deflect towards −y. The gas as a whole drifts one way, which is why a current can be driven across the field.

5 Easy tricks and shortcuts

Memorise the one cross product this question needs: î × k̂ = −ĵ. It is the one that breaks the neat cyclic pattern, and it is the reason the answer here is negative.

The structural shortcut: you never need to do the negative ion at all. Two sign reversals — charge and velocity — always cancel, so whatever the positive ion does, every ion does. Work out one and you have both.

That also tells you two of the four options are dead on sight: any option that sends positive and negative ions in opposite directions cannot be right for this arrangement.

6 Solution

All ions — positive and negative alike — deflect towards −y

7 Diagram

8 Weakness area

This is the sharpest one on the paper, because the answer chosen is correct for a very similar question. On the ILTS-523 paper the same problem appeared with B along −z, and there the answer genuinely is +y.

Flipping B from −z to +z flips the force. So this is not a misunderstood concept — it is a remembered answer applied to a changed question.

The habit to build: when a question feels familiar, circle the signs in the stem before answering. Here the single character that matters is the + in “+z”. Recognition is useful for choosing a method, but never for choosing an answer.

The underlying structure was clearly understood, since “all ions the same way” was correctly identified — that is the harder half of the question, and it was right.

Q 15Attempted · wrongGalvanometer resistance from two deflections

The galvanometer deflection, when key K₁ is closed but K₂ is open, equals θ₀. On closing K₂ also and adjusting R₂ to 5 Ω, the deflection in the galvanometer becomes θ₀/5. The resistance of the galvanometer is then (neglect the internal resistance of the battery)

Her answer12 Ω

Correct answer22 Ω

1 Given

R₁ = 220 Ω, in series with the battery.
K₂ open: full current passes through G, deflection θ₀.
K₂ closed with R₂ = 5 Ω in parallel with G.
New deflection = θ₀/5, so the galvanometer current falls to one fifth.
Battery internal resistance is negligible.

2 Asked

The galvanometer resistance G.

3 Concept

Deflection is proportional to the current through the galvanometer, so θ₀/5 means Iₛ has fallen to one fifth.

Two things change when K₂ closes, and both must be tracked:
1. R₂ shunts G, so the parallel combination has a lower resistance and the total circuit current goes up.
2. That larger total current then splits, and only a fraction R₂/(R₂+G) goes through G.

The second effect is much the stronger, so the galvanometer current falls overall.

4 Method and baby steps

K₂ open. The whole current goes through G:
I₁ = E/(R₁ + G) = E/(220 + G).
K₂ closed. The parallel combination is R₂G/(R₂+G), so the total current is
I = E / [R₁ + R₂G/(R₂+G)].
Split it. The fraction through G is R₂/(R₂+G), so
Iₛ = E R₂ / [R₁(R₂+G) + R₂G].
Apply the condition Iₛ = I₁/5:
E R₂ / [R₁(R₂+G) + R₂G] = E / [5(R₁ + G)]
Cross-multiplying and cancelling E:
5R₂(R₁ + G) = R₁(R₂ + G) + R₂G
Substitute R₁ = 220, R₂ = 5:
25(220 + G) = 220(5 + G) + 5G
5500 + 25G = 1100 + 220G + 5G
5500 + 25G = 1100 + 225G
4400 = 200G → G = 22 Ω

5 Easy tricks and shortcuts

Because R₁ = 220 Ω is far larger than both R₂ and G, the total current barely changes when K₂ closes. To a good approximation the current is fixed at E/R₁ and only the split matters:

R₂/(R₂ + G) ≈ 1/5 → R₂ + G ≈ 5R₂ → G ≈ 4R₂ = 20 Ω

That takes about ten seconds and lands within 10 per cent of the exact 22 Ω — close enough to pick the option with confidence, since the alternatives are 5, 12 and 25.

Sanity check: for the current to be cut to a fifth, most of it must be diverted, so G must be several times larger than R₂ = 5 Ω. Anything below about 15 Ω is implausible on inspection.

6 Solution

G = 22 Ω

7 Diagram

8 Weakness area

12 Ω is roughly half the correct value, and it fails the sanity check above: with G = 12 Ω the split would be 5/(5+12) ≈ 0.29, cutting the deflection to under a third rather than a fifth.

