A coil fights changes in its own current. This file measures how hard it fights (L), what that depends on, and how to scale it when a solenoid is stretched, rewound or given a core.
Self-induction. A coil carrying current threads its own field through its own turns. When the current changes, that flux changes, and the coil induces an emf in itself that fights the change.
Picture it
Push a heavy shopping trolley. It is hard to get it moving and hard to stop it once it is rolling, but keeping it rolling at a steady speed takes almost nothing. That stubbornness is inertia, and its size is the mass.
A coil treats current the same way. It resists the current starting, it resists the current stopping, but a steady current passes as if the coil were an ordinary wire. That stubbornness is self-induction, and its size is the self-inductance L. This is why Allen calls it the inertia of electricity.
In exam language
The flux linked with a coil due to its own current is proportional to that current:
Nφ ∝ I → Nφ = LI → L = Nφ/I
When the current changes, the self-induced emf is
e = −N dφ/dt = −L dI/dt
The minus sign (Lenz's law) means the emf opposes the change in current. Because of this, the current through an inductor cannot jump suddenly.
Mechanics
Electricity
Mass m resists change in velocity
Inductance L resists change in current
Force F = m dv/dt
Emf e = L dI/dt
Momentum mv
Flux linkage LI
Kinetic energy ½mv²
Magnetic energy ½LI² (Topic 06)
What L depends on
Property
Value
Nature
Scalar
SI unit
henry (H) = Wb A⁻¹ = V s A⁻¹ = Ω s
Dimensions
[M L² T⁻² A⁻²]
Depends on
Geometry (number of turns, area, length) and the medium inside (μ = μ₀μᵣ)
Does not depend on
The current I or the flux φ
Trap
L = Nφ/I looks as if L depends on I. It does not: double the current and the flux doubles too, so the ratio stays the same. L is fixed by how the coil is built, just as the capacitance of a capacitor is fixed by its plates, not by its charge.
Correction to the Allen module
Page 106 prints "Medium (μ₀ = μ₀μᵣ)". It should read μ = μ₀μᵣ: the permeability of the medium equals the permeability of free space times the relative permeability.
Emf from a current–time graph
Slope becomes emf. e = −L × (slope of the I–t graph). A rising current gives a negative emf, a falling current a positive one. The size of the current does not matter, only how steeply it changes.
This is Allen Beginner's Box 4 Q4. The same three rules from Topic 02 work, with L in place of N and current in place of flux: flat part → zero, rising → negative, falling → positive, steeper → bigger.
Trap
At the moment the current passes through zero in the middle of the graph, the emf is not zero. The current is zero, but its slope is not.
Self-inductance of a flat circular coil
For N turns of radius R carrying current I, the field at the centre is B = μ₀NI/2R. Allen takes this field as the same over the whole area of the coil:
Nφ = N × (μ₀NI/2R) × πR² = μ₀N²πRI/2 → L = μ₀N²πR/2
In exam language
This is an approximation: the field of a loop is not really uniform across its face. For NEET, use L = μ₀N²πR/2 exactly as given. The part that matters most is the pattern L ∝ N²R.
Self-inductance of a long solenoid
Inside a long solenoid, B = μ₀nI, with n = N/ℓ turns per metre. Every turn links the flux BA:
Nφ = N(μ₀nI)A = (μ₀N²A/ℓ) I
L = μ₀N²A/ℓ = μ₀n²Aℓ = μ₀n²V (V = Aℓ, the volume)
with a core of relative permeability μᵣ: L = μ₀μᵣN²A/ℓ
Allen Illustration 14. 240 turns, ℓ = 12 cm, r = 2 cm, dI/dt = 0.8 A/s. L = 4π × 10⁻⁷ × 240² × π(0.02)² / 0.12 = 7.58 × 10⁻⁴ H, so e = 7.58 × 10⁻⁴ × 0.8 = 6.1 × 10⁻⁴ V. Allen rounds this to 6 × 10⁻⁴ V. Checked: correct.
Two meanings of 'double the length'. The formulas μ₀N²A/ℓ and μ₀n²Aℓ are the same thing, but they suit different questions. Use the first when the number of turns is fixed and the second when the turns per metre are fixed.
