Everything in this chapter happens because flux changes. This file makes sure you can count flux correctly first: the right angle, the right component, the right number of turns.
Tilting hoop. Straight arrows are field lines. As the hoop turns, fewer lines pass through it. The green bar is the flux. It is largest when the hoop faces the field and zero when the hoop is edge-on.
Picture it
Hold a hula hoop in the rain when the rain is falling straight down. If the hoop lies flat, like a table top, lots of raindrops fall through it. Tilt it and fewer drops get through. Hold it upright, like a door, and almost no drops pass through at all. The rain did not change. Only the way the hoop faces the rain changed.
Magnetic field lines are like that rain. Magnetic flux is simply a count of how many field lines pass through a surface. Three things decide the count: how strong the field is (how heavy the rain is), how big the hoop is, and which way the hoop faces.
In exam language
The magnetic flux linked with a surface held in a magnetic field is the number of field lines crossing that surface. For a flat surface of area A in a uniform field B,
φ = B · A = BA cos θ
where θ is the angle between the field B and the area vectorA. The area vector points straight out of the surface (along its normal) and its length equals the area.
The angle trap: plane or normal?
Plane versus normal. The plane is tilted 30° to the field, but θ in the formula is measured to the normal, so θ = 90° − 30° = 60°.
This is the single most common way to lose marks on flux. The formula uses the angle between the field and the normal to the surface, not the angle between the field and the surface itself. Those two angles always add up to 90°.
A quick check that never fails: if the loop's plane is perpendicular to the field, the field goes straight through, so flux must be the largest, BA. If the plane is parallel to the field, the lines skim along it, so flux must be zero. Your cos or sin choice has to agree with those two cases.
Trap
"Plane inclined at 30° to the field" means θ = 60°, so φ = BA cos 60° = BA/2. Writing BA cos 30° gives 0.866 BA, which is always one of the options. Allen Illustration 1 uses exactly this wording.
In exam language
If α is the angle between the field and the plane, then θ = 90° − α and φ = BA sin α. Both forms give the same number when used correctly.
Flux with vectors: the dot product
Picking a component. The dot product keeps only the part of B that lies along the area vector. The other parts skim along the surface and add nothing.
When the field is given as B = Bxî + Byĵ + Bzk̂ and the area as a vector, multiply matching parts and add:
φ = BxAx + ByAy + BzAz
A surface lying in the x–y plane has its area vector along k̂, so only Bz matters. A surface in the y–z plane has its area vector along î, so only Bx matters. A surface in the z–x plane has its area vector along ĵ, so only By matters.
Allen Illustration 2: B = 0.3ĵ T and A = (2î + 5ĵ − 3k̂) m² give φ = 0.3 × 5 = 1.5 Wb. Illustration 4: (2î + 3ĵ + 4k̂)·(3î + 4ĵ) = 6 + 12 = 18 Wb. The k̂ part of B has no partner, so it contributes nothing.
Trap
A negative answer is not wrong. It only means the field goes through the surface opposite to the chosen area vector. In Allen Illustration 3(c), φ = −BVS means the earth's vertical field points down while the chosen area vector points up.
Flux linkage: many turns
Picture it
A coil with 100 turns is like 100 hoops stacked on top of each other. Every hoop catches the same field lines, so the total count is 100 times what one hoop catches.
In exam language
Flux linkage = Nφ = NBA cos θ
Flux through one turn is φ. Flux linkage for the whole coil is Nφ. Faraday's law uses the flux linkage, so N multiplies everything later in the chapter.
Trap
Questions sometimes say "flux linked with the coil" and give a number that is already the total, Nφ. Other times the number is per turn. When the options only work one way, that tells you which was meant. This matters in Allen Beginner's Box 6 Q2 and Illustration 43 (see the corrections section).
Units, dimensions and nature
Item
Value
How to remember it
SI unit
weber (Wb)
Named after Wilhelm Weber
Same unit written another way
T m² (tesla metre squared)
Straight from B × A
Also equal to
V s, or J/A
From e = dφ/dt, so Wb = V × s
CGS unit
maxwell (Mx)
1 Wb = 10⁸ Mx, because 1 T = 10⁴ G and 1 m² = 10⁴ cm²
Dimensional formula
[M L² T⁻² A⁻¹]
B is [M T⁻² A⁻¹] (from F = ILB); multiply by L²
Nature
scalar
It is a dot product of two vectors, and a dot product is a number
Trap
Three lookalike dimensional formulas: magnetic field [M T⁻² A⁻¹], magnetic flux [M L² T⁻² A⁻¹], and inductance (henry) [M L² T⁻² A⁻²]. Flux has A⁻¹; inductance has A⁻². The volt is [M L² T⁻³ A⁻¹], which has T⁻³.
Field lines are imaginary drawings that help us picture a field. Flux is a real, measurable quantity with units and dimensions.
When the field is not the same everywhere
Adding up strips. Near the wire the field is strong. Far away it is weak. We cut the loop into thin strips, find the flux through each, and add them all with an integral.
BA cos θ only works when B is the same over the whole surface. Next to a long straight wire, B = μ₀I/2πx gets weaker as x grows. So we slice the loop into thin strips, each so thin that B is nearly constant across it.
φ = ∫ B · dA
For a rectangle with its side ℓ parallel to the wire, stretching from distance a to distance b:
φ = ∫ₐᵇ (μ₀I / 2πx) ℓ dx = (μ₀Iℓ / 2π) ln(b/a)
Allen Illustration 6 has a square of side ℓ whose near edge is at distance t, so b = t + ℓ and φ = (μ₀Iℓ/2π) ln((t + ℓ)/t).
Trap
Using B at the middle of the loop times the area gives a close number but not the right one, and NEET options are usually written with the ln already in them.
Closed surfaces always give zero
Bubble around a magnet. Every field line that leaves the bubble must come back into it, because field lines form closed loops. Out-going and in-coming cancel.
Picture it
Imagine a soap bubble around a bar magnet. Field lines come out of N, curve round, and go back into S, then continue inside the magnet back to N. Each line is a closed loop. So each line that pokes out of the bubble has to poke back in somewhere. Count out, count in, subtract: zero.
