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Topic 13 of 14

Eddy currents

When flux changes inside a solid lump of metal, induced currents swirl around inside it. This file covers how they brake and heat, how engineers suppress them, and where they are put to work.

Gap contentAllen module pages 132–134Illustrations 34–37, Beginner's Box 8

Whirlpools of current inside metal

Eddy currents in a metal sheet entering a magnetic fieldA solid metal sheet slides into a field region. Where its flux changes, near the field's edge, swirling eddy currents appear inside the sheet. They pull back on it, slowing it down.field into pagesolid sheet moving right →eddy currents swirl at the field edgered arrow: braking force (Lenz)
Eddy currents. A solid piece of metal has no single wire to follow, so induced currents swirl in closed loops wherever the flux through the metal is changing. Like any induced current, they oppose the change.
Picture it

Where a river flows past a rock, the water curls round in little whirlpools called eddies. When the magnetic flux through a solid block of metal changes, the induced current has no wire to follow, so it curls round inside the metal in whirlpools too. That is why they are called eddy currents. They are also called Foucault currents, after the scientist who studied them.

In exam language

Eddy currents are induced currents set up in the body of a conductor when the magnetic flux through it changes, either because the field changes with time or because the conductor moves through a field.

What eddy currents do

PropertyWhy
They flow in closed loops throughout the volume of the metalThere is no single path, so they cannot be measured with a meter
Their loops lie in planes perpendicular to the fieldEach loop encloses changing flux (Allen Illustration 36: angle 90°)
They can be very largeBulk metal has very low resistance
They heat the metal, sometimes enough to melt itJoule heating, I²R
They oppose the motion of the metal (force or torque)Lenz's law
A solid plate and a slotted plate swinging through a magnetic fieldBoth plates swing through the gap between magnet poles. The solid plate carries large eddy currents and stops within a few swings. The slotted plate's eddy currents are broken up, so it swings for much longer.solid plateslotted platestops quicklykeeps swinging
Electromagnetic damping. Cutting slots in the plate breaks the big eddy current loops into small ones, which carry far less current and give far less braking.

Allen page 133 shows the same idea with a single loop and a solid sheet pulled out of a field: the sheet feels a much larger braking force than the loop, because eddy currents flow all through it.

Keeping eddy currents small

Solid core versus laminated coreCross-sections of an iron core with an alternating field along its length, out of the page here. A solid core lets one big eddy current loop form. A core made of thin insulated sheets confines the eddy currents to thin strips, so they are much weaker.solid corelaminated coreone large loop: big current, much heatthin insulated sheets: small loopsField out of the page; sheets lie parallel to the field.
Lamination. Thin sheets, insulated from each other with varnish or oxide, leave the magnetic path intact but cut the eddy current paths.

In transformers, generators and motors, eddy currents waste energy as heat. To reduce them:

MethodHow it helps
Laminated core (thin insulated sheets)Eddy loops are confined to thin sheets, greatly raising their resistance
Slots cut in metal platesBreaks up the large loops
Powdered iron or ferrite cores (high-frequency devices)Tiny insulated grains carry almost no eddy current
Trap

The sheets of a laminated core lie parallel to the magnetic field. The eddy current loops are perpendicular to the field, so the insulating layers cut straight across them.

Allen Illustration 35: to avoid eddy currents in a transformer core, a laminated core is used. Checked: correct.

Where eddy currents are useful

ApplicationHow eddy currents are used
Electromagnetic damping, dead-beat galvanometerThe coil is wound on a metal frame; eddy currents in the frame stop the needle swinging so it settles at once
Electromagnetic brakes (some trains)Magnets near the rotating wheels or rails induce eddy currents that brake smoothly
Induction furnaceA high-frequency alternating field induces huge eddy currents that melt metal
Induction motorEddy currents in the rotor are dragged along by a rotating field
SpeedometerA spinning magnet drags an aluminium drum through eddy currents; the drag sets the pointer
DiathermyDeep heating of body tissue
Undesirable effects
Energy loss as heatWastes power in transformers, motors and generators
Opposition to motionUnwanted braking
HeatingCan shorten the life of electrical devices