Change
What stays fixed
New L
Turns N doubled
Length and area
4L
Length doubled
N and area (turns spread out)
L/2
Length doubled
n and area (more turns added)
2L
Radius doubled
N and length
4L (area × 4)
All linear dimensions × 3
n
27L (A × 9, ℓ × 3)
Iron core of μᵣ inserted
Everything else
μᵣL
Formula sheet
Formula
Meaning
When to use
Watch out
Nφ = LI
Definition of L
Flux linkage from current
Nφ is total linkage
e = −L dI/dt
Self-induced emf
Current changing in a coil
Slope of I–t, not I itself
L = μ₀N²πR/2
Flat circular coil
Coil of N turns, radius R
Approximation using the centre field
L = μ₀N²A/ℓ
Solenoid, N fixed
Given total turns
L ∝ N²
L = μ₀n²Aℓ = μ₀n²V
Solenoid, n fixed
Given turns per metre
L ∝ ℓ at fixed n
L = μ₀μᵣN²A/ℓ
Solenoid with a core
Iron or other core
μ = μ₀μᵣ
1 H = 1 Wb/A = 1 V s/A = 1 Ω s
Units
Unit questions
Not V/s
[M L² T⁻² A⁻²]
Dimensions of L
Dimension questions
Flux has A⁻¹
Allen pages 105–107 checked
Item
Check
Nφ = LI, e = −L dI/dt
Correct
Unit henry, dimensions [M L² T⁻² A⁻²], scalar
Correct
L depends on geometry and medium, not on I or φ
Correct
"Medium (μ₀ = μ₀μᵣ)"
Should be μ = μ₀μᵣ
Circular coil L = μ₀N²πR/2
Correct as the standard approximation
Solenoid L = μ₀N²A/ℓ = μ₀n²V
Correct
Illustration 14: 6 × 10⁻⁴ V
Correct (6.1 × 10⁻⁴ V before rounding)
Beginner's Box 4 answer key
Q
Answer
Working
1
(3) 1.0 V
e = L ΔI/Δt = 5 × 1/5 = 1 V.
2
(2) 8 V
Current goes from +2 A to −2 A: ΔI = −4 A in 1 s. e = −L ΔI/Δt = −2 × (−4) = +8 V.
3
(1) 2 s
dI/dt = (2t − t²)e⁻ᵗ = t(2 − t)e⁻ᵗ, zero at t = 2 s (apart from the start, t = 0).
4
(2)
I rises, falls through zero, then rises back: e = −L × slope is negative, then positive, then negative.
5
(1) 4 H
L = μ₀n²Aℓ with n and A fixed: doubling ℓ doubles L.
6
(3) 27
L = μ₀n²Aℓ with n fixed: A grows 9 times and ℓ grows 3 times.
7
(4) 11.6
Nφ = LI = 2 × 5.8 = 11.6 Wb. This is the total flux linkage; the options only fit that reading.
NEET practice: 34 questions
Watch for ratio questions: they are where marks go missing in this topic. Every number here was recalculated in Python.
Q1Numerical
A coil of 500 turns has a flux of 4 mWb through each turn when it carries 2 A. Its self-inductance is
(A)0.25 H
(B)2 × 10⁻³ H
(C)4 H
(D)1 H
Show the solution
Given
N = 500, φ = 4 × 10⁻³ Wb per turn, I = 2 A
Asked
L
Concept
Self-inductance is flux linkage per unit current.
Formula
L = Nφ/I
Baby steps
Nφ = 500 × 4 × 10⁻³ = 2 Wb.
L = 2/2 = 1 H.
Answer
(D) 1 H
Why not the others
2 × 10⁻³ H forgets N. 4 H multiplies by I. 0.25 H inverts the ratio.
Shortcut
Linkage ÷ current.
Where it went wrong
Leaving out the number of turns.
Q2Numerical
The current in a 0.2 H coil falls uniformly from 3 A to 1 A in 0.1 s. The size of the self-induced emf is
(A)2 V
(B)6 V
(C)4 V
(D)0.04 V
Show the solution
Given
L = 0.2 H, ΔI = −2 A, Δt = 0.1 s
Asked
|e|
Concept
Self-induced emf depends on the rate of change of current.
