A magnet sitting beside a coil does nothing. Move it and a current appears. This file follows Faraday's experiments to the rule behind them: emf equals the rate at which flux changes.
In 1820 Hans Christian Oersted noticed that a wire carrying current made a compass needle swing. So electricity can make magnetism. Michael Faraday, in England, asked the obvious next question: can magnetism make electricity?
For years the answer seemed to be no. A magnet sitting next to a wire did nothing. In 1831 Faraday found the missing ingredient: change. A magnet makes a current only while something about it is changing. Joseph Henry in the USA found the same thing independently at about the same time, which is why NCERT calls these the experiments of Faraday and Henry.
In exam language
Electromagnetic induction is the production of an emf in a conductor when the magnetic flux linked with it changes. The discovery was the reverse of Oersted's experiment, and it is the working principle of every generator and transformer.
Trap
Oersted: current produces a magnetic field. Faraday: a changing magnetic flux produces an emf. Swapping these two names is an attribution error, which is one of the recurring mistake types in her papers.
Experiment 1: a magnet and a coil
Magnet and coil. Watch the needle. It moves only while the magnet moves, one way for coming in and the other way for going out. Stopping the magnet, even inside the coil, sends the needle back to zero.
A coil is connected to a galvanometer, a meter that shows a tiny current by swinging its needle. Then:
What you do
What the needle does
Margin shorthand on p. 99
Hold the magnet still, anywhere
Nothing
—
Push the N pole towards the coil
Swings right
N → R
Pull the N pole away
Swings left
N ← L
Push the S pole towards the coil
Swings left
S → L
Pull the S pole away
Swings right
S ← R
The pattern: changing the pole reverses the swing, and changing the direction of motion reverses the swing. Change both and you are back to the original swing, which is why "N pole away" and "S pole towards" give the same result.
The margin shorthand on page 99 matches the book exactly. The "right" and "left" belong to this particular setup. What carries over to every question is the pattern of reversals, and Topic 03 explains why using Lenz's law.
Faster motion, bigger swing
Two speeds. Both magnets travel the same distance. The faster one covers it in half the time, so the flux changes twice as fast and the needle swings twice as far.
Picture it
Think of filling a bucket. Pouring a litre in one second makes a big splash. Pouring the same litre over ten seconds is a gentle trickle. The total water is the same. What changes is how fast it arrives. Emf measures how fast flux arrives, not how much.
Allen writes this as "induced emf ∝ relative velocity". That is true when you compare the same magnet at the same position moving at different speeds. The deeper rule is Faraday's law: emf depends on the rate of change of flux.
Experiments 2 and 3: using a second coil
Experiment 2. Replace the magnet with a second coil connected to a battery. A coil carrying current behaves like a magnet. Moving it towards or away from the first coil deflects the galvanometer, just as the magnet did.
Experiment 3. Now keep both coils completely still, and put a tapping key in the battery circuit.
Switching on and off. The galvanometer kicks only at the moment the key is closed or opened. While a steady current flows, its flux is steady, so there is no emf in the second coil.
Nothing moves in Experiment 3, yet there is a current in the second coil. So motion is not the real cause. The real cause is that the current in the first coil is changing, and so the flux through the second coil is changing. Pushing a soft iron rod into both coils makes the kicks much larger, because the iron strengthens the field and so strengthens the change in flux.
Trap
A steady current in the primary gives no deflection, however large the current. Only switching on, switching off, or changing the current does something.
What all the experiments share
In exam language
Whenever the magnetic flux through a coil changes with time, an emf is induced in it. The emf lasts only as long as the flux is changing. If the coil is part of a closed circuit, the emf drives a current, and the galvanometer shows a deflection.
Every experiment is a different way of changing flux. Moving a magnet changes B at the coil. Switching a current changes B. Inserting iron changes B. Later topics add two more ways: changing the area of the loop, and turning the loop to change the angle.
Faraday's laws
First law (what happens)
Whenever the magnetic flux linked with a circuit changes, an emf is induced in it.
