NEET Physics · Targeted repair · Built from the ILTS of 9 Aug 2026
Directions & Graphs
Five of the thirteen wrong answers were direction questions and two were graph questions. Together that is 35 marks from two skills — not two topics. This pack fixes both.
Every direction question in this chapter is one of four things in disguise. Learn the four and there is nothing left to guess.
1. First, always set up axes
The commonest cause of a wrong direction answer is trying to picture it in your head. Don't. Write the axes down every single time, even when it feels unnecessary.
The standard compass setup — memorise this once and use it forever: x̂ = East · ŷ = North · ẑ = Up
Then: West = −x̂, South = −ŷ, Down = −ẑ.
Fifteen seconds of writing beats fifteen seconds of visualising. Write the axes before you read the options — this one habit is worth about 25 marks.
2. The cyclic rule — the engine of every direction answer
Once the axes are written, every direction question becomes a cross product, and every cross product follows one circle:
î × ĵ = k̂ ĵ × k̂ = î k̂ × î = ĵ
Going forwards round the circle î → ĵ → k̂ → î gives a plus. Going backwards gives a minus:
ĵ × î = −k̂ k̂ × ĵ = −î î × k̂ = −ĵ
One circle, one rule. Follow the dot's direction and you get a plus; go the other way and you pick up a minus sign. Every direction answer in the chapter comes out of this.
Worked example — the one she got wrong. A power line carries current from west to east. Which way does the field point at a spot just above it?
Step 1 — axes: x̂ = East, ŷ = North, ẑ = Up.
Step 2 — current direction: west to east means Î = +x̂ = î.
Step 3 — direction from the wire to the point: the point is above, so r̂ = +ẑ = k̂.
Step 4 — the field goes as Î × r̂ = î × k̂ = −ĵ.
Step 5 — read it off: −ĵ = South. So the field points from north towards south.
Five written lines, no mental gymnastics, no chance of error.
3. The two right-hand rules are NOT the same rule
This single confusion is worth several marks a paper. Fingers and thumb swap jobs between the two cases.
How to tell which rule you need in one second. Ask: what is going round in a circle?
If the field circles → straight wire → thumb is the current.
If the current circles → loop or solenoid → thumb is the field.
One second to decide which rule you need: ask what is going round in a circle. If the field circles it's a wire; if the current circles it's a loop. Fingers and thumb swap jobs between them.
4. Force on a charge — and the electron flip
F = q (v × B)
Work out v × B using the cyclic rule. Then — and this is the step she keeps missing — if the charge is negative, reverse the answer.
Write it as three lines, every time:
v = ? in î ĵ k̂
B = ? in î ĵ k̂
v × B = ? → then flip if the charge is negative
Work it out for a positive charge every time, then flip the answer if the particle is an electron. Doing the flip as a separate final step is far safer than trying to build it in.
5. The two cases where the answer is "nothing happens"
ILTS asked both of these and she got both wrong. They look complicated and they are actually the simplest questions on the paper.
Situation
What happens
Why
Charge moves along B (θ = 0° or 180°)
No magnetic force at all — straight line, constant speed
sin θ = 0, so v × B = 0
Electron along a solenoid's axis
Same thing — the field inside a solenoid IS along the axis
v ∥ B again, just disguised
E ∥ B, charge projected along both
Only the electric force acts — speeds up or slows down in a straight line
v × B = 0, leaving only qE
The words "along the axis" or "parallel to the field" are a gift. They mean the magnetic force is zero and the answer is almost always "continues undeflected".
Read the question for the words "along the axis" or "parallel to the field". Those phrases are a gift — they mean the magnetic force is zero and the answer is almost always "continues undeflected" or "only the speed changes".
6. Two wires — which way do the fields point between them?
This feels backwards, so check it with the grip rule every single time rather than trusting memory.
And now the reversal that catches everybody. For the fields at the midpoint, same-direction currents subtract. But for the force between the wires, same-direction currents attract. Two different quantities, two opposite-sounding rules. Never mix them up:
Same direction
Opposite directions
Field at the midpoint
Subtract (can be zero)
Add
Force between the wires
Attract
Repel
Same two wires, same currents, two opposite-sounding rules. The fields at the midpoint cancel, yet the wires attract. Both are correct — they are different quantities.
7. Particle tracks in a figure
Given a picture of four tracks and asked which is the electron, use this order:
Find the neutral one first. A neutron has no charge, so it goes straight. That identifies one track immediately.
Split by sign. Positive and negative curve to opposite sides. So the tracks divide into two groups.
Split by radius. With the same speed, r = mv/qB ∝ m/q. So heavier-per-unit-charge means a wider curve. An alpha (m/q = 2) curves more gently than a proton (m/q = 1); an electron (tiny mass) curves very tightly.
Sort in three steps: find the straight one (neutron), split by which side they bend to (sign of charge), then split by how tight the curve is (r ∝ m/q).
Quick sort for the standard four: neutron → straight · electron → tightest curve, and to the opposite side from the positives · proton → medium curve · alpha → widest curve of the positives.
Part 2 — Directions: rule sheet
Situation
Rule
Remember
Cross product
î×ĵ=k̂, ĵ×k̂=î, k̂×î=ĵ
Backwards round the circle gives a minus sign.
Compass axes
x̂ = E, ŷ = N, ẑ = Up
Write these down before anything else.
Out of / into the page
Dot ⊙ / Cross ⊗
Arrow tip coming at you / tail feathers going away.
Field of a straight wire
Grip the wire, thumb = current
Fingers curl the way the field goes.
Field of a loop or solenoid
Fingers = current, thumb = field
The reverse assignment. Thumb points to the north face.
Force on a charge
F = q(v × B)
Flip the answer for an electron.
Force on a wire
F = I(l × B)
l points along the current.
Charge along B
F = 0
Straight line, unchanged speed.
Electron along a solenoid axis
F = 0
Same rule, disguised.
E ∥ B, charge along both
Only qE acts
Electron slows down; proton speeds up.
Crossed E and B
v = E/B
Undeflected. Independent of charge and mass.
Field at the midpoint of two wires
Same dir → subtract Opposite dir → add
Opposite of the force rule.
Force between two wires
Same dir → attract Opposite dir → repel
Opposite of the rule for charges.
Biot–Savart direction
Screw rule on dl × r
Anticlockwise turn → out of the page.
Particle tracks
Neutron straight; opposite signs curve opposite ways; r ∝ m/q
Electron curves tightest.
The five-line method — use it on every direction question, without exception.
Write the axes.
Write v (or I) in î ĵ k̂.
Write B (or r̂) in î ĵ k̂.
Cross them using the cyclic rule.
Flip if the charge is negative. Then translate back to a compass word.
It takes fifteen seconds and it is the difference between 63 and 88.
Part 3 — 32 direction questions
Use the five-line method on every one: axes, first vector, second vector, cross, flip if negative. Write it out even when the answer feels obvious — that is exactly when mistakes happen.
D1Cross product
A proton moves along the +x axis in a magnetic field directed along +y. The force on it is along:
(a) +z
(b) −z
(c) +y
(d) −x
Show step-by-step solution
Givenv along +x, B along +y, charge positive
AskedDirection of F
ConceptCross product using the cyclic rule, then check the sign of the charge.