The likely cause is applying only one of the two effects — treating R₂ as a simple series addition, or using the split fraction without accounting for the changed total resistance.

The habit to build: for any shunt problem, write the two effects on separate lines before combining them: (i) what happens to the total current, and (ii) what fraction reaches the meter. Trying to hold both in one step is where these go wrong.

And do the ten-second approximation first. Landing on “about 20” before starting the algebra means an answer of 12 would immediately look wrong.

Q 18Attempted · wrongGalvanometer: the incorrect statement

Which of the following statements is incorrect?

Her answerThe deflection φ indicated on the scale of a moving coil galvanometer by a pointer attached to the spring is given by φ = (NAB/k) i

Correct answerIn voltage measurement, to keep the disturbance due to the measuring device below one per cent, a small resistance R is connected in parallel with the galvanometer

1 Given

Four statements about the moving coil galvanometer.
Exactly one of them is incorrect.

2 Asked

Which statement is wrong.

3 Concept

The whole question rests on one pair of facts that must not be swapped:

AMMETER — measures current, connected in series with the circuit. It must not impede the current, so it needs low resistance, achieved with a small shunt in PARALLEL.

VOLTMETER — measures potential difference, connected in parallel with the component. It must not draw current away, so it needs high resistance, achieved with a large resistor in SERIES.

The incorrect statement takes the ammeter recipe and labels it voltage measurement.

4 Method and baby steps

Statement 1: φ = (NAB/k) i. At equilibrium the magnetic torque NiAB is balanced by the restoring torque kφ, so φ = NABi/k. Correct.
Statement 2: the galvanometer as a detector. Its whole purpose is to show whether current is flowing, and in which direction. Correct.
Statement 3: the two sensitivities. Current sensitivity = φ/i, voltage sensitivity = φ/V. Both are standard definitions. Correct.
Statement 4: a small resistance in parallel for voltage measurement. That describes an ammeter. A voltmeter needs a large resistance in series, so that it draws negligible current and disturbs the circuit by less than one per cent. INCORRECT — this is the answer.

5 Easy tricks and shortcuts

A single sentence fixes the whole topic:

“Ammeter: series in the circuit, shunt in parallel, low resistance. Voltmeter: parallel in the circuit, resistor in series, high resistance.”

Note the pleasing symmetry — each instrument is connected one way and has its extra resistor connected the other way. That cross-over is exactly what the wrong statement gets backwards.

Reason it out rather than recall it: a voltmeter sits across a component. If it had low resistance it would short that component out. So it must be high — and you can rebuild the whole rule from that one thought.

6 Solution

The statement about a small resistance in parallel for voltage measurement is the incorrect one

7 Diagram

8 Weakness area

The statement chosen — φ = (NAB/k)i — is the standard and correct expression for galvanometer deflection. It probably drew attention because it is the only option carrying a formula, and formulas invite scrutiny.

This is the recurring pattern in “which is incorrect” questions: the eye goes to whatever looks most technical, when the false statement is usually a familiar sentence with one word altered. Here the altered words are small and parallel, in a sentence that should read large and series.

The habit to build: in a “which is incorrect” question, look for a statement you can disprove, not one that looks unusual. Test each against a rule you already hold — here, one sentence about ammeters and voltmeters settles it.

The same instinct cost marks on ILTS-04 Q178, where a correct statement about the pre-exponential factor was chosen because it sounded odd.

Q 19Attempted · wrongDeviation across a field region

A particle of mass m and charge q is projected into a region having a perpendicular uniform magnetic field B of width d. Find the angle of deviation θ of the particle as it comes out of the magnetic field.

Her answersin⁻¹(mv/dqB)

Correct answersin⁻¹(dqB/mv)

1 Given

Particle of mass m, charge q, speed v.
Uniform magnetic field B, perpendicular to the velocity.
The field occupies a slab of width d, and the particle passes through it.
Implicit: d must be less than the radius r, otherwise the particle turns back.

2 Asked

The angle θ through which the velocity is deviated on exit.

3 Concept

Inside the field the path is a circular arc of radius r = mv/qB. The particle enters travelling straight and leaves turned through the angle subtended by that arc.

The geometry is a right triangle: the radius is the hypotenuse, and the width d is the side opposite the deviation angle. So

sinθ = d / r

which is the only geometric fact needed. Everything else is substitution.

