Most marks lost in this chapter are lost to maths, not physics: a squared unit, a sine where a cosine belongs, a ratio written upside down. This file collects every tool the chapter needs, with the traps marked.
Companion file to Topics 01–14Maths tools used throughout Electromagnetic Induction
Units and powers of ten
More marks are lost here than anywhere else in this chapter. Convert everything to SI before you start.
Given
In SI
Trap
10 cm
0.1 m
—
10 cm²
10 × 10⁻⁴ = 10⁻³ m²
Not 10⁻² m²: the factor 10⁻² is squared
20 cm × 10 cm
0.2 × 0.1 = 0.02 m²
Convert first, then multiply
5 mWb
5 × 10⁻³ Wb
milli = 10⁻³
40 mH
0.04 H
—
2 μV
2 × 10⁻⁶ V
micro = 10⁻⁶
1200 rpm
20 rev/s, ω = 40π rad/s
Divide by 60, then × 2π
50 Hz
ω = 100π ≈ 314 rad/s
ω = 2πf
In exam language
1 cm = 10⁻² m 1 cm² = 10⁻⁴ m² 1 cm³ = 10⁻⁶ m³
1 Wb = 10⁸ maxwell 1 T = 10⁴ gauss
μ₀ = 4π × 10⁻⁷ ≈ 1.257 × 10⁻⁶
Trap
Size check. Before writing an answer, ask whether it is sensible. A coil's self-inductance is usually millihenry, a lab emf is usually millivolts to volts, and the earth's field is about 10⁻⁵ T. An answer of 5000 V from a hand-held magnet means a lost power of ten.
Useful numbers
Quantity
Value
Handy form
π
3.1416
≈ 22/7
π²
9.87
≈ 10 (fast estimates)
2π
6.28
One full turn in radians
μ₀/2π
2 × 10⁻⁷
Straight-wire field
μ₀/4π
10⁻⁷
Biot–Savart
e
2.718
e⁻¹ = 0.368
ln 2
0.693
Half-life factor
Angles: plane or normal?
Two angles, one tilt. α is measured from the plane of the loop, θ from the normal. They always add to 90°, so cos θ = sin α.
φ = BA cos θ (θ from the normal) = BA sin α (α from the plane), θ + α = 90°
Angle given
Use
Example
Normal makes 60° with B
BA cos 60° = BA/2
Straightforward
Plane makes 30° with B
BA sin 30° = BA/2
Same situation, different wording
Plane perpendicular to B
BA (maximum)
Field goes straight through
Plane parallel to B
0
Field skims along the surface
Angle
sin
cos
0°
0
1
30°
0.5
0.866
37°
0.6
0.8
45°
0.707
0.707
53°
0.8
0.6
60°
0.866
0.5
90°
1
0
Trap
Check your choice against the two extremes: plane perpendicular to B must give the largest flux, plane parallel must give zero. If your formula fails that test, you have used the wrong angle.
Vectors: dot for size, cross for direction
Dot product: flux
B · A = BxAx + ByAy + BzAz
Only matching components multiply. A surface in the x–y plane has its area vector along k̂, so only Bz counts. A negative answer just means the field crosses the surface opposite to the chosen area vector.
Cross product: which end is positive
Right-hand rule. Fingers along v, curl towards B, thumb along v × B. That is the way positive charge is pushed, so that end is at the higher potential.
î × ĵ = k̂, ĵ × k̂ = î, k̂ × î = ĵ (reverse the order → minus sign)
Writing the axes as x right, y up, z out of the page turns every direction question into algebra: a field into the page is −k̂, velocity to the right is +î, and î × (−k̂) = +ĵ, which is up.
Trap
Doing the cross product in the wrong order flips the answer. î × k̂ = −ĵ, not +ĵ. When a direction answer comes out backwards, this is usually why.
Differentiation: turning flux into emf
Derivative = slope. The sliding straight line is the tangent. Its steepness is the size of the emf. Where the curve levels off, the emf is zero even though the flux is largest.