FormulaF = q(v × B); î × ĵ = k̂
Solutionv = î, B = ĵ v × B = î × ĵ = k̂ Charge is positive, so no flip. F is along +z.
Answer: +z
D2Cross product
An electron moves along the +x axis in a magnetic field directed along +y. The force on it is along:
(a) +z
(b) −z
(c) +x
(d) −y
Show step-by-step solution
Givenv along +x, B along +y, charge negative
AskedDirection of F
ConceptWork it out for a positive charge first, then reverse it for the electron. This flip is the step most often forgotten.
FormulaF = q(v × B), q negative
Solutionv × B = î × ĵ = k̂, i.e. +z The electron is negative, so reverse it. F is along −z.
Answer: −z
D3Cross product
A positive charge moves along +y in a field along +z. The force is along:
(a) +x
(b) −x
(c) +z
(d) −y
Show step-by-step solution
Givenv along +y, B along +z, positive charge
AskedDirection of F
ConceptAnother turn of the same cyclic circle.
Formulaĵ × k̂ = î
Solutionv × B = ĵ × k̂ = î Positive charge, no flip. F is along +x.
Answer: +x
D4Cross product
An electron moves along +z in a magnetic field along +x. The force on it is along:
(a) +y
(b) −y
(c) +z
(d) −x
Show step-by-step solution
Givenv along +z, B along +x, negative charge
AskedDirection of F
ConceptCyclic rule then flip.
Formulak̂ × î = ĵ
Solutionv × B = k̂ × î = ĵ, i.e. +y Electron is negative ⇒ reverse. F is along −y.
Answer: −y
D5Cross product
A charge moves along +y in a magnetic field also along +y. The force on it is:
(a) along +y
(b) along −y
(c) zero
(d) perpendicular to y
Show step-by-step solution
Givenv parallel to B
AskedForce on the charge
ConceptA vector crossed with itself, or with anything parallel to it, gives zero.
FormulaF = qvB sin θ, θ = 0
Solutionv and B point the same way ⇒ θ = 0° sin 0° = 0 F = 0.
Answer: zero
D6Grip rule
A vertical wire carries current upwards. At a point due NORTH of the wire, the magnetic field points towards the:
(a) north
(b) south
(c) east
(d) west
Show step-by-step solution
GivenCurrent vertically up, point due north of the wire. Axes: x̂ = East, ŷ = North, ẑ = Up.
AskedDirection of B
ConceptField of a straight wire: use Î × r̂, or grip the wire with the thumb along the current.
FormulaB ∝ Î × r̂
SolutionCurrent is up ⇒ Î = k̂. The point is north ⇒ r̂ = ĵ. Î × r̂ = k̂ × ĵ = −î −î = West.
Answer: west
D7Grip rule
A power line carries current from WEST to EAST. At a point a short distance ABOVE it, the magnetic field is directed:
(a) north to south
(b) south to north
(c) east to west
(d) west to east
Show step-by-step solution
GivenCurrent west to east, point above the wire. Axes: x̂ = East, ŷ = North, ẑ = Up.
AskedDirection of B
ConceptThis is the ILTS question. Write the axes, then cross — do not picture it.
FormulaB ∝ Î × r̂
SolutionCurrent west to east ⇒ Î = î. Point above ⇒ r̂ = k̂. Î × r̂ = î × k̂ = −ĵ −ĵ = South, so the field runs from north towards south.
Answer: north to south
D8Grip rule
A horizontal wire carries current from NORTH to SOUTH. At a point due EAST of it, the magnetic field points:
(a) upwards
(b) downwards
(c) north
(d) south
Show step-by-step solution
GivenCurrent north to south, point due east. Axes: x̂ = East, ŷ = North, ẑ = Up.
AskedDirection of B
ConceptSame five-line method.
FormulaB ∝ Î × r̂
SolutionCurrent north to south ⇒ Î = −ĵ. Point east ⇒ r̂ = î. (−ĵ) × î = −(ĵ × î) = −(−k̂) = +k̂ +k̂ = upwards.
Answer: upwards
D9Grip rule
A long wire carries current INTO the page. At a point to the RIGHT of the wire, the field points:
(a) upwards on the page
(b) downwards on the page
(c) to the right
(d) to the left
Show step-by-step solution
GivenCurrent into the page (⊗), point to the right
AskedDirection of B
ConceptInto the page is −k̂. Take right as +î and up the page as +ĵ.
FormulaB ∝ Î × r̂
SolutionÎ = −k̂, r̂ = î (−k̂) × î = −(k̂ × î) = −ĵ −ĵ = downwards on the page. (Equivalently: current into the page ⇒ field circles clockwise.)
Answer: downwards on the page
D10Grip rule
A long wire carries current OUT of the page. The magnetic field lines around it circulate:
(a) clockwise
(b) anticlockwise
(c) radially outward
(d) radially inward
Show step-by-step solution
GivenCurrent out of the page (⊙)
AskedSense of circulation
ConceptPoint your right thumb out of the page towards yourself; your fingers curl anticlockwise.
FormulaRight-hand grip rule
SolutionThumb points out of the page (towards you). The fingers curl anticlockwise as seen from your side. So the field circulates anticlockwise. (Into the page would be clockwise.)
Answer: anticlockwise
D11Two wires
Two long parallel wires carry currents in the SAME direction. At the midpoint between them, their magnetic fields:
(a) point the same way and add
(b) point opposite ways and subtract
(c) are perpendicular
(d) are both zero
Show step-by-step solution
GivenTwo parallel wires, same current direction, point at the midpoint
AskedHow the two fields combine
ConceptApply the grip rule to each wire separately at that one point.
FormulaSuperposition of B
SolutionTake both currents out of the page. Each field circles anticlockwise. At the midpoint, the left wire's field points one way and the right wire's field points the opposite way. So they SUBTRACT, and can cancel completely if the currents are equal.
Answer: point opposite ways and subtract
D12Two wires
Two long parallel wires carry EQUAL currents in the same direction. The magnetic field exactly midway between them is:
SolutionBoth wires are the same distance away, so B₁ = B₂. Same-direction currents ⇒ the fields oppose at the midpoint. B_net = B₁ − B₂ = 0.
Answer: zero
D13Two wires
Two long conductors 8 cm apart produce a field of 300 µT at the midpoint between them. If the currents are equal, they must be:
(a) 30 A in the same direction
(b) 30 A in opposite directions
(c) 60 A in opposite directions
(d) 300 A in the same direction
Show step-by-step solution
Givend = 8 cm, midpoint 4 cm from each, B = 300 µT, equal currents
AskedMagnitude and direction of the currents
ConceptEqual same-direction currents would give ZERO at the midpoint, so they must be opposite — which means the fields add.
FormulaB = 2 × (2 × 10⁻⁷) I / r
SolutionEqual currents in the same direction would cancel ⇒ must be OPPOSITE. Opposite ⇒ fields add: B = 2 × 2×10⁻⁷ × I / 0.04 3 × 10⁻⁴ = 10⁻⁵ I ⇒ I = 30 A So 30 A in opposite directions.
Answer: 30 A in opposite directions
D14Loop rule
A circular loop lies in the plane of the page and carries current ANTICLOCKWISE as you look at it. At the centre, the field points:
(a) out of the page
(b) into the page
(c) clockwise
(d) radially outward
Show step-by-step solution
GivenLoop in the page, current anticlockwise
AskedDirection of B at the centre
ConceptLoop rule: curl the FINGERS along the current, and the THUMB gives the field.