4 Method and baby steps

Radius of the arc. For circular motion in a magnetic field, qvB = mv²/r, so r = mv/qB.
Geometry. The particle traverses a horizontal width d while turning through θ. From the circle, the horizontal distance covered is r sinθ, so
d = r sinθ → sinθ = d/r.
Substitute r. Dividing by r means multiplying by qB/mv:
sinθ = d ÷ (mv/qB) = dqB/mv.
Answer. θ = sin⁻¹(dqB/mv).
Check it behaves sensibly. A stronger field or a larger charge gives a tighter circle and therefore a bigger deviation — and indeed B and q sit in the numerator. A faster or heavier particle is harder to bend, and m and v sit in the denominator. Everything points the right way.

5 Easy tricks and shortcuts

Dimensional analysis settles this in five seconds without any geometry.

The argument of sin⁻¹ must be dimensionless. Now qB/m has the dimensions of 1/time (it is the cyclotron angular frequency), so d·qB/(mv) has dimensions of (length)(1/time)/(velocity) = 1. Dimensionless, as required.

The inverted option mv/dqB is also dimensionless, so dimensions alone do not separate them — but a physical check does:

A stronger field must bend the particle MORE. So B belongs on top. That one thought eliminates the inverted form immediately.

Limiting case: as d approaches r, sinθ approaches 1 and θ approaches 90° — the particle exits sideways. That is exactly what the correct formula predicts and the inverted one does not.

6 Solution

θ = sin⁻¹(dqB / mv)

7 Diagram

8 Weakness area

The expression was inverted — the correct ratio d/r was written as r/d. Both options contain the same four symbols, so this is not a gap in the physics but a slip in the final rearrangement.

The geometric step is where it happened: sinθ = opposite / hypotenuse = d/r, and the radius is the hypotenuse, so it goes underneath. Writing the triangle out with the two sides labelled prevents the flip.

The habit to build: once an answer is written, run the physical direction check — increase one quantity and ask whether the answer should grow or shrink. “More field means more bending” takes two seconds and catches every inversion of this kind.

This is the same family as the ratio reversals flagged on ILTS-04 Q121 and ILTS-03 Q128. It is now the most frequent single error type across the papers, and the direction check is the one remedy that addresses all of them at once.

Q 22Attempted · wrongCoaxial cable: B against distance

A coaxial cable is made up of two conductors. The inner conductor is solid, of radius R₁. The outer conductor is hollow, with inner radius R₂ and outer radius R₃. The space between them is filled with air. The two conductors carry currents of equal magnitude in opposite directions. The variation of magnetic field with distance from the axis is best plotted as

Her answerA rise to a peak, then a flat plateau, then a fall

Correct answerA linear rise to a peak at R₁, then a 1/r fall, dropping to zero at R₃

1 Given

Inner conductor: solid, radius R₁, carrying current I.
Outer conductor: hollow shell from R₂ to R₃, carrying I in the opposite direction.
Air between them — no current in the gap.
Total current through the whole cable is zero.

2 Asked

The shape of the graph of B against r.

3 Concept

Apply Ampere's law region by region and ask, each time, how much net current does my loop enclose? The key feature of a coaxial cable is that the outer conductor carries the return current, so beyond it the enclosed current is exactly zero and the field vanishes.

That is why coaxial cable is used for signal transmission: it confines its own magnetic field entirely inside the cable.

4 Method and baby steps

Region 1: r < R₁ (inside the solid core). The loop encloses the fraction r²/R₁² of I, so
B = μ₀Ir / 2πR₁² — a straight line rising from the origin, peaking at R₁.
Region 2: R₁ < r < R₂ (the air gap). The whole inner current is enclosed and nothing else, so
B = μ₀I / 2πr — a 1/r decay.
Region 3: R₂ < r < R₃ (inside the outer shell). The loop now starts enclosing the return current, which cancels part of I. The enclosed current shrinks steadily towards zero, so B falls faster and reaches zero at R₃.
Region 4: r > R₃. Enclosed current = I − I = 0, so B = 0 everywhere outside.
Shape. Linear ramp to a peak at R₁, 1/r decay to R₂, a steeper fall to zero at R₃, then flat at zero. No plateau anywhere.