In exam language
∮ B · dA = 0
This is Gauss's law for magnetism. It holds for every closed surface because isolated magnetic poles (monopoles) do not exist.
Cube in a uniform field (Allen Beginner's Box 1 Q3). The two faces perpendicular to B carry equal and opposite flux. The four faces parallel to B carry none. The total is zero, as it must be for a closed surface.
Changing the flux by turning
Turning a ring over. The flux passes through zero on the way. After a half turn it is the same size but opposite in sign, so the change is 2BA, not zero.
If a ring is turned from angle θ₁ to angle θ₂ in a steady field, the change in flux is
Δφ = BA (cos θ₂ − cos θ₁)
Turn
From → to
Size of change
Face-on to edge-on
0° → 90°
BA
Face-on to face-on the other way (half turn)
0° → 180°
2BA
Full turn
0° → 360°
0
Face-on to 60°
0° → 60°
BA/2
Correction to the Allen module
Allen Illustration 5 writes Δφ = BA(cos 0° − cos 180°) = 2BA. With φ₂ − φ₁ the order should be cos 180° − cos 0°, which gives −2BA. The size, 2BA, is correct. Keep the sign straight, because Faraday's law turns this sign into the direction of the current.
Two zero-flux cases from Allen Illustrations 7 and 8. In Illustration 7 the field runs parallel to the coil's plane, so θ = 90°. In Illustration 8 a straight wire runs along the axis of a ring. Its field lines are circles lying in the ring's own plane, so none of them cross the ring.
Rate of change of flux
The next topics are all about how fast flux changes. The average rate is Δφ/Δt = (φfinal − φinitial)/Δt. The instantaneous rate is dφ/dt. If φ = 4t² + 2t, then dφ/dt = 8t + 2, which is 18 Wb/s at t = 2 s.
Formula sheet
Formula
Meaning
When to use
Watch out
φ = BA cos θ
Flux through a flat area in a uniform field
B the same everywhere on the surface
θ is measured from the normal
φ = BA sin α
Same flux, α measured from the plane
The question gives the angle with the plane
Never mix α and θ in one line
φ = B · A
Dot product form
B and A given as î, ĵ, k̂ vectors
Only matching parts multiply
Nφ = NBA cos θ
Flux linkage of N turns
Coils and solenoids
Check whether a given value is per turn or total
φ = ∫B · dA
Flux in a non-uniform field
Loop near a wire, fields that vary
Set up the strip before integrating
φ = (μ₀Iℓ/2π) ln(b/a)
Rectangle beside a long wire
Side ℓ parallel to wire, edges at a and b
ln of far over near
∮B · dA = 0
Closed surface
Any closed surface, even around a magnet
The answer is always zero
Δφ = BA(cos θ₂ − cos θ₁)
Change on rotation
Loops turned in a steady field
Half turn gives size 2BA
1 Wb = 1 T m² = 1 V s = 10⁸ Mx
Units
Conversions
Maxwell is the CGS unit
[M L² T⁻² A⁻¹]
Dimensions of flux
Dimension questions
Inductance has A⁻²
Allen worked examples checked
Every illustration on pages 95–98 was recalculated.
Illustration
Allen's answer
Check
1. Plane at 30° to field, A = 0.06 m², B = 1.2 T
0.036 Wb
Correct (θ = 60°)
2. B = 0.3ĵ, A = 2î + 5ĵ − 3k̂
1.5 Wb
Correct
3. Earth's field through xy, yz, zx planes
0, BHS, −BVS
Correct
4. (2î + 3ĵ + 4k̂)·(3î + 4ĵ)
18 Wb
Correct
5. Ring turned 180°
2BA
Size correct; the sign line is written backwards (Δφ = −2BA)
6. Square beside a long wire
(μ₀Iℓ/2π) ln((t + ℓ)/t)
Correct
7. Field parallel to coil plane
0
Correct
8. Wire along the axis of a ring
0
Correct
Beginner's Box 1 answer key
Q
Answer
Working
1
5 × 10⁻³ Wb
The coil's plane is at 30° to the field, so θ = 60°. Nφ = 100 × 0.2 × 5 × 10⁻⁴ × cos 60° = 5 × 10⁻³ Wb. (Flux through one turn is 5 × 10⁻⁵ Wb.)
2
NBA per second
Flux linkage goes from +NBA to −NBA, a change of 2NBA, in 2 s. Average rate = 2NBA/2 = NBA Wb/s.
3
EFGH: −Ba², ABCD: +Ba², the other four faces: 0
B runs from face EFGH to face ABCD. Take outward normals. Total over the closed cube is 0.
4
0.02 Wb
φ = BA = 0.2 × 0.1.
5
−0.1 Wb (size 0.1 Wb)
Loop flat in the xy plane, so A = 0.1k̂. φ = (4î − k̂)·(0.1k̂) = −0.1 Wb. The minus sign means the field crosses opposite to +k̂.
6
≈ 2.96 × 10⁻⁵ Wb
The field is inside the solenoid only, so only its 2 cm radius counts, not the disc's 10 cm. B = μ₀nI = 4π × 10⁻⁷ × 1250 × 15 = 0.0236 T. φ = B × π(0.02)² = 2.96 × 10⁻⁵ Wb.
NEET practice: 40 questions
Every number here was recalculated in Python. Try each question before opening its solution. The solutions follow the same nine parts every time, so you always know where to look.
Q1Numerical
A loop of area 0.05 m² is placed in a uniform field of 0.4 T so that its normal makes 60° with the field. The flux through the loop is
(A)0.01 Wb
(B)0.0173 Wb
(C)0.02 Wb
(D)0.005 Wb
Show the solution
Given
A = 0.05 m², B = 0.4 T, angle between normal and field = 60°
Asked
Flux φ
Concept
Flux in a uniform field depends on the angle between B and the area vector (the normal).
Formula
φ = BA cos θ
Baby steps
The angle is already measured from the normal, so θ = 60°.
BA = 0.4 × 0.05 = 0.02.
cos 60° = 0.5.
φ = 0.02 × 0.5 = 0.01 Wb.
Answer
(A) 0.01 Wb
Why not the others
0.0173 Wb uses sin 60°. 0.02 Wb ignores the angle. 0.005 Wb halves the answer twice.
Shortcut
Normal given, so use cos directly: half of BA.