Second law (how much)
The size of the induced emf equals the rate of change of flux linkage.
|e| = dφ/dt for N turns: |e| = N dφ/dt
average emf: e = −N Δφ/Δt
The minus sign comes from Lenz's law, which fixes the direction. That is Topic 03. For now, whenever a question asks only for the size of the emf, you can ignore the sign.
Three rings. The same changing flux passes through each. Every ring gets an emf. Only the metal ring has free electrons that can move, so only it carries a current.
Emf induced?
Current flows?
Closed metal ring
Yes
Yes, I = e/R
Wooden or plastic ring
Yes
No, no free charges
Metal ring with a cut in it
Yes, across the cut
No, the circuit is open
Trap
"Emf is induced irrespective of the coil's material" (Allen p. 100). NEET turns this into "for which coil is no emf induced: copper, wood, iron, or none?" The answer is none. Emf needs changing flux, not a conductor. Current needs a closed conductor.
Reading emf from a flux–time graph
Slope becomes emf. As the dot moves along the flux graph, the lower dot shows the emf. A rising line gives a negative emf, a flat line gives zero, a falling line gives a positive emf. A steeper line gives a bigger emf.
Because e = −dφ/dt, the emf at any moment is minus the slope of the φ–t graph at that moment. Three rules settle every graph question:
Flux graph
Emf
Flat (constant flux, however large)
Zero
Straight sloping line
Constant emf; size = size of slope
Steeper line
Bigger emf
Rising line
Negative emf (with the minus sign)
Curve such as φ = kt²
Emf changes; e = −2kt
Trap
A large flux does not mean a large emf. At the top of the flux graph (the flat part) the emf is zero. What matters is how steep the graph is.
Only relative motion matters
Moving the magnet towards a still coil and moving the coil towards a still magnet at the same speed give the same emf. If both move in the same direction at the same speed, nothing changes between them, so there is no emf at all.
Allen Beginner's Box 3 Q2 uses this: a coil and magnet approach at 5 m/s and 3 m/s (relative speed 8 m/s) and give 16 mV. Moving in the same direction, the relative speed is 5 − 3 = 2 m/s, a quarter as much, so the emf is 4 mV.
Formula sheet
Formula
Meaning
When to use
Watch out
|e| = dφ/dt
Instantaneous emf, one turn
φ given as a function of t
Differentiate before substituting t
|e| = N dφ/dt
Emf in N turns
Coils
N may already be inside a given linkage
e = −N Δφ/Δt
Average emf
Flux changes over an interval
Δφ = final − initial
e = −(slope of φ–t)
Emf from a graph
Graph questions
Flat part gives zero
I = e/R
Induced current
Closed conducting coil
No current in an open or non-conducting loop
e ∝ vrelative
Same geometry, different speeds
Magnet and coil both moving
Same direction: subtract speeds
Allen pages 98–100 checked
Statement on the page
Check
Faraday demonstrated the reverse effect of Oersted's experiment
Correct
N pole towards → right; away → left; S pole reverses both
Correct for the drawn setup
Deflection is more when the magnet moves faster
Correct
Margin notes N→R, N←L, S→L, S←R
Match the book
Margin note: deflection ∝ velocity of incoming magnet
Correct at a given position; the exact rule is emf ∝ rate of change of flux
First law: emf is induced when a conductor is in a varying field
Fine for exams; strictly the flux linked must change, and no conductor is needed for the emf itself
Second law: |e| = dφ/dt
Correct (multiply by N for a coil)
Emf induced irrespective of material; current only if conducting
Correct
The four handwritten answers under Illustration 9(a) on page 100 (C, AC, ACW, C) are checked in Topic 03. All four are right.
NEET practice: 32 questions
This topic is mostly ideas rather than calculation, so it has fewer questions than Topic 01. Every number was recalculated in Python.
Q1Concept
A bar magnet is held at rest inside a coil connected to a galvanometer. The galvanometer shows
(A)a steady deflection to the left
(B)a steady deflection to the right
(C)no deflection
(D)a deflection that grows slowly
Show the solution
Given
Magnet at rest inside the coil
Asked
Galvanometer reading
Concept
Emf is induced only while flux is changing.
