Change the current in one coil and a coil next to it answers. This file measures how strongly (M), how it depends on turns, size, distance and angle, and corrects one Allen illustration.
Mutual induction. Change the current in the primary and the secondary feels a changing flux. It answers with an emf that opposes the change. Steady current: silence.
Picture it
Imagine two neighbours with a shared garden hose running under their fence. When one opens or closes her tap, the hose jerks on the other side. While the water flows steadily, nothing jerks. The neighbour only notices changes. Mutual induction is the secondary coil noticing changes in the primary's current.
In exam language
The flux linkage of the secondary is proportional to the current in the primary:
N₂φ₂ = M I₁ → M = N₂φ₂ / I₁
e₂ = −N₂ dφ₂/dt = −M dI₁/dt
M is the mutual inductance. It has the same unit (henry) and dimensions [M L² T⁻² A⁻²] as self-inductance, and it is a scalar.
M depends on
M does not depend on
Number of turns and size of both coils
The current in either coil
Distance between the coils
The flux value at any moment
Relative orientation
Which coil is called primary
Medium or core (μᵣ)
Mutual inductance is a shared property: M₁₂ = M₂₁. A changing current in coil 2 induces the same emf in coil 1 as the same change in coil 1 would induce in coil 2.
Two coaxial solenoids
A short coil of N₂ turns is wound over the middle of a long solenoid of N₁ turns and length ℓ. The solenoid's field B₁ = μ₀(N₁/ℓ)I₁ passes through every turn of the outer coil, over the solenoid's cross-section A:
N₂φ₂ = N₂ B₁ A = μ₀N₁N₂A I₁ / ℓ → M = μ₀N₁N₂A/ℓ
with a core of relative permeability μᵣ: M = μ₀μᵣN₁N₂A/ℓ
Allen Illustration 23. N₁ = 2000, ℓ = 0.3 m, A = 1.2 × 10⁻³ m², N₂ = 300: M = 4π × 10⁻⁷ × 2000 × 300 × 1.2 × 10⁻³/0.3 = 3.0 × 10⁻³ H. A 2 A current reversed in 0.25 s gives e = 3 × 10⁻³ × 4/0.25 = 48 mV. Checked: correct.
Trap
Use the cross-section of the inner coil, because only there is the field. Winding the outer coil wider does not change M.
Two concentric loops
A small loop of radius r sits at the centre of a large coplanar loop of radius R ≫ r. The large loop's centre field μ₀N₁I₁/2R is nearly uniform over the small loop:
N₂φ₂ = N₂ (μ₀N₁I₁/2R) πr² → M = μ₀N₁N₂πr²/2R
So M ∝ r²/R. This is Allen Beginner's Box 6 Q7.
Correction to the Allen module
Illustration 21(b) (p. 118): a small square loop of side ℓ at the centre of a large square loop of side L. The field at the centre of a square is 2√2 μ₀I/πL, not √2 μ₀I/πL: each of the four sides gives μ₀I/(4π · L/2) × (sin 45° + sin 45°) = √2 μ₀I/2πL, and four sides make 2√2 μ₀I/πL. So the correct answer is M = 2√2 μ₀ℓ²/πL, twice Allen's value. Part (a), M = μ₀πr²/2R for circles, is correct.
Coupling factor
Orientation. Turning the secondary side-on lets the primary's field lines skim past instead of through it. M falls towards zero.
Not all the flux of one coil reaches the other. The fraction that does is the coupling factor K:
K = φ₂/φ₁ M = K√(L₁L₂), 0 ≤ K ≤ 1
Arrangement
K
One coil wound tightly over the other
≈ 1 (M = √(L₁L₂))
Coils side by side on the same axis
Between 0 and 1
Coils far apart
Close to 0
Axes at right angles
≈ 0
Illustration 22: L₁ = 2 H, L₂ = 8 H, K = 0.5 gives M = 0.5 × √16 = 2 H. Illustration 24: two 0.1 H coils wound one over the other, K = 1, M = √(0.1 × 0.1) = 0.1 H. Illustration 25: M decreases with distance, decreases when the turns are reduced, and increases with an iron core. All checked: correct.