Quick reference

IdeaRuleWatch out
OriginChanging flux through bulk metalMoving or time-varying field
DirectionLenz's lawOppose the change or the motion
Plane of loopsPerpendicular to BAngle with field lines = 90°
SizeLarge, because resistance is lowCannot be measured with a meter
ReductionLamination, slots, ferriteSheets parallel to B
UsesDamping, brakes, furnace, motor, speedometer, diathermyResistive heaters are not eddy-current devices
Rotating coil power (Illus. 34)Pavg = e₀²/2RSource of the power: the external turning force

Allen pages 132–134 checked

ItemCheck
Definition, applications and undesirable effectsCorrect
Figure: loop, disc, loop leaving field, sheet (Floop ≪ Fsheet)Correct
Illustration 34: e₀ = 0.603 V, I₀ = 0.0603 A, P = 0.018 WCorrect numbers
Illustration 34: "resistance 10 W"Misprint: should be 10 Ω
Illustration 35: laminated coreCorrect
Illustration 36: plane of eddy currents at 90° to the fieldCorrect
Illustration 37: dynamo works on electromagnetic inductionCorrect

Beginner's Box 8 answer key

These two questions sit on page 134 but test the rotating coil from Topic 12.

QAnswerWorking
1(4) Parallel to the magnetic fieldEmf is largest when the flux through the coil is passing through zero, which happens with the plane along B.
2(3) 113 VN = 60, A = 0.20 × 0.10 = 0.02 m², B = 0.5 T, 1800 rpm = 30 rev/s so ω = 60π rad/s. e₀ = 60 × 0.5 × 0.02 × 60π = 36π ≈ 113 V.

NEET practice: 25 questions

This topic is short and mostly conceptual, so it has fewer questions than the others. It is gap content: the rationalised NCERT book dropped eddy currents, but they appear in Allen and in older NEET papers.