FormulaRight-hand rule for a loop
SolutionCurl your right fingers anticlockwise in the plane of the page. Your thumb then points towards you — out of the page. (Note this is the opposite assignment to a straight wire.)
Answer: out of the page
D15Loop rule
For a current-carrying circular loop, which face behaves as the NORTH pole?
(a) the face the current appears clockwise from
(b) the face the current appears anticlockwise from
(c) both faces
(d) neither face
Show step-by-step solution
GivenCurrent loop acting as a magnetic dipole
AskedWhich face is the north pole
ConceptField lines leave the north pole, and the loop rule tells you which way they leave.
FormulaRight-hand rule for a loop
SolutionCurl the fingers along the current; the thumb gives the field direction inside the loop. Field lines emerge from that face, so it is the north pole. Viewed from that face the current appears ANTICLOCKWISE.
Answer: the face the current appears anticlockwise from
D16Solenoid rule
Looking into one end of a solenoid, the current appears to flow clockwise. That end is:
(a) a north pole
(b) a south pole
(c) neither pole
(d) both, alternately
Show step-by-step solution
GivenCurrent appears clockwise when viewed from one end
AskedNature of that end
ConceptSame loop rule. Clockwise as seen from your side means the thumb points away from you.
FormulaRight-hand rule for a loop
SolutionCurl the right fingers clockwise as seen from your position. The thumb then points away from you, into the solenoid. Field lines enter at that face ⇒ it is a SOUTH pole.
Answer: a south pole
D17Zero force
An electron is projected with uniform velocity ALONG the axis of a current-carrying long solenoid. It will:
(a) move in a circle about the axis
(b) follow a helical path
(c) continue with uniform velocity along the axis
(d) be accelerated along the axis
Show step-by-step solution
GivenElectron projected along the axis of a solenoid
AskedResulting motion
ConceptThe field inside a solenoid points ALONG the axis. So the electron is moving parallel to B — the disguised version of θ = 0.
FormulaF = qvB sin θ, θ = 0
SolutionInside a solenoid, B is directed along the axis. The electron also moves along the axis ⇒ v ∥ B ⇒ θ = 0°. sin 0° = 0, so there is no magnetic force at all. It continues in a straight line with unchanged velocity.
Answer: continue with uniform velocity along the axis
D18Zero force
A uniform electric field and a uniform magnetic field point in the SAME direction. An electron is projected along that direction. The electron will:
(a) turn to the right
(b) turn to the left
(c) slow down
(d) speed up
Show step-by-step solution
GivenE ∥ B, electron projected along both
AskedWhat happens to the electron
Conceptv ∥ B kills the magnetic force entirely, leaving only the electric force — which opposes the motion for a negative charge.
FormulaF = q(E + v × B), with v × B = 0
Solutionv is parallel to B ⇒ v × B = 0 ⇒ no magnetic force. Only qE remains, acting along the line of motion. The electron is negative, so the force opposes E and therefore opposes v. The electron slows down while travelling in a straight line.
Answer: slow down
D19Combined E and B
An ionised gas contains positive and negative ions. An electric field is applied along +x and a magnetic field along −z. The ions will:
(a) all deflect towards +y
(b) all deflect towards −y
(c) positive ions towards +y and negative towards −y
(d) positive ions towards −y and negative towards +y
Show step-by-step solution
GivenE along +x, B along −z, both signs of ion present
AskedDirection of deflection
ConceptThe electric field starts the ions moving in OPPOSITE directions. Then both the velocity and the charge sign flip, and two flips cancel.
FormulaF = q(v × B)
SolutionE along +x drives positive ions along +x and negative ions along −x. Positive: v × B = î × (−k̂) = +ĵ. Charge positive ⇒ force +y. Negative: v × B = (−î) × (−k̂) = −ĵ. Charge negative ⇒ flip ⇒ force +y. Both flips cancel, so ALL ions deflect towards +y.
Answer: all deflect towards +y
D20Combined E and B
In a velocity selector, the electric and magnetic forces on a charge must be:
(a) parallel and equal
(b) antiparallel and equal
(c) perpendicular and equal
(d) both zero
Show step-by-step solution
GivenCharge passing through undeflected
AskedRelation between the two forces
ConceptUndeflected means the net force is zero, so the two forces must cancel exactly.
FormulaqE = qvB, forces opposite
SolutionFor no deflection the net force must vanish. So the electric and magnetic forces must be equal in magnitude and opposite in direction. This gives qE = qvB, hence v = E/B.
Answer: antiparallel and equal
D21Particle tracks
A neutron, a proton, an electron and an alpha particle enter a region of uniform field (into the page) with the same velocity. Which one travels in a STRAIGHT line?
(a) the proton
(b) the electron
(c) the neutron
(d) the alpha particle
Show step-by-step solution
GivenFour particles, same velocity, uniform field into the page
AskedWhich follows a straight track
ConceptStart every track question by finding the neutral particle — it is the free mark.
FormulaF = qvB sin θ
SolutionA neutron carries no charge, so q = 0. F = qvB = 0 for it. With no force it goes straight through undeflected.
Answer: the neutron
D22Particle tracks
In the same figure, the electron and the proton curve:
(a) the same way, electron more tightly
(b) opposite ways, electron more tightly
(c) opposite ways, proton more tightly
(d) the same way, proton more tightly
Show step-by-step solution
GivenElectron and proton, same velocity, same field
AskedRelative directions and radii of the tracks
ConceptTwo separate facts: opposite charge sign gives opposite bending, and r ∝ m/q gives the relative tightness.
FormulaF = q(v × B); r = mv/qB
SolutionOpposite charges ⇒ forces in opposite directions ⇒ they bend to opposite sides. r ∝ m/q. The electron's mass is about 1836 times smaller than the proton's. So the electron has a far smaller radius — it curves much more tightly.
Answer: opposite ways, electron more tightly
D23Particle tracks
A proton and an alpha particle enter with the same velocity. Which has the WIDER curved track?
(a) the proton
(b) the alpha particle
(c) both the same
(d) depends on the field
Show step-by-step solution
GivenSame velocity; proton (m, q), alpha (4m, 2q)
AskedWhich track is wider
ConceptWith speed fixed, the radius depends on m/q.
Formular = mv/qB ⇒ r ∝ m/q
SolutionProton: m/q = m/q. Alpha: 4m/2q = 2m/q. The alpha has twice the m/q ratio. So the alpha particle traces the wider curve.
Answer: the alpha particle
D24Earth's field
A positive charge falls vertically downwards where the Earth's horizontal field runs from south to north. It is deflected towards the:
(a) north
(b) south
(c) east
(d) west
Show step-by-step solution
Givenv downwards, B from south to north, positive charge. Axes: x̂ = East, ŷ = North, ẑ = Up.
AskedDirection of deflection
ConceptWrite the axes and cross — intuition is unreliable here.
FormulaF = q(v × B)
Solutionv is downwards ⇒ v = −k̂. B is northwards ⇒ B = ĵ. v × B = (−k̂) × ĵ = −(k̂ × ĵ) = −(−î) = +î Positive charge ⇒ no flip. +î = East.