Formula
|e| = L |ΔI|/Δt
Baby steps
|ΔI|/Δt = 2/0.1 = 20 A/s.
|e| = 0.2 × 20 = 4 V.
Answer
(C) 4 V
Why not the others
6 V uses 3 A. 2 V uses 1 A. 0.04 V multiplies by Δt.
Shortcut
L × change ÷ time.
Where it went wrong
Using a current value instead of the change.
Q3Numerical
A 0.5 H coil carries 4 A. The current is reversed uniformly in 0.2 s. The size of the average self-induced emf is
(A)10 V
(B)20 V
(C)zero
(D)40 V
Show the solution
Given
L = 0.5 H, I: +4 A → −4 A, Δt = 0.2 s
Asked
|e|
Concept
Reversal means the change is twice the current.
Formula
|e| = L |ΔI|/Δt
Baby steps
ΔI = −4 − 4 = −8 A.
|e| = 0.5 × 8/0.2 = 20 V.
Answer
(B) 20 V
Why not the others
10 V uses ΔI = 4 A. Zero treats +4 and −4 as the same. 40 V doubles again.
Shortcut
Reversal: ΔI = 2I.
Where it went wrong
Taking the change in a reversal as I, or as zero.
Q4Numerical
The current in a 2 mH coil varies as I = (3t² + 2t) A. The size of the self-induced emf at t = 1 s is
(A)10 mV
(B)16 mV
(C)12 mV
(D)8 mV
Show the solution
Given
L = 2 × 10⁻³ H, I = 3t² + 2t, t = 1 s
Asked
|e|
Concept
Emf depends on dI/dt.
Formula
|e| = L dI/dt
Baby steps
dI/dt = 6t + 2 = 8 A/s at t = 1 s.
|e| = 2 mH × 8 A/s = 16 mV.
Answer
(B) 16 mV
Why not the others
10 mV uses I = 5 A as if it were dI/dt. 12 mV drops the +2. 8 mV forgets L's factor of 2.
Shortcut
Differentiate I, then multiply by L.
Where it went wrong
Using I instead of dI/dt.
Q5Numerical
An air-cored solenoid has 1000 turns, length 0.5 m and cross-sectional area 10 cm². Its self-inductance is about
(A)5.0 mH
(B)1.3 mH
(C)2.5 H
(D)2.5 mH
Show the solution
Given
N = 1000, ℓ = 0.5 m, A = 1 × 10⁻³ m²
Asked
L
Concept
Long solenoid with a fixed number of turns.
Formula
L = μ₀N²A/ℓ
Baby steps
N² = 10⁶.
μ₀N²A = 4π × 10⁻⁷ × 10⁶ × 10⁻³ = 1.257 × 10⁻³.
÷ 0.5 → 2.51 × 10⁻³ H ≈ 2.5 mH.
Answer
(D) 2.5 mH
Why not the others
1.3 mH forgets to divide by ℓ = 0.5 m. 2.5 H uses 10 cm² = 10⁻³ m² wrongly as 1 m². 5.0 mH doubles.
Shortcut
4π × 10⁻⁴ ÷ 0.5.
Where it went wrong
Converting cm² to m² with 10⁻² instead of 10⁻⁴.
Q6Numerical
The solenoid of the previous question (L ≈ 2.5 mH) is filled with an iron core of relative permeability 500. Its self-inductance becomes about
(A)1.26 H
(B)2.5 mH
(C)5 × 10⁻⁶ H
(D)0.126 H
Show the solution
Given
L₀ = 2.51 × 10⁻³ H, μᵣ = 500
Asked
L
Concept
A core multiplies L by μᵣ.
Formula
L = μᵣL₀
Baby steps
L = 500 × 2.51 × 10⁻³ = 1.26 H.
Answer
(A) 1.26 H
Why not the others
2.5 mH ignores the core. 5 × 10⁻⁶ H divides by μᵣ. 0.126 H loses a factor of 10.
Shortcut
× μᵣ.