Q1Concept
Eddy currents are also known as
  1. (A)Foucault currents
  2. (B)displacement currents
  3. (C)drift currents
  4. (D)Faraday currents
Show the solution
Given
Other name
Asked
Name
Concept
Named after Léon Foucault.
Formula
—
Baby steps
  1. Eddy currents = Foucault currents.
Answer
(A) Foucault currents
Why not the others
Displacement current is Maxwell's term for capacitors. Drift current is about semiconductors. 'Faraday currents' is not a standard name.
Shortcut
Foucault.
Where it went wrong
Attribution error.
Q2Concept
Eddy currents are produced in a piece of metal when
  1. (A)it carries a steady current
  2. (B)the magnetic flux through it changes
  3. (C)it is placed in a steady magnetic field and kept at rest
  4. (D)it is heated strongly
Show the solution
Given
Conditions
Asked
When eddy currents appear
Concept
They are induced currents.
Formula
e = −dφ/dt
Baby steps
  1. Changing flux induces emf.
  2. Bulk metal lets current swirl.
Answer
(B) the magnetic flux through it changes
Why not the others
A steady current or a steady field at rest gives no change. Heating does not induce currents.
Shortcut
No change, no eddy.
Where it went wrong
Thinking any field makes eddy currents.
Q3Concept
The plane in which eddy currents circulate makes an angle with the magnetic field lines of
  1. (A)180°
  2. (B)0°
  3. (C)45°
  4. (D)90°
Show the solution
Given
Allen Illustration 36
Asked
Angle
Concept
Each loop must enclose changing flux.
Formula
φ = BA cos θ
Baby steps
  1. A loop encloses the most flux when its plane is perpendicular to B.
  2. So eddy loops lie at 90° to the field.
Answer
(D) 90°
Why not the others
0° or 180° would enclose no flux. 45° has no reason.
Shortcut
Loops perpendicular to B.
Where it went wrong
Imagining currents flowing along the field.
Q4Concept
To reduce eddy current losses, the core of a transformer is
  1. (A)laminated
  2. (B)made of solid copper
  3. (C)made thicker
  4. (D)made of a single solid block of iron
Show the solution
Given
Transformer core
Asked
Remedy
Concept
Thin insulated sheets cut eddy paths.
Formula
—
Baby steps
  1. Laminations confine eddy currents to thin sheets.
  2. Resistance of each path rises, current falls.
Answer
(A) laminated
Why not the others
Copper has even lower resistance, so larger eddy currents. Thicker or solid cores make the loops bigger.
Shortcut
Laminate.
Where it went wrong
Choosing a better conductor.
Q5Concept
In a laminated transformer core, the thin sheets are arranged
  1. (A)perpendicular to the magnetic field, insulated from one another
  2. (B)parallel to the magnetic field, insulated from one another
  3. (C)parallel to the field, welded together
  4. (D)at 45° to the field
Show the solution
Given
Lamination
Asked
Orientation
Concept
Eddy loops are perpendicular to B; insulating layers must cut across them.
Formula
—
Baby steps
  1. Sheets parallel to B leave the magnetic path unbroken.
  2. Insulation between sheets cuts each eddy loop.
Answer
(B) parallel to the magnetic field, insulated from one another
Why not the others
Sheets perpendicular to B would break the magnetic path. Welded sheets would conduct between them. 45° is a half measure.
Shortcut
Parallel to B, insulated.
Where it went wrong
Putting the sheets across the field.
Q6Concept
A dead-beat galvanometer settles quickly because
  1. (A)its magnet is very weak
  2. (B)its spring is very stiff
  3. (C)its needle is very light
  4. (D)eddy currents in the metal frame of its coil damp the motion
Show the solution
Given
Dead-beat galvanometer
Asked
Why it settles fast
Concept
Electromagnetic damping.
Formula
—
Baby steps
  1. The coil is wound on a conducting frame.
  2. As it swings, eddy currents in the frame oppose the motion.
Answer
(D) eddy currents in the metal frame of its coil damp the motion
Why not the others
Stiffness and mass change the swing but do not remove it. A weak magnet would reduce damping.
Shortcut
Metal frame → damping.
Where it went wrong
Missing the frame's role.
Q7Concept
An induction furnace melts metals by
  1. (A)eddy currents induced by a high-frequency alternating field
  2. (B)passing a direct current through the metal
  3. (C)burning fuel around the crucible
  4. (D)a steady magnetic field
Show the solution
Given
Induction furnace
Asked
Principle
Concept
Large eddy currents cause Joule heating.
Formula
H = I²Rt
Baby steps
  1. High-frequency field → rapidly changing flux → huge eddy currents.
  2. The metal heats up and melts.
Answer
(A) eddy currents induced by a high-frequency alternating field
Why not the others
Direct current is resistive heating, not induction. Fuel is combustion. A steady field induces nothing.
Shortcut
Alternating field → eddy heating.
