One flux change, five consequences. This file sorts out which ones depend on how fast the change happens (emf, current, heat) and which one does not (charge).
The chain. Emf and current are 'right now' quantities: they vanish as soon as the flux stops changing. Charge and heat are totals: once delivered, they stay delivered.
Picture it
Think of a water tank connected to a pipe with a small water wheel in it. Tilting the tank (changing the flux) creates a push (emf). The push makes water flow (current). A bucket at the end collects water (charge). The wheel rubbing in the pipe gets warm (heat). Stop tilting and the push and the flow stop at once, but the water already in the bucket and the warmth already made do not disappear.
In exam language
Allen lists the induced parameters as emf e, current I, charge q, electric field Ein (Topic 11) and heat H. For a coil of N turns and total resistance R whose flux per turn changes by Δφ in time Δt:
average emf e = −N Δφ/Δt instantaneous e = −N dφ/dt
current I = e/R = −(N/R) dφ/dt
charge dq = I dt = −(N/R) dφ → q = N|Δφ|/R
heat H = ∫ I²R dt = ∫ (e²/R) dt
Charge does not care how fast; heat does
Slow versus fast. The charge is the area under the current–time graph. Halving the time doubles the current, so the area stays the same. Heat goes as I²t, so it doubles.
Look at the charge formula again. The time Δt cancels out:
q = I Δt = (N Δφ / R Δt) × Δt = N Δφ / R
So pulling a coil out of a field slowly or quickly sends the same charge round the circuit. Only the change in flux and the resistance matter. This is one of NEET's favourite ideas in this chapter.
If the same flux change happens in half the time
Factor
Emf e = NΔφ/Δt
× 2
Current I = e/R
× 2
Charge q = NΔφ/R
× 1 (unchanged)
Heat H = I²RΔt (uniform change)
× 2
In exam language
For a uniform change, H = (NΔφ)² / (R Δt). Faster change, more heat; the charge stays the same.
Trap
Allen writes dq = −dφ/R for a single loop. For a coil use q = NΔφ/R. Forgetting N is the usual slip. If a question gives the change in flux linkage directly, do not multiply by N again.
Four ways the flux can change
Four types. Flux φ = NBA cos θ can change through B, through the area (or its radius), or through the angle. Each gives its own formula for the emf.
Since φ = NBA cos θ, the flux can change in only a few ways. Hold the rest constant and differentiate only what moves:
Type
What changes
Emf
Example
1
Field B
e = −NA cos θ (dB/dt)
A magnet approaching; a current being switched
2
Area A
e = −NB cos θ (dA/dt)
A rod sliding on rails
3
Radius r of a circular loop
e = −NB cos θ (2πr dr/dt)
A ring shrinking or expanding
4
Angle θ
e = −NBA d(cos θ)/dt = NBAω sin ωt
A coil turning in a field (generator)
Allen Illustration 11. A loop in 0.04 T shrinks at 2 mm/s; at r = 2 cm, e = 2πrB(dr/dt) = 2π(0.02)(0.04)(0.002) = 3.2π × 10⁻⁶ V = 3.2π μV. Checked: correct.
Trap
In type 3 the emf depends on the radius at that moment. A ring shrinking at a steady rate gives an emf that gets smaller as the ring gets smaller.