Where it went wrong
Using sin because 60° 'looks like' a plane angle. Read which line the angle is measured from.
Q2Numerical
A flat coil of area 0.2 m² has its plane inclined at 30° to a uniform magnetic field of 0.5 T. The flux through the coil is
(A)0.05 Wb
(B)0.0866 Wb
(C)0.1 Wb
(D)zero
Show the solution
Given
A = 0.2 m², B = 0.5 T, angle between plane and field = 30°
Asked
Flux φ
Concept
θ in BA cos θ is the angle with the normal, which is 90° minus the plane angle.
Formula
φ = BA cos θ, θ = 90° − 30°
Baby steps
Plane at 30° to field, so the normal is at 60°.
BA = 0.5 × 0.2 = 0.1.
φ = 0.1 × cos 60° = 0.05 Wb.
Answer
(A) 0.05 Wb
Why not the others
0.0866 Wb uses cos 30° (the plane angle). 0.1 Wb treats the field as perpendicular. Zero would need the plane parallel to the field.
Shortcut
Plane angle α given: φ = BA sin α = 0.1 × sin 30° = 0.05 Wb.
Where it went wrong
The angle-trap: putting the plane angle straight into cos.
Q3Numerical
A circular coil of 200 turns and radius 5 cm is held with its plane perpendicular to a uniform field of 0.02 T. The flux linkage of the coil is
(A)0.126 Wb
(B)1.57 × 10⁻⁴ Wb
(C)6.28 × 10⁻³ Wb
(D)3.14 × 10⁻² Wb
Show the solution
Given
N = 200, r = 5 cm = 0.05 m, B = 0.02 T, plane perpendicular to B (θ = 0°)
Asked
Flux linkage Nφ
Concept
Each turn catches the same flux, so the linkage is N times the flux through one turn.
Multiplying magnitudes instead of taking the dot product.
Q5Numerical
A field B = (2î + 3ĵ − k̂) T passes through a flat area of 5 m² lying in the y–z plane. The size of the flux through it is
(A)15 Wb
(B)10 Wb
(C)5 Wb
(D)20 Wb
Show the solution
Given
B = 2î + 3ĵ − k̂ T, area 5 m² in the y–z plane
Asked
|φ|
Concept
A surface in the y–z plane has its normal along the x-axis.
Formula
φ = B · A with A = 5î
Baby steps
Area vector = 5î.
Only Bx = 2 T counts.
φ = 2 × 5 = 10 Wb.
Answer
(B) 10 Wb
Why not the others
15 Wb uses By (a z–x plane). 5 Wb uses Bz (an x–y plane). 20 Wb is not produced by any single component.
Shortcut
Name the missing axis of the plane: y–z plane → x-component of B.
Where it went wrong
Picking a component that lies in the plane instead of the one along the normal.
Q6Numerical
A square loop of side 10 cm is in a uniform field of 0.2 T. The normal to the loop makes 37° with the field (cos 37° = 0.8). The flux is
(A)1.6 × 10⁻³ Wb
(B)1.2 × 10⁻³ Wb
(C)2.0 × 10⁻³ Wb
(D)1.6 × 10⁻² Wb
Show the solution
Given
side 0.1 m, B = 0.2 T, θ = 37°
Asked
φ
Concept
Uniform field; θ is measured from the normal.
Formula
φ = BA cos θ
Baby steps
A = 0.1 × 0.1 = 0.01 m².
BA = 0.2 × 0.01 = 2 × 10⁻³.
φ = 2 × 10⁻³ × 0.8 = 1.6 × 10⁻³ Wb.
Answer
(A) 1.6 × 10⁻³ Wb
Why not the others
1.2 × 10⁻³ Wb uses sin 37° = 0.6. 2.0 × 10⁻³ Wb ignores the angle. 1.6 × 10⁻² Wb takes the side as 10 cm² instead of 100 cm².
Shortcut
BA × 0.8.
Where it went wrong
Squaring 10 cm to 10 cm² then converting wrongly.
Q7Concept
A flux of 5 × 10⁻⁴ Wb expressed in maxwell is
(A)5 × 10⁻⁸ Mx
(B)5 × 10⁻¹² Mx
(C)5 × 10⁴ Mx
(D)50 Mx
Show the solution
Given
φ = 5 × 10⁻⁴ Wb
Asked
φ in maxwell
Concept
The maxwell is the CGS unit of flux.
Formula
1 Wb = 10⁸ Mx
Baby steps
Multiply by 10⁸.
5 × 10⁻⁴ × 10⁸ = 5 × 10⁴ Mx.
Answer
(C) 5 × 10⁴ Mx
Why not the others
5 × 10⁻¹² Mx divides by 10⁸ (conversion upside down). 5 × 10⁻⁸ Mx and 50 Mx use the wrong power.
Shortcut
Maxwell is the smaller unit, so the number must get bigger.
Where it went wrong
Ratio reversal: dividing when you should multiply.
Q8Concept
The dimensional formula of magnetic flux is
(A)[M T⁻² A⁻¹]
(B)[M L² T⁻² A⁻¹]
(C)[M L² T⁻³ A⁻¹]
(D)[M L² T⁻² A⁻²]
Show the solution
Given
Flux = B × area
Asked
Dimensions of φ
Concept
Build from B = F/(IL), then multiply by L².
Formula
[B] = [M L T⁻²]/([A][L]) = [M T⁻² A⁻¹]
Baby steps
[B] = [M T⁻² A⁻¹].
[area] = [L²].
[φ] = [M L² T⁻² A⁻¹].
Answer
(B) [M L² T⁻² A⁻¹]
Why not the others
[M T⁻² A⁻¹] is the field B. [M L² T⁻³ A⁻¹] is the volt. [M L² T⁻² A⁻²] is the henry.
Shortcut
Flux = volt × second: take the volt's T⁻³ and add one T.
Where it went wrong
Mixing up flux with inductance (one extra A⁻¹).
Q9Concept
One weber is the same as
(A)1 V s
(B)1 V s⁻¹
(C)1 T m⁻²
(D)1 N A⁻¹
Show the solution
Given
Units of flux
Asked
An equivalent of 1 Wb
Concept
Faraday's law e = dφ/dt links flux to volts and seconds.