When the primary current is alternating
Sine in, cosine out. The secondary emf follows the slope of the primary current, so it is largest when the current passes through zero.
I₁ = I₀ sin ωt → e₂ = −M dI₁/dt = −MI₀ω cos ωt, peak = MI₀ω = 2πf MI₀
Beginner's Box 6 Q4: M = 1.5 H, I₀ = 1 A, f = 50 Hz gives a peak emf of 1.5 × 1 × 100π = 150π V. This is the idea behind the transformer in Topic 14.
Formula sheet
Formula
Meaning
When to use
Watch out
N₂φ₂ = MI₁
Definition of M
Linkage of one coil due to the other's current
N₂φ₂ is total linkage
e₂ = −M dI₁/dt
Mutually induced emf
Primary current changing
Secondary current = e₂/R₂
M = μ₀N₁N₂A/ℓ
Coaxial solenoids
Coil wound on a long solenoid
A of the inner coil
M = μ₀N₁N₂πr²/2R
Concentric coplanar loops, R ≫ r
Small loop at the centre of a big one
M ∝ r²/R
M = 2√2 μ₀ℓ²/πL
Small square inside a large square
Allen Illustration 21(b), corrected
Allen prints √2
M = K√(L₁L₂)
Coupling factor
Given L₁, L₂ and K
Mmax = √(L₁L₂)
e₂,peak = MI₀ω
Sinusoidal primary
AC in the primary
ω = 2πf
M₁₂ = M₂₁
Reciprocity
Swapping primary and secondary
Holds even for unequal coils
Allen pages 116–120 checked
Item
Check
N₂φ₂ = MI₁, e₂ = −M dI₁/dt, unit and dimensions
Correct
Coaxial solenoids M = μ₀N₁N₂A/ℓ
Correct
Concentric loops M = μ₀N₁N₂πr²/2R
Correct
M = K√(L₁L₂), 0 ≤ K ≤ 1
Correct
Illustration 21(a)
Correct
Illustration 21(b): B = √2 μ₀I/πL
Should be 2√2 μ₀I/πL; M = 2√2 μ₀ℓ²/πL
Illustrations 22, 23, 24, 25
Correct
Box 6 Q2 answer
Works only if "flux linked with B" means total linkage N₂φ₂
Beginner's Box 6 answer key
Q
Answer
Working
1
(1) 4.0 A/s
Needed secondary emf = I₂R₂ = 0.4 × 5 = 2 V. dI₁/dt = e₂/M = 2/0.5 = 4 A/s. The primary's resistance is not needed.
2
(2) 3 × 10⁻⁵ H
M = (flux linked with B)/I_A = 9.0 × 10⁻⁵/3.0 = 3 × 10⁻⁵ H, taking 9.0 × 10⁻⁵ Wb as the total linkage of B. If it were the flux per turn, M = 600 × 9.0 × 10⁻⁵/3 = 1.8 × 10⁻² H, which is not an option.
3
(2) e = −MI₀ω cos ωt
e = −M d(I₀ sin ωt)/dt.
4
(2) 150π V
Peak = MI₀ω = 1.5 × 1 × 2π × 50.
5
(3) 25 H
M ∝ N₁N₂, and 5 × 10 = 10 × 5.
6
(1) 4.8 × 10⁻² V
Same numbers as Illustration 23: M = 3 × 10⁻³ H, ΔI = 4 A in 0.25 s.
7
(4) R₂²/R₁
Small loop R₂ at the centre of the large loop R₁: M = μ₀πR₂²/2R₁.