Function
Derivative
Used for
tⁿ
n tⁿ⁻¹
φ = 4t² + 2t → dφ/dt = 8t + 2
constant
0
Steady flux gives no emf
sin ωt
ω cos ωt
Rotating coils
cos ωt
−ω sin ωt
φ = NBA cos ωt → e = NBAω sin ωt
ekt
k ekt
LR circuits
t e⁻ᵗ (product rule)
(1 − t)e⁻ᵗ
Box 4 Q3: emf zero at t = 1 s
t² e⁻ᵗ
t(2 − t)e⁻ᵗ
Zero at t = 2 s
In exam language
Product rule: d(uv)/dt = u(dv/dt) + v(du/dt).
Trap
Differentiate first, substitute after. For φ = 4t² + 2t at t = 2 s, the emf is 8(2) + 2 = 18 V. Putting t = 2 into φ gives 20, which is the flux, not the emf. This single habit fixes a whole family of errors.
Maximum and minimum
A quantity is largest or smallest where its derivative is zero. The current in Box 4 Q3 peaks where dI/dt = 0, and that is exactly where the induced emf vanishes.
Integration: adding up strips
Integral
Result
Used for
∫₀ˡ x dx
ℓ²/2
Rotating rod: e = ∫Bωx dx = ½Bωℓ²
∫ₐᵇ dx/x
ln(b/a)
Flux near a long wire: (μ₀Iℓ/2π) ln(b/a)
∫ I dt
charge q
Area under an I–t graph
∫ I²R dt
heat H
Energy lost
∫₀ᴵ LI dI
½LI²
Energy stored in an inductor
Area = charge. Integrating a current over time gives the charge that has flowed. For straight-sided graphs, just find the area of the triangles and rectangles.
In exam language
For a graph made of straight lines, no calculus is needed:
rectangle: q = I × t triangle: q = ½ × base × height
Exponentials and logarithms
Decay by equal factors. Each time constant multiplies what is left by 0.368. The curve approaches zero but never gets there, so it is never a straight line down to zero.
t
e−t/τ (what is left)
1 − e−t/τ (what has grown)
0
1
0
0.693τ
0.50
0.50
τ
0.368
0.632
2τ
0.135
0.865
3τ
0.050
0.950
5τ
0.007
0.993
In exam language
To find the time to reach a given value, take logarithms:
I = I₀e−t/τ → t = τ ln(I₀/I)
I = I₀(1 − e−t/τ) → t = τ ln[I₀/(I₀ − I)]
For growth, the log uses the remaining fraction. To reach 75% of the final current, one quarter remains, so t = τ ln 4, not τ ln(4/3).
Ratios and scaling
Many NEET questions change one quantity and ask what happens. Write the formula, mark which letters change, and multiply the factors.
Relation
Double the input
Where it appears
y ∝ x
× 2
e ∝ v for a sliding rod
y ∝ x²
× 4
L ∝ N², U ∝ I², e ∝ ℓ² for a rotating rod
y ∝ 1/x
× ½
Iₛ ∝ 1/Nₛ, L ∝ 1/ℓ at fixed N
y ∝ 1/x²
× ¼
Line loss ∝ 1/V²
In exam language
Worked example. A solenoid's turns are doubled and its length halved, with the area unchanged. L = μ₀N²A/ℓ: N² gives × 4, and 1/ℓ gives × 2, so L becomes 8 times larger.
Trap
The ratio reversal check. After writing a ratio, ask which side should be bigger and confirm your fraction agrees. In a transformer, the coil with more turns has the higher voltage but the smaller current, so Vₛ/Vₚ = Nₛ/Nₚ while Iₛ/Iₚ = Nₚ/Nₛ. Writing both the same way is the single most common slip in Topic 14.
