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Moving Charges and Magnetism

One chapter, twenty-six questions

Every physics question lost on this paper, rebuilt in full. Each one carries what was given, what was asked, the concept behind it, the formula, the steps written out, the fastest route through, and an animated figure wherever seeing the thing settles the answer. Where an answer was attempted and missed, there is also a note on exactly where it went wrong.

26Questions lost
13Attempted, wrong
13Left blank
117Marks at stake

How this paper divided up

All twenty-six come from a single chapter, so they are grouped here by sub-topic within Moving Charges and Magnetism rather than by chapter. The split between the two failure modes is exactly even — thirteen attempted and missed, thirteen not attempted at all — and on NEET marking the thirteen blanks were worth 52 marks while the thirteen wrong answers cost 52 more plus 13 in negatives.

What stands out is that the two halves fail for different reasons, and so they need different remedies. The blanks are not concept gaps: Q108 is a single dot product, Q107 is one determinant, Q117 and Q119 are two lines each once the right formula is written down. The wrong answers are not knowledge gaps either — in almost every case the physics was right and a sign, a factor, or a direction was not.

Contents

A. Biot–Savart Law: Arcs, Polygons and Rotating Charge3 questions · 3 wrong · 0 blank
Q92Q94Q123
B. Long Straight Wires: Fields, Null Points and Direction4 questions · 2 wrong · 2 blank
Q93Q95Q96Q113
C. Ampere's Circuital Law: Cylinders, Tubes and Toroids7 questions · 4 wrong · 3 blank
Q98Q100Q101Q112Q126Q127Q128
D. Lorentz Force: When It Acts and Which Way5 questions · 2 wrong · 3 blank
Q105Q107Q108Q131Q133
E. Circular and Helical Motion in a Magnetic Field6 questions · 2 wrong · 4 blank
Q106Q109Q116Q117Q118Q119
F. The Theory Behind Biot–Savart's Law1 question · 0 wrong · 1 blank
Q120
Red = attempted and missed · Amber = left blank

A. Biot–Savart Law: Arcs, Polygons and Rotating Charge

3 wrong · 0 blank

Three questions, all of them about turning a shape into a field. The recurring demand is to see what fraction of a full circle you are looking at, or how a straight side sits relative to the point you care about.

Q 92Attempted · wrongArc · field at the centre

A piece of wire carrying a current of 6 A is bent in the form of a circular arc of radius 10 cm and it subtends an angle of 120° at the centre. Find the magnetic field B due to this piece of wire at the centre.

  1. 0.5 × 10−5 T
  2. Correct1.26 × 10−5 T
  3. Her answer2.4 × 10−5 T
  4. 3.2 × 10−5 T
Given
  • Current I = 6 A.
  • Radius of the arc R = 10 cm = 0.10 m.
  • The arc subtends 120° at the centre.
Asked

The magnetic field B at the centre of the arc.

Concept to use

An arc is simply a fraction of a full circle. A complete loop of radius R gives μ0I/2R at its centre. An arc that sweeps θ out of the whole 360° gives that same field cut down by the fraction θ/360°. The one thing the formula insists on is that θ be in radians when you use the μ0Iθ/4πR version.

Diagram
120°R = 10 cmB at centreI = 6 AThe lit arc is one third of the dashed full circle — not two thirds.
AnimatedThe 120° arc against the full circle it belongs to.
Formula to useB = μ0Iθ / 4πR (θ in radians) = (θ/360°) × μ0I/2R
Baby steps
  1. Convert the angle: 120° = 120 × π/180 = 2π/3 = 2.094 radians.
  2. Use μ0/4π = 10−7 directly — never write out 4π × 10−7 when the 10−7 form is available.
  3. B = 10−7 × 6 × 2.094 / 0.10
  4. B = 10−7 × 125.66 = 1.26 × 10−5 T
Answer
B = 1.26 × 10−5 T
Shortcut
Skip radians altogether. 120° out of 360° is one third. A full loop would give μ0I/2R = (4π × 10−7 × 6)/(2 × 0.1) = 3.77 × 10−5 T. One third of that is 1.26 × 10−5 T. Two lines, and no chance of a radian slip.
Where it went wrong
2.4 × 10−5 T is exactly twice the right answer, and that doubling has a specific cause: the fraction was taken as 120/180 = 2/3 instead of 120/360 = 1/3. The arc was measured against a semicircle rather than against the whole circle. Whenever an arc answer comes out as a clean multiple of another option, the fraction is the first thing to re-check.
Q 94Attempted · wrongPolygon · field at the centroid

The magnetic induction field at the centroid of an equilateral triangle of side 'ℓ' and carrying a current 'i' is

  1. 2√2 μ0i / πℓ
  2. Correct9μ0i / 2πℓ
  3. 4μ0i / πℓ
  4. Her answer3√3 μ0i / πℓ
Given
  • An equilateral triangle of side ℓ carrying current i.
  • The field point is the centroid.
Asked

The magnetic induction at the centroid.

Concept to use

This is three identical finite straight wires, not three infinite ones. Two pieces of geometry do all the work: the perpendicular distance from the centroid to each side (the apothem) is ℓ/2√3, and from the foot of that perpendicular each end of the side is seen at 60°. All three fields point the same way out of the page, so they add arithmetically.

Diagram
d = ℓ/2√360°centroidside ℓEach side is a FINITE wire — half-angles 60°, not 90°.
AnimatedThe apothem and the 60° half-angles at the centroid.
Formula to useB_one = (μ0i / 4πd)(sin θ₁ + sin θ₂), with d = ℓ/2√3 and θ₁ = θ₂ = 60°
Baby steps
  1. Apothem: for an equilateral triangle the centroid sits d = ℓ/(2√3) from every side.
  2. Half-angles: sin θ₁ + sin θ₂ = 2 sin 60° = √3.
  3. One side: B = (μ0i/4π)(2√3/ℓ)(√3) = (μ0i/4π)(6/ℓ) = 3μ0i / 2πℓ.
  4. Three sides, all pointing the same way: B = 3 × 3μ0i/2πℓ = 9μ0i / 2πℓ.
Answer
9μ0i / 2πℓ
Shortcut
There is one formula that covers every regular polygon of n sides of length a: B = nμ0i tan(π/n) sin(π/n) / πa. For n = 3 it gives 9μ0i/2πℓ; for n = 4 it gives 2√2 μ0i/πa — which is sitting right there as option A, because the square is the other polygon NEET likes. Learn both results and you never derive either one under exam pressure.
Where it went wrong
3√3 μ0i/πℓ is precisely what you get by treating each side as an infinite wire: B = μ0i/2πd with d = ℓ/2√3 gives √3 μ0i/πℓ per side, and three of those is 3√3 μ0i/πℓ. So the apothem was found correctly — that is the hard part — and then the (sin θ₁ + sin θ₂) bracket was dropped. A triangle side is short; it can never behave like an infinite wire.
Q 123Attempted · wrongRotating charge as a current

A ring of radius 'r' is uniformly charged with a charge 'q'. If the ring is rotated about its own axis with an angular frequency 'ω' then the magnetic induction field at the centre is

  1. Correct(μ0/4π)(qω/r)
  2. Her answer(μ0/4π)(r/qω)
  3. (μ0/4π)(q/rω)
  4. (μ0/4π)(ω/qr)
Given
  • A ring of radius r carrying total charge q, spread uniformly.
  • The ring spins about its own axis at angular frequency ω.
Asked

The magnetic induction at the centre of the ring.