8
(a) opposite to the primary current; (b) no current; (c) same direction as the primary current
Just closed: primary flux grows, so the secondary opposes it. Steady: no change. Just opened: primary flux collapses, so the secondary supports it. This holds for all three circuits (i), (ii), (iii); the scan does not show the cell polarity clearly enough to name clockwise or anticlockwise.
NEET practice: 31 questions
Every number was recalculated in Python. Q14 checks the corrected Illustration 21(b) value.
Q1Numerical
The mutual inductance of two coils is 0.5 H. The current in the primary changes at 4 A s⁻¹. The emf induced in the secondary is
(A)0.125 V
(B)8 V
(C)2 V
(D)4.5 V
Show the solution
Given
M = 0.5 H, dI₁/dt = 4 A/s
Asked
|e₂|
Concept
Mutually induced emf.
Formula
|e₂| = M dI₁/dt
Baby steps
0.5 × 4 = 2 V.
Answer
(C) 2 V
Why not the others
8 V divides. 0.125 V inverts. 4.5 V adds.
Shortcut
M × rate.
Where it went wrong
Ratio reversal.
Q2Numerical
A current of 3 A in coil P produces a total flux linkage of 0.012 Wb in a nearby coil S. The mutual inductance is
(A)4 mH
(B)36 mH
(C)0.25 H
(D)12 mH
Show the solution
Given
I₁ = 3 A, N₂φ₂ = 0.012 Wb
Asked
M
Concept
Definition of M.
Formula
M = N₂φ₂/I₁
Baby steps
0.012/3 = 0.004 H = 4 mH.
Answer
(A) 4 mH
Why not the others
36 mH multiplies. 0.25 H inverts. 12 mH forgets to divide.
Shortcut
Linkage per ampere.
Where it went wrong
Multiplying by the current.
Q3Numerical
A long solenoid of length 0.4 m and 1000 turns has a cross-section of 2 cm². A 200-turn coil is wound tightly over its middle. The mutual inductance is about
6.28 × 10⁻⁵ H is half the correct value. 1.26 × 10⁻² H uses 2 cm² = 2 × 10⁻² m². 5.03 × 10⁻⁴ H is ten times μ₀N₁N₂A, with no division by ℓ.
Shortcut
4π × 10⁻⁷ × N₁N₂A/ℓ.
Where it went wrong
Converting cm² with 10⁻² instead of 10⁻⁴.
Q4Concept
In the previous arrangement, the solenoid is given an iron core of relative permeability 100. The mutual inductance becomes
(A)100 times smaller
(B)unchanged
(C)100 times larger
(D)10 times larger
Show the solution
Given
μᵣ = 100
Asked
Effect on M
Concept
M = μ₀μᵣN₁N₂A/ℓ.
Formula
M ∝ μᵣ
Baby steps
The core multiplies B, so it multiplies M by μᵣ.
Answer
(C) 100 times larger
Why not the others
Unchanged ignores the core. Smaller reverses. 10 uses √μᵣ.
Shortcut
Core → × μᵣ.
Where it went wrong
Thinking a core shields the second coil.
Q5Numerical
A single-turn loop of radius 1 cm lies at the centre of a coplanar single-turn loop of radius 50 cm. The mutual inductance is about
(A)7.9 × 10⁻¹⁰ H
(B)3.9 × 10⁻¹⁰ H
(C)3.9 × 10⁻⁸ H
(D)1.3 × 10⁻⁶ H
Show the solution
Given
r = 0.01 m, R = 0.5 m
Asked
M
Concept
Centre field of the big loop is nearly uniform over the small one.
Formula
M = μ₀πr²/2R
Baby steps
πr² = 3.14 × 10⁻⁴ m².
μ₀/2R = 4π × 10⁻⁷/1 = 1.26 × 10⁻⁶.
M = 1.26 × 10⁻⁶ × 3.14 × 10⁻⁴ = 3.9 × 10⁻¹⁰ H.