Where it went wrong
Dividing by μᵣ.
Q7Numerical
A flat circular coil of 100 turns has radius 10 cm. Using L = μ₀N²πR/2, its self-inductance is about
(A)2 × 10⁻³ H
(B)4 × 10⁻³ H
(C)2 × 10⁻⁵ H
(D)6.3 × 10⁻⁴ H
Show the solution
Given
N = 100, R = 0.1 m
Asked
L
Concept
Flat coil, field at the centre taken as uniform.
Formula
L = μ₀N²πR/2
Baby steps
μ₀N² = 4π × 10⁻⁷ × 10⁴ = 4π × 10⁻³.
× πR/2 = × 0.05π.
L = 0.2π² × 10⁻³ ≈ 1.97 × 10⁻³ H ≈ 2 × 10⁻³ H.
Answer
(A) 2 × 10⁻³ H
Why not the others
4 × 10⁻³ H forgets the ½. 2 × 10⁻⁵ H uses N instead of N². 6.3 × 10⁻⁴ H drops one factor of π.
Shortcut
π² ≈ 10 speeds up the arithmetic.
Where it went wrong
Using N instead of N².
Q8Concept
The number of turns of a solenoid is doubled while its length and cross-sectional area stay the same. Its self-inductance becomes
(A)unchanged
(B)twice as large
(C)half as large
(D)four times as large
Show the solution
Given
N → 2N; ℓ, A fixed
Asked
New L
Concept
L ∝ N² when ℓ and A are fixed.
Formula
L = μ₀N²A/ℓ
Baby steps
(2N)² = 4N².
L × 4.
Answer
(D) four times as large
Why not the others
Twice forgets the square. Half and unchanged have no basis.
Shortcut
Turns count twice: once for the field, once for the linkage.
Where it went wrong
Treating L as proportional to N.
Q9Concept
A solenoid is cut to half its length, keeping the same number of turns per metre and the same area. The self-inductance of the remaining piece is
(A)half the original
(B)double the original
(C)a quarter of the original
(D)unchanged
Show the solution
Given
ℓ → ℓ/2, n fixed, A fixed
Asked
New L
Concept
With n fixed, L = μ₀n²Aℓ ∝ ℓ.
Formula
L = μ₀n²Aℓ
Baby steps
n and A unchanged.
ℓ halves, so L halves.
Answer
(A) half the original
Why not the others
Double comes from using L ∝ 1/ℓ, which only holds when N is fixed. A quarter squares the factor.
Shortcut
Fixed n → L ∝ length.
Where it went wrong
Using μ₀N²A/ℓ when n, not N, is fixed.
Q10Concept
A solenoid's turns are spread out so that its length doubles, with the same total number of turns and the same area. Its self-inductance becomes
(A)four times as large
(B)twice as large
(C)half as large
(D)unchanged
Show the solution
Given
ℓ → 2ℓ, N fixed, A fixed
Asked
New L
Concept
With N fixed, L = μ₀N²A/ℓ ∝ 1/ℓ.
Formula
L = μ₀N²A/ℓ
Baby steps
N and A unchanged.
ℓ doubles, so L halves.
Answer
(C) half as large
Why not the others
Twice is the fixed-n case. Four times and unchanged have no basis.
Shortcut
Fixed N → L ∝ 1/length.
Where it went wrong
Answering the fixed-n version of the question.
Q11Concept
The radius of a solenoid is doubled, with the number of turns and the length unchanged. Its self-inductance becomes
(A)twice as large
(B)four times as large
(C)unchanged
(D)half as large
Show the solution
Given
r → 2r; N, ℓ fixed
Asked
New L
Concept
L ∝ A ∝ r².
Formula
L = μ₀N²(πr²)/ℓ
Baby steps
Area × 4.
L × 4.
Answer
(B) four times as large
Why not the others
Twice treats L ∝ r. The others ignore the area.
Shortcut
Radius doubled → area × 4.
Where it went wrong
Forgetting that area goes as r².