Formula
e = −N dφ/dt
Baby steps
Flux through the coil is large but constant.
dφ/dt = 0.
e = 0, so no current and no deflection.
Answer
(C) no deflection
Why not the others
A steady deflection would need a steady rate of change, but nothing changes. A growing deflection would need an increasing rate.
Shortcut
Still magnet → zero, wherever it is.
Where it went wrong
Thinking large flux means large emf.
Q2Concept
In a coil–galvanometer experiment, pushing the N pole of a magnet towards the coil makes the needle swing right. Which of these also makes it swing right?
(A)Pulling the S pole away from the coil
(B)Pulling the N pole away from the coil
(C)Pushing the S pole towards the coil
(D)Holding the N pole still inside the coil
Show the solution
Given
N pole towards → right
Asked
Another action giving a right swing
Concept
Reversing the pole reverses the swing; reversing the motion also reverses it.
Formula
Two reversals cancel
Baby steps
S pole instead of N: one reversal.
Away instead of towards: a second reversal.
Two reversals → same swing as the original: right.
Answer
(A) Pulling the S pole away from the coil
Why not the others
N pole away and S pole towards each have one reversal: left. A still magnet gives no swing.
Shortcut
Count reversals: even number → same direction.
Where it went wrong
Changing only one thing and expecting the same swing.
Q3Concept
In the same experiment (N pole towards → right), the S pole of the magnet is pushed towards the coil. The needle
(A)swings right, then left
(B)swings right
(C)does not move
(D)swings left
Show the solution
Given
N towards → right; now S towards
Asked
Direction of swing
Concept
Changing the pole reverses the induced current.
Formula
One reversal
Baby steps
Motion is the same (towards).
Pole is reversed.
One reversal → left.
Answer
(D) swings left
Why not the others
Right would need no reversal or two. No movement would need no change of flux. Right-then-left describes a magnet passing through.
Shortcut
S → L (margin shorthand).
Where it went wrong
Forgetting that the pole matters.
Q4Concept
A magnet is pushed into a coil first slowly and then quickly, through the same distance. Compared with the slow push, the quick push gives
(A)a smaller deflection
(B)the same deflection for a shorter time
(C)a larger deflection for a shorter time
(D)the same deflection for the same time
Show the solution
Given
Same distance, less time
Asked
Size and duration of deflection
Concept
Emf depends on the rate of change of flux.
Formula
|e| = N Δφ/Δt
Baby steps
Total Δφ is the same both times.
Δt is smaller for the quick push.
So e (and the current) is larger, but lasts for less time.
Answer
(C) a larger deflection for a shorter time
Why not the others
Same deflection ignores the rate. Smaller deflection reverses the rule. Same time is impossible if the push is quicker.
Shortcut
Faster → bigger and briefer.
Where it went wrong
Thinking only the total change matters.
Q5Concept
A coil and a magnet move in the same direction with the same velocity. The emf induced in the coil is
(A)zero
(B)maximum
(C)proportional to their common speed
(D)equal to that when only the magnet moves at this speed
Show the solution
Given
Same velocity for both
Asked
Induced emf
Concept
Only relative motion changes the flux through the coil.
Formula
vrelative = 0
Baby steps
Their separation never changes.
The flux through the coil stays constant.
e = 0.
Answer
(A) zero
Why not the others
Each other option assumes some change in flux, but the relative speed is zero.
Shortcut
No relative motion, no emf.
Where it went wrong
Adding the two speeds.
Q6Concept
A wooden ring is placed where the magnetic flux through it is changing. Which is correct?
(A)An emf is induced, but no current flows
(B)Neither emf nor current is induced
(C)Both emf and current are induced
(D)A current flows, but there is no emf
Show the solution
Given
Non-conducting ring, changing flux
Asked
Emf and current
Concept
Emf depends on changing flux; current needs free charges to move.
Formula
e = −dφ/dt; I = e/R with R → ∞
Baby steps
The flux changes, so e ≠ 0.