Two ways "double the length" can mean different things
Fixed quantity
Formula to use
Effect of doubling ℓ
Total turns N
L = μ₀N²A/ℓ
L halves
Turns per metre n
L = μ₀n²Aℓ
L doubles
Reading graphs
What you want
What to read
Example
Emf from a flux graph
Slope (and flip the sign)
Rising flux → negative emf
Charge from a current graph
Area under the curve
Triangle: ½ × base × height
Where the emf is zero
Flat parts of the flux graph
Loop fully inside a field region
Which segment has the biggest emf
The steepest segment
Not the tallest one
Growth or decay
Curve bending towards a level value
LR circuits, terminal velocity
Trap
Two graph traps run through this chapter. First, a flat graph at a high value means zero emf, not a large one. Second, do not copy the shape of the input graph into the answer: the emf is the slope of the flux, not the flux itself.
Recognising shapes
Shape
Relation
Straight line through the origin
y ∝ x
Parabola through the origin
y ∝ x²
Hyperbola falling away
y ∝ 1/x
Rising curve that levels off
y = y₀(1 − e−x/a)
Falling curve approaching zero
y = y₀e−x/a
Rises, peaks, then falls as 1/x
Induced electric field against r
One-page summary
Tool
Key fact
Chapter use
Unit conversion
1 cm² = 10⁻⁴ m²; rpm ÷ 60 × 2π
Every numerical question
Angles
θ (normal) + α (plane) = 90°
Flux, Topic 01
Dot product
Only matching components multiply
Flux from vectors
Cross product
î × ĵ = k̂; reverse → minus
Which end is positive, Topics 09–10
Differentiation
Differentiate first, then substitute
Emf from flux or current
d(sin ωt)/dt
ω cos ωt
Rotating coils, Topic 12
∫x dx = ℓ²/2
Average of 0 and ℓ is ℓ/2
Rotating rod, Topic 10
∫dx/x = ln(b/a)
Non-uniform field
Loop near a wire
Area under I–t
Charge
Topic 04
e⁻¹ = 0.368, ln 2 = 0.693
One time constant; half-life
LR circuits, Topic 07
Scaling
Square the factor for x² relations
L ∝ N², U ∝ I²
Ratio check
Ask which side should be bigger
Transformers, Topic 14
Practice: 30 questions
These test only the maths, not new physics. If a question here is slow, that is the tool to drill before going back to the topic files.
Q1Numerical
An area of 25 cm² expressed in square metres is
(A)0.25 m²
(B)2.5 × 10⁻³ m²
(C)2.5 × 10⁻² m²
(D)2.5 × 10⁻⁵ m²
Show the solution
Given
A = 25 cm²
Asked
A in m²
Concept
The centimetre-to-metre factor is squared for areas.
Formula
1 cm² = 10⁻⁴ m²
Baby steps
25 × 10⁻⁴ = 2.5 × 10⁻³ m².
Answer
(B) 2.5 × 10⁻³ m²
Why not the others
0.25 m² and 2.5 × 10⁻² m² use 10⁻² instead of 10⁻⁴. 2.5 × 10⁻⁵ m² over-corrects.
Shortcut
Square the 10⁻².
Where it went wrong
Using the length factor for an area.
Q2Numerical
A coil turning at 900 rpm has an angular speed of
(A)30π rad/s
(B)900 rad/s
(C)15π rad/s
(D)60π rad/s
Show the solution
Given
900 rpm
Asked
ω
Concept
Convert to revolutions per second, then to radians.
Formula
ω = 2π × rpm/60
Baby steps
900/60 = 15 rev/s.
ω = 15 × 2π = 30π rad/s ≈ 94.2 rad/s.
Answer
(A) 30π rad/s
Why not the others
900 rad/s uses rpm directly. 15π rad/s forgets the factor 2. 60π rad/s forgets to divide by 60 and instead uses 1800 rpm.
Shortcut
rpm ÷ 60 × 2π.
Where it went wrong
Putting rpm straight into a formula that needs rad/s.
Q3Numerical
A loop's plane makes 60° with a uniform field. The angle to use in φ = BA cos θ is
(A)120°
(B)60°
(C)90°
(D)30°
Show the solution
Given
Plane at 60° to B
Asked
θ (from the normal)
Concept
The two angles add to 90°.
Formula
θ = 90° − α
Baby steps
α = 60° from the plane.