5 Easy tricks and shortcuts

Check the outside first. Equal and opposite currents mean the total enclosed current beyond the cable is zero, so B must be zero for r > R₃. Any graph that still shows a tail beyond R₃ is wrong immediately — and that single check eliminates most of the options.

Then check the origin. The core is solid, so B must start at zero and rise linearly. A graph that begins at a non-zero value belongs to a hollow core.

Why there is never a plateau: a flat section would require the enclosed current to grow in exact proportion to r, which happens nowhere in this geometry. B is constant only inside a solenoid, not around a conductor.

6 Solution

Linear rise to a maximum at R₁, 1/r fall to R₂, decreasing to zero at R₃ and staying zero beyond

7 Diagram

8 Weakness area

The graph chosen has a flat plateau between R₁ and R₂. In that region the enclosed current is constant at I while r keeps increasing, so B = μ₀I/2πr must fall — a constant enclosed current gives a 1/r decay, never a flat line.

It is worth being clear about what a plateau would mean: B constant while r grows requires the enclosed current to grow in proportion to r, which no cylindrical geometry produces. Constant B belongs to the inside of a solenoid, not to the region around a wire.

The habit to build: for any multi-region Ampere's law graph, write the enclosed current for each region first, in words, before looking at any option:
“grows as r² · constant at I · shrinking to zero · zero.”
Then match a shape to that sequence. Choosing between four similar-looking curves by eye is far harder than deriving the sequence and matching it.

Q 23Attempted · wrongCircle and square of the same wire

Two wires of the same length are made into a circle and a square respectively. Currents are passed in them such that their magnetic moments are equal. Then the ratio of the magnetic field at their respective centres (circle : square) is

Her answer√2 π³ / 16

Correct answer√2 π³ / 64 ≈ 0.685 (the key says /32 — see the note above)

The paper's answer key appears to be wrong hereThe key marks √2 π³/32, but working the problem through gives √2 π³/64 ≈ 0.685. The full derivation is below and the numerical check is unambiguous. Your answer was also incorrect, so a mark was lost either way — but it is worth knowing the method is sound.

1 Given

Both wires have the same length L.
Circle: circumference L, so radius r = L/2π.
Square: perimeter L, so side a = L/4.
The two magnetic moments are equal: I_cA_c = I_sA_s.

2 Asked

The ratio B_{centre of circle} : B_{centre of square}.

3 Concept

Three separate results have to be combined, which is what makes this the hardest question on the paper:

1. Areas from the fixed length — a circle encloses more area than a square of the same perimeter.
2. The equal-moment condition M = IA then fixes the ratio of the currents, which are not equal.
3. Field at each centre — μ₀I/2r for the loop, and four finite wires summed for the square.

The currents differ, so you cannot simply compare the two field formulas.

4 Method and baby steps

Areas. Circle: r = L/2π, so A_c = πr² = L²/4π.
Square: a = L/4, so A_s = L²/16.
Current ratio from equal moments.
I_c(L²/4π) = I_s(L²/16)
I_c/I_s = 4π/16 = π/4 ≈ 0.785
So the circle carries the smaller current — it encloses more area, so it needs less current for the same moment.
Field at the centre of the circle.
B_c = μ₀I_c/2r = μ₀I_c/(2 · L/2π) = πμ₀I_c/L.
Field at the centre of the square. Each side is a finite wire at perpendicular distance a/2, subtending 45° on each side:
one side: μ₀I_s/(4π · a/2) × 2sin45° = √2 μ₀I_s/(2πa)
four sides: B_s = 4 × that = 2√2 μ₀I_s/(πa), and with a = L/4 this is 8√2 μ₀I_s/(πL).
Take the ratio.
B_c/B_s = [πμ₀I_c/L] ÷ [8√2 μ₀I_s/(πL)] = π² I_c / (8√2 I_s)
Substitute the current ratio I_c/I_s = π/4:
= π² · (π/4) / (8√2) = π³ / (32√2)
Rationalising: = √2 π³ / 64 ≈ 0.685.

5 Easy tricks and shortcuts

Estimate numerically before choosing. All four options are of the form √2π³/n, and π³ ≈ 31, so √2π³ ≈ 43.9. Then:

/8 ≈ 5.5 · /16 ≈ 2.74 · /32 ≈ 1.37 · /64 ≈ 0.685

So the four options are just four numbers, and you only need to know roughly how big the answer should be.