Q12Concept
A fixed length of wire is wound into a close-packed solenoid of fixed length ℓ. It is rewound with half the radius, still close-packed over the same length, so that the number of turns doubles. The self-inductance
(A)halves
(B)doubles
(C)becomes four times
(D)stays the same
Show the solution
Given
Same wire length; r → r/2 so N → 2N; ℓ fixed
Asked
New L
Concept
L = μ₀N²A/ℓ: N² goes up while A goes down.
Formula
L ∝ N²r²
Baby steps
Each turn uses half the wire, so N doubles: N² × 4.
Radius halves: A × ¼.
Net factor 4 × ¼ = 1.
Answer
(D) stays the same
Why not the others
Doubles or four times ignore the smaller area. Halves ignores the extra turns.
Shortcut
Nr is fixed by the wire length, so N²r² is fixed.
Where it went wrong
Changing only one of N and r.
Q13Concept
The dimensional formula of self-inductance is
(A)[M L² T⁻³ A⁻²]
(B)[M L² T⁻² A⁻¹]
(C)[M L² T⁻² A⁻²]
(D)[M L T⁻² A⁻²]
Show the solution
Given
L = Nφ/I
Asked
Dimensions of L
Concept
Divide the dimensions of flux by current.
Formula
[L] = [φ]/[A]
Baby steps
[φ] = [M L² T⁻² A⁻¹].
÷ [A] → [M L² T⁻² A⁻²].
Answer
(C) [M L² T⁻² A⁻²]
Why not the others
[M L² T⁻² A⁻¹] is flux. [M L² T⁻³ A⁻²] is resistance. [M L T⁻² A⁻²] is μ₀.
Shortcut
Flux over current.
Where it went wrong
Stopping at flux.
Q14Concept
One henry is equivalent to
(A)1 Ω s⁻¹
(B)1 Ω s
(C)1 Wb A
(D)1 V A s⁻¹
Show the solution
Given
Units of L
Asked
Equivalent unit
Concept
From e = L dI/dt: L = e/(dI/dt), in V s/A.
Formula
H = V s A⁻¹ = Ω s
Baby steps
Unit of L = V ÷ (A/s) = V s/A.
V/A = Ω, so H = Ω s.
Answer
(B) 1 Ω s
Why not the others
Ω s⁻¹ and V A s⁻¹ put the second on the wrong side. Wb A should be Wb A⁻¹.
Shortcut
Rearrange e = L dI/dt in units.
Where it went wrong
Writing L = e × dI/dt.
Q15Numerical
A coil of self-inductance 40 mH carries a current of 2.5 A. The flux linkage of the coil is
(A)0.1 Wb
(B)16 Wb
(C)0.016 Wb
(D)100 Wb
Show the solution
Given
L = 0.04 H, I = 2.5 A
Asked
Nφ
Concept
Flux linkage = LI.
Formula
Nφ = LI
Baby steps
Nφ = 0.04 × 2.5 = 0.1 Wb.
Answer
(A) 0.1 Wb
Why not the others
16 Wb divides by 2.5 and forgets mH. 0.016 Wb divides L by I. 100 Wb uses 40 H.
Shortcut
LI.
Where it went wrong
Converting mH wrongly.
Q16Numerical
An emf of 12 V is induced in a coil when the current through it changes at 3 A s⁻¹. The self-inductance of the coil is
(A)0.25 H
(B)36 H
(C)4 H
(D)15 H
Show the solution
Given
|e| = 12 V, dI/dt = 3 A/s
Asked
L
Concept
Rearrange the self-induced emf formula.
Formula
L = |e|/(dI/dt)
Baby steps
L = 12/3 = 4 H.
Answer
(C) 4 H
Why not the others
36 H multiplies. 0.25 H inverts. 15 H adds.
Shortcut
Volts per (amps per second).
Where it went wrong
Ratio reversal.
Q17Concept
Two solenoids have the same cross-sectional area. Solenoid P has N turns and length 2ℓ; solenoid Q has 2N turns and length ℓ. The ratio LP : LQ is
(A)1 : 4
(B)8 : 1
(C)1 : 2
(D)1 : 8
Show the solution
Given
P: N, 2ℓ. Q: 2N, ℓ. Same A
Asked
LP : LQ
Concept
L ∝ N²/ℓ at fixed A.