θ = 90° − 60° = 30°.
Answer
(D) 30°
Why not the others
60° is the plane angle itself. 90° and 120° do not follow.
Shortcut
Flip it before using cos.
Where it went wrong
Putting the plane angle straight into cos.
Q4Numerical
For B = (2î − 3ĵ + 5k̂) T and an area vector A = 4ĵ m², the flux B · A is
(A)8 Wb
(B)12 Wb
(C)20 Wb
(D)−12 Wb
Show the solution
Given
B = 2î − 3ĵ + 5k̂, A = 4ĵ
Asked
φ
Concept
Only matching components multiply.
Formula
φ = BxAx + ByAy + BzAz
Baby steps
Only the ĵ parts pair up.
φ = (−3)(4) = −12 Wb.
Answer
(D) −12 Wb
Why not the others
12 Wb drops the minus sign. 20 Wb uses Bz. 8 Wb uses Bx.
Shortcut
Pick out the matching component.
Where it went wrong
Ignoring the sign, or picking the wrong component.
Q5Concept
A negative value of magnetic flux means that
(A)the flux is decreasing
(B)the field passes through the surface opposite to the chosen area vector
(C)an error has been made, since flux cannot be negative
(D)the field is zero
Show the solution
Given
φ < 0
Asked
Meaning
Concept
Flux is a dot product, so it carries a sign.
Formula
φ = BA cos θ
Baby steps
cos θ is negative for θ greater than 90°.
That means the field crosses the surface the other way.
Answer
(B) the field passes through the surface opposite to the chosen area vector
Why not the others
A negative value is not the same as a decreasing value. Flux can certainly be negative. Zero field gives zero flux, not negative.
Shortcut
Sign = which way through.
Where it went wrong
Treating the sign as an error.
Q6Numerical
If φ = (6t² − 4t + 3) Wb, the size of dφ/dt at t = 2 s is
(A)24 Wb/s
(B)19 Wb/s
(C)20 Wb/s
(D)8 Wb/s
Show the solution
Given
φ = 6t² − 4t + 3, t = 2 s
Asked
|dφ/dt|
Concept
Differentiate, then substitute.
Formula
d(tⁿ)/dt = n tⁿ⁻¹
Baby steps
dφ/dt = 12t − 4.
At t = 2: 24 − 4 = 20 Wb/s.
Answer
(C) 20 Wb/s
Why not the others
19 Wb/s is φ at t = 2 s. 24 Wb/s drops the −4. 8 Wb/s uses t = 1.
Shortcut
Differentiate first.
Where it went wrong
Substituting into φ instead of its derivative.
Q7Numerical
If I = 5 sin(200t) A, the size of dI/dt at t = 0 is
(A)1000 A/s
(B)5 A/s
(C)200 A/s
(D)zero
Show the solution
Given
I = 5 sin 200t, t = 0
Asked
|dI/dt|
Concept
Differentiate a sine.
Formula
d(sin ωt)/dt = ω cos ωt
Baby steps
dI/dt = 5 × 200 cos 200t = 1000 cos 200t.
At t = 0, cos 0 = 1, so 1000 A/s.
Answer
(A) 1000 A/s
Why not the others
5 A/s forgets ω. 200 A/s forgets the amplitude. Zero is the value of I at t = 0, not its rate.
Shortcut
Amplitude × ω.
Where it went wrong
Confusing where I is zero with where its rate is zero.
Q8Numerical
If φ = 0.4 cos(50t) Wb, the peak value of the induced emf in a single loop is
(A)0.4 V
(B)20 V
(C)50 V
(D)125 V
Show the solution
Given
φ = 0.4 cos 50t
Asked
e₀
Concept
Differentiate a cosine.
Formula
e = −dφ/dt = 0.4 × 50 sin 50t
Baby steps
dφ/dt = −20 sin 50t.
Peak emf = 20 V.
Answer
(B) 20 V
Why not the others
0.4 V is the peak flux. 50 V is ω. 125 V divides instead of multiplying.
Shortcut
Amplitude × ω.