Static, dynamic and periodic EMI (Allen p. 105)
Name
Held constant
Changing
Where it happens
Static EMI
A and θ
B (usually because a current changes)
Coils at rest: self-induction and mutual induction
Dynamic EMI
B and θ
Area
A straight wire or rod moving through a field
Periodic EMI
A and B
Angle θ
A rotating coil
Formula sheet
Formula
Meaning
When to use
Watch out
eavg = −N Δφ/Δt
Average emf
Flux given at two instants
Δφ = final − initial
e = −N dφ/dt
Instantaneous emf
φ given as a function of t
Differentiate first
I = e/R
Induced current
Closed circuit of resistance R
R is the whole circuit's resistance
q = N|Δφ|/R
Induced charge
Total charge for a flux change
Independent of time
q = area under I–t
Charge from a graph
Current pulse given
Then Δφ = qR/N
H = ∫ e²/R dt
Heat
Any change
Depends on how fast
H = (NΔφ)²/(RΔt)
Heat for a uniform change
Steady rate of change
Halve Δt → double H
e = −NA cos θ dB/dt
Type 1: B changes
Field ramps, B = B₀ sin ωt
Fixed area and angle
e = −NB cos θ dA/dt
Type 2: area changes
Sliding rods, stretching loops
Topic 09 develops this
e = NB(2πr)(dr/dt)
Type 3: radius changes
Shrinking or growing rings
Use r at that instant
e = NBAω sin ωt
Type 4: angle changes
Rotating coils
Maximum NBAω (Topic 12)
Allen pages 103–105 checked
Item
Check
Average and instantaneous emf, I = e/R
Correct
dq = −dφ/R
Correct for one loop; a coil of N turns needs N dφ
H = ∫ I²R dt = ∫ (e²/R) dt
Correct
Types 1–4 formulas
Correct
Illustration 11: 3.2π μV
Correct
Illustration 12: 1180 J into a 1 kg ring, s = 236 J/kg°C, Δθ = 5°C
Correct (covered in Topic 03)
Static, dynamic and periodic EMI table
Correct
Beginner's Box 3 answer key
Q
Answer
Working
1
(4) None
Emf needs only a changing flux. Copper, wood and iron coils all get an emf; only the current depends on the material.
2
(2) 4 mV
Approaching: relative speed 5 + 3 = 8 m/s gives 16 mV. Same direction: 5 − 3 = 2 m/s, so 16 × 2/8 = 4 mV.
3
(4) 2.0 V
dφ/dt = 10t − 4 = −2 at t = 0.2 s, so e = −dφ/dt = +2 V. (The current would be 0.2 A.)
4
(3)
Rising flux gives a negative emf, the flat part gives zero, the falling part gives a positive emf: negative block, gap, positive block.
e = A dB/dt = (0.04)² × 0.4 = 6.4 × 10⁻⁴ V; I = 6.4 × 10⁻⁴ / 2 × 10⁻³ = 0.32 A.
7
(1) 4.0 m
The ring opposes the fall, so a < g and the distance is less than ½g(1)² ≈ 4.9 m. Only 4.0 m is smaller.
8
(2) 4.0 C
q = Δφ/R = (10 − 2)/2 = 4 C.
9
(2) 0.02 V/m
E(2πr) = πr² dB/dt, so E = (r/2)(dB/dt) = 0.01 × 2 = 0.02 V/m. This uses the induced electric field (Topic 11).
NEET practice: 33 questions
Most NEET numericals in this chapter are one of these shapes. Every number here was recalculated in Python.
Q1Numerical
A coil of 100 turns and area 0.01 m² lies perpendicular to a field that rises uniformly from 0.2 T to 0.8 T in 0.3 s. The average emf induced is
(A)2 V
(B)0.02 V
(C)2.67 V
(D)0.67 V
Show the solution
Given
N = 100, A = 0.01 m², ΔB = 0.6 T, Δt = 0.3 s, θ = 0°
Asked
eavg
Concept
Type 1: only B changes.
Formula
|e| = NA ΔB/Δt
Baby steps
ΔB/Δt = 0.6/0.3 = 2 T/s.
e = 100 × 0.01 × 2 = 2 V.
Answer
(A) 2 V
Why not the others
0.02 V forgets N. 2.67 V uses the final field 0.8 T. 0.67 V uses the initial field 0.2 T as if it were the change.
Shortcut
NA × (ΔB/Δt).
Where it went wrong
Using a field value instead of the change in field.
Q2Numerical
The flux through a closed circuit of resistance 5 Ω varies as φ = (4t² − 6t + 2) Wb. The induced current at t = 0.5 s is
(A)2 A
(B)0.4 A
(C)0 A
(D)0.8 A
Show the solution
Given
φ = 4t² − 6t + 2, R = 5 Ω, t = 0.5 s
Asked
I
Concept
Instantaneous emf from the derivative, then Ohm's law.
Formula
e = −dφ/dt, I = e/R
Baby steps
dφ/dt = 8t − 6 = 4 − 6 = −2 Wb/s.
e = +2 V.
I = 2/5 = 0.4 A.