Formula
Wb = V × s = T × m²
Baby steps
From e = dφ/dt, the unit of φ = volt × second.
So 1 Wb = 1 V s.
Answer
(A) 1 V s
Why not the others
V s⁻¹ is the unit of a rate of change of emf. T m⁻² should be T m². N A⁻¹ is T m, not T m².
Shortcut
Rearrange e = dφ/dt in units.
Where it went wrong
Writing T m⁻² because the field is 'per area'.
Q10Numerical
A ring of area 0.04 m² lies with its plane perpendicular to a uniform field of 0.5 T. It is turned through 180° about a diameter. The size of the change in flux is
(A)zero
(B)0.04 Wb
(C)0.02 Wb
(D)0.08 Wb
Show the solution
Given
A = 0.04 m², B = 0.5 T, θ₁ = 0°, θ₂ = 180°
Asked
|Δφ|
Concept
After a half turn the same field enters from the other face, so the flux changes sign.
Formula
Δφ = BA(cos θ₂ − cos θ₁)
Baby steps
φ₁ = BA cos 0° = +0.02 Wb.
φ₂ = BA cos 180° = −0.02 Wb.
Δφ = −0.02 − 0.02 = −0.04 Wb, size 0.04 Wb.
Answer
(B) 0.04 Wb
Why not the others
Zero treats the start and end as identical. 0.02 Wb is the flux at one instant. 0.08 Wb doubles twice.
Shortcut
Half turn: |Δφ| = 2BA = 2 × 0.5 × 0.04.
Where it went wrong
Thinking a half turn brings the flux back to where it was. Only a full turn does.
Q11Numerical
A 50-turn coil of area 0.02 m² is turned from a position where its plane is perpendicular to a 0.1 T field to a position where its plane is parallel to the field. The size of the change in flux linkage is
(A)0.002 Wb
(B)0.2 Wb
(C)zero
(D)0.1 Wb
Show the solution
Given
N = 50, A = 0.02 m², B = 0.1 T, θ goes from 0° to 90°
Asked
|Δ(Nφ)|
Concept
Plane perpendicular to B means maximum flux; plane parallel means zero flux.
Formula
Δ(Nφ) = NBA(cos 90° − cos 0°)
Baby steps
Initial linkage = 50 × 0.1 × 0.02 = 0.1 Wb.
Final linkage = 0.
Change = 0.1 Wb.
Answer
(D) 0.1 Wb
Why not the others
0.2 Wb is the half-turn value. Zero ignores the turn. 0.002 Wb forgets N.
Shortcut
Quarter turn from face-on: change = NBA.
Where it went wrong
Reading 'plane parallel' as maximum flux.
Q12Numerical
A loop of area 0.5 m² starts with its normal along a 0.4 T field and is turned until the normal is at 60° to the field. The size of the change in flux is
(A)0.173 Wb
(B)0.2 Wb
(C)0.1 Wb
(D)zero
Show the solution
Given
A = 0.5 m², B = 0.4 T, θ₁ = 0°, θ₂ = 60°
Asked
|Δφ|
Concept
Change = final flux − initial flux.
Formula
Δφ = BA(cos 60° − cos 0°)
Baby steps
BA = 0.2 Wb.
Initial φ = 0.2 Wb; final φ = 0.2 × 0.5 = 0.1 Wb.
Change = 0.1 − 0.2 = −0.1 Wb, size 0.1 Wb.
Answer
(C) 0.1 Wb
Why not the others
0.2 Wb is the starting flux. 0.173 Wb uses sin 60°. Zero ignores the turn.
Shortcut
BA(1 − cos θ) = 0.2 × 0.5.
Where it went wrong
Giving the final flux as the change.
Q13Concept
A cube of side a is placed in a uniform magnetic field B. The total flux through all six faces is
(A)Ba²
(B)6Ba²
(C)zero
(D)2Ba²
Show the solution
Given
Closed cube in uniform B
Asked
Total flux through the closed surface
Concept
For any closed surface, flux going in equals flux coming out.
Formula
∮B · dA = 0
Baby steps
One face perpendicular to B has flux −Ba² (field enters).
The opposite face has +Ba² (field leaves).
Four side faces are parallel to B: zero.
Total = 0.
Answer
(C) zero
Why not the others
6Ba² counts every face as face-on. Ba² counts one face. 2Ba² adds sizes and ignores the sign of entering flux.
Shortcut
Closed surface → zero, no calculation needed.
Where it went wrong
Adding magnitudes of in-going and out-going flux.
Q14Concept
A hemispherical bowl of radius R sits in a uniform field B that is parallel to its axis. The flux through the curved surface of the bowl is
(A)4πR²B
(B)2πR²B
(C)zero
(D)πR²B
Show the solution
Given
Hemisphere radius R, B along the axis
Asked
Flux through the curved surface
Concept
The curved surface and the flat base together form a closed surface, so their fluxes are equal and opposite.
Formula
∮B · dA = 0 → φ(curved) = −φ(base)
Baby steps
Flux through the flat circular base = B × πR².
Closed surface total must be zero.
So the curved surface carries the same size of flux, πR²B.
Answer
(D) πR²B
Why not the others
2πR²B multiplies B by the curved area, but the field is not perpendicular to most of it. Zero confuses the open bowl with a closed surface. 4πR²B is a full sphere's area.
Shortcut
Every line through the curved surface also goes through the base.
Where it went wrong
Using the curved area 2πR² instead of the projected area πR².
Q15Concept
A closed surface completely encloses a strong bar magnet. The net magnetic flux through the surface is
(A)μ₀ times the pole strength
(B)zero
(C)positive, leaving from the N side
(D)negative, entering at the S side
Show the solution
Given
Closed surface around a whole magnet
Asked
Net flux
Concept
Gauss's law for magnetism: net flux through any closed surface is zero.
Formula
∮B · dA = 0
Baby steps
Lines leave from N and return to S, then continue inside the magnet.
Every line leaving also enters.
Net flux = 0.
Answer
(B) zero
Why not the others
μ₀ × pole strength would be true only if isolated poles existed. The other two describe parts of the surface, not the net value.