Where it went wrong
Thinking any magnetic field heats metal.
Q8Concept
Which of these does not make use of eddy currents?
  1. (A)An electric heater with a resistance coil
  2. (B)An electromagnetic brake
  3. (C)A speedometer
  4. (D)An induction motor
Show the solution
Given
Applications
Asked
Odd one out
Concept
A heater coil uses Joule heating from a supplied current.
Formula
—
Baby steps
  1. Brakes, speedometers and induction motors all rely on induced currents.
  2. A resistance heater does not.
Answer
(A) An electric heater with a resistance coil
Why not the others
Each of the other three is listed by Allen as an eddy current application.
Shortcut
Supplied current ≠ induced current.
Where it went wrong
Treating all heating as eddy heating.
Q9Concept
A solid metal plate and a similar plate with slots cut in it swing as pendulums through the gap of a strong magnet. Which is true?
  1. (A)The slotted plate stops much sooner
  2. (B)The solid plate stops much sooner
  3. (C)Both stop at the same time
  4. (D)Neither is affected by the magnet
Show the solution
Given
Solid versus slotted plate
Asked
Damping
Concept
Slots break up eddy current loops.
Formula
—
Baby steps
  1. Solid plate: large loops, large eddy currents, strong braking.
  2. Slotted plate: small loops, weak braking.
Answer
(B) The solid plate stops much sooner
Why not the others
The slotted plate is damped less, not more. Their damping differs. Both are affected by the magnet.
Shortcut
Slots reduce damping.
Where it went wrong
Thinking slots let field through more easily.
Q10Concept
A strong magnet falls through a vertical aluminium pipe and, separately, through a glass pipe of the same size. Compared with the glass pipe, in the aluminium pipe it falls
  1. (A)at the same rate
  2. (B)faster
  3. (C)more slowly
  4. (D)faster at first, then more slowly
Show the solution
Given
Conducting versus insulating pipe
Asked
Speed of fall
Concept
Eddy currents in aluminium oppose the motion.
Formula
—
Baby steps
  1. Aluminium conducts: eddy currents are induced in the pipe walls.
  2. They oppose the magnet's fall.
  3. Glass has none.
Answer
(C) more slowly
Why not the others
Faster would create energy. The same rate ignores the eddy currents. There is no reason for early speeding up.
Shortcut
Conductor slows it.
Where it went wrong
Thinking aluminium is not affected because it is not magnetic.
Q11Concept
A spinning aluminium disc slows down quickly when a magnet is brought near it. This is because
  1. (A)the magnet increases friction
  2. (B)aluminium is attracted to the magnet
  3. (C)eddy currents induced in the disc oppose its rotation
  4. (D)the disc becomes magnetised
Show the solution
Given
Magnet near a spinning disc
Asked
Reason for slowing
Concept
Lenz's law acting on eddy currents.
Formula
—
Baby steps
  1. The rotating disc cuts field lines.
  2. Eddy currents form and oppose the motion.
Answer
(C) eddy currents induced in the disc oppose its rotation
Why not the others
Aluminium is not noticeably attracted to magnets. Friction is not changed. Aluminium does not become a magnet.
Shortcut
Electromagnetic braking.
Where it went wrong
Thinking it is magnetic attraction.
Q12Concept
Eddy currents in a metal block are usually large because
  1. (A)the resistance of the bulk metal is very low
  2. (B)the magnetic field inside metals is very large
  3. (C)metals store charge
  4. (D)eddy currents do not obey Ohm's law
Show the solution
Given
Size of eddy currents
Asked
Reason
Concept
I = e/R with small R.
Formula
I = e/R
Baby steps
  1. The emf may be modest.
  2. The resistance of thick metal paths is tiny.
  3. So the current is large.
Answer
(A) the resistance of the bulk metal is very low
Why not the others
The field need not be large. Metals do not store charge. Eddy currents obey Ohm's law.
Shortcut
Low R, big I.
Where it went wrong
Blaming the field.
Q13Concept
In a transformer, eddy currents in the core are undesirable mainly because they
  1. (A)increase the output voltage
  2. (B)waste energy as heat
  3. (C)reverse the direction of the flux
  4. (D)make the transformer work on DC
Show the solution
Given
Transformer core
Asked
Why eddy currents are unwanted
Concept
Joule heating in the core is a loss.
Formula
—
Baby steps
  1. Alternating flux in the core induces eddy currents.
  2. They heat the core, wasting input energy.
Answer
(B) waste energy as heat
Why not the others
They do not raise voltage, reverse flux, or allow DC operation.
Shortcut
Eddy loss is part of iron loss.
Where it went wrong
Thinking losses change the voltage ratio.
Q14Concept
The force or torque that eddy currents exert on a moving metal body
  1. (A)always helps its motion