Concept to use

A spinning charge is a current. In one complete revolution the whole charge q sweeps past any fixed point exactly once, so the current is simply charge divided by the time for one turn. Once you have that current, this is an ordinary circular loop and nothing new is needed.

Diagram
++++++++++Brcharge q, spun at ωOne full turn every 2π/ω seconds → I = qω/2π.
AnimatedA spinning ring of charge is an ordinary current loop.
Formula to useT = 2π/ω ; I = q/T = qω/2π ; B = μ0I/2r
Baby steps
  1. Time for one revolution: T = 2π/ω.
  2. Equivalent current: I = q/T = qω/2π.
  3. Field at the centre of a loop: B = μ0I/2r = μ0 qω / (4π r).
  4. Written in the option's style: B = (μ0/4π)(qω/r).
Answer
B = (μ0/4π)(qω/r)
Shortcut
A pure reasoning check settles this without any algebra. More charge must mean more field. Faster spin must mean more field. A bigger ring must mean less field. Only one option has q and ω upstairs and r downstairs. Three options die on inspection.
Where it went wrong
The option chosen is the exact reciprocal of the right one — r on top, qω underneath. Read as English it says a bigger ring gives a stronger field and a faster spin gives a weaker one, and both of those are backwards. When two options in a list are reciprocals of each other, one of them is the planted trap; deciding which way each quantity ought to push takes about five seconds and rules it out.

B. Long Straight Wires: Fields, Null Points and Direction

2 wrong · 2 blank

Four questions where the directions do more work than the arithmetic. Three of them can be half-answered before any number is written, just by asking whether the currents run the same way or opposite ways.

Q 93Left blankNull point between two wires

Two parallel long straight conductors are placed at right angles to the metre scale at the 2 cm and 6 cm marks as shown in the figure. If they carry currents i and 3i respectively in the same direction, then they will produce zero magnetic field at

  1. 1 cm mark
  2. 5 cm mark
  3. Correct3 cm mark
  4. 7 cm mark
Given
  • Wire 1 at the 2 cm mark carries current i.
  • Wire 2 at the 6 cm mark carries current 3i, in the same direction.
  • Separation between the wires = 4 cm.
Asked

The mark on the scale where the resultant magnetic field is zero.

Concept to use

Where the null point lives is decided by the directions, before any arithmetic. With currents in the same direction, the two fields point opposite ways only in the region between the wires — so that is the only place they can cancel. And since the field of a wire weakens with distance, the cancelling point has to sit closer to the smaller current.

Diagram
0123456789101112131415i3i4 cmB = 0 hereBetween the wires, closer to the SMALLER current.
AnimatedSliding the probe across the gap: the field cancels at 3 cm.
Formula to useμ0i / 2πx = μ0(3i) / 2π(4 − x) → i/x = 3i/(4 − x)
Baby steps
  1. Let the null point be x cm from the 2 cm wire, so 4 − x cm from the 6 cm wire.
  2. Set the two field magnitudes equal: i/x = 3i/(4 − x). The i cancels straight away.
  3. Cross-multiply: 4 − x = 3x, so 4x = 4 and x = 1 cm.
  4. x is measured from the 2 cm wire, so the point is at 2 + 1 = the 3 cm mark.
Answer
The 3 cm mark
Shortcut
For straight wires B ∝ 1/distance, so the gap splits in the same ratio as the currents: 1 : 3. That means one quarter of the way across from the weaker wire. One quarter of 4 cm is 1 cm, giving the 3 cm mark — no equation written down at all. Worth remembering the companion rule too: if the currents were opposite, the null would fall outside the pair, beyond the smaller current.
Q 95Attempted · wrongDirection of the field of a wire

A current carrying power line carries current from west to east. The direction of magnetic field at a short distance, above it, is

  1. CorrectNorth to South
  2. Her answerSouth to North
  3. East to West
  4. West to East
Given
  • A long horizontal power line with current running west → east.
  • The field point is a short distance directly above the line.
Asked

The direction of the magnetic field at that point.

Concept to use

The field of a straight wire wraps around it in circles. Rather than trying to picture that rotation in three dimensions, write the axes down and take a cross product. The field direction goes as Î × r̂, where Î points along the current and r̂ points from the wire out to the field point. Written this way there is no mental rotation to get wrong.

Diagram
East (x̂)North (ŷ)Up (ẑ)current W → EPB points SOUTHCirculation runs Up → South → Down → North.
AnimatedThe field circles the wire; above it, the tangent points south.
Formula to useB ∝ Î × r̂ with x̂ = East, ŷ = North, ẑ = Up
Baby steps
  1. Fix the axes on paper: x̂ = East, ŷ = North, ẑ = Up.
  2. Current runs west to east, so Î = +x̂ = î.
  3. The point is above the wire, so r̂ = +ẑ = k̂.
  4. î × k̂ = −ĵ. (Going î → ĵ → k̂ is forwards and gives a plus; î × k̂ runs backwards, so it picks up the minus.)
  5. −ĵ is South. The field at that point runs from north towards south.
Answer
North to South
Shortcut
The five lines above are the shortcut. Mental rotation of a three-dimensional field pattern is the thing that fails under time pressure; writing î × k̂ = −ĵ does not. Fix the habit of writing the axis triad in the margin the moment a compass direction appears in a stem.
Where it went wrong
South to North is the exact reverse, which means the sign of the cross product was lost rather than the geometry being misread. The cyclic order is î → ĵ → k̂ → î: forwards gives a plus, backwards a minus. Here î × k̂ runs backwards and must be −ĵ. Note the companion fact, since it is the one being confused with: k̂ × î = +ĵ. Same two vectors, opposite order, opposite answer.
Q 96Attempted · wrongField at the mid-point of two wires

Two long current carrying conductors are placed parallel to each other at a distance of 8 cm between them. The magnitude of magnetic field produced at mid-point between the two conductors due to current flowing in them is 300 μT. The equal current flowing in the two conductors is:

  1. 30 A in the same direction.
  2. Correct30 A in the opposite direction.
  3. Her answer60 A in the opposite direction.
  4. 300 A in the opposite direction.
Given
  • Two long parallel conductors, 8 cm apart.
  • Equal currents I in each.
  • Field at the mid-point = 300 μT = 300 × 10−6 T.
Asked

The size of the current, and whether the two currents run the same way or opposite ways.

Concept to use

At the mid-point each wire is the same distance away, so each contributes the same size of field. The directions then decide everything. Currents in the same direction produce fields that oppose each other at the mid-point and cancel to zero. Currents in opposite directions produce fields that point the same way and add. A non-zero reading therefore settles the direction half of the answer before any arithmetic happens.