Shortcut
A magnet always has both poles inside the surface.
Where it went wrong
Carrying Gauss's law of electrostatics (q/ε₀) over to magnetism.
Q16Concept
The statement ∮B · dA = 0 for every closed surface tells us that
(A)isolated magnetic poles do not exist
(B)magnetic field lines are always straight
(C)magnetic flux is a vector
(D)the magnetic field is zero inside every closed surface
Show the solution
Given
Gauss's law for magnetism
Asked
Physical meaning
Concept
A net outward flux would need a net 'magnetic charge' inside.
Formula
∮B · dA = 0
Baby steps
Zero net flux for every surface means there is no source or sink of field lines.
A source of field lines would be a monopole.
So monopoles do not exist.
Answer
(A) isolated magnetic poles do not exist
Why not the others
Field lines are loops, not straight. Flux is a scalar. The field can be large inside; only the net flux is zero.
Shortcut
Magnetic Gauss law = no monopoles.
Where it went wrong
Reading 'net flux zero' as 'field zero'.
Q17Concept
A long straight wire carries current I. A square loop of side a lies in the same plane, with its near edge a distance a from the wire and two sides parallel to the wire. The flux through the loop is
(A)(μ₀Ia/2π) ln 2
(B)(μ₀Ia/2π) ln(3/2)
(C)μ₀Ia/4π
(D)μ₀Ia/2π
Show the solution
Given
Near edge at a, far edge at 2a, side parallel to wire = a
Asked
φ
Concept
B = μ₀I/2πx varies across the loop, so integrate over strips.
Formula
φ = (μ₀Iℓ/2π) ln(b/a)
Baby steps
Strip at distance x, width dx: dφ = (μ₀I/2πx)(a dx).
Integrate from x = a to x = 2a.
φ = (μ₀Ia/2π) ln(2a/a) = (μ₀Ia/2π) ln 2.
Answer
(A) (μ₀Ia/2π) ln 2
Why not the others
ln(3/2) takes the far edge at 3a. μ₀Ia/4π uses B at 2a times area a². μ₀Ia/2π uses B at a times area a².
Shortcut
ln(far/near) = ln(2a/a).
Where it went wrong
Using one value of B for the whole loop.
Q18Numerical
The flux through each turn of a 500-turn coil is 2 mWb. The flux linkage of the coil is
(A)0.25 Wb
(B)2 mWb
(C)1 Wb
(D)10 Wb
Show the solution
Given
N = 500, φ = 2 × 10⁻³ Wb
Asked
Nφ
Concept
Linkage = number of turns × flux per turn.
Formula
Nφ
Baby steps
500 × 2 × 10⁻³ = 1 Wb.
Answer
(C) 1 Wb
Why not the others
2 mWb forgets N. 0.25 Wb divides instead of multiplying. 10 Wb converts mWb wrongly.
Shortcut
500 × 2 mWb = 1000 mWb.
Where it went wrong
Leaving the answer per turn.
Q19Concept
Magnetic flux is
(A)a scalar only in a uniform field
(B)a vector along B
(C)a vector along the area vector
(D)a scalar, because it is the dot product of two vectors
Show the solution
Given
φ = B · A
Asked
Nature of flux
Concept
The dot product of two vectors is a scalar.
Formula
φ = B · A
Baby steps
B and A are vectors.
Their dot product is a number with sign but no direction.
So flux is a scalar in every field.
Answer
(D) a scalar, because it is the dot product of two vectors
Why not the others
Flux has a sign, not a direction, so neither vector option works. It stays scalar in non-uniform fields too (∫B · dA).
Shortcut
Dot product → scalar.
Where it went wrong
Treating the + and − sign of flux as a direction.
Q20Numerical
At a place the earth's field has a horizontal component 3 × 10⁻⁵ T and a vertical component 4 × 10⁻⁵ T. The flux through a horizontal table top of area 2 m² is
(A)8 × 10⁻⁵ Wb
(B)6 × 10⁻⁵ Wb
(C)1 × 10⁻⁴ Wb
(D)zero
Show the solution
Given
BH = 3 × 10⁻⁵ T, BV = 4 × 10⁻⁵ T, horizontal area 2 m²
Asked
φ through the table
Concept
A horizontal surface has a vertical normal, so only the vertical component crosses it.
Formula
φ = BVA
Baby steps
Normal to the table is vertical.
BH skims along the table: no flux.
φ = 4 × 10⁻⁵ × 2 = 8 × 10⁻⁵ Wb.
Answer
(A) 8 × 10⁻⁵ Wb
Why not the others
6 × 10⁻⁵ Wb uses the horizontal component. 1 × 10⁻⁴ Wb uses the total field 5 × 10⁻⁵ T. Zero ignores the vertical part.
Shortcut
Horizontal surface → vertical component.
Where it went wrong
Choosing the component that lies along the surface.
Q21Numerical
A vertical loop of area 0.5 m² stands with its plane perpendicular to the magnetic meridian. The earth's horizontal field is 3 × 10⁻⁵ T and vertical field 4 × 10⁻⁵ T. The flux through the loop is
(A)zero
(B)2.0 × 10⁻⁵ Wb
(C)2.5 × 10⁻⁵ Wb
(D)1.5 × 10⁻⁵ Wb
Show the solution
Given
Vertical loop, plane perpendicular to meridian, A = 0.5 m²
Asked
φ
Concept
Plane perpendicular to the meridian means the normal lies along the meridian (horizontal).
Formula
φ = BHA
Baby steps
Normal is horizontal and along BH.
BV lies in the vertical plane of the loop: no flux.
φ = 3 × 10⁻⁵ × 0.5 = 1.5 × 10⁻⁵ Wb.
Answer
(D) 1.5 × 10⁻⁵ Wb
Why not the others
2.0 × 10⁻⁵ Wb uses BV. 2.5 × 10⁻⁵ Wb uses the total field. Zero would need the plane along the meridian.
Shortcut
Find the normal first, then the component along it.
Where it went wrong
Confusing 'plane perpendicular to the meridian' with 'plane along the meridian'.