Answer
(B) 0.4 A
Why not the others
2 A is the emf in volts given as the current. 0 A puts t = 0.5 into φ (which is 0). 0.8 A uses 4 V.
Shortcut
Differentiate, substitute, divide by R.
Where it went wrong
Substituting into φ instead of dφ/dt.
Q3Numerical
The flux per turn through a 50-turn coil changes by 4 mWb. The total resistance of the circuit is 10 Ω. The charge that flows is
(A)4 × 10⁻⁴ C
(B)0.02 C
(C)0.2 C
(D)2 C
Show the solution
Given
N = 50, Δφ = 4 × 10⁻³ Wb per turn, R = 10 Ω
Asked
q
Concept
Induced charge depends only on the change in flux linkage and the resistance.
Formula
q = NΔφ/R
Baby steps
Change in linkage = 50 × 4 × 10⁻³ = 0.2 Wb.
q = 0.2/10 = 0.02 C.
Answer
(B) 0.02 C
Why not the others
4 × 10⁻⁴ C forgets N. 0.2 C forgets to divide by R. 2 C multiplies by R.
Shortcut
Linkage change ÷ R.
Where it went wrong
Asking for a time that the question never gives.
Q4Numerical
A 200-turn coil of area 0.05 m² lies perpendicular to a 0.1 T field. The circuit resistance is 20 Ω. The coil is pulled completely out of the field. The charge that flows is
(A)1 C
(B)0.1 C
(C)0.05 C
(D)2.5 × 10⁻⁴ C
Show the solution
Given
N = 200, A = 0.05 m², B = 0.1 T, R = 20 Ω; flux goes from BA to 0
Asked
q
Concept
Removing the coil changes the flux from BA to zero.
Formula
q = NBA/R
Baby steps
NBA = 200 × 0.1 × 0.05 = 1 Wb.
q = 1/20 = 0.05 C.
Answer
(C) 0.05 C
Why not the others
0.1 C doubles the change (that would be a half turn). 1 C forgets R. 2.5 × 10⁻⁴ C forgets N.
Shortcut
Out of the field: Δ(linkage) = NBA.
Where it went wrong
Asking 'how fast was it pulled?' The speed does not matter.
Q5Numerical
A 100-turn coil of area 0.02 m² lies perpendicular to a 0.5 T field and is turned through 180° about a diameter. The circuit resistance is 25 Ω. The charge that flows is
(A)0.08 C
(B)0.04 C
(C)zero
(D)2 C
Show the solution
Given
N = 100, A = 0.02 m², B = 0.5 T, R = 25 Ω, half turn
Asked
q
Concept
A half turn changes the linkage by 2NBA.
Formula
q = 2NBA/R
Baby steps
NBA = 100 × 0.5 × 0.02 = 1 Wb.
Change = 2 Wb.
q = 2/25 = 0.08 C.
Answer
(A) 0.08 C
Why not the others
0.04 C uses NBA (a quarter turn). Zero treats the end flux as equal to the start. 2 C forgets R.
Shortcut
Half turn → 2NBA/R.
Where it went wrong
Forgetting that the flux changes sign.
Q6Concept
A coil is pulled out of a magnetic field once slowly and once quickly. Compared with the slow pull, the quick pull gives
(A)the same charge but more heat
(B)more charge and more heat
(C)the same charge and the same heat
(D)less charge but more heat
Show the solution
Given
Same flux change, less time
Asked
Charge and heat
Concept
q = NΔφ/R has no time in it; heat H = (NΔφ)²/(RΔt) does.
Formula
q = NΔφ/R, H ∝ 1/Δt
Baby steps
Charge: same Δφ and R → same q.
Heat: smaller Δt → larger H.
Answer
(A) the same charge but more heat
Why not the others
More charge assumes q depends on speed. Same heat ignores the larger current. Less charge has no basis.
Shortcut
Charge: speed-proof. Heat: speed-sensitive.
Where it went wrong
Treating charge like emf.
Q7Numerical
A uniform flux-linkage change of 0.2 Wb happens in 0.1 s in a circuit of resistance 2 Ω. The heat produced is
(A)0.4 J
(B)0.2 J
(C)0.1 J
(D)2 J
Show the solution
Given
Δ(Nφ) = 0.2 Wb, Δt = 0.1 s, R = 2 Ω, uniform rate
Asked
H
Concept
Constant emf during the change, so H = I²RΔt.