Wood has no free charges, so R is effectively infinite.
I = e/R = 0.
Answer
(A) An emf is induced, but no current flows
Why not the others
'Neither' ignores Faraday's law. 'Both' needs a conductor. A current with no emf is impossible.
Shortcut
Emf: needs changing flux. Current: also needs a closed conductor.
Where it went wrong
Believing emf needs a metal.
Q7Concept
Electromagnetic induction was discovered by
(A)Hans Christian Oersted
(B)Michael Faraday, and independently by Joseph Henry
(C)André-Marie Ampère
(D)Heinrich Lenz
Show the solution
Given
History of EMI
Asked
Discoverer
Concept
Faraday (1831) and Henry found induced currents; Oersted found the opposite effect.
Formula
—
Baby steps
Oersted (1820): current → magnetic field.
Faraday and Henry (early 1830s): changing flux → emf.
Lenz later gave the rule for direction.
Answer
(B) Michael Faraday, and independently by Joseph Henry
Why not the others
Oersted showed current makes a field. Ampère worked out forces between currents. Lenz added the direction rule.
Shortcut
Faraday = induction. Lenz = direction.
Where it went wrong
Attribution swap between Oersted and Faraday.
Q8Concept
Oersted's experiment showed that
(A)a changing magnetic flux produces an emf
(B)an electric current produces a magnetic field
(C)the induced current opposes the change that causes it
(D)magnetic monopoles do not exist
Show the solution
Given
Oersted, 1820
Asked
What it showed
Concept
A current-carrying wire deflects a nearby compass.
Formula
—
Baby steps
Current in a wire → compass needle swings.
So a current produces a magnetic field.
Answer
(B) an electric current produces a magnetic field
Why not the others
Changing flux → emf is Faraday's result. Opposition to change is Lenz's law. No monopoles is Gauss's law for magnetism.
Shortcut
Oersted: electricity to magnetism. Faraday: magnetism back to electricity.
Where it went wrong
Reversing the two discoveries.
Q9Concept
Two coils are placed side by side and neither moves. A key in the battery circuit of the first coil is pressed and held down. The galvanometer in the second coil
(A)kicks only when the key is held for a long time
(B)shows a steady deflection while the key is held
(C)shows no deflection at any time
(D)kicks briefly when the key is pressed, then returns to zero
Show the solution
Given
Two stationary coils; key pressed and held
Asked
Galvanometer behaviour
Concept
Flux through coil 2 changes only while the current in coil 1 is changing.
Formula
e₂ ∝ dI₁/dt
Baby steps
Pressing the key: I₁ rises from 0 → flux in coil 2 changes → kick.
Holding the key: I₁ steady → no change → zero.
So: a brief kick, then zero.
Answer
(D) kicks briefly when the key is pressed, then returns to zero
Why not the others
A steady deflection needs a steady rate of change. 'No deflection' ignores the switching moment. Time held has no effect.
Shortcut
Kicks at switching moments only.
Where it went wrong
Expecting a reading for as long as current flows.
Q10Concept
In the two-coil experiment, the key is released after being held down. Compared with the kick when it was pressed, the galvanometer now
(A)kicks both ways equally
(B)kicks the same way
(C)does not kick
(D)kicks the other way
Show the solution
Given
Key released after steady current
Asked
Direction of kick
Concept
Pressing increases flux; releasing decreases it. Opposite changes give opposite emfs.
Formula
sign of e follows sign of −dφ/dt
Baby steps
Pressing: flux in coil 2 increases.
Releasing: flux in coil 2 decreases.
Opposite changes → opposite kicks.
Answer
(D) kicks the other way
Why not the others
The same way would need the same kind of change. No kick ignores the fall of current. 'Both ways' is not a single change.
Shortcut
On and off give mirror-image kicks.
Where it went wrong
Assuming every kick goes the same way.