Q22Concept
A straight wire carrying current passes along the axis of a circular ring, perpendicular to the ring's plane. The flux through the ring due to the wire's field is
(A)zero
(B)μ₀I/2
(C)μ₀IR/2
(D)maximum
Show the solution
Given
Wire along the ring's axis
Asked
φ through the ring
Concept
Field lines of a straight wire are circles around it, lying in planes perpendicular to the wire.
Formula
φ = BA cos 90°
Baby steps
The ring's plane is perpendicular to the wire.
The field circles lie in that same plane, so B is parallel to the ring's surface.
θ between B and the ring's normal = 90°, so φ = 0.
Answer
(A) zero
Why not the others
The other options would need field lines crossing the ring, but they only run around within its plane.
Shortcut
Straight-wire field lines never pierce a coaxial ring.
Where it went wrong
Picturing the wire's field as pointing along the wire.
Q23Numerical
A large circular loop of radius 0.5 m carries 5 A. A small loop of radius 1 cm lies at its centre in the same plane. The flux through the small loop is about (take B uniform over the small loop)
(A)6.28 × 10⁻⁶ Wb
(B)3.95 × 10⁻⁹ Wb
(C)1.97 × 10⁻⁹ Wb
(D)9.87 × 10⁻¹⁰ Wb
Show the solution
Given
R = 0.5 m, I = 5 A, r = 0.01 m, coplanar and concentric
Asked
φ through the small loop
Concept
The small loop is tiny, so the big loop's centre field is nearly uniform over it.
Formula
B = μ₀I/2R, φ = Bπr²
Baby steps
B = 4π × 10⁻⁷ × 5/(2 × 0.5) = 6.28 × 10⁻⁶ T.
Area = π(0.01)² = 3.14 × 10⁻⁴ m².
φ = 6.28 × 10⁻⁶ × 3.14 × 10⁻⁴ = 1.97 × 10⁻⁹ Wb.
Answer
(C) 1.97 × 10⁻⁹ Wb
Why not the others
3.95 × 10⁻⁹ Wb forgets the 2 in μ₀I/2R. 6.28 × 10⁻⁶ is B, not flux. 9.87 × 10⁻¹⁰ Wb halves the correct value.
Shortcut
μ₀Iπr²/2R.
Where it went wrong
Using the field of a straight wire, μ₀I/2πR.
Q24Numerical
A long solenoid of radius 1 cm has 1000 turns per metre and carries 2 A. A flat disc of radius 5 cm is centred on its axis and perpendicular to it, inside the solenoid's middle. Take the field outside the solenoid as zero. The flux through the disc is about
(A)2.0 × 10⁻⁵ Wb
(B)7.9 × 10⁻⁷ Wb
(C)2.5 × 10⁻³ Wb
(D)zero
Show the solution
Given
n = 1000 m⁻¹, I = 2 A, solenoid radius 0.01 m, disc radius 0.05 m
Asked
φ through the disc
Concept
Only the part of the disc inside the solenoid has field through it.
Formula
B = μ₀nI, φ = B × π(rsolenoid)²
Baby steps
B = 4π × 10⁻⁷ × 1000 × 2 = 2.51 × 10⁻³ T.
Useful area = π(0.01)² = 3.14 × 10⁻⁴ m².
φ = 2.51 × 10⁻³ × 3.14 × 10⁻⁴ = 7.9 × 10⁻⁷ Wb.
Answer
(B) 7.9 × 10⁻⁷ Wb
Why not the others
2.0 × 10⁻⁵ Wb uses the disc's 5 cm radius. 2.5 × 10⁻³ is B in tesla, not flux. Zero would need no field at all.
Shortcut
Area = the smaller of the two circles.
Where it went wrong
Using the bigger disc area. The same trap is Allen Beginner's Box 1 Q6.
Q25Numerical
The flux through a coil increases from 0.02 Wb to 0.08 Wb in 0.3 s. The average rate of change of flux is
(A)0.018 Wb s⁻¹
(B)0.33 Wb s⁻¹
(C)0.067 Wb s⁻¹
(D)0.2 Wb s⁻¹
Show the solution
Given
φ₁ = 0.02 Wb, φ₂ = 0.08 Wb, Δt = 0.3 s
Asked
Δφ/Δt
Concept
Average rate = change divided by time.
Formula
Δφ/Δt = (φ₂ − φ₁)/Δt
Baby steps
Δφ = 0.08 − 0.02 = 0.06 Wb.
0.06/0.3 = 0.2 Wb s⁻¹.
Answer
(D) 0.2 Wb s⁻¹
Why not the others
0.33 uses 0.1 Wb (added instead of subtracted). 0.067 uses φ₁ alone. 0.018 multiplies by Δt.
Shortcut
6 hundredths ÷ 3 tenths = 0.2.
Where it went wrong
Adding the two flux values.
Q26Numerical
The flux through a loop varies as φ = (4t² + 2t) Wb. The rate of change of flux at t = 2 s is
(A)20 Wb s⁻¹
(B)18 Wb s⁻¹
(C)10 Wb s⁻¹
(D)16 Wb s⁻¹
Show the solution
Given
φ = 4t² + 2t, t = 2 s
Asked
dφ/dt at t = 2 s
Concept
Instantaneous rate = derivative.
Formula
d(t²)/dt = 2t
Baby steps
dφ/dt = 8t + 2.
At t = 2: 16 + 2 = 18 Wb s⁻¹.
Answer
(B) 18 Wb s⁻¹
Why not the others
20 is φ itself at t = 2 s. 10 is the average rate over 0–2 s. 16 drops the 2t term.
Shortcut
Differentiate first, substitute second.
Where it went wrong
Substituting t before differentiating.
Q27Concept
A loop is in a uniform field. At what angle between its normal and the field is the flux half of its maximum value?
A circle and a square are made from wires of the same length and placed with their planes perpendicular to the same uniform field. The ratio of flux through the circle to flux through the square is
(A)π : 4
(B)4 : π
(C)1 : 1
(D)16 : π
Show the solution
Given
Same perimeter L for both shapes, same B, both face-on
Asked
φcircle : φsquare
Concept
Flux ∝ area in the same field. Compare areas for the same perimeter.
Formula
Acircle = L²/4π, Asquare = L²/16
Baby steps
Circle: 2πr = L, so A = πr² = L²/4π.