Formula
H = (NΔφ)²/(RΔt)
Baby steps
e = 0.2/0.1 = 2 V.
I = 2/2 = 1 A.
H = 1² × 2 × 0.1 = 0.2 J.
Answer
(B) 0.2 J
Why not the others
0.4 J is the answer if the change took 0.05 s. 0.1 J halves it. 2 J forgets the time 0.1 s.
Shortcut
(0.2)²/(2 × 0.1) = 0.2 J.
Where it went wrong
Leaving out Δt in I²Rt.
Q8Concept
The same flux change is made to happen uniformly in half the time. The heat produced in the circuit becomes
(A)unchanged
(B)four times as large
(C)half as large
(D)twice as large
Show the solution
Given
Δt → Δt/2, same Δφ and R
Asked
New H / old H
Concept
H = (NΔφ)²/(RΔt) for a uniform change.
Formula
H ∝ 1/Δt
Baby steps
Current doubles, so I² is 4 times.
Time halves.
4 × ½ = 2.
Answer
(D) twice as large
Why not the others
Four times forgets that the time is halved. Half reverses the rule. Unchanged is the charge, not the heat.
Shortcut
I² × t: 4 × ½.
Where it went wrong
Squaring the time factor as well.
Q9Numerical
A loop lies perpendicular to a uniform 0.5 T field. Its area decreases steadily at 0.02 m² s⁻¹. The induced emf is
(A)0.025 V
(B)0.04 V
(C)0.01 V
(D)0.1 V
Show the solution
Given
B = 0.5 T, dA/dt = −0.02 m² s⁻¹
Asked
|e|
Concept
Type 2: area changes.
Formula
|e| = B dA/dt
Baby steps
|e| = 0.5 × 0.02 = 0.01 V.
Answer
(C) 0.01 V
Why not the others
0.04 V divides 0.02 by 0.5. 0.025 V divides 0.5 by 20. 0.1 V multiplies by 10 somewhere.
Shortcut
B × (rate of area change).
Where it went wrong
Dividing when the formula multiplies.
Q10Numerical
A circular loop is perpendicular to a 0.05 T field. Its radius shrinks at 1 mm/s. When the radius is 5 cm, the induced emf is
(A)10π μV
(B)2.5π μV
(C)5π μV
(D)0.5π μV
Show the solution
Given
B = 0.05 T, dr/dt = 1 × 10⁻³ m/s, r = 0.05 m
Asked
|e|
Concept
Type 3: radius changes (Allen Illustration 11).
Formula
|e| = B(2πr)(dr/dt)
Baby steps
2πr = 2π × 0.05 = 0.1π m.
e = 0.05 × 0.1π × 10⁻³ = 5π × 10⁻⁶ V = 5π μV.
Answer
(C) 5π μV
Why not the others
2.5π μV forgets the 2 in 2πr. 10π μV doubles the correct value. 0.5π μV loses a factor of 10.
Shortcut
d(πr²)/dt = 2πr dr/dt.
Where it went wrong
Writing d(πr²) = πr dr (dropping the 2).
Q11Concept
A circular loop in a steady perpendicular field shrinks so that its radius falls at a constant rate. As the loop gets smaller, the induced emf
(A)decreases in proportion to the area
(B)stays constant
(C)increases
(D)decreases in proportion to the radius
Show the solution
Given
dr/dt constant, B constant
Asked
How e changes
Concept
e = B(2πr)(dr/dt): with dr/dt fixed, e ∝ r.
Formula
e ∝ r
Baby steps
B and dr/dt are constants.
So e is proportional to r.
Smaller r → smaller e.
Answer
(D) decreases in proportion to the radius
Why not the others
Constant emf would need dA/dt constant, not dr/dt. It cannot increase. e ∝ r, not r².
Shortcut
Steady dr/dt → e ∝ r.
Where it went wrong
Assuming a steady rate always means a steady emf.