  2. (B)always opposes its motion
  3. (C)is always zero
  4. (D)is perpendicular to its motion and does no work
Show the solution
Given
Eddy force
Asked
Direction
Concept
Lenz's law.
Formula
—
Baby steps
  1. Eddy currents oppose the change that causes them.
  2. The cause is the motion, so the force opposes the motion.
Answer
(B) always opposes its motion
Why not the others
Helping would create energy. They are not zero while flux changes. The force does negative work (braking).
Shortcut
Eddy = brake.
Where it went wrong
Thinking magnetic forces never do work here.
Q15Concept
Diathermy, a medical treatment, uses eddy currents to
  1. (A)heat tissues deep inside the body
  2. (B)take images of bones
  3. (C)measure blood pressure
  4. (D)stop bleeding by freezing
Show the solution
Given
Diathermy
Asked
Purpose
Concept
Induced currents heat tissue.
Formula
—
Baby steps
  1. A high-frequency field induces eddy currents in tissue.
  2. Joule heating warms it from inside.
Answer
(A) heat tissues deep inside the body
Why not the others
Imaging, pressure measurement and freezing are unrelated.
Shortcut
Deep heat treatment.
Where it went wrong
Guessing an imaging technique.
Q16Numerical
A 10-turn circular coil of radius 10 cm rotates at 100 rad/s in a 0.02 T field, with its axis of rotation perpendicular to the field. Its ends are joined through a total resistance of 5 Ω. The average power dissipated as heat is about
  1. (A)0.63 W
  2. (B)0.079 W
  3. (C)0.020 W
  4. (D)0.039 W
Show the solution
Given
N = 10, r = 0.1 m, ω = 100 rad/s, B = 0.02 T, R = 5 Ω
Asked
Pavg
Concept
Peak emf, then average power for a sinusoidal emf (Allen Illustration 34).
Formula
e₀ = NBπr²ω, Pavg = e₀²/2R
Baby steps
  1. e₀ = 10 × 0.02 × 0.0314 × 100 = 0.628 V.
  2. e₀² = 0.395.
  3. P = 0.395/10 = 0.039 W.
Answer
(D) 0.039 W
Why not the others
0.079 W forgets the ½ (it is the peak power). 0.020 W halves once too often. 0.63 W is the emf in volts.
Shortcut
Average of sin² is ½.
Where it went wrong
Using the peak power.
Q17Concept
In Allen Illustration 34, a coil rotating in a field dissipates power as heat. The source of this energy is
  1. (A)the resistance of the coil
  2. (B)the magnetic field, which gradually weakens
  3. (C)the coil's own kinetic energy, even at constant speed
  4. (D)the external agent that keeps the coil turning
Show the solution
Given
Rotating coil at constant speed
Asked
Energy source
Concept
The induced current produces a torque opposing rotation.
Formula
Pmech = Pheat
Baby steps
  1. The opposing torque must be balanced by an external torque.
  2. That external work becomes electrical energy, then heat.
Answer
(D) the external agent that keeps the coil turning
Why not the others
A permanent magnet's field does not run down. Constant speed means no loss of kinetic energy. Resistance converts energy; it is not a source.
Shortcut
Work in, heat out.
Where it went wrong
Thinking the field supplies the energy.
Q18Numerical
A rectangular coil of 50 turns, 20 cm by 10 cm, rotates at 1500 rpm about an axis perpendicular to a 0.4 T field. The maximum emf is about
  1. (A)600 V
  2. (B)63 V
  3. (C)126 V
  4. (D)31 V
Show the solution
Given
N = 50, A = 0.02 m², 1500 rpm, B = 0.4 T
Asked
e₀
Concept
Convert rpm, then e₀ = NBAω (Beginner's Box 8 Q2 pattern).
Formula
ω = 2π × rpm/60
Baby steps
  1. 1500 rpm = 25 rev/s → ω = 50π rad/s.
  2. e₀ = 50 × 0.4 × 0.02 × 50π = 20π ≈ 63 V.
Answer
(B) 63 V
Why not the others
600 V uses ω = 1500. 126 V doubles. 31 V halves.
Shortcut
rpm ÷ 60 × 2π.
Where it went wrong
Using rpm as ω.
Q19Graph
A solid metal plate and an identical slotted plate are set swinging through a strong magnetic field. Which graph compares their amplitudes against time (the faster-falling curve is the solid plate)?
  1. (A)Graph of amp against tampt
  2. (B)Graph of amp against tampt
  3. (C)Graph of amp against tampt
  4. (D)Graph of amp against tampt
Show the solution
Given
Solid versus slotted plate
Asked
Amplitude against time
Concept
Eddy damping is much stronger in the solid plate.
Formula
—
Baby steps
  1. Both amplitudes decay.
  2. The solid plate's decays far faster.
Answer
(C) the graph in option C
Why not the others
Identical curves ignore the slots. Flat lines ignore damping. Identical straight falls ignore both the difference and the gradual decay.
Shortcut
Two decays, one much faster.
Where it went wrong
Thinking slots make no difference.
Q20Graph
An aluminium disc spinning freely between the poles of a strong magnet is left to itself. Which graph shows its angular speed against time?
  1. (A)Graph of ω against tωt
  2. (B)Graph of ω against tωt
  3. (C)Graph of ω against tωt
  4. (D)Graph of ω against tωt
Show the solution
Given
Eddy current braking