Answer: east
D25Earth's field
An electron enters a magnetic field acting vertically DOWNWARDS with velocity from east to west. It is deflected towards the:
(a) north
(b) south
(c) up
(d) down
Show step-by-step solution
Givenv from east to west, B vertically downwards, electron. Axes: x̂ = East, ŷ = North, ẑ = Up.
AskedDirection of deflection
ConceptCross product then flip for the negative charge.
FormulaF = q(v × B)
Solutionv east to west ⇒ v = −î. B downwards ⇒ B = −k̂. v × B = (−î) × (−k̂) = î × k̂ = −ĵ (south) The electron is negative ⇒ flip ⇒ +ĵ = North.
Answer: north
D26Force on a wire
A horizontal wire carries current towards the EAST in a magnetic field pointing NORTH. The force on the wire is directed:
(a) upwards
(b) downwards
(c) west
(d) south
Show step-by-step solution
GivenI towards east, B towards north. Axes: x̂ = East, ŷ = North, ẑ = Up.
AskedDirection of the force
ConceptSame cyclic rule, with the length vector in place of the velocity.
FormulaF = I(l × B)
Solutionl = î (east), B = ĵ (north) l × B = î × ĵ = k̂ k̂ = upwards.
Answer: upwards
D27Force on a wire
A vertical wire carries current UPWARDS in a magnetic field pointing EAST. The force on it is directed:
(a) north
(b) south
(c) up
(d) down
Show step-by-step solution
GivenI upwards, B towards east. Axes: x̂ = East, ŷ = North, ẑ = Up.
AskedDirection of the force
ConceptCross product again — note the order matters: it is l × B, not B × l.
FormulaF = I(l × B)
Solutionl = k̂ (up), B = î (east) l × B = k̂ × î = ĵ ĵ = North.
Answer: north
D28Two wires
Two long parallel wires carry currents in the same direction. The force between them is:
(a) attractive
(b) repulsive
(c) zero
(d) attractive only if the currents are equal
Show step-by-step solution
GivenParallel currents in the same direction
AskedNature of the force
ConceptLearn this as a fact, because reasoning by analogy with charges gives the wrong answer.
Formulaf = μ₀I₁I₂/2πd
SolutionSame-direction (parallel) currents ATTRACT. Opposite-direction (antiparallel) currents repel. This is the reverse of the rule for like charges, which repel.
Answer: attractive
D29Two wires
Beams of electrons and protons travel parallel to each other, in the same direction, at low speed. They:
(a) attract each other
(b) repel each other
(c) neither attract nor repel
(d) attract only at high speed
Show step-by-step solution
GivenElectron beam and proton beam, parallel, same direction
AskedNet interaction
ConceptTwo effects compete: an electric one and a magnetic one. At ordinary speeds the electric one wins by a huge margin.
FormulaCoulomb attraction vs magnetic force
SolutionElectrically, opposite charges attract — a strong effect. Magnetically, the beams are equivalent to currents in OPPOSITE directions (since the charges are opposite), which repel — a much weaker effect. At speeds well below c the electrostatic attraction dominates, so they attract.
Answer: attract each other
D30Loop near a wire
A rectangular loop lies beside a long straight wire, with its nearest arm parallel to the wire and carrying current in the SAME direction as the wire. The loop is:
(a) pushed away from the wire
(b) pulled towards the wire
(c) not acted upon
(d) rotated but not moved
Show step-by-step solution
GivenCoplanar loop beside a long wire, near arm's current parallel to the wire's
AskedNet force on the loop
ConceptThe field of the wire is non-uniform, so the near and far arms feel different forces and do not cancel.
Formulaf ∝ 1/d
SolutionThe near arm carries current parallel to the wire ⇒ attracted. The far arm carries it antiparallel ⇒ repelled. Force falls off as 1/d, so the nearer arm wins. The loop is pulled towards the wire.
Answer: pulled towards the wire
D31Screw rule
According to the right-hand screw rule applied to dl × r, if the turn from dl to r is ANTICLOCKWISE, the resulting field at that point is:
(a) towards the observer
(b) away from the observer
(c) along dl
(d) along r
Show step-by-step solution
GivenBiot–Savart direction convention
AskedDirection of dB
ConceptImagine turning a screwdriver from the first vector towards the second.
FormuladB ∝ dl × r
SolutionTurning a screw anticlockwise brings it OUT towards you. So the resultant points towards the observer (a dot, ⊙). A clockwise turn would send it away (a cross, ⊗).
Answer: towards the observer
D32Convention
In magnetism diagrams, the symbol ⊗ represents a quantity that is:
(a) coming out of the page
(b) going into the page
(c) lying in the plane of the page
(d) equal to zero
Show step-by-step solution
GivenStandard notation
AskedMeaning of the cross symbol
ConceptThink of an arrow: you see the tip coming at you, or the tail feathers going away.
FormulaDot = out, cross = in
SolutionA dot ⊙ is the tip of an arrow pointed at you ⇒ out of the page. A cross ⊗ is the feathered tail of an arrow flying away ⇒ into the page.
Answer: going into the page
Part 4 — Graphs: the concepts
A graph question is not asking you to plot anything. It is asking one thing: which formula applies in which region. Get that and the shape draws itself.
1. The three cylinders — the family she lost both marks on
ILTS asked two of these three, and the same wrong shape was chosen both times. They differ only in how much current the Amperian loop encloses.
SOLID wire, current spread through itHOLLOW pipe, current on the wallsTHIN wire, no inside at all
Type
Inside
Outside
Shape
Solid conductor, radius a
Loop catches only part of the current ⇒ B = μ₀Ir/2πa², so B ∝ r
B = μ₀I/2πr
Straight rise from zero, peak at the surface, then 1/r fall
Hollow pipe, radius R
Loop catches no current ⇒ B = 0
B = μ₀I/2πr
Flat at zero, jump at the surface, then 1/r fall
Thin wire
There is no inside
B = μ₀I/2πr
Pure 1/r hyperbola
The exact mistake to unlearn. On both ILTS graph questions the option chosen was flat at a high value inside, then falling. That shape is wrong for every case in this chapter. There is no configuration where the field is constant and non-zero inside and then falls off outside.
Inside a solid wire the field rises from zero. Inside a hollow pipe it is zero. Never flat-and-high.
The one question to ask yourself. Put your finger at a point inside the conductor and ask: how much current is inside my circle?
Solid wire, halfway in → a quarter of the current (area ratio) → small but growing field.
Hollow pipe, anywhere inside → none of it → zero field.
That question alone tells you which of the two shapes to draw.
The growing loop keeps catching more current, so the field climbs. Past the surface it has caught it all, so only distance matters from then on and the field falls.
Same loop, same growth — but here it encloses nothing until it crosses the wall. That single difference is what separates the two graphs, and it is the one to get right.
2. Reading a ratio off the graph
ILTS also asked for the ratio of the fields at a/3 and 2a — a question she left blank. It is two substitutions:
Inside at a/3: B = μ₀I(a/3)/2πa² = μ₀I/6πaOutside at 2a: B = μ₀I/2π(2a) = μ₀I/4πa
Ratio = (1/6) ÷ (1/4) = 2/3. The only skill is noticing that one point is inside and the other outside, so two different formulas are needed. The graph is what tells you that.