Diagram
IIup the pagedown the page8 cmmid-pointB₁B₂B₁+B₂Opposite currents → at the mid-point the two fields POINT THE SAME WAY.
AnimatedOpposite currents: at the mid-point the two fields reinforce.
Formula to useB_mid = 2 × μ0I / 2πd, with d = 4 cm = 0.04 m
Baby steps
  1. The reading is not zero, so the currents cannot be in the same direction. They must be opposite.
  2. Each wire is d = 4 cm = 0.04 m from the mid-point.
  3. One wire alone: B₁ = 2 × 10−7 × I / 0.04 = 5 × 10−6 I.
  4. Opposite currents → the two add: B = 2 × 5 × 10−6 I = 10−5 I.
  5. Set equal to the reading: 10−5 I = 300 × 10−6, giving I = 30 A.
Answer
30 A, flowing in opposite directions
Shortcut
Read the direction off the question first, then halve. Non-zero at the mid-point means opposite currents, which means the total is double one wire's field. So each wire supplies 150 μT, and I = Bd/(2 × 10−7) = (150 × 10−6 × 0.04)/(2 × 10−7) = 30 A.
Where it went wrong
60 A is exactly double the right current, and doubling always points at the same cause here: the whole 300 μT was set equal to one wire's field instead of to the sum of two. The direction was read correctly — the harder half — and then the second wire was left out of the arithmetic. If opposite currents make the fields add, the 300 μT has to be split into two equal shares of 150 μT.
Q 113Left blankBeams of charges side by side

Beams of electrons and protons move parallel to each other in the same direction. They

  1. CorrectAttract each other
  2. Repel each other
  3. Neither attract nor repel
  4. Force of attraction or repulsion depends upon speed of beams
Given
  • An electron beam and a proton beam, side by side.
  • Both travelling in the same direction.
Asked

Whether they attract, repel, or neither.

Concept to use

Two effects act at once and they pull in opposite directions. Electrically, electrons and protons carry opposite charges, so they attract — and that force is strong. Magnetically, a beam of electrons moving right counts as a conventional current pointing left (the carriers are negative), while the proton beam is a current pointing right; antiparallel currents repel — but that force is weaker by roughly (v/c)². At any speed a question like this intends, the electric effect wins outright.

Diagram
electronsprotonselectric: ATTRACT(strong)magnetic: repel(weak)Net result: they attract
AnimatedElectric attraction against magnetic repulsion — and which one wins.
Formula to useF_magnetic / F_electric ≈ v² / c²
Baby steps
  1. Electric: opposite charges → attraction, and this is the strong effect.
  2. Magnetic: the electron beam is equivalent to a current in the opposite sense to the proton beam, so antiparallel currents → repulsion.
  3. Compare their sizes: the magnetic force is smaller by about (v/c)², which is vanishingly small unless the beams are near light speed.
  4. Net result: they attract.
Answer
They attract each other
Shortcut
Whenever a question pits an electric effect against a magnetic one between charged particles, the electric one wins — unless the stem explicitly says the speed is comparable to the speed of light. Note the contrast that makes this a favourite trap: two wires carrying current the same way attract, because a wire is electrically neutral and only the magnetic effect survives. Beams are not neutral, so the rule flips.

C. Ampere's Circuital Law: Cylinders, Tubes and Toroids

4 wrong · 3 blank

The largest cluster in the paper — seven questions, and every one of them answered by the same single question: how much current does my loop enclose? Two of the seven are the graph pair that were answered identically.

Q 98Attempted · wrongAmpere's law · line integral

A current of 2 A is flowing in a long straight conductor. The line integral of magnetic induction around a closed path enclosing the current carrying conductor is

  1. 4π × 10−7 weber / metre
  2. Correct8π × 10−7 weber / metre
  3. 2π × 10−7 weber / metre
  4. Her answerzero
Given
  • A long straight conductor carrying I = 2 A.
  • A closed path that encircles the conductor.
Asked

The value of ∫B·dl around that closed path.

Concept to use

Ampere's circuital law. The line integral of B around any closed loop depends on exactly one thing: the net current threading the loop. It does not care about the loop's shape, its size, or where inside it the wire happens to sit. That indifference is the whole content of the law.

Diagram
I = 2 A∫B·dl adds up all the way round — it is NOT zero.It depends only on the current threading the loop, not on its shape.
AnimatedAny closed path, however lumpy, gives the same μ0I.
Formula to use∫B·dl = μ0 Ienclosed
Baby steps
  1. Current threading the loop: I = 2 A.
  2. ∫B·dl = μ0 × 2 = 4π × 10−7 × 2.
  3. = 8π × 10−7 weber/metre.
Answer
8π × 10−7 Wb/m
Shortcut
Leave μ0 as 4π × 10−7 and simply double it. The options are all written in π, so converting to decimals only creates work and then forces you to convert back.
Where it went wrong
“Zero” is the correct answer to a different law. For a static electric field, ∫E·dl = 0 around any closed loop, because that field is conservative — go all the way round and you gain nothing. The magnetic field is not conservative, and Ampere's law is the precise statement of that difference. The two integrals look nearly identical on the page and behave in opposite ways, which is exactly why this option is offered.
Q 100Left blankHollow tube · statement set

A hollow tube is carrying an electric current along its length distributed uniformly over its surface.
I. The magnetic field is constant inside the tube and equal to zero.
II. The magnetic field outside the tube decreases with distance.
III. The magnetic field at the surface is maximum.
Which of the statements are correct?

  1. I, II only
  2. II, III only
  3. I, III only
  4. CorrectI, II, III
Given
  • A hollow tube carrying current along its length.
  • The current is spread uniformly over the surface only.
Asked

Which of the three statements are correct.

Concept to use

One tool answers all three parts: draw an Amperian loop and ask how much current it encloses. Inside the hollow region the loop encloses nothing, so B is zero there. Outside, the loop encloses all of it, so B = μ0I/2πr, which thins out as 1/r. The largest value therefore occurs at the smallest radius at which the full current is already enclosed — that is, right at the surface.

Diagram
no current hereB = 0RAmperian loops (green) enclose ALL the current once you are outside.
AnimatedInside a hollow tube an Amperian loop encloses nothing.
Formula to useB = 0 for r < R ; B = μ0I / 2πr for r > R
Baby steps
  1. Statement I. Inside, the enclosed current is zero, so B = 0 everywhere in the cavity — and a field that is zero everywhere is certainly constant. Correct.
  2. Statement II. Outside, B = μ0I/2πr falls as 1/r as you move away. Correct.
  3. Statement III. B is zero inside and falling outside, so its greatest value is at r = R. Correct.
  4. All three stand together.
Answer
I, II and III
Shortcut
Sketch the graph once instead of reading three statements separately: flat along the axis at zero, a vertical jump at R, then a 1/r tail. Every one of the three statements can be read straight off that single picture in a few seconds. Note the connection to Q128, which asks for that very graph.
Q 101Left blankSolid wire · inside vs outside

A long, straight wire of radius 'a' carries a current distributed uniformly over its cross-section. The ratio of the magnetic fields due to the wire at distance a/3 and 2a respectively from the axis of the wire is:

  1. 1/2
  2. 3/2
  3. Correct2/3
  4. 2
Given
  • A solid wire of radius a, current spread uniformly across the cross-section.
  • Point 1 at r = a/3 (inside the wire).
  • Point 2 at r = 2a (outside the wire).
Asked

The ratio of the magnetic fields at those two points.

Concept to use

Two different formulas for two different regions, and the first job is to notice which point is where. Inside, an Amperian loop catches only the share of the current lying within it; that share grows as r²/a², and the two effects combine to give B ∝ r. Outside, the loop catches the whole current, so B ∝ 1/r.