Q12Numerical
A coil of 10 turns and area 0.01 m² is fixed perpendicular to a field B = 0.2 sin(100t) T. The maximum emf induced is
(A)20 V
(B)0.2 V
(C)0.02 V
(D)2 V
Show the solution
Given
N = 10, A = 0.01 m², B₀ = 0.2 T, ω = 100 rad/s
Asked
emax
Concept
Type 1 with a sinusoidal field.
Formula
e = −NA dB/dt = −NAB₀ω cos ωt
Baby steps
NAB₀ω = 10 × 0.01 × 0.2 × 100.
= 2 V.
Answer
(D) 2 V
Why not the others
0.2 V and 20 V are a power of ten out. 0.02 V forgets the ω that differentiation brings out.
Shortcut
Maximum = NAB₀ω.
Where it went wrong
Forgetting the ω that differentiation brings out.
Q13Numerical
A single loop of area 0.1 m² is perpendicular to a field B = 0.02t² T. The emf at t = 5 s is
(A)0.02 V
(B)0.05 V
(C)0.01 V
(D)0.1 V
Show the solution
Given
A = 0.1 m², B = 0.02t², t = 5 s
Asked
|e|
Concept
Differentiate B, then multiply by A.
Formula
|e| = A dB/dt
Baby steps
dB/dt = 0.04t = 0.2 T/s at t = 5 s.
e = 0.1 × 0.2 = 0.02 V.
Answer
(A) 0.02 V
Why not the others
0.05 V uses B itself (0.5 T) × A. 0.01 V drops the factor 2 from differentiating t². 0.1 V forgets A.
Shortcut
d(t²)/dt = 2t.
Where it went wrong
Using B instead of dB/dt.
Q14Numerical
A 20-turn square coil of side 10 cm lies perpendicular to a field that decreases at 0.5 T s⁻¹. The coil's resistance is 2 Ω. The induced current is
(A)0.05 A
(B)0.1 A
(C)5 A
(D)0.005 A
Show the solution
Given
N = 20, side 0.1 m, dB/dt = 0.5 T/s, R = 2 Ω
Asked
I
Concept
Type 1, then Ohm's law.
Formula
I = NA(dB/dt)/R
Baby steps
A = 0.01 m².
e = 20 × 0.01 × 0.5 = 0.1 V.
I = 0.1/2 = 0.05 A.
Answer
(A) 0.05 A
Why not the others
0.1 A is the emf taken as current. 5 A squares 10 without converting cm to m. 0.005 A drops a factor of 10.
Shortcut
e first, then ÷ R.
Where it went wrong
Not converting 10 cm to 0.1 m before squaring.
Q15Numerical
In the circuit of the previous question (e = 0.1 V, R = 2 Ω), the power dissipated as heat is
(A)2.5 × 10⁻³ W
(B)0.05 W
(C)0.2 W
(D)5 × 10⁻³ W
Show the solution
Given
e = 0.1 V, R = 2 Ω
Asked
P
Concept
Joule heating at a constant rate.
Formula
P = e²/R
Baby steps
e² = 0.01.
P = 0.01/2 = 5 × 10⁻³ W.
Answer
(D) 5 × 10⁻³ W
Why not the others
0.05 W is e/R (a current, not power). 0.2 W is eR. 2.5 × 10⁻³ W halves twice.
Shortcut
e²/R.
Where it went wrong
Using eR.
Q16Numerical
A 20 cm × 10 cm rectangular loop of resistance 0.02 Ω lies perpendicular to a field B = 0.4t T. The induced current is
(A)0.4 A
(B)8 × 10⁻³ A
(C)4 A
(D)0.04 A
Show the solution
Given
A = 0.02 m², dB/dt = 0.4 T/s, R = 0.02 Ω
Asked
I
Concept
Type 1.
Formula
I = A(dB/dt)/R
Baby steps
e = 0.02 × 0.4 = 8 × 10⁻³ V.
I = 8 × 10⁻³/0.02 = 0.4 A.
Answer
(A) 0.4 A
Why not the others
8 × 10⁻³ is the emf in volts. 4 A uses area 0.2 m². 0.04 A uses R = 0.2 Ω.
Shortcut
Area 200 cm² = 0.02 m².
Where it went wrong
Converting 20 cm × 10 cm to 2 m².