Where it went wrong
Reporting the flux amplitude as the emf.
Q9Numerical
The current in a coil is I = t²e⁻ᵗ A. The induced emf is zero (after the start) at
(A)t = 2 s
(B)t = 1 s
(C)t = 0.5 s
(D)t = e s
Show the solution
Given
I = t²e⁻ᵗ
Asked
t where dI/dt = 0
Concept
Use the product rule.
Formula
d(uv) = u dv + v du
Baby steps
dI/dt = 2te⁻ᵗ − t²e⁻ᵗ = t(2 − t)e⁻ᵗ.
Zero when t = 0 or t = 2.
So t = 2 s.
Answer
(A) t = 2 s
Why not the others
t = 1 s is the answer for I = te⁻ᵗ. 0.5 s and e s do not make the bracket zero.
Shortcut
Factorise: t(2 − t).
Where it went wrong
Solving I = 0 instead of dI/dt = 0.
Q10Numerical
The value of ∫₀^0.4 x dx is
(A)0.4
(B)0.16
(C)0.08
(D)0.032
Show the solution
Given
∫₀^0.4 x dx
Asked
Value
Concept
Standard integral.
Formula
∫₀ˡ x dx = ℓ²/2
Baby steps
0.4²/2 = 0.16/2 = 0.08.
Answer
(C) 0.08
Why not the others
0.16 forgets the ½. 0.4 forgets the square. 0.032 divides by 5.
A triangular current pulse rises from 0 to 6 A and back to 0 over 0.4 s. The charge that flows is
(A)2.4 C
(B)1.2 C
(C)0.6 C
(D)15 C
Show the solution
Given
Triangle: base 0.4 s, height 6 A
Asked
q
Concept
Charge is the area under the I–t graph.
Formula
q = ½ × base × height
Baby steps
½ × 0.4 × 6 = 1.2 C.
Answer
(B) 1.2 C
Why not the others
2.4 C forgets the ½. 0.6 C halves twice. 15 C divides instead of multiplying.
Shortcut
Half base times height.
Where it went wrong
Using the rectangle's area.
Q13Numerical
The value of e⁻² to three decimal places is
(A)0.865
(B)0.368
(C)0.135
(D)0.050
Show the solution
Given
e⁻²
Asked
Value
Concept
Each factor of e⁻¹ multiplies by 0.368.
Formula
e⁻² = (e⁻¹)²
Baby steps
0.368 × 0.368 = 0.135.
Answer
(C) 0.135
Why not the others
0.368 is e⁻¹. 0.865 is 1 − e⁻², the growth value. 0.050 is e⁻³.
Shortcut
Square 0.368.
Where it went wrong
Using the growth fraction for a decay.
Q14Numerical
A quantity decays as y = y₀e−t/τ. The time at which it falls to one quarter of y₀ is
(A)4 τ
(B)0.25 τ
(C)0.69 τ
(D)1.39 τ
Show the solution
Given
y = y₀/4
Asked
t
Concept
Take logarithms.
Formula
t = τ ln(y₀/y)
Baby steps
y₀/y = 4.
t = τ ln 4 = 2 × 0.693 τ = 1.39 τ.
Answer
(D) 1.39 τ
Why not the others
0.25τ treats the decay as linear. 0.69τ is the half-life. 4τ leaves 1.8% remaining.
Shortcut
Two half-lives.
Where it went wrong
Reading a quarter as 0.25τ.
Q15Numerical
A growing quantity follows y = y₀(1 − e−t/τ). It reaches 75% of y₀ at
(A)τ ln 4
(B)τ ln(4/3)
(C)0.75 τ
(D)τ ln 3
Show the solution
Given
y = 0.75 y₀
Asked
t
Concept
Solve for the remaining fraction.
Formula
e−t/τ = 1 − 0.75
Baby steps
e−t/τ = 0.25.
t = τ ln 4 ≈ 1.39 τ.