Diagram
dBRB ∝ dB ∝ 1/da/32aRises inside, peaks exactly at the surface, then falls away.
AnimatedThe two field points marked on the one curve.
Formula to useinside: B = μ0Ir / 2πa² outside: B = μ0I / 2πr
Baby steps
  1. a/3 is less than a, so point 1 is inside. 2a is greater than a, so point 2 is outside.
  2. Inside at r = a/3: B₁ = μ0I(a/3)/(2πa²) = μ0I/6πa.
  3. Outside at r = 2a: B₂ = μ0I/(2π × 2a) = μ0I/4πa.
  4. Ratio B₁ : B₂ = (1/6) : (1/4) = 4 : 6 = 2 : 3.
Answer
2/3
Shortcut
Work in units of the surface value B_s = μ0I/2πa, which is the peak. Inside at a/3 the field is (1/3)B_s, since B ∝ r. Outside at 2a it is (1/2)B_s, since B ∝ 1/r. The ratio is (1/3) ÷ (1/2) = 2/3, and μ0 never appears anywhere.
Q 112Attempted · wrongSolid conductor · B against distance
Paired with Q128The identical graph was chosen again in Q128, which asks about a hollow cylinder. One picture is being used for both cases, and it is the right answer to neither. See the note on Q128.

A cylindrical conductor of radius R is carrying a constant current uniformly distributed across its cross-section. The plot of the magnitude of the magnetic field, B with the distance, d from the centre of the conductor, is correctly represented by the figure

  1. A straight rise from the origin up to R, then flat and constant beyond R
  2. Her answerConstant and non-zero from the axis out to R, then falling away
  3. Zero out to R, then a jump at R followed by a falling curve
  4. CorrectA straight rise from the origin to a peak at R, then a 1/d fall
Given
  • A solid cylindrical conductor of radius R.
  • Current spread uniformly across the whole cross-section.
  • B is plotted against distance d from the centre.
Asked

Which graph correctly represents B against d.

Concept to use

The same two-region argument as Q101, now read as a picture. Inside, the enclosed current grows as d² while the loop circumference grows as d, and the two together give B ∝ d — a straight line out of the origin. Outside, the whole current is enclosed and B ∝ 1/d. The two pieces meet at d = R, which is where B reaches its largest value.

Diagram
dBRB ∝ dB ∝ 1/dRises inside, peaks exactly at the surface, then falls away.
AnimatedSolid conductor: rise, peak at R, then 1/d decay.
Formula to used < R: B = μ0Id / 2πR² d > R: B = μ0I / 2πd
Baby steps
  1. At d = 0, right on the axis, an infinitesimal loop encloses no current at all, so B = 0. The graph must start at the origin.
  2. For d < R the enclosed current is I(d²/R²), giving B = μ0Id/2πR² — a straight rising line.
  3. At d = R the entire current is enclosed and B is at its peak.
  4. Beyond R, B = μ0I/2πd falls as 1/d, approaching but never reaching zero.
  5. Rise, peak at R, then decay — only one graph has that shape.
Answer
The straight rise to a peak at R followed by a 1/d fall
Shortcut
Test the two ends and ignore the middle. Does the curve start at zero at d = 0? Does it fall away far from the wire? Only one of the four graphs passes both tests, and both tests take a second each.
Where it went wrong
The graph chosen is flat and non-zero inside, then falls. A constant field inside would mean the enclosed current stays the same no matter how small the loop is drawn — but shrink the loop towards the axis and the enclosed current shrinks with it, all the way down to zero. The field cannot be non-zero at d = 0.
Q 126Left blankToroid · relative permeability

The current in the windings on a toroid is 2 A. There are 400 turns and the mean radius is 40 cm. If the magnetic field inside is 1 T, the relative permeability of the core of toroid is

  1. 2000
  2. Correct2500
  3. 1000
  4. 1500
Given
  • I = 2 A, N = 400 turns.
  • Mean radius r = 40 cm = 0.40 m.
  • Measured field inside the core B = 1 T.
Asked

The relative permeability μr of the core.

Concept to use

A toroid's field is μ0nI where n is the turns per unit length measured along the mean circle, so n = N/2πr. Fill the core with a magnetic material and the field is multiplied by μr. So the method is: work out what the same winding would give with air inside, then see how many times bigger the actual field is.

Diagram
r (mean)N turnsB lives inside the core only, and it thins out as the ring gets bigger.
AnimatedTurns per unit length are counted along the mean circle.
Formula to useB = μr μ0 N I / 2πr → μr = B × 2πr / (μ0 N I)
Baby steps
  1. Mean circumference: 2πr = 2π × 0.4 = 2.513 m.
  2. Field the winding would give in air: B_air = μ0NI/2πr = (4π × 10−7 × 400 × 2) / 2.513.
  3. = 1.0053 × 10−3 / 2.513 = 4.0 × 10−4 T.
  4. μr = B / B_air = 1 / (4 × 10−4) = 2500.
Answer
μr = 2500
Shortcut
Write it as one fraction and cancel π before touching a calculator: μr = 2πrB/(μ0NI) = (2 × 0.4 × 1) / (4 × 10−7 × 800). The π cancels top and bottom, leaving 0.8/(3.2 × 10−4) = 2500. Cancelling π first removes every decimal from the working.
Q 127Attempted · wrongTwo toroids compared

Two toroids with number of turns 400 and 200 have average radii respectively 30 cm and 60 cm. If they carry the same current, the ratio of magnetic fields in these two toroids is

  1. Her answer2 : 1
  2. 1 : 4
  3. 2 : 3
  4. Correct4 : 1
Given
  • Toroid 1: N₁ = 400 turns, mean radius r₁ = 30 cm.
  • Toroid 2: N₂ = 200 turns, mean radius r₂ = 60 cm.
  • Both carry the same current.
Asked

The ratio B₁ : B₂.

Concept to use

B = μ0NI/2πr, so with the current fixed B depends on N/r — the turns and the radius together. Turns alone is only half the story: packing the same number of turns onto a bigger ring spreads them out and weakens the field.

Diagram
r (mean)N turnsB lives inside the core only, and it thins out as the ring gets bigger.
AnimatedMore turns and a smaller mean radius both push the field up.
Formula to useB ∝ N/r → B₁/B₂ = (N₁/N₂) × (r₂/r₁)
Baby steps
  1. B₁ ∝ 400/30 = 13.33.
  2. B₂ ∝ 200/60 = 3.33.
  3. Ratio = 13.33 / 3.33 = 4.
  4. B₁ : B₂ = 4 : 1.
Answer
4 : 1
Shortcut
Take the two changes separately and multiply. Toroid 1 has twice the turns (×2) and half the radius (×2 again, since r is downstairs). Two doublings give 4. Whenever two quantities both change, the answer is the product of the two ratios, never just one of them.
Where it went wrong
2 : 1 is the ratio of the turns alone — the radius was left out of the comparison. Toroid 1 wins twice over: more turns and a tighter ring. This is worth flagging as a habit rather than a one-off, because Q101 and Q112 in this same paper also turn on holding two competing quantities in view at once. When a stem hands you two numbers that both changed, neither one on its own is the answer.
Q 128Attempted · wrongHollow cylinder · B against distance
Paired with Q112Both graph questions in this paper were answered with the same curve. They have different answers, and the difference is one word in the stem. Treat “solid or hollow?” as the first thing you read in any B-against-r question.