Answer
(B) 3.9 × 10⁻¹⁰ H
Why not the others
7.9 × 10⁻¹⁰ H forgets the 2 in 2R. 3.9 × 10⁻⁸ H uses r = 10 cm. 1.3 × 10⁻⁶ H is B per ampere, not M.
Shortcut
μ₀πr²/2R.
Where it went wrong
Using the big loop's area.
Q6Concept
For a small loop of radius r at the centre of a large coplanar loop of radius R (R ≫ r), both r and R are doubled. The mutual inductance becomes
(A)twice as large
(B)four times as large
(C)unchanged
(D)half as large
Show the solution
Given
r → 2r, R → 2R
Asked
New M
Concept
M ∝ r²/R.
Formula
M = μ₀πr²/2R
Baby steps
r² × 4.
÷ R × 2.
Net × 2.
Answer
(A) twice as large
Why not the others
Four times ignores R. Unchanged treats M ∝ r/R. Half inverts.
Shortcut
Square the small radius, divide by the big one.
Where it went wrong
Treating M ∝ r/R.
Q7Numerical
Two coils have self-inductances 4 mH and 9 mH and mutual inductance 3 mH. The coupling factor is
(A)0.25
(B)0.5
(C)0.75
(D)2
Show the solution
Given
L₁ = 4 mH, L₂ = 9 mH, M = 3 mH
Asked
K
Concept
Coupling factor.
Formula
K = M/√(L₁L₂)
Baby steps
√(4 × 9) = 6 mH.
K = 3/6 = 0.5.
Answer
(B) 0.5
Why not the others
0.25 divides by 12. 0.75 divides by 4. 2 inverts (and K cannot exceed 1).
Shortcut
√(L₁L₂) first.
Where it went wrong
Dividing by L₁ + L₂ or by the product.
Q8Numerical
Two coils of self-inductance 2 H and 8 H have a coupling factor of 0.8. Their mutual inductance is
(A)4.0 H
(B)3.2 H
(C)8.0 H
(D)12.8 H
Show the solution
Given
L₁ = 2 H, L₂ = 8 H, K = 0.8
Asked
M
Concept
M = K√(L₁L₂).
Formula
M = K√(L₁L₂)
Baby steps
√16 = 4 H.
M = 0.8 × 4 = 3.2 H.
Answer
(B) 3.2 H
Why not the others
4.0 H assumes K = 1. 8.0 H uses L₂. 12.8 H multiplies without the root.
Shortcut
K × geometric mean.
Where it went wrong
Forgetting the square root.
Q9Numerical
The largest possible mutual inductance between coils of self-inductance 0.2 H and 0.8 H is
(A)0.5 H
(B)1.0 H
(C)0.16 H
(D)0.4 H
Show the solution
Given
L₁ = 0.2 H, L₂ = 0.8 H
Asked
Mmax
Concept
K ≤ 1.
Formula
Mmax = √(L₁L₂)
Baby steps
√(0.16) = 0.4 H.
Answer
(D) 0.4 H
Why not the others
1.0 H adds. 0.16 H forgets the root. 0.5 H averages.
Shortcut
Geometric mean.
Where it went wrong
Using the arithmetic mean.
Q10Numerical
An alternating current of peak 2 A and frequency 50 Hz flows in a primary coil. The mutual inductance is 0.2 H. The peak emf in the secondary is about
(A)63 V
(B)20 V
(C)0.4 V
(D)126 V
Show the solution
Given
M = 0.2 H, I₀ = 2 A, f = 50 Hz
Asked
e₂,peak
Concept
e₂ = −MI₀ω cos ωt.
Formula
e₂,peak = MI₀(2πf)
Baby steps
ω = 100π ≈ 314 rad/s.
0.2 × 2 × 314 = 126 V.
Answer
(D) 126 V
Why not the others
20 V uses f = 50 instead of ω. 0.4 V forgets ω. 63 V uses πf.
Shortcut
40π V.