Asked
ω against t
Concept
Braking torque ∝ angular speed.
Formula
dω/dt ∝ −ω
Baby steps
  1. Torque proportional to speed → exponential decay.
Answer
(D) the graph in option D
Why not the others
Constant speed ignores the braking. The rising curve needs a drive. A sudden stop needs an infinite torque.
Shortcut
Same shape as F = −bv.
Where it went wrong
Drawing no braking because aluminium is non-magnetic.
Q21Graph
A strong magnet is dropped from rest down a long vertical copper pipe. Which graph shows its speed against time?
  1. (A)Graph of v against tvt
  2. (B)Graph of v against tvt
  3. (C)Graph of v against tvt
  4. (D)Graph of v against tvt
Show the solution
Given
Magnet in a copper pipe
Asked
v against t
Concept
Eddy braking grows with speed until it balances gravity.
Formula
mg = bvT
Baby steps
  1. Starts from rest, speeds up.
  2. Levels off at a terminal speed.
Answer
(C) the graph in option C
Why not the others
The straight line is free fall. The falling curve starts at the wrong value. Constant speed from rest is impossible.
Shortcut
Terminal speed.
Where it went wrong
Choosing free fall.
Q22Assertion–reason
Assertion (A): The core of a transformer is laminated.
Reason (R): Lamination greatly reduces eddy currents in the core.
  1. (A)Both A and R are true, and R is the correct explanation of A.
  2. (B)Both A and R are true, but R is not the correct explanation of A.
  3. (C)A is true, but R is false.
  4. (D)A is false, but R is true.
Show the solution
Given
A: laminated core. R: reduces eddy currents.
Asked
Truth and link
Concept
Purpose of lamination.
Formula
—
Baby steps
  1. A is true.
  2. R is true.
  3. R is the reason for A.
Answer
(A) Both A and R are true, and R is the correct explanation of A.
Why not the others
(B) denies a direct link. (C) and (D) need a false statement.
Shortcut
Purpose explains practice.
Where it went wrong
Choosing (B) by habit.
Q23Assertion–reason
Assertion (A): A magnet falls more slowly through a copper pipe than through a glass pipe.
Reason (R): Copper is strongly attracted by magnets.
  1. (A)Both A and R are true, and R is the correct explanation of A.
  2. (B)Both A and R are true, but R is not the correct explanation of A.
  3. (C)A is true, but R is false.
  4. (D)A is false, but R is true.
Show the solution
Given
A: slower in copper. R: copper is attracted.
Asked
Truth and link
Concept
Copper is not ferromagnetic; eddy currents cause the braking.
Formula
—
Baby steps
  1. A is true.
  2. R is false.
  3. A true, R false.
Answer
(C) A is true, but R is false.
Why not the others
(A) and (B) need R true. (D) needs A false.
Shortcut
Eddy currents, not attraction.
Where it went wrong
Blaming magnetic attraction.
Q24Assertion–reason
Assertion (A): Eddy currents circulate in planes parallel to the magnetic field.
Reason (R): Eddy currents obey Lenz's law.
  1. (A)Both A and R are true, and R is the correct explanation of A.
  2. (B)Both A and R are true, but R is not the correct explanation of A.
  3. (C)A is true, but R is false.
  4. (D)A is false, but R is true.
Show the solution
Given
A: loops parallel to B. R: Lenz's law.
Asked
Truth and link
Concept
Loops are perpendicular to B.
Formula
—
Baby steps
  1. R is true.
  2. A is false: they are perpendicular.
  3. A false, R true.
Answer
(D) A is false, but R is true.
Why not the others
(A), (B) and (C) need A true.
Shortcut
90°, not 0°.
Where it went wrong
Confusing the sheets' orientation with the loops'.
Q25Two statements
Statement I: Electromagnetic brakes and induction furnaces both use eddy currents.
Statement II: Eddy currents can be measured directly with an ammeter placed in the metal.
  1. (A)Both Statement I and Statement II are true.
  2. (B)Both Statement I and Statement II are false.
  3. (C)Statement I is true, but Statement II is false.
  4. (D)Statement I is false, but Statement II is true.
Show the solution
Given
Applications; measurement
Asked
Which are true
Concept
Eddy currents have no single path.
Formula
—
Baby steps
  1. Statement I is true.
  2. Statement II is false: they flow throughout the volume, so they cannot be measured with a meter.
Answer
(C) Statement I is true, but Statement II is false.
Why not the others
Both-true fails on II. Both-false fails on I. I-false-II-true reverses them.
Shortcut
No single path to measure.
Where it went wrong
Assuming every current can be metered.

Answer key

1 A
2 B
3 D
4 A
5 B
6 D
7 A
8 A
9 B
10 C
11 C
12 A
13 B
14 B
15 A
16 D
17 D
18 B
19 C
20 D
21 C
22 A
23 C
24 D
25 C

Spread across letters: A 7, B 6, C 6, D 6. No letter repeats more than twice in a row. Question mix: Concept 16, Numerical 2, Graph 3, Assertion–reason 3, Two statements 1. Balancing seed 4000.