Formula
L = μ₀N²A/ℓ
Baby steps
LP ∝ N²/2ℓ.
LQ ∝ 4N²/ℓ.
Ratio = (1/2)/4 = 1/8.
Answer
(D) 1 : 8
Why not the others
8 : 1 is the right numbers upside down. 1 : 2 ignores the length. 1 : 4 ignores the length too.
Shortcut
P has fewer turns and is longer: it must be smaller.
Where it went wrong
Ratio reversal.
Q18Numerical
A current I = 2 sin(100t) A flows through a 10 mH coil. The maximum self-induced emf is
(A)2 V
(B)0.2 V
(C)20 V
(D)0.02 V
Show the solution
Given
L = 0.01 H, I₀ = 2 A, ω = 100 rad/s
Asked
emax
Concept
dI/dt = I₀ω cos ωt has a maximum of I₀ω.
Formula
emax = LI₀ω
Baby steps
LI₀ω = 0.01 × 2 × 100 = 2 V.
Answer
(A) 2 V
Why not the others
0.2 V and 20 V shift a power of ten. 0.02 V forgets ω.
Shortcut
L × I₀ × ω.
Where it went wrong
Forgetting ω.
Q19Numerical
The current in a coil varies as I = t e⁻ᵗ A. The self-induced emf is zero at
(A)t = 2 s
(B)t = 1 s
(C)t = 0.5 s
(D)t = e s
Show the solution
Given
I = t e⁻ᵗ
Asked
t where e = 0 (other than never)
Concept
Emf is zero when dI/dt = 0.
Formula
e = −L dI/dt
Baby steps
dI/dt = e⁻ᵗ − t e⁻ᵗ = (1 − t)e⁻ᵗ.
Zero when t = 1 s.
Answer
(B) t = 1 s
Why not the others
t = 2 s is the answer for I = t²e⁻ᵗ (Beginner's Box 4 Q3). 0.5 s and e s do not make (1 − t) zero.
Shortcut
Maximum current ↔ zero emf.
Where it went wrong
Setting I = 0 instead of dI/dt = 0.
Q20Concept
The current through a coil is doubled. Its self-inductance
(A)halves
(B)doubles
(C)stays the same
(D)becomes four times
Show the solution
Given
I → 2I
Asked
New L
Concept
L depends on geometry and medium only.
Formula
L = Nφ/I with φ ∝ I
Baby steps
Doubling I doubles φ.
The ratio Nφ/I stays the same.
Answer
(C) stays the same
Why not the others
Each other option makes L depend on I.
Shortcut
L is a property of the coil, like mass of a body.
Where it went wrong
Reading L = Nφ/I as L ∝ 1/I.
Q21Concept
While the current through a coil is increasing, the self-induced emf
(A)is in the same direction as the current
(B)opposes the current
(C)is zero
(D)is in the same direction only if the coil has an iron core
Show the solution
Given
dI/dt > 0
Asked
Direction of self-induced emf
Concept
Lenz's law: the emf opposes the change (here, the increase).
Formula
e = −L dI/dt
Baby steps
dI/dt positive → e negative.
The emf acts against the current, slowing its rise. It is called a back emf.
Answer
(B) opposes the current
Why not the others
Along the current is the falling-current case. Zero needs a steady current. The core changes size, not direction.
Shortcut
Rising → back emf.
Where it went wrong
Thinking the emf always acts against the current. When I falls, it acts along it.
Q22Concept
While the current through a coil is decreasing, the self-induced emf
(A)acts against the current
(B)acts in the same direction as the current
(C)is zero
(D)reverses the current immediately
Show the solution
Given
dI/dt < 0
Asked
Direction
Concept
Opposing a decrease means trying to keep the current going.
Formula
e = −L dI/dt
Baby steps
dI/dt negative → e positive.
The emf supports the current, slowing its fall.
Answer
(B) acts in the same direction as the current
Why not the others
Against the current is the rising case. Zero needs a steady current. An inductor never reverses current suddenly.
Shortcut
Falling → emf helps.
Where it went wrong
Direction reversal: using the rising-current rule.