Where it went wrong
Using f instead of 2πf.
Q11Numerical
The primary current varies as I₁ = 2t² A and the mutual inductance is 0.1 H. The secondary emf at t = 3 s is
(A)1.2 V
(B)1.8 V
(C)0.6 V
(D)3.6 V
Show the solution
Given
I₁ = 2t², M = 0.1 H, t = 3 s
Asked
|e₂|
Concept
Differentiate I₁.
Formula
|e₂| = M dI₁/dt
Baby steps
dI₁/dt = 4t = 12 A/s.
e₂ = 0.1 × 12 = 1.2 V.
Answer
(A) 1.2 V
Why not the others
1.8 V uses I₁ = 18 A. 0.6 V drops the factor 2. 3.6 V uses 36 A/s.
Shortcut
Rate, not value.
Where it went wrong
Using I₁ instead of dI₁/dt.
Q12Numerical
Two coils have M = 0.2 H. The primary current changes at 10 A s⁻¹. The secondary circuit has resistance 4 Ω. The secondary current is
(A)0.5 A
(B)2 A
(C)8 A
(D)0.08 A
Show the solution
Given
M = 0.2 H, dI₁/dt = 10 A/s, R₂ = 4 Ω
Asked
I₂
Concept
Emf, then Ohm's law.
Formula
I₂ = M(dI₁/dt)/R₂
Baby steps
e₂ = 2 V.
I₂ = 2/4 = 0.5 A.
Answer
(A) 0.5 A
Why not the others
2 A is the emf in volts. 8 A multiplies by R. 0.08 A divides by 25.
Shortcut
Emf first.
Where it went wrong
Stopping at the emf.
Q13Numerical
Two coils have M = 5 mH. A current of 3 A in the primary is reversed uniformly in 0.1 s. The emf induced in the secondary is
(A)0.6 V
(B)0.15 V
(C)zero
(D)0.3 V
Show the solution
Given
M = 5 × 10⁻³ H, ΔI₁ = 6 A, Δt = 0.1 s
Asked
|e₂|
Concept
Reversal changes the current by 2I.
Formula
|e₂| = M ΔI₁/Δt
Baby steps
ΔI₁ = 6 A.
e₂ = 5 × 10⁻³ × 60 = 0.3 V.
Answer
(D) 0.3 V
Why not the others
0.15 V uses ΔI = 3 A. Zero treats +3 and −3 as equal. 0.6 V doubles again.
Shortcut
Reversal: 2I.
Where it went wrong
Using I instead of 2I.
Q14Numerical
A small square loop of side 1 cm lies at the centre of a large coplanar square loop of side 1 m, with sides parallel. Taking the large loop's centre field as uniform over the small loop, the mutual inductance is about
(A)1.13 × 10⁻¹⁰ H
(B)5.66 × 10⁻¹¹ H
(C)4.0 × 10⁻¹¹ H
(D)2.26 × 10⁻¹⁰ H
Show the solution
Given
L = 1 m, ℓ = 0.01 m
Asked
M
Concept
Field at the centre of a square: four finite wires at distance L/2.
Formula
B = 2√2 μ₀I/πL, M = Bℓ²/I
Baby steps
Each side: μ₀I/(4π · 0.5) × (sin 45° + sin 45°) = √2 μ₀I/2π.
Four sides: B = 2√2 μ₀I/π (for L = 1 m).
M = 2√2 × 4π × 10⁻⁷ × 10⁻⁴ / π = 1.13 × 10⁻¹⁰ H.
Answer
(A) 1.13 × 10⁻¹⁰ H
Why not the others
5.66 × 10⁻¹¹ H uses Allen's printed √2 μ₀I/πL, which is half the true field. 4.0 × 10⁻¹¹ H drops the √2 altogether. 2.26 × 10⁻¹⁰ H doubles again.
Shortcut
B(square centre) = 2√2 μ₀I/πL.