Q11Concept
In the two-coil experiment, a soft iron rod is pushed through both coils. The kicks when the key is pressed become
(A)much larger
(B)smaller
(C)zero
(D)unchanged
Show the solution
Given
Iron rod inserted along the common axis
Asked
Effect on induced emf
Concept
Iron greatly increases the magnetic field for the same current.
Formula
e ∝ dφ/dt, φ ∝ μ
Baby steps
The same switching now changes a much larger flux.
A larger change in the same time gives a larger emf.
Answer
(A) much larger
Why not the others
Iron does not weaken or block the field. It strengthens it, so the kick cannot stay the same.
Shortcut
Iron core → stronger coupling → bigger kicks (the idea behind a transformer core).
Where it went wrong
Thinking iron shields the second coil.
Q12Concept
According to Faraday's second law, the magnitude of the induced emf equals
(A)the rate of change of flux linkage
(B)the total flux linkage
(C)the change in flux linkage
(D)the product of flux linkage and time
Show the solution
Given
Faraday's second law
Asked
What emf equals
Concept
Emf is a rate, measured in volts = webers per second.
Formula
|e| = d(Nφ)/dt
Baby steps
Units check: Wb/s = V.
So emf = rate of change of flux linkage.
Answer
(A) the rate of change of flux linkage
Why not the others
Total flux or change in flux alone are in webers, not volts. Flux × time is in Wb s.
Shortcut
Check units: only Wb/s gives volts.
Where it went wrong
Dropping the 'per time'.
Q13Numerical
The flux linked with a single loop changes from 5 Wb to 1 Wb in 0.2 s. The size of the average induced emf is
(A)0.8 V
(B)25 V
(C)30 V
(D)20 V
Show the solution
Given
φ₁ = 5 Wb, φ₂ = 1 Wb, Δt = 0.2 s, N = 1
Asked
|eavg|
Concept
Average emf = change in flux over time.
Formula
|e| = |Δφ|/Δt
Baby steps
Δφ = 1 − 5 = −4 Wb.
|e| = 4/0.2 = 20 V.
Answer
(D) 20 V
Why not the others
25 V uses 5 Wb as the change. 30 V adds 5 and 1. 0.8 V multiplies by Δt.
Shortcut
4 ÷ 0.2 = 20.
Where it went wrong
Using the starting flux instead of the change.
Q14Numerical
A coil of 100 turns has the flux through each turn changed by 2 mWb in 0.01 s. The size of the induced emf is
(A)20 V
(B)0.2 V
(C)2000 V
(D)2 V
Show the solution
Given
N = 100, Δφ = 2 × 10⁻³ Wb per turn, Δt = 0.01 s
Asked
|e|
Concept
Each turn adds the same emf.
Formula
|e| = N Δφ/Δt
Baby steps
Per turn: 2 × 10⁻³/0.01 = 0.2 V.
Coil: 100 × 0.2 = 20 V.
Answer
(A) 20 V
Why not the others
0.2 V is one turn. 2000 V uses 2 Wb instead of 2 mWb. 2 V divides by 10 somewhere.
Shortcut
N × Δφ/Δt = 100 × 0.2.
Where it went wrong
Forgetting N.
Q15Numerical
The flux through a loop is φ = (3t² + 2t) Wb. The size of the emf at t = 1 s is
(A)6 V
(B)5 V
(C)8 V
(D)2 V
Show the solution
Given
φ = 3t² + 2t, t = 1 s
Asked
|e| at t = 1 s
Concept
Instantaneous emf is the derivative of flux.
Formula
|e| = dφ/dt
Baby steps
dφ/dt = 6t + 2.
At t = 1: 6 + 2 = 8 V.
Answer
(C) 8 V
Why not the others
5 V is φ at t = 1 s. 6 V drops the 2t term. 2 V drops the 3t² term.
Shortcut
Differentiate, then put t = 1.
Where it went wrong
Putting t = 1 into φ instead of dφ/dt.