A long thin hollow metallic cylinder of radius R has a current 'i' ampere. The magnetic induction B away from the axis at a distance 'r' from the axis varies as shown in

  1. CorrectZero out to R, then a jump at R followed by a 1/r fall
  2. Her answerConstant and non-zero from the axis out to R, then falling away
  3. A straight rise from the origin to a peak at R, then a fall
  4. A straight rise from the origin up to R, then dropping to nothing
Given
  • A hollow metallic cylinder of radius R.
  • It carries current i, which therefore lies entirely on the surface.
  • B is plotted against distance r from the axis.
Asked

Which graph correctly represents B against r.

Concept to use

Hollow means every bit of the current sits on the wall. Any Amperian loop drawn inside the cavity therefore encloses nothing at all, and B is exactly zero right up to the wall. Step outside and the loop suddenly catches the whole current, so B = μ0i/2πr, falling as 1/r.

Diagram
rBRB = 0B ∝ 1/rFlat ON the axis line at zero — no current is enclosed inside a hollow tube.
AnimatedHollow cylinder: flat at zero, then a jump, then 1/r.
Formula to useB = 0 for r < R ; B = μ0i / 2πr for r > R
Baby steps
  1. Inside: Ienclosed = 0, so B = 0 for every r < R. The graph lies flat along the r-axis.
  2. At r = R the entire current is enclosed at once, so B jumps to its peak value μ0i/2πR.
  3. Outside: B ∝ 1/r, a falling curve.
  4. Flat at zero, a vertical jump at R, then a 1/r tail.
Answer
Zero along the axis until R, then a jump and a 1/r decay
Shortcut
Ask one question and one only: how much current is inside my loop? For a hollow tube the answer is “none” until you cross the wall. That single sentence picks the graph without any formula at all.
Where it went wrong
The same graph — constant inside, then falling — was chosen here as in Q112, and it is the right answer to neither question. Q112 is a solid conductor, where B rises from zero and then falls. Q128 is hollow, where B is zero and then falls. One remembered picture is being applied to both. The word to hunt for in the stem is “solid” or “hollow”; it decides the entire left-hand half of the graph.

D. Lorentz Force: When It Acts and Which Way

2 wrong · 3 blank

Five questions on the force itself. Two turn on the sign of a cross product, one on the fact that a perpendicular force means a zero dot product, and two on the angle between v and B being zero — which switches the magnetic force off entirely.

Q 105Attempted · wrongE and B together · ion drift

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

  1. Positive ions deflect towards +y direction and negative ions towards −y direction.
  2. CorrectAll ions deflect towards +y direction
  3. Her answerAll ions deflect towards −y direction
  4. Positive ions deflect towards −y direction and negative ions towards +y direction
Given
  • An ionised gas holding both positive and negative ions.
  • Electric field E along +x.
  • Magnetic field B along −z (into the page).
Asked

Which way each kind of ion is deflected.

Concept to use

Two forces act, and the order matters. The electric force is what sets the ions moving: positive ions get pushed along +x, negative ions along −x. Only then does the magnetic force q(v × B) act. Now notice what happens: for the negative ion the charge sign flips and the velocity direction flips. Two flips cancel, so both kinds of ion end up pushed the same way.

Diagram
B into page (along −z)E along +x+−+y+yThe ion signs flip TWICE — once in v, once in q — so both end up going the same way.
AnimatedBoth signs flip together, so both ions drift the same way.
Formula to useF = qE + q(v × B)
Baby steps
  1. Positive ion. E pushes it along +x, so v = +v x̂.
  2. v × B = (v x̂) × (−B ẑ) = −vB (x̂ × ẑ) = −vB(−ŷ) = +vB ŷ. Charge positive → force along +y.
  3. Negative ion. E pushes it along −x, so v = −v x̂.
  4. v × B = (−v x̂) × (−B ẑ) = +vB (x̂ × ẑ) = −vB ŷ. Multiply by the negative charge → force along +y again.
  5. Both kinds of ion deflect towards +y.
Answer
All ions deflect towards the +y direction
Shortcut
You only ever need to do one of the two calculations. The sign appears twice — once in v and once in q — and a double flip is no flip at all. So whenever an electric field is what sets the ions moving, every ion deflects the same way regardless of its charge. Do the positive one and copy the answer across.
Where it went wrong
−y is the right axis with the wrong sign, so the physics of the double flip was understood — the answer chosen does say all ions go the same way, which is the insight the question is testing. What was lost was a single minus sign: x̂ × ẑ = −ŷ, because x̂ × ẑ runs backwards round the cycle. Getting the concept and losing the sign is a different failure from not knowing the concept, and it is fixed by writing the cyclic triangle in the margin, not by re-reading the chapter.
Q 107Left blankAcceleration from v × B

A proton of velocity (3î + 2ĵ) ms−1 enters a field of magnetic induction (2ĵ + 3k̂) tesla. The acceleration (in m/s²) produced in the proton is (charge to mass ratio of proton = 0.96 × 108 C kg−1)

  1. 2.8 × 108 (2î − 3ĵ)
  2. Correct2.88 × 108 (2î − 3ĵ + 2k̂)
  3. 2.8 × 108 (2î + 3k̂)
  4. 2.88 × 108 (î − 3ĵ + 2k̂)
Given
  • v = (3î + 2ĵ) m/s.
  • B = (2ĵ + 3k̂) T.
  • q/m = 0.96 × 108 C/kg.
Asked

The acceleration produced in the proton.

Concept to use

The magnetic force is F = q(v × B), so a = F/m = (q/m)(v × B). The fact that the question hands you the charge-to-mass ratio ready-made is the hint that q and m are never needed separately — work out the cross product and multiply once.

Diagram
x̂ŷẑforward = +x̂×ŷ=ẑ   ŷ×ẑ=x̂   ẑ×x̂=ŷ  —  backwards picks up a minus.
AnimatedThe cyclic circle that fixes every cross-product sign.
Formula to usea = (q/m)(v × B)
Baby steps
  1. Write the components: v = (3, 2, 0) and B = (0, 2, 3).
  2. î component: (2)(3) − (0)(2) = 6.
  3. ĵ component: −[(3)(3) − (0)(0)] = −9. (Remember the middle term of a determinant carries a minus.)
  4. k̂ component: (3)(2) − (2)(0) = 6.
  5. v × B = 6î − 9ĵ + 6k̂ = 3(2î − 3ĵ + 2k̂).
  6. a = 0.96 × 108 × 3 × (2î − 3ĵ + 2k̂) = 2.88 × 108 (2î − 3ĵ + 2k̂).
Answer
a = 2.88 × 108 (2î − 3ĵ + 2k̂) m/s²
Shortcut
The number in front decides it before the vector does. 0.96 × 3 = 2.88, and only two options carry 2.88 at all. Between those two, check just the î component: it is 6, which is 2 after the 3 is pulled out, not 1. One multiplication and one component and you are done.