Q17Numerical
The flux linked with a coil of resistance 4 Ω changes from 2 Wb to 10 Wb. The charge that flows through the coil is
(A)3 C
(B)2 C
(C)0.5 C
(D)32 C
Show the solution
Given
Δ(linkage) = 8 Wb, R = 4 Ω
Asked
q
Concept
Charge does not need the time.
Formula
q = Δ(Nφ)/R
Baby steps
q = 8/4 = 2 C.
Answer
(B) 2 C
Why not the others
3 C uses 12 Wb (sum). 0.5 C inverts. 32 C multiplies.
Shortcut
Change in linkage over R.
Where it went wrong
Adding the two flux values.
Q18Numerical
A triangular current pulse, rising from 0 to 4 A and back to 0 in a total of 0.5 s, flows in a circuit of resistance 10 Ω because of a change in flux linkage. The size of that change is
(A)1 Wb
(B)20 Wb
(C)10 Wb
(D)5 Wb
Show the solution
Given
Triangle: base 0.5 s, height 4 A; R = 10 Ω
Asked
Δ(Nφ)
Concept
Charge = area under I–t; then Δ(Nφ) = qR.
Formula
q = ½ × base × height, Δ(Nφ) = qR
Baby steps
q = ½ × 0.5 × 4 = 1 C.
Δ(Nφ) = 1 × 10 = 10 Wb.
Answer
(C) 10 Wb
Why not the others
20 Wb uses the rectangle 0.5 × 4 (forgets ½). 1 Wb is the charge. 5 Wb halves again.
Shortcut
Area of the pulse × R.
Where it went wrong
Taking the area of a triangle as base × height.
Q19Concept
A coil is rewound from the same kind of wire with twice as many turns of the same radius, so its resistance doubles. For the same change in field, the charge that flows is
(A)doubled
(B)unchanged
(C)halved
(D)four times as large
Show the solution
Given
N → 2N, R → 2R, same B change and area
Asked
New q / old q
Concept
q = NΔφ/R.
Formula
q ∝ N/R
Baby steps
N doubles: q × 2.
R doubles: q × ½.
Net: unchanged.
Answer
(B) unchanged
Why not the others
Doubled ignores the extra resistance. Halved ignores the extra turns. Four times doubles twice.
Shortcut
Both N and R double: they cancel.
Where it went wrong
Changing only one of N and R.
Q20Concept
A flat loop of area A and resistance R is perpendicular to a field B. The field is reduced uniformly to zero in time t. The heat produced is
(A)BA²/Rt
(B)B²A²t/R
(C)BA/R
(D)B²A²/Rt
Show the solution
Given
Uniform change from BA to 0 in time t
Asked
H
Concept
Constant emf e = BA/t for time t.
Formula
H = (e²/R)t
Baby steps
e = BA/t.
H = (BA/t)² × t/R = B²A²/(Rt).
Answer
(D) B²A²/Rt
Why not the others
B²A²t/R puts t on top. BA/R is the charge. BA²/Rt has wrong dimensions.
Shortcut
Heat has 1/t; charge has no t.
Where it went wrong
Squaring e but forgetting it contains 1/t.
Q21Concept
A rectangular loop rotates at constant angular speed in a uniform magnetic field. This is an example of
(A)dynamic EMI
(B)static EMI
(C)periodic EMI
(D)self-induction
Show the solution
Given
Rotating loop, steady field
Asked
Classification
Concept
Periodic EMI: A and B fixed, θ changes.
Formula
e = NBAω sin ωt
Baby steps
B constant, A constant, θ = ωt changing.
That is periodic EMI.
Answer
(C) periodic EMI
Why not the others
Static EMI needs B to change with the coil at rest. Dynamic EMI is a straight conductor moving. Self-induction is a type of static EMI.
Shortcut
Rotation → periodic.
Where it went wrong
Calling any motion 'dynamic'.
Q22Concept
A rod slides along two parallel rails in a steady magnetic field, changing the area of the circuit. This is
(A)static EMI
(B)periodic EMI
(C)dynamic EMI
(D)mutual induction
Show the solution
Given
Sliding rod, steady B, fixed angle
Asked
Classification
Concept
Dynamic EMI: B and θ fixed, area changes.
Formula
e = B dA/dt
Baby steps
Only the area changes.