The only skill here is noticing that the two points sit in different regions, so two different formulas are needed. The graph is what tells you that — which is why it is worth drawing before calculating.
3. The motion graphs — what depends on speed and what does not
r ∝ v — a straight lineT does NOT depend on vKE never changes
Straight from the formulas:
r = mv/qB contains v ⇒ r vs v is a straight line through the origin.
T = 2πm/qB has no v ⇒ T vs v is a horizontal line.
Magnetic force does no work ⇒ KE vs time is a horizontal line.
Watch the two dots stay in step lap after lap. Bigger circle, higher speed — the two cancel exactly, which is why the T graph is a horizontal line.
4. How to tell 1/r from 1/r² from 1/r³ by eye
All three fall away, but at very different rates. The higher the power, the more brutally it collapses near the start and the flatter it lies later.
B ∝ 1/r — long straight wireE ∝ 1/r² — point chargeB ∝ 1/x³ — dipole, far from a loop
The trick that removes the guesswork. A curve is hard to identify by eye, but a straight line is not. So re-plot against the reciprocal:
If B vs 1/r is a straight line through the origin ⇒ B ∝ 1/r ⇒ straight wire.
If B vs 1/r² is a straight line ⇒ inverse square.
If B vs 1/x³ is a straight line ⇒ dipole.
Hard to tell apart by eye at a glance — which is why the reliable method is to plot against the reciprocal and look for a straight line instead.
5. The sine-shaped graphs
τ = mB sin θ — zero, peak, zeroU = −mB cos θ — lowest at 0°
Torque starts at zero when the moment is aligned with the field, peaks at 90°, and returns to zero at 180°. Potential energy does the opposite: lowest at 0° (stable), highest at 180° (unstable). The same sine factor governs F = qvB sin θ for a charge.
One rotation, two curves. Torque starts at zero when aligned and peaks at 90°; energy is lowest when aligned and highest when flipped. They are 90° out of step with each other.
6. Along a solenoid
Plot the field along the axis of a long solenoid and you get a flat plateau in the middle that drops to half at each end and dies away outside. The plateau is the whole point of a solenoid — it is the only easy way to make a genuinely uniform field.
Uniform right through the middle, exactly half at each end, and essentially zero outside. That flat plateau is the entire reason solenoids are useful.
Part 5 — Graphs: shape sheet
Every graph in this chapter is one of these. Learn to draw each from its formula rather than recognising it from a picture.
Graph
Formula behind it
Shape
Solid conductor: B vs r
μ₀Ir/2πa² then μ₀I/2πr
Rise from zero → peak at r = a → 1/r fall
Hollow pipe: B vs r
0 then μ₀I/2πr
Zero inside → jump at r = R → 1/r fall
Thin wire: B vs r
μ₀I/2πr
Pure hyperbola, no inside region
Thin wire: B vs 1/r
μ₀I/2π × (1/r)
Straight line through the origin
Coil: B vs x along the axis
μ₀NIR²/2(x²+R²)^{3/2}
Max at the centre, symmetric fall, → 1/x³ far away
Solenoid: B vs position
μ₀nI
Flat plateau, half value at the ends, ≈0 outside
Solenoid: B vs radius
μ₀nI
Horizontal line — no radius in the formula
Any config: B vs I
B ∝ I
Straight line through the origin
Coil centre: B vs N
μ₀NI/2R
Straight line through the origin
Coil centre: B vs R
μ₀NI/2R
Hyperbola (1/R)
Charge: r vs v
r = mv/qB
Straight line through the origin
Charge: r vs 1/B
r = mv/qB
Straight line through the origin
Charge: r vs √K
r = √(2mK)/qB
Straight line through the origin
Charge: T vs v
T = 2πm/qB
Horizontal line
Charge: KE vs t
W = 0
Horizontal line
Loop: τ vs θ
τ = mB sin θ
Sine hump: 0 → max at 90° → 0
Loop: U vs θ
U = −mB cos θ
Minimum at 0°, maximum at 180°
Two wires: f vs d
μ₀I₁I₂/2πd
Hyperbola (1/d), not 1/d²
Ampere: ∮B·dl vs Ienc
μ₀ Ienc
Straight line, slope μ₀
The three-question checklist for any graph MCQ.
Where does it start? At zero, or at a finite value? (Solid wire starts at zero; hollow starts at zero; thin wire starts at infinity.)
Is there a kink or a jump? If the question mentions a radius a or R, there must be a change of behaviour at that point.
Does it fall as 1/r or die faster? Straight wire 1/r, point charge 1/r², dipole 1/x³.
Answer those three and only one option will survive.
Part 6 — 32 graph questions
For each one, run the three-question checklist before looking at the options: where does it start, is there a kink, and how fast does it fall?
G1Cylinder graphs
A solid cylindrical conductor of radius R carries a steady current distributed uniformly over its cross-section. Which plot of B against distance d from the axis is correct?
(a)
(b)
(c)
(d)
Show step-by-step solution
GivenSolid conductor of radius R, uniform current
AskedCorrect B versus d graph
ConceptInside, the Amperian loop encloses only part of the current, so B grows with d. Outside, it encloses all of it, so B falls as 1/d.
FormulaInside B = μ₀Id/2πR²; outside B = μ₀I/2πd
SolutionAt d = 0 the loop encloses no current ⇒ B = 0. So the graph must START AT ZERO. Inside, enclosed current ∝ d² and circumference ∝ d, giving B ∝ d — a straight rise. At d = R the field is maximum. Outside, B ∝ 1/d — a hyperbolic fall. So: straight rise from zero, peak at R, then 1/d decay — option (b).
Answer: (b) rise from zero to a peak at R, then 1/d fall
G2Cylinder graphs
A long thin HOLLOW metallic cylinder of radius R carries current i. Which plot shows B against distance r from the axis?
(a)
(b)
(c)
(d)
Show step-by-step solution
GivenHollow cylinder of radius R carrying current i
AskedCorrect B versus r graph
ConceptInside a hollow pipe the Amperian loop encloses NO current at all, so the field there is exactly zero.
FormulaInside B = 0; outside B = μ₀i/2πr
SolutionAll the current flows along the walls. An Amperian loop drawn inside the cavity encloses zero current ⇒ B = 0 throughout the inside. At the surface the field jumps to its maximum, μ₀i/2πR. Outside it falls as 1/r. So: flat at ZERO inside, then 1/r decay — option (c).
Answer: (c) zero inside, then 1/r fall from R
G3Cylinder graphs
For a long THIN straight wire, which plot of B against r is correct?
(a)
(b)
(c)
(d)
Show step-by-step solution
GivenLong thin straight wire
AskedCorrect B versus r graph
ConceptA thin wire has no interior region, so there is only one formula and one shape.
FormulaB = μ₀I/2πr
SolutionThere is no 'inside' to worry about. B = μ₀I/2πr everywhere outside the wire. That is a pure hyperbola with no kink and no flat portion — option (a).
Answer: (a) a pure 1/r hyperbola
G4Cylinder graphs
Inside a solid conductor carrying a uniformly distributed current, the magnetic field varies with distance r from the axis as:
(a) B ∝ 1/r
(b) B ∝ 1/r²
(c) B ∝ r
(d) B ∝ r²
Show step-by-step solution
GivenPoint inside a solid conductor
AskedDependence of B on r
ConceptThe enclosed current grows as the area, but the loop's circumference grows too.