Answer
(A) τ ln 4
Why not the others
τ ln(4/3) solves e−t/τ = 0.75, which is the wrong fraction. 0.75τ treats growth as linear. τ ln 3 has no basis.
Shortcut
Use what is left, not what is reached.
Where it went wrong
Putting the reached fraction into the exponential.
Q16Concept
A quantity y is proportional to x². If x is increased by a factor of 3, y becomes
(A)1/9 as large
(B)3 times as large
(C)6 times as large
(D)9 times as large
Show the solution
Given
y ∝ x², x → 3x
Asked
New y
Concept
Square the factor.
Formula
y ∝ x²
Baby steps
3² = 9.
Answer
(D) 9 times as large
Why not the others
3 times treats y ∝ x. 6 times doubles the factor. 1/9 inverts.
Shortcut
Square the multiplier.
Where it went wrong
Forgetting to square.
Q17Concept
A solenoid's number of turns is doubled and its length is halved, with the area unchanged. Its self-inductance becomes
(A)2 times as large
(B)4 times as large
(C)8 times as large
(D)unchanged
Show the solution
Given
N → 2N, ℓ → ℓ/2, A fixed
Asked
New L
Concept
Mark which letters change and multiply the factors.
Formula
L = μ₀N²A/ℓ
Baby steps
N²: × 4.
1/ℓ: × 2.
Total: × 8.
Answer
(C) 8 times as large
Why not the others
4 times uses only the turns. 2 times uses only the length. Unchanged assumes they cancel.
Shortcut
Multiply the factors.
Where it went wrong
Changing only one quantity.
Q18Concept
In an ideal transformer, if the secondary has 5 times as many turns as the primary, then
(A)Vₛ = 5Vₚ and Iₛ = Iₚ/5
(B)Vₛ = 5Vₚ and Iₛ = 5Iₚ
(C)Vₛ = Vₚ/5 and Iₛ = 5Iₚ
(D)Vₛ = Vₚ/5 and Iₛ = Iₚ/5
Show the solution
Given
Nₛ = 5Nₚ
Asked
Vₛ and Iₛ
Concept
Voltage follows the turns, current goes against them.
Formula
Vₛ/Vₚ = Nₛ/Nₚ = Iₚ/Iₛ
Baby steps
More turns → higher voltage: Vₛ = 5Vₚ.
Power is unchanged, so the current falls by 5.
Answer
(A) Vₛ = 5Vₚ and Iₛ = Iₚ/5
Why not the others
Raising both would create power. The last two step the voltage down.
Shortcut
Check: does the power stay the same?
Where it went wrong
Ratio reversal on the current.
Q19Concept
A quantity y is inversely proportional to x. Its graph against x is
(A)a horizontal line
(B)a straight line through the origin
(C)a parabola
(D)a hyperbola falling towards zero
Show the solution
Given
y ∝ 1/x
Asked
Graph shape
Concept
Recognising standard shapes.
Formula
y = k/x
Baby steps
Large y for small x, falling steeply then flattening, never reaching zero.
Answer
(D) a hyperbola falling towards zero
Why not the others
The straight line is y ∝ x. The parabola is y ∝ x². A flat line means y does not depend on x.
Shortcut
1/x → hyperbola.
Where it went wrong
Drawing a straight fall to zero.
Q20Numerical
A straight-line graph of flux against time passes through (1 s, 3 Wb) and (4 s, 12 Wb). The size of the induced emf in a single loop is
(A)15 V
(B)9 V
(C)3 V
(D)0.33 V
Show the solution
Given
Two points on a straight φ–t graph
Asked
|e|
Concept
Emf is the slope.
Formula
slope = Δφ/Δt
Baby steps
Δφ = 12 − 3 = 9 Wb.
Δt = 4 − 1 = 3 s.
slope = 3 Wb/s, so |e| = 3 V.
Answer
(C) 3 V
Why not the others
9 V is Δφ alone. 15 V adds the flux values. 0.33 V inverts the slope.
Shortcut
Rise over run.
Where it went wrong
Using one point's coordinates instead of the differences.