So it is dynamic EMI.
Answer
(C) dynamic EMI
Why not the others
Periodic needs rotation. Static and mutual induction need B to change with conductors at rest.
Shortcut
Moving straight conductor → dynamic.
Where it went wrong
Mixing up the three names.
Q23Concept
Two coils are at rest and the current in one of them is changing. The emf induced in the other is an example of
(A)dynamic EMI
(B)static EMI (mutual induction)
(C)periodic EMI
(D)motional emf
Show the solution
Given
Both coils at rest, changing current
Asked
Classification
Concept
Static EMI: A and θ fixed, B changes because a current changes.
Formula
dI/dt → dB/dt → dφ/dt
Baby steps
Nothing moves.
B changes because I changes.
Static EMI; between two coils it is mutual induction.
Answer
(B) static EMI (mutual induction)
Why not the others
Dynamic and motional both need moving conductors. Periodic needs rotation.
Shortcut
At rest + changing current → static.
Where it went wrong
Thinking EMI always needs motion.
Q24Concept
A short bar magnet is released from rest along the axis of a fixed horizontal metal ring (g = 9.8 m s⁻²). The distance it falls in the first second could be
(A)4.5 m
(B)4.9 m
(C)5.2 m
(D)9.8 m
Show the solution
Given
Magnet falling through a closed ring
Asked
Possible distance in 1 s
Concept
Lenz's law: the ring opposes the fall, so a < g.
Formula
s < ½g t²
Baby steps
Free fall would give ½ × 9.8 × 1 = 4.9 m.
The ring slows it, so s < 4.9 m.
Only 4.5 m fits.
Answer
(A) 4.5 m
Why not the others
4.9 m is free fall. 5.2 m and 9.8 m need a > g.
Shortcut
Any closed ring → less than free fall.
Where it went wrong
Picking free fall because 'the magnet only falls'.
Q25Graph
The flux through a coil varies as φ = φ₀ sin ωt. Which graph shows the induced emf (with its sign) against time?
(A)
(B)
(C)
(D)
Show the solution
Given
φ = φ₀ sin ωt
Asked
e against t
Concept
e = −dφ/dt.
Formula
e = −φ₀ω cos ωt
Baby steps
dφ/dt = φ₀ω cos ωt.
e = −φ₀ω cos ωt.
At t = 0 it starts at its most negative value.
Answer
(B) the graph in option B
Why not the others
The sine graph copies φ. The +cos graph forgets the minus sign. The −sin graph differentiates sin to sin.
Shortcut
Differentiate sin → cos, then flip.
Where it went wrong
Dropping the minus sign.
Q26Graph
A circular loop in a steady perpendicular field shrinks with its radius falling at a constant rate until it vanishes. Which graph shows the size of the induced emf against time?
(A)
(B)
(C)
(D)
Show the solution
Given
dr/dt constant; r falls linearly to zero
Asked
|e| against t
Concept
e = B(2πr)(dr/dt) ∝ r.
Formula
|e| ∝ r ∝ (r₀ − vt)
Baby steps
r falls in a straight line.
e ∝ r, so e also falls in a straight line.
It reaches zero when the loop vanishes.
Answer
(D) the graph in option D
Why not the others
The flat line needs constant dA/dt. The curved graph is the area πr², not the emf. The rising line gets the trend backwards.
Shortcut
e follows r, not r².
Where it went wrong
Drawing the area instead of the emf.
Q27Graph
The flux linked with a closed coil changes at a steady rate for a time T and then stays constant. Which graph shows the total charge that has flowed against time?
(A)
(B)
(C)
(D)
Show the solution
Given
Steady rate for time T, then no change
Asked
q against t
Concept
Steady rate → steady current → charge grows linearly; then no current.
Formula
q = It
Baby steps
While changing: I constant, q rises in a straight line.
After T: I = 0, q stays at its total NΔφ/R.
Answer
(B) the graph in option B
Why not the others
The straight line forever ignores that the change stops. The step shape is the current graph, not the charge. The parabola needs a growing current.
Shortcut
Charge is a running total: ramp, then flat.
Where it went wrong
Drawing the current instead of the charge.
Q28Graph
The same change in flux linkage is carried out uniformly in different times Δt. Which graph shows the heat produced against Δt?