Q16Numerical
A 50-turn coil of area 0.02 m² and resistance 5 Ω lies with its plane perpendicular to a field that rises uniformly from 0.1 T to 0.5 T in 0.2 s. The induced current is
(A)2 A
(B)0.4 A
(C)0.008 A
(D)1 A
Show the solution
Given
N = 50, A = 0.02 m², R = 5 Ω, ΔB = 0.4 T, Δt = 0.2 s
Asked
I
Concept
Only B changes; area and angle are fixed.
Formula
|e| = NA ΔB/Δt, I = e/R
Baby steps
ΔB/Δt = 0.4/0.2 = 2 T/s.
e = 50 × 0.02 × 2 = 2 V.
I = 2/5 = 0.4 A.
Answer
(B) 0.4 A
Why not the others
2 A is the emf in volts, reported as if it were the current. 0.008 A forgets N. 1 A uses the final field 0.5 T as the change.
Shortcut
e = 2 V, then ÷ 5 Ω.
Where it went wrong
Stopping at the emf and giving it as the current.
Q17Numerical
A magnet and a coil approach each other with speeds 4 m/s and 2 m/s, and the emf is 12 mV. If instead both move in the same direction with these speeds (the faster one behind), at the same separation the emf is
(A)24 mV
(B)12 mV
(C)4 mV
(D)zero
Show the solution
Given
Approach: relative speed 6 m/s gives 12 mV. Same direction: relative speed 2 m/s
Asked
New emf
Concept
At the same separation, emf ∝ relative speed.
Formula
e ∝ vrel
Baby steps
Approaching: vrel = 4 + 2 = 6 m/s.
Same direction: vrel = 4 − 2 = 2 m/s.
e = 12 × 2/6 = 4 mV.
Answer
(C) 4 mV
Why not the others
12 mV ignores the change in relative speed. 24 mV adds instead of subtracting. Zero needs equal speeds.
Shortcut
Ratio of relative speeds: 2/6.
Where it went wrong
Using 4 + 2 in both cases.
Q18Numerical
Pushing a magnet into a coil at 2 cm/s produces 3 mV at a certain position. Pushing it at 6 cm/s past the same position produces
(A)1 mV
(B)9 mV
(C)3 mV
(D)18 mV
Show the solution
Given
v₁ = 2 cm/s → 3 mV; v₂ = 6 cm/s
Asked
e₂
Concept
At the same position, the rate of flux change is proportional to speed.
Formula
e ∝ v
Baby steps
Speed is 3 times larger.
e₂ = 3 × 3 = 9 mV.
Answer
(B) 9 mV
Why not the others
1 mV divides instead of multiplying. 3 mV ignores speed. 18 mV squares the factor.
Shortcut
Three times the speed, three times the emf.
Where it went wrong
Ratio reversal (6/2 upside down).
Q19Numerical
A 20-turn coil of area 0.05 m² starts with its plane perpendicular to a uniform 0.1 T field and is turned through 180° in 0.5 s. The average emf induced is
(A)0.1 V
(B)0.2 V
(C)zero
(D)0.4 V
Show the solution
Given
N = 20, A = 0.05 m², B = 0.1 T, half turn, Δt = 0.5 s
Asked
|eavg|
Concept
A half turn reverses the flux, so the change is 2NBA.
Formula
|e| = 2NBA/Δt
Baby steps
NBA = 20 × 0.1 × 0.05 = 0.1 Wb.
Change = 2 × 0.1 = 0.2 Wb.
e = 0.2/0.5 = 0.4 V.
Answer
(D) 0.4 V
Why not the others
0.2 V uses NBA as the change. Zero treats the start and end flux as equal. 0.1 V is NBA itself.
Shortcut
Half turn: 2NBA over the time.
Where it went wrong
Forgetting that the flux changes sign.
Q20Numerical
The flux linkage of a coil falls steadily from 12 Wb to 2 Wb in 2 s. The size of the induced emf is
(A)1 V
(B)6 V
(C)5 V
(D)7 V
Show the solution
Given
Nφ: 12 Wb → 2 Wb in 2 s
Asked
|e|
Concept
Emf = rate of change of flux linkage.
Formula
|e| = |Δ(Nφ)|/Δt
Baby steps
Change = 12 − 2 = 10 Wb.
e = 10/2 = 5 V.