FormulaB(2πr) = μ₀ I r²/a²
SolutionEnclosed current = I × (πr²/πa²) = I r²/a² B × 2πr = μ₀ I r²/a² B = μ₀ I r / 2πa², so B ∝ r.
Answer: B ∝ r
G5Cylinder graphs
For the same solid conductor, at which distance is the magnetic field MAXIMUM?
(a) at the axis
(b) at the surface
(c) at twice the radius
(d) far away
Show step-by-step solution
GivenSolid conductor of radius a
AskedPosition of maximum B
ConceptThe field rises inside and falls outside, so the peak must be at the join.
FormulaInside B ∝ r; outside B ∝ 1/r
SolutionInside, B increases with r — so the maximum is not at the axis. Outside, B decreases with r — so it is not beyond the surface. The two behaviours meet at r = a, which is where the field peaks.
Answer: at the surface
G6Reading a graph
A long straight wire of radius a carries a uniformly distributed current. The ratio of the magnetic fields at distances a/3 and 2a from the axis is:
(a) 1/2
(b) 3/2
(c) 2/3
(d) 2
Show step-by-step solution
GivenWire of radius a; points at r = a/3 (inside) and r = 2a (outside)
AskedRatio of the two fields
ConceptThe two points lie in DIFFERENT regions, so two different formulas are needed. Spotting that is the whole question.
FormulaInside B = μ₀Ir/2πa²; outside B = μ₀I/2πr
SolutionAt r = a/3 (inside): B₁ = μ₀I(a/3)/2πa² = μ₀I/6πa At r = 2a (outside): B₂ = μ₀I/2π(2a) = μ₀I/4πa Ratio = (1/6) ÷ (1/4) = 4/6 = 2/3.
Answer: 2/3
G7Reading a graph
For a solid conductor of radius a, the field at r = a/2 equals the field at which outside distance?
(a) r = a
(b) r = 2a
(c) r = 4a
(d) no such point exists
Show step-by-step solution
GivenSolid conductor of radius a
AskedOutside point with the same field as r = a/2
ConceptSet the inside expression equal to the outside expression and solve.
Formulaμ₀Ir/2πa² = μ₀I/2πr′
SolutionInside at a/2: B = μ₀I(a/2)/2πa² = μ₀I/4πa Outside at r′: B = μ₀I/2πr′ Setting equal: 2πr′ = 4πa ⇒ r′ = 2a. A point inside and a point outside CAN have the same field — the graph shows this clearly.
Answer: r = 2a
G8Motion graphs
For a charged particle moving perpendicular to a fixed magnetic field, which graph shows radius r against speed v?
(a)
(b)
(c)
(d)
Show step-by-step solution
GivenFixed q, m and B; varying speed
AskedShape of the r versus v graph
ConceptRead the proportionality straight from the formula.
Formular = mv/qB ⇒ r ∝ v
SolutionWith m, q, B fixed, r = (m/qB) × v. This is y = kx. So the graph is a straight line through the origin — option (a).
Answer: (a) straight line through the origin
G9Motion graphs
For the same particle, which graph shows time period T against speed v?
(a)
(b)
(c)
(d)
Show step-by-step solution
GivenFixed q, m and B; varying speed
AskedShape of the T versus v graph
ConceptCheck whether v appears in the formula at all — it does not.
FormulaT = 2πm/qB
SolutionT depends only on m, q and B. There is no v anywhere in the expression. So T is the same at every speed — a horizontal line, option (a).
Answer: (a) a horizontal line
G10Motion graphs
Which graph shows the kinetic energy of a charged particle against time, while it moves in a uniform magnetic field?
(a)
(b)
(c)
(d)
Show step-by-step solution
GivenCharged particle moving in a uniform magnetic field
AskedShape of the KE versus time graph
ConceptThe magnetic force is always perpendicular to the velocity, so it does no work.
FormulaW = 0 ⇒ KE constant
SolutionNo work is done by a magnetic force. So the kinetic energy cannot change at any moment. The graph is a horizontal line — option (a).
Answer: (a) a horizontal line
G11Motion graphs
Which graph correctly shows the radius r of the circular path against 1/B, at fixed speed?
(a)
(b)
(c)
(d)
Show step-by-step solution
GivenFixed m, v and q; varying B
AskedShape of the r versus 1/B graph
ConceptRewrite the formula so that 1/B appears as the variable.
Formular = (mv/q) × (1/B)
Solutionr = mv/qB = (mv/q) × (1/B) With mv/q constant, r is directly proportional to 1/B. So the graph is a straight line through the origin — option (a).
Answer: (a) straight line through the origin
G12Motion graphs
At fixed field, the graph of radius r against √K (K = kinetic energy) is:
(a) a horizontal line
(b) a straight line through the origin
(c) a hyperbola
(d) a parabola
Show step-by-step solution
GivenFixed q, m and B; varying kinetic energy
AskedShape of the r versus √K graph
ConceptExpress the radius in terms of kinetic energy first.
Formular = √(2mK)/qB
Solutionr = √(2mK)/qB = (√(2m)/qB) × √K The bracket is constant, so r ∝ √K. Plotted against √K this is a straight line through the origin.
Answer: a straight line through the origin
G13Motion graphs
The graph of angular frequency ω against speed v for a charge in a uniform magnetic field is:
(a) a straight line through the origin
(b) a horizontal line
(c) a hyperbola
(d) a parabola
Show step-by-step solution
GivenFixed q, m and B
AskedShape of the ω versus v graph
ConceptSame reasoning as for the time period — no v in the formula.
Formulaω = qB/m
Solutionω = qB/m contains no v. So ω is unchanged however fast the particle goes. The graph is a horizontal line.
Answer: a horizontal line
G14Distance laws
Which graph represents the axial magnetic field of a small current loop at large distances?
(a)
(b)
(c)
(d)
Show step-by-step solution
GivenSmall current loop, axial point with x ≫ R
AskedShape of the B versus x graph
ConceptThe dipole field collapses as 1/x³ — much faster than a wire's 1/x.
FormulaB = (μ₀/4π)(2m/x³)
SolutionFar from the loop, B ∝ 1/x³. This falls away far more steeply than 1/x and flattens sooner. Option (a) shows that steep collapse.
Answer: (a) the steeply falling 1/x³ curve
G15Distance laws
A graph of B against 1/r for a long straight wire is:
(a)
(b)
(c)
(d)
Show step-by-step solution
GivenLong straight wire, B plotted against 1/r
AskedShape of the graph
ConceptThis is the standard trick for identifying a power law: plot against the reciprocal and look for a straight line.
FormulaB = (μ₀I/2π) × (1/r)
SolutionB = μ₀I/2πr can be written as B = (μ₀I/2π)(1/r). With I fixed, B is directly proportional to 1/r. So the graph is a straight line through the origin — option (a).
Answer: (a) straight line through the origin
G16Distance laws
Three fields fall off as 1/r, 1/r² and 1/r³. These correspond respectively to:
(a) point charge, straight wire, dipole
(b) straight wire, point charge, dipole
(c) dipole, straight wire, point charge
(d) straight wire, dipole, point charge
Show step-by-step solution
GivenThree different distance dependences
AskedWhich source gives which
ConceptThese three are the only distance laws in the chapter, and NEET mixes them up deliberately.