And to check the finished answer in five seconds: a must be perpendicular to both v and B. Test a·B = (2)(0) + (−3)(2) + (2)(3) = 0 ✓ and a·v = (2)(3) + (−3)(2) + (2)(0) = 0 ✓. Any answer failing either dot product is wrong on the spot.
Q 108Left blankPerpendicularity as an equation

A charged particle has acceleration a⃗ = 2î + xĵ in a magnetic field B⃗ = −3î + 2ĵ − 4k̂. Find the value of 'x'.

  1. x = 2
  2. Correctx = 3
  3. x = 4
  4. x = −3
Given
  • a⃗ = 2î + xĵ (the k component is zero).
  • B⃗ = −3î + 2ĵ − 4k̂.
Asked

The value of x.

Concept to use

The magnetic force is always perpendicular to B — that is guaranteed by the cross product itself, for any v whatsoever. Acceleration points along the force, so a must be perpendicular to B too. And perpendicular means the dot product is zero. That one fact turns the whole question into a single linear equation.

Diagram
x̂ŷẑforward = +x̂×ŷ=ẑ   ŷ×ẑ=x̂   ẑ×x̂=ŷ  —  backwards picks up a minus.
AnimatedPerpendicular means the dot product vanishes — no cross product needed.
Formula to usea⃗ · B⃗ = 0
Baby steps
  1. a·B = (2)(−3) + (x)(2) + (0)(−4).
  2. = −6 + 2x + 0 = −6 + 2x.
  3. Set it to zero: −6 + 2x = 0.
  4. x = 3.
Answer
x = 3
Shortcut
No cross product is needed anywhere in this question, and no velocity either. “Find the missing component of the acceleration of a charge in a magnetic field” always means a·B = 0. Recognising the pattern turns a two-minute question into a fifteen-second one — which is exactly the kind of question a paper cannot afford to leave blank.
Q 131Left blankE parallel to B

A uniform electric field and a uniform magnetic field are acting along the same direction in a certain region. If an electron is projected in the region such that its velocity is pointed along the direction of fields, then the electron:

  1. will turn towards right of direction of motion
  2. will turn towards left of direction of motion
  3. Correctspeed will decrease
  4. speed will increase
Given
  • E and B both point along the same direction.
  • An electron is projected along that same direction.
Asked

What happens to the electron.

Concept to use

The magnetic force needs the velocity to have a component across B. Here v is exactly along B, so the angle between them is zero, sin 0° = 0, and the magnetic force vanishes completely — the electron cannot be turned at all. That leaves only the electric force. Since the electron is negative, the electric force on it points opposite to E, which is backwards along its own motion.

Diagram
EBv (parallel to both)e−force on the electron: BACKWARDSv ∥ B → magnetic force is zero. Only E acts — and it pulls the electron back.
Animatedv along B kills the magnetic force; only E is left.
Formula to useF = qE + qvB sin θ, with θ = 0 → F = qE only
Baby steps
  1. v is parallel to B, so sin θ = 0 and the magnetic force is zero. Turning left or right is off the table immediately.
  2. The electric force on a charge q is qE. For an electron q is negative, so the force points opposite to E.
  3. E points along the direction of motion, so the force points against the motion.
  4. A force opposing the motion slows the electron down: the speed decreases.
Answer
The speed will decrease
Shortcut
Two of the four options are about turning, and the magnetic force here is zero, so both die in one stroke. That leaves speed up or speed down, and the electron's negative sign settles it. Half the option list can be eliminated before any thinking about magnitudes at all.
Q 133Attempted · wrongElectron along a solenoid axis

An electron is projected with uniform velocity along the axis of a current carrying long solenoid. Which of the following is true?

  1. The electron will be accelerated along the axis.
  2. Her answerThe electron path will be circular about the axis.
  3. The electron will experience a force at 45° to the axis and hence execute a helical path.
  4. CorrectThe electron will continue to move with uniform velocity along the axis of the solenoid.
Given
  • A long solenoid carrying a steady current.
  • An electron projected with uniform velocity along the axis.
Asked

What the electron does.

Concept to use

Inside a long solenoid the field is uniform and directed along the axis. The electron is also travelling along the axis. So v and B are parallel, the angle between them is zero, and the magnetic force is exactly zero. With no force acting, Newton's first law finishes the question.

Diagram
Be−v is along the axis, and so is B. The angle between them is zero.sin 0° = 0 → no force → it sails straight through unchanged.
AnimatedStraight down the axis, with nothing to bend it.
Formula to useB = μ0nI along the axis ; F = qvB sin θ with θ = 0 → F = 0
Baby steps
  1. Field inside a long solenoid: B = μ0nI, directed along the axis.
  2. The electron's velocity is also along the axis, so θ = 0.
  3. F = qvB sin 0° = 0.
  4. No force means no acceleration, so it keeps the same speed in the same straight line.
Answer
The electron continues to move with uniform velocity along the axis
Shortcut
The stem says “along the axis” and the field is along the axis. That is the same sentence written twice, and parallel means zero force. Any option describing a circle, a helix or an acceleration is dead the moment you notice it.
Where it went wrong
A circular path requires a force pointing towards a centre, and here there is no force at all. The remembered picture — “charged particle in a magnetic field goes in a circle” — was applied without checking the one condition that makes it true: v must be across B, not along it. It is the same missing check as in Q131, where the same parallel arrangement makes the magnetic force vanish. Two marks in this paper turn on that single question: what is the angle between v and B?

E. Circular and Helical Motion in a Magnetic Field

2 wrong · 4 blank

Six questions built on r = mv/qB and T = 2πm/qB. Four of the six were left blank, and three of those four are two-line calculations once the right formula is on the page.

Q 106Left blankWhat stays constant

Which of the following quantities remains constant for a charged particle moving inside a magnetic field?

  1. Velocity
  2. CorrectKinetic energy
  3. Acceleration
  4. Force
Given
  • A charged particle moving inside a magnetic field.
Asked

Which quantity remains constant.

Concept to use

The magnetic force is always at right angles to the velocity, because that is what a cross product produces. A force perpendicular to the motion can never do work, so the energy cannot change — and with the mass fixed, neither can the speed. But the direction of the velocity keeps turning, which means the velocity vector, the acceleration and the force are all changing even though their sizes stay put.

Diagram
vFspeed unchangedF stays at 90° to v the whole way round, so it never does any work.Energy is fixed; the velocity VECTOR is turning every instant.
AnimatedThe force never stops turning — and never does any work.
Formula to useW = F·d = 0 when F ⊥ v → kinetic energy constant
Baby steps
  1. F = q(v × B) is perpendicular to v by the nature of the cross product.
  2. Work done per second = F·v = 0, always.
  3. No work done → kinetic energy unchanged → speed unchanged.
  4. But the direction of v keeps turning, so the velocity vector is not constant; neither is the acceleration nor the force, both of which keep swinging round to point at the centre.
Answer
Kinetic energy
Shortcut
“Speed” and “velocity” are different words in physics and examiners lean on it constantly. Speed is constant here; velocity is not. If an option says velocity, it is wrong — and that single reading habit is worth marks across the whole chapter.
Q 109Attempted · wrongIdentifying tracks in a field

A neutron, a proton, an electron and an alpha-particle enter a region of uniform magnetic field with the same velocities. The magnetic field is perpendicular and directed into the plane of the paper. The tracks of the particles are labelled in the figure. The electron follows the track

  1. A
  2. Her answerB
  3. C
  4. CorrectD
Given
  • Four particles entering together with the same velocity: neutron, proton, electron, alpha.
  • B is uniform, perpendicular to the page and directed into it.
  • Four tracks are drawn, labelled A, B, C and D.
Asked

Which track the electron follows.