(A)
(B)
(C)
(D)
Show the solution
Given
Fixed Δ(Nφ) and R; varying Δt
Asked
H against Δt
Concept
H = (NΔφ)²/(RΔt).
Formula
H ∝ 1/Δt
Baby steps
H is inversely proportional to Δt.
That is a hyperbola: large for quick changes, small for slow ones.
Answer
(D) the graph in option D
Why not the others
A rising line reverses the relationship. The flat line is the charge graph. A straight falling line would reach zero at a finite time.
Shortcut
Heat ∝ 1/time.
Where it went wrong
Choosing the flat line by confusing heat with charge.
Q29Assertion–reason
Assertion (A): The charge that flows through a coil when a magnet is pushed in does not depend on how quickly it is pushed. Reason (R): The induced charge is q = NΔφ/R.
(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: charge independent of speed. R: q = NΔφ/R.
Asked
Truth and link
Concept
Time cancels in q = IΔt.
Formula
q = NΔφ/R
Baby steps
A is true.
R is true.
R has no time in it, which is exactly why A holds.
Answer
(A) Both A and R are true, and R is the correct explanation of A.
Why not the others
(B) denies the direct link. (C) and (D) need a false statement.
Shortcut
No t in the formula → no dependence on speed.
Where it went wrong
Picking (B) out of habit.
Q30Assertion–reason
Assertion (A): More heat is produced in a coil when a magnet is pushed in quickly than when it is pushed in slowly. Reason (R): More charge flows when the magnet is pushed in quickly.
(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: faster → more heat. R: faster → more charge.
Asked
Truth and link
Concept
Heat depends on rate; charge does not.
Formula
H ∝ 1/Δt; q = NΔφ/R
Baby steps
A is true.
R is false: the charge is the same.
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
Heat: yes. Charge: no.
Where it went wrong
Assuming heat and charge rise together.
Q31Assertion–reason
Assertion (A): A coil rotating in a steady magnetic field is an example of periodic EMI. Reason (R): In periodic EMI the field and the area stay constant while the angle between them changes.
(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: rotating coil is periodic. R: definition of periodic EMI.
Asked
Truth and link
Concept
Classification by what changes.
Formula
e = NBA d(cos θ)/dt
Baby steps
A is true.
R is true.
R is the definition that makes A true.
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
Definition explains the example.
Where it went wrong
Choosing (B) when R is a definition.
Q32Assertion–reason
Assertion (A): For a circular loop whose radius shrinks at a steady rate in a steady field, the induced emf stays constant. Reason (R): The emf is e = B(2πr)(dr/dt).
(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: constant emf. R: e = 2πrB dr/dt.
Asked
Truth and link
Concept
With dr/dt constant, e ∝ r.
Formula
e ∝ r
Baby steps
R is true.
By R, e falls as r falls, 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) all need A true.
Shortcut
Read R carefully: it contains r.
Where it went wrong
Assuming 'steady rate' means 'steady emf'.
Q33Two statements
Statement I: The induced current in a circuit depends on the circuit's resistance. Statement II: The induced emf in a circuit depends on the circuit's resistance.
(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
Current versus emf
Asked
Which are true
Concept
e = −N dφ/dt has no R; I = e/R does.
Formula
I = e/R
Baby steps
Statement I is true.
Statement II is false.
Answer
(C) Statement I is true, but Statement II is false.
Why not the others
Both-true would put R into Faraday's law. Both-false rejects Ohm's law. I-false-II-true reverses them.
Shortcut
Emf: flux only. Current: flux and R.
Where it went wrong
Treating emf and current alike.
Answer key
1 A
2 B
3 B
4 C
5 A
6 A
7 B
8 D
9 C
10 C
11 D
12 D
13 A
14 A
15 D
16 A
17 B
18 C
19 B
20 D
21 C
22 C
23 B
24 A
25 B
26 D
27 B
28 D
29 A
30 C
31 A
32 D
33 C
Spread across letters: A 9, B 8, C 8, D 8. No letter repeats more than twice in a row. Question mix: Numerical 15, Concept 9, Graph 4, Assertion–reason 4, Two statements 1. Balancing seed 0.