Where it went wrong
Copying the misprinted field from Illustration 21(b).
Q15Concept
The number of turns of the primary coil is doubled, keeping everything else the same. The mutual inductance
(A)halves
(B)becomes four times
(C)doubles
(D)stays the same
Show the solution
Given
N₁ → 2N₁
Asked
New M
Concept
M ∝ N₁N₂.
Formula
M = μ₀N₁N₂A/ℓ
Baby steps
Only N₁ doubles.
M doubles.
Answer
(C) doubles
Why not the others
Four times would need both N₁ and N₂ doubled (like L ∝ N²). Halves reverses. Same ignores N₁.
Shortcut
M ∝ N₁N₂, not N².
Where it went wrong
Carrying L ∝ N² over to M.
Q16Concept
The current in the primary coil is doubled. The mutual inductance of the pair
(A)doubles
(B)stays the same
(C)halves
(D)becomes four times
Show the solution
Given
I₁ → 2I₁
Asked
New M
Concept
M depends on geometry, separation, orientation and medium only.
Formula
M = N₂φ₂/I₁
Baby steps
Doubling I₁ doubles φ₂.
The ratio is unchanged.
Answer
(B) stays the same
Why not the others
The other options make M depend on current.
Shortcut
Like L, M is fixed by construction.
Where it went wrong
Reading M = N₂φ₂/I₁ as M ∝ 1/I₁.
Q17Concept
Coil A has 100 turns and coil B has 500 turns. Compared with the mutual inductance M_AB (current in A, emf in B), M_BA (current in B, emf in A) is
(A)25 times larger
(B)5 times larger
(C)5 times smaller
(D)equal
Show the solution
Given
Unequal coils; swap primary and secondary
Asked
M_BA vs M_AB
Concept
Reciprocity: mutual inductance is a shared property.
Formula
M₁₂ = M₂₁
Baby steps
M ∝ N_A N_B, which is symmetric.
So M_BA = M_AB.
Answer
(D) equal
Why not the others
The others treat M as belonging to one coil.
Shortcut
Swap the roles, same M.
Where it went wrong
Thinking the bigger coil has the bigger M.
Q18Concept
The distance between two coaxial coils is increased. The mutual inductance
(A)increases
(B)decreases
(C)stays the same
(D)first increases, then decreases
Show the solution
Given
Separation increased
Asked
Effect on M
Concept
Less of one coil's flux reaches the other (Allen Illustration 25).
Formula
M = K√(L₁L₂), K falls
Baby steps
The field of a coil weakens with distance.
So the flux linkage per ampere falls: M decreases.
Answer
(B) decreases
Why not the others
Increase and 'same' ignore the weakening field. Nothing makes it rise first.
Shortcut
Farther → weaker coupling.
Where it went wrong
Confusing M with L of each coil (which does not change).
Q19Concept
Two circular coils are placed with their centres close together. The mutual inductance is nearly zero when
(A)they share an iron core
(B)they are coaxial and touching
(C)one is wound over the other
(D)their axes are at right angles
Show the solution
Given
Relative orientation
Asked
Arrangement with M ≈ 0
Concept
Coupling depends on how much of one coil's flux passes through the other.
Formula
K ≈ 0
Baby steps
With axes at right angles, one coil's field lines run along the other's plane.
Almost no flux passes through, so M ≈ 0.
Answer
(D) their axes are at right angles
Why not the others
The other three give strong coupling.
Shortcut
Side-on → no linkage.
Where it went wrong
Thinking closeness alone decides M.
Q20Concept
The maximum value of the coupling factor K between two coils is
(A)√2
(B)0.5
(C)1
(D)infinite
Show the solution
Given
Coupling factor
Asked
Kmax
Concept
K is the fraction of flux linked.
Formula
K = φ₂/φ₁ ≤ 1
Baby steps
A fraction cannot exceed 1.