Square: side L/4, so A = L²/16.
Ratio = (1/4π)/(1/16) = 16/4π = 4/π.
Answer
(B) 4 : π
Why not the others
π : 4 is the same ratio upside down. 1 : 1 assumes equal areas. 16 : π forgets the 4 in 4π.
Shortcut
A circle encloses the most area for a given length, so the circle's number must be bigger. 4 > π, so 4 : π.
Where it went wrong
Ratio reversal: writing the answer upside down.
Q29Concept
The radius of a circular loop is doubled and the uniform field through it is halved. The flux through the loop becomes
(A)half as large
(B)twice as large
(C)unchanged
(D)four times as large
Show the solution
Given
r → 2r, B → B/2
Asked
New φ / old φ
Concept
φ = Bπr², so φ ∝ Br².
Formula
φ ∝ Br²
Baby steps
Doubling r multiplies area by 4.
Halving B multiplies by 1/2.
Net factor = 4 × 1/2 = 2.
Answer
(B) twice as large
Why not the others
Half forgets that area goes as r². Unchanged treats doubling and halving as cancelling. Four times ignores the halved field.
Shortcut
Factor = (2)² × ½.
Where it went wrong
Treating area as proportional to r.
Q30Graph
A flat loop is in a uniform field. The angle α between the plane of the loop and the field is increased from 0° to 90°. Which graph shows the flux φ against α?
(A)
(B)
(C)
(D)
Show the solution
Given
α = angle with the plane, from 0° to 90°
Asked
Shape of φ against α
Concept
With the plane angle, φ = BA sin α.
Formula
φ = BA sin α
Baby steps
At α = 0° the plane is along the field: φ = 0.
At α = 90° the plane is perpendicular: φ = BA.
Between them φ follows sin α: rises quickly, then flattens near the top.
Answer
(D) the graph in option D
Why not the others
The falling cosine is the graph against θ (the normal angle), not α. A straight line would need φ ∝ α. A flat line would mean the angle has no effect.
Shortcut
Check the two ends: 0 at α = 0°, maximum at α = 90°.
Where it went wrong
Drawing the cos θ graph when the angle is measured from the plane.
Q31Graph
A square loop of side a moves at constant speed into, through and out of a region of uniform field that is 2a wide. Which graph shows the flux φ against the position x of the loop's front edge (x = 0 as the front edge enters)?
(A)
(B)
(C)
(D)
Show the solution
Given
Loop side a, field region width 2a, constant speed
Asked
φ against x
Concept
Flux = B × (area of loop inside the field).
Formula
φ = Ba × (length inside)
Baby steps
x = 0 to a: area inside grows steadily, so φ rises in a straight line.
x = a to 2a: loop fully inside, φ = Ba² stays constant.
x = 2a to 3a: loop leaves, φ falls in a straight line.
After 3a: φ = 0.
Answer
(D) the graph in option D
Why not the others
The triangle has no flat part, but the loop is fully inside for a distance a. The rise-then-flat graph never lets the loop leave. The rectangle jumps instantly, but the area changes gradually.
Shortcut
Rise for a, flat for (width − a), fall for a.
Where it went wrong
Forgetting the fully-inside stretch.
Q32Graph
A small square loop lies in the plane of a long straight current-carrying wire. It is moved slowly away from the wire. Which graph shows the flux φ through the loop against the distance d of its near edge from the wire?
(A)
(B)
(C)
(D)
Show the solution
Given
Coplanar loop moving away from a long wire
Asked
φ against d
Concept
B = μ₀I/2πx falls with distance, and faster near the wire.
Formula
φ = (μ₀Ia/2π) ln((d + a)/d)
Baby steps
Close to the wire the flux is large and drops sharply.
Far away it keeps falling but more and more slowly.
It never quite reaches zero, and it never rises.
Answer
(B) the graph in option B
Why not the others
The straight line would need B to fall evenly with distance. The rising curve has flux growing as the loop goes away. A constant line ignores the weakening field.
Shortcut
Steep then gentle, always falling.
Where it went wrong
Drawing a straight line because 'further means less'.
Q33Graph
A long solenoid of radius R carries a steady current. A flat disc of radius r, centred on the axis and perpendicular to it, sits inside the solenoid's middle. Taking the field outside the solenoid as zero, which graph shows the flux through the disc as r increases from 0 to beyond R?
(A)
(B)
(C)
(D)
Show the solution
Given
Uniform field inside radius R, zero outside
Asked
φ against r
Concept
Flux = B × (area of the disc that lies inside the solenoid).
Formula
φ = Bπr² for r ≤ R; φ = BπR² for r ≥ R
Baby steps
For r less than R, flux grows as r²: a parabola.
At r = R the whole field region is covered.
For r greater than R, extra disc area catches no field: flat.
Answer
(A) the graph in option A
Why not the others
The parabola that keeps rising ignores that there is no field outside. The straight-line rise would need φ ∝ r. The falling tail would need flux to decrease, but adding area never removes flux here.
Shortcut
Parabola up to R, then flat.
Where it went wrong
Continuing to use the disc's full area beyond R.
Q34Assertion–reason
Assertion (A): The net magnetic flux through a closed surface enclosing a bar magnet is zero. Reason (R): Magnetic field lines are imaginary.
(A)Both A and R are true, and R is the correct explanation of A.
(B)Both A and R are true, but R is not the correct explanation of A.
(C)A is true, but R is false.
(D)A is false, but R is true.
Show the solution
Given
A: net flux around a magnet is zero. R: field lines are imaginary.
Asked
Truth of A, truth of R, and whether R explains A
Concept
Gauss's law for magnetism.
Formula
∮B · dA = 0
Baby steps
A is true: no monopoles, so net flux is zero.
R is true: field lines are a drawing aid.
But A is true because monopoles do not exist, not because lines are imaginary.
So both true, R does not explain A.
Answer
(B) Both A and R are true, but R is not the correct explanation of A.
Why not the others
(A) needs R to be the reason. (C) and (D) need one of the statements to be false, but both are true.
Shortcut
Ask: 'A is true because R?' Here that sentence does not make sense.
Where it went wrong
Picking (A) whenever both statements are true. Always test the 'because' link.