SolutionA long straight wire gives B = μ₀I/2πr ⇒ 1/r. A point charge gives E ∝ 1/r². A current loop far away (a dipole) gives B ∝ 1/x³. So the order is: straight wire, point charge, dipole.
Answer: straight wire, point charge, dipole
G17Coil graphs
The graph of the field at the centre of a circular coil against the number of turns N is:
(a) a horizontal line
(b) a straight line through the origin
(c) a hyperbola
(d) a parabola
Show step-by-step solution
GivenCircular coil, fixed I and R, varying N
AskedShape of the B versus N graph
ConceptN sits on top of the formula, to the first power.
FormulaB = μ₀NI/2R
SolutionB = (μ₀I/2R) × N The bracket is constant, so B ∝ N. A straight line through the origin.
Answer: a straight line through the origin
G18Coil graphs
The graph of the field at the centre of a circular coil against its radius R is:
(a)
(b)
(c)
(d)
Show step-by-step solution
GivenCircular coil, fixed N and I, varying R
AskedShape of the B versus R graph
ConceptR sits in the denominator, to the first power.
FormulaB = μ₀NI/2R
SolutionB = (μ₀NI/2) × (1/R) So B ∝ 1/R — a hyperbola. Option (a).
Answer: (a) a 1/R hyperbola
G19Solenoid graphs
The graph of the field inside a long solenoid against its radius is:
(a)
(b)
(c)
(d)
Show step-by-step solution
GivenLong solenoid, fixed n and I, varying radius
AskedShape of the B versus radius graph
ConceptThere is no radius term anywhere in the solenoid formula — this is tested precisely because students expect one.
FormulaB = μ₀nI
SolutionB = μ₀nI contains only n and I. The radius does not appear at all. So B is unchanged as the radius varies — a horizontal line, option (a).
Answer: (a) a horizontal line
G20Solenoid graphs
Plotted along the axis from one end of a long solenoid to the other, the field is:
(a) zero in the middle and maximum at the ends
(b) uniform in the middle and half that value at the ends
(c) uniform everywhere including outside
(d) increasing steadily from one end to the other
Show step-by-step solution
GivenLong solenoid, field plotted along the axis
AskedShape of the profile
ConceptThe 'very long on both sides' assumption holds in the middle but fails at the ends.
FormulaMiddle B = μ₀nI; end B = μ₀nI/2
SolutionIn the middle, turns extend far in both directions ⇒ B = μ₀nI, and it is uniform. At each end only one side contributes ⇒ B = μ₀nI/2. So: a flat plateau in the middle dropping to half at the ends.
Answer: uniform in the middle and half that value at the ends
G21Torque graphs
Which graph shows the torque on a current loop against the angle θ between its magnetic moment and the field?
(a)
(b)
(c)
(d)
Show step-by-step solution
GivenCurrent loop in a uniform field, θ from 0° to 180°
AskedShape of the τ versus θ graph
ConceptThe torque carries a sine factor, so it starts and ends at zero with a peak in between.
Formulaτ = mB sin θ
SolutionAt θ = 0°: sin 0 = 0 ⇒ τ = 0. At θ = 90°: sin 90° = 1 ⇒ τ is maximum. At θ = 180°: sin 180° = 0 ⇒ τ = 0 again. That is a sine hump — option (a).
Answer: (a) a sine hump, zero at both ends
G22Torque graphs
The potential energy of a magnetic dipole in a uniform field, plotted against θ from 0° to 180°:
(a) is maximum at 0° and minimum at 180°
(b) is minimum at 0° and maximum at 180°
(c) is zero throughout
(d) is maximum at 90°
Show step-by-step solution
GivenMagnetic dipole in a uniform field
AskedShape of the U versus θ graph
ConceptU follows a negative cosine, so it is lowest where the torque is zero and the dipole is aligned.
FormulaU = −mB cos θ
SolutionAt θ = 0°: U = −mB, the minimum — stable equilibrium. At θ = 90°: U = 0. At θ = 180°: U = +mB, the maximum — unstable equilibrium.
Answer: is minimum at 0° and maximum at 180°
G23Force graphs
The force on a charge moving in a magnetic field, plotted against the angle θ between v and B, is:
(a) constant
(b) a sine curve, zero at 0° and 180°
(c) a cosine curve, maximum at 0°
(d) a straight line
Show step-by-step solution
GivenCharge moving at angle θ to the field
AskedShape of the F versus θ graph
ConceptSame sine factor as the torque — zero when aligned, maximum when perpendicular.
FormulaF = qvB sin θ
SolutionAt θ = 0° and 180° the force is zero (motion along the field). At θ = 90° it is maximum, qvB. So it traces a sine curve.
Answer: a sine curve, zero at 0° and 180°
G24Two wires
The graph of the force per unit length between two parallel wires against their separation d is:
(a)
(b)
(c)
(d)
Show step-by-step solution
GivenTwo long parallel current-carrying wires
AskedShape of the f versus d graph
ConceptThis is a 1/d law, not an inverse square — a very common slip.
Formulaf = μ₀I₁I₂/2πd
Solutiond appears once in the denominator, so f ∝ 1/d. That is a hyperbola — option (a). Option (b) is the inverse-square shape, which would be wrong here.
Answer: (a) a 1/d hyperbola
G25Ampere's law
The graph of ∮B·dl around a loop against the current enclosed by it is:
(a) a horizontal line
(b) a straight line through the origin of slope μ₀
(c) a hyperbola
(d) a parabola
Show step-by-step solution
GivenAmperian loop with varying enclosed current
AskedShape of the graph
ConceptAmpere's law is a simple direct proportion between the two quantities.
Formula∮B·dl = μ₀ I_enclosed
SolutionThe relation is a direct proportion. So the graph is a straight line through the origin. Its slope is μ₀.
Answer: a straight line through the origin of slope μ₀
G26Coaxial cable
For a coaxial cable carrying equal and opposite currents in its inner and outer conductors, the field OUTSIDE the whole cable is:
(a) μ₀I/2πr
(b) twice μ₀I/2πr
(c) zero
(d) μ₀I/4πr
Show step-by-step solution
GivenCoaxial cable, equal and opposite currents, point outside both conductors
AskedField outside
ConceptAmpere's law counts the ALGEBRAIC sum of enclosed currents.
Formula∮B·dl = μ₀ I_enclosed(net)
SolutionA loop drawn outside encloses both conductors. The two currents are equal and opposite, so the net enclosed current is zero. Hence B = 0 outside the cable.
Answer: zero
G27Statements
For a long hollow current-carrying tube, consider: (I) the field inside is constant and equal to zero; (II) the field outside decreases with distance; (III) the field at the surface is maximum. Which are correct?
(a) I and II only
(b) II and III only
(c) I and III only
(d) I, II and III
Show step-by-step solution
GivenLong hollow tube carrying current
AskedWhich statements are correct
ConceptCheck each against the hollow-cylinder graph: zero inside, peak at the surface, 1/r fall outside.
FormulaInside B = 0; outside B = μ₀i/2πr
Solution(I) An Amperian loop inside encloses no current ⇒ B = 0 — correct. (II) Outside, B = μ₀i/2πr decreases with r — correct. (III) The field is greatest right at the surface, where r is smallest among the outside points — correct. All three are correct.