K = 1 when all of one coil's flux links the other.
Answer
(C) 1
Why not the others
0.5 is an example value. √2 and infinite exceed a fraction's limit.
Shortcut
K is a fraction.
Where it went wrong
Forgetting the square root sets the upper bound on M, not on K.
Q21Concept
The primary coil carries a large steady current. The emf induced in the secondary is
(A)maximum
(B)M times the current
(C)zero
(D)equal to the primary's voltage
Show the solution
Given
dI₁/dt = 0
Asked
e₂
Concept
Emf needs a changing primary current.
Formula
e₂ = −M dI₁/dt
Baby steps
Steady current → dI₁/dt = 0.
e₂ = 0.
Answer
(C) zero
Why not the others
MI₁ is flux linkage, not emf. Maximum and 'equal' have no basis.
Shortcut
No change, no emf.
Where it went wrong
Confusing MI₁ with the emf.
Q22Graph
The primary current rises steadily, stays constant, then falls steadily at half the rate at which it rose. Which graph shows the secondary emf (with sign)?
(A)
(B)
(C)
(D)
Show the solution
Given
Rise (fast), steady, fall (half the rate)
Asked
e₂ against t
Concept
e₂ = −M × slope of I₁.
Formula
e₂ = −M dI₁/dt
Baby steps
Rise: negative emf.
Steady: zero.
Fall at half the rate: positive emf, half the size, lasting twice as long.
Answer
(B) the graph in option B
Why not the others
The mirrored graph forgets the minus sign. The equal-heights graph ignores the slower fall. The trapezoid copies I₁.
Shortcut
Sign from −, height from slope.
Where it went wrong
Copying the primary current.
Q23Graph
Two coaxial coils are moved apart. Which graph shows their mutual inductance against separation?
(A)
(B)
(C)
(D)
Show the solution
Given
Separation increasing
Asked
M against d
Concept
Coupling weakens with distance.
Formula
K falls with d
Baby steps
Largest when close.
Falls off steeply, then slowly, towards zero.
Answer
(A) the graph in option A
Why not the others
Rising M gets the trend backwards. Constant M ignores distance. Rise-then-fall has no physical reason.
Shortcut
Always falling.
Where it went wrong
Choosing constant because L₁ and L₂ do not change.
Q24Graph
An alternating current of fixed peak value flows in the primary. Its frequency is varied. Which graph shows the peak emf in the secondary against frequency?
(A)
(B)
(C)
(D)
Show the solution
Given
I₀ fixed, f varies
Asked
e₂,peak against f
Concept
e₂,peak = MI₀(2πf).
Formula
e ∝ f
Baby steps
M and I₀ fixed.
Peak emf ∝ f: straight line through the origin.
Answer
(C) the graph in option C
Why not the others
Constant ignores that faster changes give bigger emfs. The hyperbola inverts the relation. The parabola squares f.
Shortcut
Double the frequency, double the emf.
Where it went wrong
Thinking only the current's size matters.
Q25Graph
A secondary coil at the same centre is turned so that the angle between its axis and the primary's axis goes from 0° to 90°. Which graph shows the mutual inductance against this angle?
(A)
(B)
(C)
(D)
Show the solution
Given
Angle between axes 0° → 90°
Asked
M against angle
Concept
Flux through a tilted coil ∝ cos of the angle between field and its axis.
Formula
M ∝ cos θ
Baby steps
0°: coaxial, maximum.
90°: side-on, zero.
Falls like a cosine.
Answer
(B) the graph in option B
Why not the others
The rising sine has maximum side-on. The flat line ignores orientation. The V shape rises again at 90°.
Shortcut
Same as φ = BA cos θ from Topic 01.
Where it went wrong
Measuring the angle from the plane instead of the axis.
Q26Assertion–reason
Assertion (A): The mutual inductance of two coils is the same whichever coil carries the changing current. Reason (R): Mutual inductance depends on the geometry of both coils and their relative position.