Q35Assertion–reason
Assertion (A): The flux through a flat loop is zero when the loop's plane is parallel to a uniform field. Reason (R): When the plane is parallel to the field, the field lines run along the surface and none of them cross it.
(A)Both A and R are true, and R is the correct explanation of A.
(B)Both A and R are true, but R is not the correct explanation of A.
(C)A is true, but R is false.
(D)A is false, but R is true.
Show the solution
Given
A: zero flux for plane parallel to B. R: lines do not cross the surface.
Asked
Truth and link
Concept
Flux counts lines crossing a surface.
Formula
φ = BA cos 90° = 0
Baby steps
A is true.
R is true.
R is exactly why A is true: no crossings means zero flux.
Answer
(A) Both A and R are true, and R is the correct explanation of A.
Why not the others
(B) would deny the link, but R is the direct reason. (C) and (D) need a false statement.
Shortcut
Flux = line count, so 'no lines cross' explains 'zero flux'.
Where it went wrong
Choosing (B) out of caution when the link is direct.
Q36Assertion–reason
Assertion (A): A flat loop of area A has its plane inclined at 30° to a uniform field B, so the flux through it is BA cos 30°. Reason (R): Flux is BA cos θ, where θ is the angle between the field and the area vector.
(A)Both A and R are true, and R is the correct explanation of A.
(B)Both A and R are true, but R is not the correct explanation of A.
(C)A is true, but R is false.
(D)A is false, but R is true.
Show the solution
Given
A uses cos of the plane angle. R states the correct formula.
Asked
Truth and link
Concept
θ is measured from the normal, not the plane.
Formula
θ = 90° − 30° = 60°
Baby steps
R is the correct definition: true.
The normal is at 60°, so the flux is BA cos 60°, not BA cos 30°.
A is false, R is true.
Answer
(D) A is false, but R is true.
Why not the others
(A) and (B) need A to be true. (C) would make the correct formula false.
Shortcut
Whenever an angle is given with the plane, flip it before using cos.
Where it went wrong
Accepting A because the formula 'looks right'.
Q37Assertion–reason
Assertion (A): The flux linkage of a 100-turn coil is 100 times the flux through one of its turns. Reason (R): Flux linkage depends on the material the coil wire is made of.
(A)Both A and R are true, and R is the correct explanation of A.
(B)Both A and R are true, but R is not the correct explanation of A.
(C)A is true, but R is false.
(D)A is false, but R is true.
Show the solution
Given
A: linkage = Nφ. R: linkage depends on wire material.
Asked
Truth and link
Concept
Flux linkage counts field lines through all turns; wire material does not change the count.
Formula
Nφ
Baby steps
A is true: each turn links the same flux.
R is false: copper, aluminium or even a non-conducting loop links the same flux.
A true, R false.
Answer
(C) A is true, but R is false.
Why not the others
(A) and (B) need R to be true. (D) would make the definition of linkage false.
Shortcut
Flux is about the field and the area, never the wire.
Where it went wrong
Mixing up flux (which needs no conductor) with current (which does).
Q38Two statements
Statement I: The dimensional formula of magnetic flux is [M L² T⁻² A⁻¹]. Statement II: The SI unit of magnetic flux is T m⁻².
(A)Both Statement I and Statement II are true.
(B)Both Statement I and Statement II are false.
(C)Statement I is true, but Statement II is false.
(D)Statement I is false, but Statement II is true.
(C) Statement I is true, but Statement II is false.
Why not the others
Both-true and I-false options fail on the dimension check; both-false fails because I is correct.
Shortcut
Flux multiplies by area, so the metres are squared on top.
Where it went wrong
Reading 'field per area' and writing m⁻².
Q39Concept
Which change will not alter the flux through a flat loop in a uniform magnetic field?
(A)Squeezing the loop into a smaller area
(B)Turning the loop about a diameter
(C)Increasing the field strength
(D)Sliding the loop sideways, keeping it fully inside the field and its orientation fixed
Show the solution
Given
Uniform field; four possible changes
Asked
The change that leaves φ the same
Concept
φ = BA cos θ changes only if B, A or θ changes.
Formula
φ = BA cos θ
Baby steps
Sliding within a uniform field keeps B, A and θ the same.
Turning changes θ; a stronger field changes B; squeezing changes A.
Answer
(D) Sliding the loop sideways, keeping it fully inside the field and its orientation fixed
Why not the others
Each of the other three changes one of B, A or θ.
Shortcut
Uniform field + no turning + same area = same flux.
Where it went wrong
Thinking any motion changes flux. Only motion that changes B, A or θ does.
Q40Concept
A magnetic field line is described as imaginary while magnetic flux is described as real because
(A)field lines can be seen with iron filings, so they are real and flux is not
(B)flux exists only in uniform fields
(C)flux is a measurable quantity with units and dimensions, while lines are a way of drawing the field
(D)flux is a vector but field lines are not
Show the solution
Given
Allen note (1) on page 95
Asked
Why flux is 'real'
Concept
A physical quantity can be measured and has units.
Formula
φ in weber
Baby steps
Flux has a unit (Wb) and dimensions [M L² T⁻² A⁻¹].
Field lines are a picture: the number we draw is our choice of scale.
So flux is real; lines are imaginary.
Answer
(C) flux is a measurable quantity with units and dimensions, while lines are a way of drawing the field
Why not the others
Flux exists in any field. Iron filings show the pattern, not actual lines. Flux is a scalar.
Shortcut
Units and dimensions → physical quantity.
Where it went wrong
Believing iron filings prove lines are physical objects.
Answer key
1 A
2 A
3 D
4 C
5 B
6 A
7 C
8 B
9 A
10 B
11 D
12 C
13 C
14 D
15 B
16 A
17 A
18 C
19 D
20 A
21 D
22 A
23 C
24 B
25 D
26 B
27 C
28 B
29 B
30 D
31 D
32 B
33 A
34 B
35 A
36 D
37 C
38 C
39 D
40 C
Spread across letters: A 10, B 10, C 10, D 10. No letter repeats more than twice in a row. Question mix: Numerical 16, Concept 15, Graph 4, Assertion–reason 4, Two statements 1. Balancing seed 0.