Answer: I, II and III
G28Cylinder graphs
Which of the following B versus r shapes does NOT occur for any configuration in this chapter?
(a) rise from zero, then 1/r fall
(b) zero, then 1/r fall
(c) constant non-zero inside, then 1/r fall
(d) pure 1/r fall
Show step-by-step solution
GivenThe standard cylindrical configurations
AskedThe shape that never occurs
ConceptThis is the exact wrong option chosen twice on the ILTS — worth learning as an impossible shape.
FormulaAmpere's law applied inside each configuration
Solution'Rise then fall' is the solid conductor — real. 'Zero then fall' is the hollow pipe — real. 'Pure 1/r' is the thin wire — real. 'Constant non-zero inside, then falling' occurs for NONE of them. Inside a solid wire the field rises from zero; inside a hollow pipe it is zero.
Answer: constant non-zero inside, then 1/r fall
G29Coil graphs
Along the axis of a circular coil, plotting B against the axial distance x on both sides of the centre gives:
(a) a maximum at the centre falling symmetrically on both sides
(b) a minimum at the centre
(c) a constant value
(d) a straight line
Show step-by-step solution
GivenCircular coil, field plotted along its axis
AskedShape of the profile
ConceptPut x = 0 into the axial formula and compare with larger x.
FormulaB = μ₀NIR²/2(x²+R²)^{3/2}
SolutionAt x = 0 the denominator is smallest, so B is greatest — the centre value μ₀NI/2R. As |x| grows the denominator grows and B falls. The formula depends on x², so it is symmetric about the centre. A peak at the centre falling away on both sides.
Answer: a maximum at the centre falling symmetrically on both sides
G30Recognition
A graph of B against r starts at a non-zero value at r = 0 and decreases smoothly. This corresponds to:
(a) a solid conductor
(b) a hollow pipe
(c) no configuration in this chapter
(d) a solenoid
Show step-by-step solution
GivenGraph starting at a finite non-zero value at r = 0
AskedWhich configuration it matches
ConceptAsk what the field is exactly on the axis in each case.
FormulaInside solid B ∝ r; inside hollow B = 0
SolutionOn the axis of a solid conductor, B = 0 (no enclosed current). Inside a hollow pipe, B = 0 as well. A thin wire's field is undefined at r = 0, not finite. So no configuration in this chapter gives a finite non-zero value at the axis.
Answer: no configuration in this chapter
G31Recognition
Which quantity, plotted against the current I, always gives a straight line through the origin?
(a) the field at the centre of a coil
(b) the time period of a charge in a field
(c) the radius of a charge's path
(d) the angular frequency of a charge
Show step-by-step solution
GivenFour quantities from the chapter
AskedThe one proportional to I
ConceptLook for I appearing to the first power on top, with nothing else varying.
FormulaB = μ₀NI/2R
SolutionField at a coil's centre: B ∝ I ⇒ straight line through the origin. Time period T = 2πm/qB has no I in it. The radius r = mv/qB does not involve the coil current. ω = qB/m likewise.
Answer: the field at the centre of a coil
G32Recognition
A student plots B against r for some configuration and obtains a straight line rising from the origin that then stops abruptly. The most likely explanation is:
(a) the field really does stop
(b) only the inside of a solid conductor has been plotted
(c) it is a hollow pipe
(d) it is a solenoid
Show step-by-step solution
GivenGraph showing only a linear rise, ending at some r
AskedInterpretation
ConceptA linear rise is the signature of the inside of a solid conductor; the outside region is simply missing.
FormulaInside B = μ₀Ir/2πa²
SolutionA rise proportional to r happens only inside a solid conductor. The plot ends where the conductor's surface is. The 1/r portion outside has just not been drawn. This is a distractor option in several NEET graph questions.
Answer: only the inside of a solid conductor has been plotted
G33Comparing
Two graphs are drawn for the same solid conductor: one for a current I and one for 2I. The second graph:
(a) has the same shape but is twice as tall everywhere
(b) has a different shape
(c) peaks at twice the radius
(d) is identical
Show step-by-step solution
GivenSolid conductor, current doubled
AskedEffect on the graph
ConceptBoth the inside and outside formulas are linear in I, so doubling I scales the whole curve.
FormulaInside μ₀Ir/2πa²; outside μ₀I/2πr
SolutionI appears to the first power in both regions. So every value of B doubles. The peak stays at r = a; only the height changes. Same shape, twice as tall.
Answer: has the same shape but is twice as tall everywhere
G34Comparing
For a solid conductor of radius a, the field at r = 2a compared with the field at the surface is:
(a) the same
(b) half
(c) one quarter
(d) double
Show step-by-step solution
GivenSolid conductor of radius a; compare r = a and r = 2a
AskedRatio of the fields
ConceptBoth points are outside or on the surface, so the same 1/r formula applies to both.
FormulaB = μ₀I/2πr for r ≥ a
SolutionAt the surface: B = μ₀I/2πa At r = 2a: B = μ₀I/4πa The second is half the first.
Answer: half
G35Recognition
Which graph shape would you expect for the magnetic field inside a solenoid plotted against the current I?
(a)
(b)
(c)
(d)
Show step-by-step solution
GivenLong solenoid, fixed n, varying current
AskedShape of the B versus I graph
ConceptI appears to the first power on top.
FormulaB = μ₀nI
SolutionB = (μ₀n) × I with μ₀n constant. So B ∝ I. A straight line through the origin — option (a).
Answer: (a) straight line through the origin
G36Checklist
When identifying a B versus r graph, the FIRST thing to check is:
(a) how steeply it falls at large r
(b) whether it starts at zero or at a finite value
(c) the units on the axes
(d) the position of the label R
Show step-by-step solution
GivenAny B versus r graph question
AskedThe most useful first check
ConceptThe starting value distinguishes the solid conductor and hollow pipe from the thin wire immediately, and it eliminates the commonest wrong option.
FormulaThree-question checklist
SolutionStart value: zero for solid and hollow; unbounded for a thin wire. Then look for a kink or jump at r = R. Then check the rate of fall (1/r, 1/r² or 1/r³). The start value alone rules out the 'flat and high inside' distractor every time.
Answer: whether it starts at zero or at a finite value
How to use this pack
Day 1 — directions. Read Part 1, copy the rule sheet by hand, then do D1–D16. Write the five lines on every single question, even the easy ones.
Day 2 — directions. D17–D32, then redo any from Day 1 that needed a second attempt.
Day 3 — graphs. Read Part 4. Then, before touching the questions, draw all three cylinder graphs from scratch on blank paper — solid, hollow, thin — straight from the formulas. Repeat until they come out right without looking.
Day 4 — graphs. G1–G32.
Day 5. Re-attempt the five direction questions and two graph questions from the ILTS itself. She should now get all seven.
The habit that actually fixes this
Direction errors do not come from not knowing the rules — she knows them. They come from trying to do the rotation in her head under time pressure. The fix is mechanical: write the axes before reading the options. Fifteen seconds of writing beats fifteen seconds of visualising, every time.
What this is worth. On the 9 August paper: five direction errors and two graph errors. Converting a wrong answer to a right one is a five-mark swing each under +4/−1 marking. Seven questions × 5 = 35 marks, taking 63 to about 98 without learning a single new formula.