Q21Concept
On a flux–time graph, a horizontal section at a large value of flux means that during that time the emf is
(A)large and constant
(B)zero
(C)large and increasing
(D)equal to the flux value
Show the solution
Given
Flat section, high flux
Asked
Emf
Concept
Emf is the slope, not the height.
Formula
e = −dφ/dt
Baby steps
A horizontal line has zero slope.
So the emf is zero, however high the line is.
Answer
(B) zero
Why not the others
The other options confuse the height of the graph with its steepness.
Shortcut
Flat → zero.
Where it went wrong
Reading the value instead of the slope.
Q22Direction
Taking x to the right, y up and z out of the page, a rod moves to the right (+î) in a field into the page (−k̂). The direction of v × B is
(A)to the left (−î)
(B)down (−ĵ)
(C)up (+ĵ)
(D)out of the page (+k̂)
Show the solution
Given
v = +î, B = −k̂
Asked
Direction of v × B
Concept
Use the standard cross products.
Formula
î × k̂ = −ĵ
Baby steps
î × (−k̂) = −(î × k̂) = −(−ĵ) = +ĵ.
So v × B points up.
Answer
(C) up (+ĵ)
Why not the others
Down comes from missing a sign. Left and out of the page are not perpendicular to both vectors.
Shortcut
î × ĵ = k̂ and cycle round.
Where it went wrong
Getting î × k̂ = +ĵ instead of −ĵ.
Q23Concept
Reversing the order of a cross product
(A)doubles its size
(B)leaves it unchanged
(C)makes it zero
(D)reverses its direction
Show the solution
Given
a × b versus b × a
Asked
Effect
Concept
The cross product is anti-commutative.
Formula
b × a = −(a × b)
Baby steps
Swapping the order flips the sign.
The size stays the same.
Answer
(D) reverses its direction
Why not the others
Unchanged would make it like a dot product. It is only zero if the vectors are parallel. Its size never doubles.
Shortcut
Swap → minus.
Where it went wrong
Treating the cross product like multiplication of numbers.
Q24Numerical
A solenoid has n = 1000 turns per metre, area 4 cm² and length 0.5 m. The value of μ₀n²Aℓ is about
(A)6.3 × 10⁻⁴ H
(B)2.5 × 10⁻² H
(C)2.5 × 10⁻⁴ H
(D)2.5 × 10⁻⁶ H
Show the solution
Given
n = 1000 m⁻¹, A = 4 × 10⁻⁴ m², ℓ = 0.5 m
Asked
L
Concept
Substitute carefully, converting the area first.
Formula
L = μ₀n²Aℓ
Baby steps
n² = 10⁶.
μ₀n² = 4π × 10⁻⁷ × 10⁶ = 1.257.
× A × ℓ = × 4 × 10⁻⁴ × 0.5 = 2.5 × 10⁻⁴ H.
Answer
(C) 2.5 × 10⁻⁴ H
Why not the others
2.5 × 10⁻² H uses 4 cm² = 4 × 10⁻² m². 6.3 × 10⁻⁴ H forgets to multiply by ℓ = 0.5. 2.5 × 10⁻⁶ H loses two powers of ten.
Shortcut
Convert cm² with 10⁻⁴.
Where it went wrong
Area conversion.
Q25Numerical
Using π² ≈ 10, a quick estimate of 4π² × 10⁻⁵ is about
(A)4 × 10⁻⁴
(B)4 × 10⁻⁵
(C)1.3 × 10⁻⁴
(D)4 × 10⁻³
Show the solution
Given
4π² × 10⁻⁵
Asked
Estimate
Concept
π² is close to 10, which shifts the power of ten by one.
Formula
π² ≈ 10
Baby steps
4 × 10 × 10⁻⁵ = 4 × 10⁻⁴.
Answer
(A) 4 × 10⁻⁴
Why not the others
4 × 10⁻⁵ drops the π². 1.3 × 10⁻⁴ uses 4π instead of 4π². 4 × 10⁻³ shifts too far.
Shortcut
π² ≈ 10 makes the arithmetic quick.