Concept to use

Two separate questions, and they must be answered in order. First: which side does it bend to? Positive and negative charges bend opposite ways, and an uncharged particle does not bend at all. Second: how tightly? r = mv/qB, so a larger m/q means a wider curve. Answering these in the wrong order is what makes this question hard.

Diagram
B into pageABDCall four enter here, moving upC straight = neutron · A, B bend LEFT = positive · D bends RIGHT = electron
AnimatedThe four tracks, redrawn: side first, tightness second.
Formula to useF = q(v × B) fixes the side ; r = mv/qB fixes the tightness
Baby steps
  1. The particles enter moving up the page with B into the page. For a positive charge, v × B points to the left, so positive particles curve left.
  2. Tracks A and B both bend left, so those two are the proton and the alpha particle.
  3. Track C is straight, so it carries no charge: that is the neutron.
  4. That leaves D — the only track bending right — for the negatively charged electron.
  5. The tightness confirms it. The electron's m/q is thousands of times smaller than any of the others, so its radius is by far the smallest, and D is the sharpest turn on the figure.
Answer
Track D
Shortcut
Find the straight track first. It is always the neutron, and it splits the figure down the middle. The electron is then the lone track on the opposite side from everything else, because it is the only negatively charged particle in the list. Two glances, no formula.
Where it went wrong
Track B bends left, which is the positive side — B is in fact the alpha particle, drawn as the widest of the left-hand curves because its m/q is twice the proton's. So the tightness was being judged before the side was settled. Deciding left-or-right first, and only then asking how tight the curve is, keeps these two steps from blurring together. This is the same left-or-right step that cost the mark in Q95 and the sign in Q105.
Q 116Left blankTwo particles · avoiding a collision

Two identical particles having the same mass m and charges +q and −q separated by a distance d enter in a uniform magnetic field B directed perpendicular to paper inwards with speeds v₁ and v₂ as shown in figure. The particles will not collide if

  1. d > (m/qB)(v₁ + v₂)
  2. d < (m/qB)(v₁ + v₂)
  3. Correctd > (2m/qB)(v₁ + v₂)
  4. v₁ = v₂
Given
  • Two particles of the same mass m, charges +q and −q.
  • Separated by a distance d, both moving horizontally.
  • B is uniform, perpendicular to the paper and directed inwards.
  • Speeds v₁ (upper particle) and v₂ (lower particle).
Asked

The condition on d for the particles not to collide.

Concept to use

Each particle travels a circle of radius r = mv/qB. Opposite charges bend opposite ways, and in this arrangement that means they curve towards each other. Now the key point: a particle's furthest reach across the gap is not r but 2r. Its circle's centre lies a distance r into the gap, and it swings a full r beyond that centre before coming back — so it reaches the point diametrically opposite its start.

Diagram
−+v₁v₂d2r₁2r₂They miss each other only if d beats the two DIAMETERS, not the two radii.
AnimatedEach particle reaches a full diameter into the gap.
Formula to user = mv/qB ; safe if d > 2r₁ + 2r₂
Baby steps
  1. Radii: r₁ = mv₁/qB and r₂ = mv₂/qB.
  2. The upper particle curves down and reaches 2r₁ into the gap — its diameter, not its radius.
  3. The lower particle curves up and reaches 2r₂ into the gap.
  4. They stay apart provided d > 2r₁ + 2r₂.
  5. d > 2mv₁/qB + 2mv₂/qB = (2m/qB)(v₁ + v₂).
Answer
d > (2m/qB)(v₁ + v₂)
Shortcut
When two options differ only by a factor of 2, the question is testing radius against diameter and nothing else. Sketch the two circles: each particle's deepest point into the gap is directly opposite where it entered, which is a whole diameter away. Then pick the option with the 2 in it.
Q 117Left blankForce from an energy in MeV

A 2 MeV proton is moving perpendicular to a uniform magnetic field of 2.5 T. The force on the proton is

  1. 2.5 × 10−10 N
  2. 8 × 10−11 N
  3. 2.5 × 10−11 N
  4. Correct8 × 10−12 N
Given
  • A proton with kinetic energy 2 MeV.
  • B = 2.5 T, and the motion is perpendicular to B.
  • mp = 1.67 × 10−27 kg, e = 1.6 × 10−19 C.
Asked

The magnetic force on the proton.

Concept to use

F = qvB needs the speed, but the question gives you an energy instead. So there are two conversions before any force appears: MeV into joules, then joules into speed through KE = ½mv². Recognising that this is really an energy-conversion question wearing a magnetism costume is most of the battle.

Formula to useKE = ½mv² → v = √(2KE/m) ; F = qvB
Baby steps
  1. Energy in joules: 2 MeV = 2 × 106 × 1.6 × 10−19 = 3.2 × 10−13 J.
  2. Speed: v = √(2 × 3.2 × 10−13 / 1.67 × 10−27) = √(3.83 × 1014) = 1.96 × 107 m/s.
  3. Force: F = qvB = 1.6 × 10−19 × 1.96 × 107 × 2.5.
  4. = 7.8 × 10−12 ≈ 8 × 10−12 N.
Answer
About 8 × 10−12 N
Shortcut
Commit one number to memory: a 1 MeV proton travels at about 1.4 × 107 m/s. Energies scale as v², so a 2 MeV proton moves at √2 times that, roughly 1.96 × 107 m/s — which skips the whole first half of the working. MeV protons recur often enough across magnetism and modern physics to make this worth the memory.
Q 118Attempted · wrongReading momentum off a radius

Two particles A and B of masses mA and mB respectively and having the same charge are moving in a plane. A uniform magnetic field exists perpendicular to this plane. The speeds of the particles are vA and vB respectively and the trajectories are as shown in fig. Then

  1. Her answermAvA < mBvB
  2. CorrectmAvA > mBvB
Given
  • Particles A and B, same charge, in the same uniform perpendicular field.
  • From the figure, A travels the outer (wider) arc and B the inner one.
  • The screenshot captured two of the four options.
Asked

The relationship between mAvA and mBvB.

Concept to use

The radius of the circular path is r = mv/qB. With the charge and the field the same for both particles, the radius depends on mv — the momentum — and nothing else. So reading which curve is wider is reading which momentum is bigger. You cannot separate m from v here, and you are not asked to.