(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: reciprocity. R: M depends on geometry of both.
Asked
Truth and link
Concept
Reciprocity is a deeper result; depending on both coils does not by itself make the two values equal.
Formula
M₁₂ = M₂₁
Baby steps
A is true.
R is true.
Something can depend on both coils without being symmetric: R does not prove A.
Answer
(B) Both A and R are true, but R is not the correct explanation of A.
Why not the others
(A) claims R proves A; it does not. (C) and (D) need a false statement.
Shortcut
Test: does R alone force equality? No.
Where it went wrong
Accepting any related true R as the reason.
Q27Assertion–reason
Assertion (A): Placing a soft iron core inside two coaxial coils increases their mutual inductance. Reason (R): Mutual inductance is proportional to the relative permeability of the medium.
(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: iron increases M. R: M ∝ μᵣ.
Asked
Truth and link
Concept
M = μ₀μᵣN₁N₂A/ℓ.
Formula
M ∝ μᵣ
Baby steps
A is true.
R is true.
Iron has a large μᵣ, so 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
Formula explains the fact.
Where it went wrong
Choosing (B) by habit.
Q28Assertion–reason
Assertion (A): No emf is induced in the secondary while the primary carries a steady current. Reason (R): The mutual inductance becomes zero when the primary current is steady.
(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: e₂ = 0 for steady I₁. R: M = 0 then.
Asked
Truth and link
Concept
e₂ = 0 because dI₁/dt = 0; M does not change.
Formula
e₂ = −M dI₁/dt
Baby steps
A is true.
R is false: M depends on geometry, not on whether the current changes.
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
Zero emf comes from zero rate, not zero M.
Where it went wrong
Blaming M instead of dI/dt.
Q29Assertion–reason
Assertion (A): Two coils with their axes at right angles have almost zero mutual inductance. Reason (R): Almost none of the flux of one coil passes through the other.
(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: perpendicular → M ≈ 0. R: flux does not pass through.
Asked
Truth and link
Concept
M measures flux linkage per ampere.
Formula
M = N₂φ₂/I₁
Baby steps
A is true.
R is true.
No linked flux means no mutual inductance: 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
Linkage is what M counts.
Where it went wrong
Choosing (B) out of caution.
Q30Two statements
Statement I: The mutual inductance of a coil wound over a long solenoid is μ₀N₁N₂A/ℓ. Statement II: Mutual inductance does not depend on the medium between the coils.
(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; medium
Asked
Which are true
Concept
A core multiplies M by μᵣ.
Formula
M = μ₀μᵣN₁N₂A/ℓ
Baby steps
Statement I is true (air core).
Statement II is false.
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
Medium matters.
Where it went wrong
Forgetting μᵣ.
Q31Two statements
Statement I: The mutual inductance of two coils depends on the current in the primary coil. Statement II: The SI unit of mutual inductance is the henry.
(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
Dependence on current; unit
Asked
Which are true
Concept
M is fixed by construction.
Formula
—
Baby steps
Statement I is false.
Statement II is true.
Answer
(D) Statement I is false, but Statement II is true.
Why not the others
Both-true fails on I. Both-false fails on II. I-true-II-false reverses them.
Shortcut
M ignores current; unit henry.
Where it went wrong
Reading M = N₂φ₂/I₁ as dependence on I₁.
Answer key
1 C
2 A
3 D
4 C
5 B
6 A
7 B
8 B
9 D
10 D
11 A
12 A
13 D
14 A
15 C
16 B
17 D
18 B
19 D
20 C
21 C
22 B
23 A
24 C
25 B
26 B
27 A
28 C
29 A
30 C
31 D
Spread across letters: A 8, B 8, C 8, D 7. No letter repeats more than twice in a row. Question mix: Numerical 12, Concept 9, Graph 4, Assertion–reason 4, Two statements 2. Balancing seed 0.