Q23Concept
In an inductor, which quantity cannot change suddenly (in zero time)?
(A)The current through it
(B)The voltage across it
(C)The rate of change of current
(D)The emf across it
Show the solution
Given
Ideal inductor
Asked
Quantity that cannot jump
Concept
A sudden jump in current would need dI/dt → ∞, so an infinite emf.
Formula
e = −L dI/dt
Baby steps
Finite emf → finite dI/dt.
So the current must change continuously.
Answer
(A) The current through it
Why not the others
The voltage and the emf can jump (they do at switching). dI/dt can jump too.
Shortcut
Inductor: current is continuous. (Capacitor: voltage is continuous.)
Where it went wrong
Swapping the inductor and capacitor rules.
Q24Concept
A soft iron rod is inserted into a coil. Its self-inductance
(A)decreases
(B)increases
(C)stays the same
(D)becomes zero
Show the solution
Given
Iron core added
Asked
Effect on L
Concept
Iron has μᵣ ≫ 1.
Formula
L = μ₀μᵣN²A/ℓ
Baby steps
μᵣ of soft iron is large.
L is multiplied by μᵣ.
Answer
(B) increases
Why not the others
Iron strengthens the field, so L cannot decrease, stay the same or vanish.
Shortcut
Core → × μᵣ.
Where it went wrong
Thinking a solid rod blocks the field.
Q25Concept
Self-inductance is called the inertia of electricity because
(A)a coil opposes any change in the current through it
(B)a coil opposes the current itself
(C)the coil's current keeps increasing on its own
(D)inductance is measured in kilograms
Show the solution
Given
Analogy with mass
Asked
Reason for the name
Concept
Mass resists change in motion; L resists change in current.
Formula
F = m dv/dt ↔ e = L dI/dt
Baby steps
Inertia means resisting change, not resisting motion.
An inductor passes a steady current freely, but resists changes.
Answer
(A) a coil opposes any change in the current through it
Why not the others
An inductor does not oppose a steady current. Current does not grow on its own. The unit is the henry.
Shortcut
Opposes change, not the current.
Where it went wrong
Thinking an inductor reduces a steady current.
Q26Graph
The current through a 1 H coil rises steadily from zero and then stays constant. Which graph shows the self-induced emf (with its sign) against time?
The positive block forgets the minus sign. The ramp-and-flat graph copies the current. A constant emf forever ignores the flat part.
Shortcut
Slope, then flip the sign.
Where it went wrong
Copying the current graph.
Q27Graph
Which graph shows the self-inductance of an air-cored coil against the current through it?
(A)
(B)
(C)
(D)
Show the solution
Given
Air core; current varied
Asked
L against I
Concept
L depends only on geometry and medium.
Formula
L = μ₀N²A/ℓ
Baby steps
Nothing about the coil's shape or core changes.
So L is the same for every current: a flat line.
Answer
(D) the graph in option D
Why not the others
The rising line and the parabola make L depend on I. The hyperbola misreads L = Nφ/I.
Shortcut
Property of the coil → horizontal line.
Where it went wrong
Reading L = Nφ/I as L ∝ 1/I.
Q28Graph
Which graph shows the flux linkage of a coil against the current through it?
(A)
(B)
(C)
(D)
Show the solution
Given
Nφ = LI with L constant
Asked
Nφ against I
Concept
Flux linkage is proportional to current.
Formula
Nφ = LI
Baby steps
L is constant.
Nφ = LI is a straight line through the origin.
Its slope is L.
Answer
(C) the graph in option C
Why not the others
A flat line would mean current makes no field. A parabola would need L ∝ I. A hyperbola would have flux falling as current rises.
Shortcut
Straight line, slope L.
Where it went wrong
Mixing up this graph with L against I.
Q29Graph
Solenoids of the same length and area are wound with different numbers of turns N. Which graph shows L against N?
(A)
(B)
(C)
(D)
Show the solution
Given
ℓ and A fixed; N varies
Asked
L against N
Concept
L ∝ N².
Formula
L = μ₀N²A/ℓ
Baby steps
Doubling N gives four times L.
That is a parabola through the origin.