Where it went wrong
Losing track of the power of ten.
Q26Graph
Which graph shows a quantity that is inversely proportional to the square of x?
(A)
(B)
(C)
(D)
Show the solution
Given
y ∝ 1/x²
Asked
Graph shape
Concept
Recognising standard shapes.
Formula
y = k/x²
Baby steps
Falls faster than 1/x and flattens sooner.
Answer
(B) the graph in option B
Why not the others
The gentler fall is 1/x. The rising curve is x². The straight fall reaches zero at a finite x, which 1/x² never does.
Shortcut
1/x² falls off fastest.
Where it went wrong
Confusing 1/x with 1/x².
Q27Graph
A quantity grows from zero and levels off at a steady value. Which graph shows it?
(A)
(B)
(C)
(D)
Show the solution
Given
Growth to a limit
Asked
Shape
Concept
Standard growth curve.
Formula
y = y₀(1 − e−t/τ)
Baby steps
Steepest at the start, then flattening towards a limit.
Answer
(D) the graph in option D
Why not the others
The falling curve is decay. The straight line and the parabola keep rising without limit.
Shortcut
Rise and level off.
Where it went wrong
Choosing a straight ramp.
Q28Graph
The graph shows a quantity y against x. Which relation does it represent?
(A)y ∝ x
(B)y ∝ x²
(C)y ∝ 1/x
(D)y ∝ √x
Show the solution
Given
Curve through the origin, getting steeper
Asked
Relation
Concept
Recognising standard shapes.
Formula
—
Baby steps
It passes through the origin, so y = 0 when x = 0.
It gets steeper as x grows, which is a square law.
Answer
(B) y ∝ x²
Why not the others
y ∝ x is a straight line. y ∝ 1/x falls. y ∝ √x flattens as x grows instead of steepening.
Shortcut
Through the origin and steepening → x².
Where it went wrong
Confusing x² with √x.
Q29Concept
The best first step when a numerical answer comes out a thousand times too large is to check
(A)whether the formula exists
(B)the unit conversions, especially areas and prefixes
(C)the answer key
(D)whether the question is solvable
Show the solution
Given
Answer far too large
Asked
First check
Concept
Powers of ten are the usual cause.
Formula
1 cm² = 10⁻⁴ m²
Baby steps
Areas, millihenry, milliweber and micro-prefixes are the usual culprits.
Check them before doubting the formula.
Answer
(B) the unit conversions, especially areas and prefixes
Why not the others
The formula is rarely the problem when the digits are right but the power of ten is wrong.
Shortcut
Digits right, power wrong → units.
Where it went wrong
Redoing the whole solution instead of checking units.
Q30Concept
In φ = BA cos θ, the safest way to confirm you have used the right angle is to
(A)check that the plane perpendicular to B gives the maximum flux and the plane parallel gives zero
(B)check that θ is less than 45°
(C)always use cos
(D)always use the angle printed in the question
Show the solution
Given
Choosing between sin and cos
Asked
Best check
Concept
Test the formula at the two extreme positions.
Formula
—
Baby steps
Plane ⊥ B must give BA.
Plane ∥ B must give 0.
If your formula fails either test, the angle is wrong.
Answer
(A) check that the plane perpendicular to B gives the maximum flux and the plane parallel gives zero
Why not the others
There is no rule that θ is small. Using cos blindly fails when the plane angle is given. The printed angle may be measured from either line.
Shortcut
Test the extremes.
Where it went wrong
Trusting the printed number without checking what it is measured from.
Answer key
1 B
2 A
3 D
4 D
5 B
6 C
7 A
8 B
9 A
10 C
11 A
12 B
13 C
14 D
15 A
16 D
17 C
18 A
19 D
20 C
21 B
22 C
23 D
24 C
25 A
26 B
27 D
28 B
29 B
30 A
Spread across letters: A 8, B 8, C 7, D 7. No letter repeats more than twice in a row. Question mix: Numerical 17, Concept 9, Direction 1, Graph 3. Balancing seed 6005.