And you can see that it must be less than 1. For the same perimeter the circle encloses more area, so at equal magnetic moment it carries less current; and its centre is also further from the wire than the square's is. Both effects push the same way, so the circle's field is the smaller and the ratio is below 1. Only one option qualifies.

6 Solution

B_circle : B_square = π³/(32√2) = √2 π³ / 64 ≈ 0.685

7 Diagram

8 Weakness area

The option chosen, √2π³/16 ≈ 2.74, is four times the correct value — and, more tellingly, it is greater than 1, which the physical argument rules out before any algebra.

The check that costs nothing: for a fixed length of wire the circle encloses the largest possible area. At equal magnetic moment it therefore carries the smallest current, and its centre is furthest from the conductor. Both effects reduce its field, so the ratio must be less than one. That single observation eliminates three of the four options.

The habit to build: in a multi-step ratio problem, decide which way the answer should go before starting, and check the final number against that expectation. Three chained results give three chances to drop a factor, and the sanity check is the only thing that catches it.

Do also note the flagged key discrepancy above — the paper's stated answer does not survive checking.

Q 24Attempted · wrongFlux of a dipole through a plane

Consider a circular coil of wire carrying constant current I, forming a magnetic dipole. The magnetic flux through an infinite plane that contains the circular coil and excludes the circular coil area is given by φ_i. The magnetic flux through the area of the circular coil is given by φ₀. Which of the following options is correct?

Her answerφ_i < φ₀

Correct answerφ_i = −φ₀

1 Given

A circular current loop, acting as a magnetic dipole.
An infinite plane containing the loop.
φ₀ = flux through the coil's own area.
φ_i = flux through the rest of that infinite plane.

2 Asked

The relationship between φ_i and φ₀.

3 Concept

Magnetic field lines are always closed loops — there are no magnetic monopoles, so no line ever starts or ends anywhere. This is Gauss's law for magnetism: ∮B·dA = 0 over any closed surface.

For a current loop, every field line that passes upward through the coil must curve around and come back downward somewhere outside it. The infinite plane catches all of them.

So the total flux through the entire infinite plane is exactly zero, and the two parts must be equal and opposite.

4 Method and baby steps

Split the plane. The infinite plane consists of exactly two parts: the coil's own area, and everything else. So
φ_total = φ₀ + φ_i.
Find the total. Every field line leaving through the coil returns through the plane outside it, so every line crosses the plane an equal number of times upward and downward.
φ_total = 0.
Therefore φ₀ + φ_i = 0, which gives
φ_i = −φ₀.
Interpretation. The two fluxes are equal in magnitude and opposite in sign: strong field over a small area inside the coil, weak field over an infinite area outside, and the two exactly balance.

5 Easy tricks and shortcuts

Recognise the question type. Any question about total magnetic flux through a closed surface — or through an infinite plane, which catches every returning line — has the answer zero, because magnetic monopoles do not exist. Once you spot that, the rest is one line of algebra.

Eliminate by sign. Three of the four options (>, <, =) compare magnitudes and all imply the two fluxes have the same sign. But the lines go up inside and down outside, so the signs must differ. Only one option carries a minus sign, and that alone identifies the answer.

The contrast with electric flux: for a charge, Gauss's law gives q/ε₀ — non-zero, because charges are monopoles. For magnetism the right-hand side is always zero. That difference is the single most important structural fact separating the two fields.

6 Solution

φ_i = −φ₀ — equal in magnitude, opposite in sign

7 Diagram

8 Weakness area

The answer chosen compares the two fluxes as though both were positive quantities of different size. It misses that they have opposite signs — and the magnitudes are in fact equal, not different.

The intuition behind “φ_i < φ₀” is reasonable in itself: the field is much weaker outside the coil. But it is spread over an infinite area, and weak field × infinite area comes to exactly the same total. Assuming a weaker field means a smaller flux is what goes wrong.

The habit to build: when a question offers three magnitude comparisons and one signed relation, ask first whether the two quantities can have opposite signs. Flux is a signed quantity, and here the direction reverses between the two regions — so the odd option out is very likely the answer.

More generally: magnetic flux through any closed surface is zero. Adding that single sentence to the notes converts this from a reasoning problem into a one-line lookup.

Moving Charges and Magnetism

Eleven questions, every one attempted and missed

Contents

The shape of this set