Answer
(A) the graph in option A
Why not the others
The straight line treats L ∝ N. The square-root curve treats N ∝ L². The flat line ignores N.
Shortcut
N² → parabola.
Where it went wrong
Treating L ∝ N.
Q30Assertion–reason
Assertion (A): The self-inductance of a coil does not change when the current through it is doubled. Reason (R): Self-inductance depends only on the geometry of the coil and the medium inside 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: L independent of I. R: L depends on geometry and medium.
Asked
Truth and link
Concept
L is a property of construction.
Formula
L = μ₀μᵣN²A/ℓ
Baby steps
A is true.
R is true.
Since I is not in R's list, L cannot depend on I. R explains A.
Answer
(A) Both A and R are true, and R is the correct explanation of A.
Why not the others
(B) denies a direct link. (C) and (D) need a false statement.
Shortcut
R lists what L depends on; I is not on it.
Where it went wrong
Picking (B) by habit.
Q31Assertion–reason
Assertion (A): The henry is the SI unit of self-inductance. Reason (R): Self-inductance is a scalar quantity.
(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: unit henry. R: L is scalar.
Asked
Truth and link
Concept
Both are facts about L, but one does not cause the other.
Formula
—
Baby steps
A is true.
R is true.
Being a scalar has nothing to do with which unit is used. 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 a causal link. (C) and (D) need a false statement.
Shortcut
'The unit is henry because L is a scalar' makes no sense.
Where it went wrong
Linking two true facts from the same bullet list.
Q32Assertion–reason
Assertion (A): An ideal inductor carrying a steady current has no potential difference across it. Reason (R): The self-induced emf is proportional to the current itself.
(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: no p.d. for steady current. R: emf ∝ I.
Asked
Truth and link
Concept
Emf ∝ dI/dt, not I.
Formula
e = −L dI/dt
Baby steps
A is true: dI/dt = 0 and there is no resistance.
R is false: it is proportional to the rate of change.
A true, R false.
Answer
(C) A is true, but R is false.
Why not the others
(A) and (B) need R true. (D) needs A false.
Shortcut
Steady current → zero slope → zero emf.
Where it went wrong
Confusing I with dI/dt.
Q33Assertion–reason
Assertion (A): If the number of turns of a solenoid is doubled, keeping its length and area the same, its self-inductance doubles. Reason (R): The self-inductance of a solenoid is proportional to the square of its number of turns.
(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: L doubles. R: L ∝ N².
Asked
Truth and link
Concept
L = μ₀N²A/ℓ.
Formula
L ∝ N²
Baby steps
R is true.
By R, L becomes four times, so A is false.
A false, R true.
Answer
(D) A is false, but R is true.
Why not the others
(A), (B) and (C) need A true.
Shortcut
Apply R to test A.
Where it went wrong
Accepting 'doubles' without squaring.
Q34Two statements
Statement I: The self-inductance of a long solenoid is μ₀n²Aℓ, where n is the number of turns per unit length. Statement II: The self-inductance of a solenoid does not depend on the material of its core.
(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.
Show the solution
Given
Solenoid formula; core dependence
Asked
Which are true
Concept
L = μ₀μᵣn²Aℓ with a core.
Formula
μ = μ₀μᵣ
Baby steps
Statement I is true (air core).
Statement II is false: an iron core multiplies L by μᵣ.
Answer
(C) Statement I is true, but Statement II is false.
Why not the others
Both-true fails on II. Both-false fails on I. I-false-II-true reverses them.
Shortcut
Core matters: × μᵣ.
Where it went wrong
Forgetting the medium in 'geometry and medium'.
Answer key
1 D
2 C
3 B
4 B
5 D
6 A
7 A
8 D
9 A
10 C
11 B
12 D
13 C
14 B
15 A
16 C
17 D
18 A
19 B
20 C
21 B
22 B
23 A
24 B
25 A
26 D
27 D
28 C
29 A
30 A
31 B
32 C
33 D
34 C
Spread across letters: A 9, B 9, C 8, D 8. No letter repeats more than twice in a row. Question mix: Numerical 11, Concept 14, Graph 4, Assertion–reason 4, Two statements 1. Balancing seed 1.