Diagram
B out of pageABrArBSame charge, same B → a bigger circle means a bigger mv.
AnimatedSame charge, same field: the wider arc carries more momentum.
Formula to user = mv/qB → mv = qBr → mv ∝ r
Baby steps
  1. r = mv/qB, and q and B are identical for the two particles.
  2. Rearranged: mv = qBr, so momentum is directly proportional to radius.
  3. The figure shows A on the outer arc, so rA > rB.
  4. Therefore mAvA > mBvB.
Answer
mAvA > mBvB
Shortcut
Bigger circle = bigger momentum. That is the entire question, and it needs no algebra whatsoever. A fast, heavy particle is harder to bend, so it traces a wider arc — the intuition and the formula agree, which makes it safe to trust under time pressure.
Where it went wrong
The inequality is stated the wrong way round — the physics was right and the direction of the < sign was not. This is the same failure as the ratio errors flagged in earlier papers, and the fix is the same: say the sentence out loud before choosing the symbol. “A is the wider curve, so A is harder to bend, so A has the bigger momentum, so mAvA is the larger one.” Spoken sentences do not invert; remembered symbols do.
Q 119Left blankPitch of a helical path

A beam of protons with speed 4 × 105 ms−1 enters a uniform magnetic field of 0.3 T at an angle of 60° to the magnetic field. The pitch of the resulting helical path of protons is close to: (Mass of the proton 1.67 × 10−27 kg, charge of the proton 1.69 × 10−19 C)

  1. 12 cm
  2. Correct4 cm
  3. 2 cm
  4. 5 cm
Given
  • v = 4 × 105 m/s, entering at 60° to B.
  • B = 0.3 T.
  • m = 1.67 × 10−27 kg, q = 1.69 × 10−19 C.
Asked

The pitch of the helical path.

Concept to use

Split the velocity into two independent pieces. The part across the field, v sin θ, is what the magnetic force acts on, and it drives the circular motion. The part along the field, v cos θ, feels no magnetic force at all and simply carries the proton forward at a steady rate. The pitch is how far that forward drift takes it during exactly one complete circle.

Diagram
Bpitch = v∥ × Tv at 60°v⊥ makes the circle; v∥ slides it forward. One turn = one pitch.
AnimatedOne full turn advances the proton by exactly one pitch.
Formula to usepitch = (v cos θ) × T, with T = 2πm/qB
Baby steps
  1. Time for one turn: T = 2πm/qB = 2π × 1.67 × 10−27 / (1.69 × 10−19 × 0.3).
  2. = 1.049 × 10−26 / 5.07 × 10−20 = 2.07 × 10−7 s.
  3. Forward speed: v cos 60° = 4 × 105 × 0.5 = 2 × 105 m/s.
  4. Pitch = 2 × 105 × 2.07 × 10−7 = 4.14 × 10−2 m ≈ 4 cm.
Answer
About 4 cm
Shortcut
The period T contains no speed and no angle — only m, q and B. Work it out first and set it aside, and all that remains is a single multiplication. Then guard the one trap in the question: the pitch uses cos θ, the component along B. Using sin 60° here would give about 7 cm, which is why 5 cm and 12 cm are on the list.

F. The Theory Behind Biot–Savart's Law

0 wrong · 1 blank

A single statement question, and one that rewards reading rather than calculating.

Q 120Left blankBiot-Savart law · statements

Statement I: Biot-Savart's law gives us the expression for the magnetic field strength of an infinitesimal current element (Idl) of a current carrying conductor only.
Statement II: Biot-Savart's law is analogous to Coulomb's inverse square law of charge q, with the former being related to the field produced by a scalar source, Idl while the latter being produced by a vector source, q.
In light of above statements choose the most appropriate answer from the options given below:

  1. Both Statement I and Statement II are incorrect
  2. CorrectStatement I is correct and statement II is incorrect
  3. Statement I is incorrect and statement II is correct
  4. Both Statement I and Statement II are correct
Given
  • Statement I: the law gives the field of an infinitesimal element Idl only.
  • Statement II: Idl is described as a scalar source and q as a vector source.
Asked

Which of the two statements are correct.

Concept to use

Statement I is simply the honest definition of the law: it gives dB for one infinitesimal element, and anything larger has to be built up by integration. Statement II makes a real comparison with Coulomb's law but gets the two source types the wrong way round. In Coulomb's law the source q is a scalar — a plain number with no direction. In the Biot-Savart law the source Idl is a vector — it points along the wire.

Diagram
Coulombqq is a SCALAR — just a numberno cross product in the lawBiot–SavartI dlI dl is a VECTOR — it has a directionwhich is why the law needs dl × r̂Statement II swaps these two labels round. That is the whole error.
AnimatedA scalar source against a vector source, side by side.
Formula to usedB = (μ0/4π)(I dl × r̂)/r² compared with E = (1/4πε0)(q/r²) r̂
Baby steps
  1. Statement I. The law is written for an element Idl and must be integrated along the conductor to give the field of anything real. Correct as stated.
  2. Statement II. It claims Idl is the scalar source and q the vector one.
  3. But charge q is a plain number — a scalar. And a current element Idl carries a direction — a vector. The statement has swapped them.
  4. So Statement I is correct and Statement II is incorrect.
Answer
Statement I is correct and Statement II is incorrect
Shortcut
The giveaway is the cross product. Biot-Savart contains Idl × r̂; Coulomb's law does not contain any cross product at all. A law can only have a cross product if its source has a direction — so Idl must be the vector and q the scalar. Read Statement II only for which word is attached to which source, because that swap is the only thing it changes.

What the twenty-six have in common

Reading the paper as a whole

The thirteen wrong answers fall into four families

FamilyQuestionsWhat actually happened
Sign and direction95, 105, 109, 118 The physics was right and the direction was reversed. Q95 and Q105 are both a single minus sign in a cross product; Q109 chose a track on the positive side for a negative particle; Q118 wrote a correct comparison with the inequality the wrong way round.
A clean factor of two92, 96, 127 Q92 came out exactly double, from comparing the arc against a semicircle rather than a full circle. Q96 came out exactly double, from forgetting the second wire. Q127 used the turns and left the radius out. In all three the method was sound and one ingredient was missing.
One picture used for two cases112, 128 The same graph was chosen for a solid conductor and for a hollow one. It is the right answer to neither. A single stored image is being applied without checking the word in the stem that decides between them.
The wrong rule imported94, 98, 123, 133 Q94 used the infinite-wire formula on a triangle's side. Q98 used the electrostatic result ∫E·dl = 0 on a magnetic integral. Q123 inverted a formula. Q133 applied “charge in a field goes in a circle” without checking that v must be across B.

Notice what is not on this list: there is no question here that was missed because the underlying idea was unknown. Every one of the thirteen was attempted with the right general approach.

The thirteen blanks are mostly short questions

This is the more expensive half of the paper, and the more fixable. Timed honestly, most of these are under ninety seconds:

Only Q116 really needs thinking time, and even that turns on noticing that the answer uses diameters rather than radii.

The one habit that would recover the most marks

Four of the thirteen wrong answers — Q95, Q105, Q109 and Q118 — are a direction or a sign, and nothing else. Together they cost 16 marks in unearned scores plus 4 in negatives. Every one of them is prevented by the same thirty-second discipline: write the axes down in the margin, then write the cross product out symbolically before choosing.

For Q95 that is five lines: x̂ = East, ŷ = North, ẑ = Up; Î = î; r̂ = k̂; î × k̂ = −ĵ; −ĵ = South. Written like that the answer cannot come out backwards. Held in the head, it comes out backwards roughly half the time — which is what this paper shows.

The companion habit, for the ratio and inequality questions, is to say the comparison as a sentence before writing the symbol. “A is the wider curve, so A is harder to bend, so A has the bigger momentum.” Spoken sentences do not invert. Remembered symbols do.

Three things worth carrying into the next paper