Answer
(C) 5 V
Why not the others
6 V uses 12 Wb. 1 V uses 2 Wb. 7 V averages 12 and 2.
Shortcut
10 over 2.
Where it went wrong
Dividing one of the flux values by the time.
Q21Concept
A flat loop moves sideways at constant speed, staying entirely inside a large region of uniform magnetic field, without turning. The induced emf is
(A)increasing with time
(B)constant and non-zero
(C)zero, because the flux does not change
(D)maximum, because the loop is moving
Show the solution
Given
Uniform field, loop fully inside, pure sideways motion
Asked
Induced emf
Concept
Motion matters only if it changes flux.
Formula
φ = BA cos θ
Baby steps
B is the same everywhere the loop goes.
A and θ do not change.
So φ is constant and e = 0.
Answer
(C) zero, because the flux does not change
Why not the others
Each other option assumes motion alone creates emf, but the flux is unchanged.
Shortcut
Uniform field + fully inside + no turning → zero.
Where it went wrong
Thinking any motion induces an emf.
Q22Concept
In Experiment 2, a coil carrying a steady current is moved towards a second coil connected to a galvanometer. The galvanometer
(A)does not deflect, because the current is steady
(B)deflects while the coils approach
(C)deflects only if the current is switched off
(D)deflects only if an iron core is used
Show the solution
Given
Current-carrying coil moving towards a second coil
Asked
Galvanometer behaviour
Concept
A current-carrying coil acts like a magnet. Moving it changes the flux through the second coil.
Formula
e ∝ dφ/dt
Baby steps
The steady current makes a steady field around coil 1.
Moving coil 1 closer increases the field at coil 2.
Changing flux → emf → deflection.
Answer
(B) deflects while the coils approach
Why not the others
'Steady current' is steady only in coil 1; the flux in coil 2 still changes with distance. Switching and iron are not required.
Shortcut
Moving a current coil = moving a magnet.
Where it went wrong
Assuming a steady current can never induce anything.
Q23Graph
The flux through a coil changes with time as shown. Which graph shows the induced emf e (including its sign) against time?
(A)
(B)
(C)
(D)
Show the solution
Given
φ rises slowly, stays flat, then falls in half the time it took to rise
Asked
e against t
Concept
e = −(slope of φ–t).
Formula
e = −dφ/dt
Baby steps
Rising part: positive slope → negative emf.
Flat part: zero slope → zero emf.
Falling part: the same drop in half the time → slope twice as steep, negative → positive emf twice as large.
Answer
(A) the graph in option A
Why not the others
The mirror-image graph forgets the minus sign. The equal-height graph ignores that the fall is steeper. The trapezoid just copies the flux graph.
Shortcut
Sign from the minus; height from the steepness.
Where it went wrong
Copying the flux graph, or dropping the minus sign.
Q24Graph
The flux through a coil varies as φ = kt², where k is a positive constant. Which graph shows the magnitude of the induced emf against time?
(A)
(B)
(C)
(D)
Show the solution
Given
φ = kt²
Asked
|e| against t
Concept
Emf is the derivative of flux.
Formula
|e| = d(kt²)/dt = 2kt
Baby steps
|e| = 2kt.
That is a straight line through the origin.
It rises steadily with time.
Answer
(B) the graph in option B
Why not the others
The parabola is the flux graph itself. A flat line would need φ to rise linearly. A falling line would need φ to rise more and more slowly.
Shortcut
Differentiate once: the power drops by one.
Where it went wrong
Drawing the flux instead of its slope.
Q25Graph
The emf induced in a coil has a constant, non-zero magnitude for 5 s. Which flux–time graph could produce this?
(A)
(B)
(C)
(D)
Show the solution
Given
|e| constant and non-zero
Asked
Matching φ–t graph
Concept
Constant emf means constant slope.
Formula
|e| = |dφ/dt| = constant
Baby steps
Constant slope means a straight sloping line.
The line may rise or fall; either gives a constant size of emf.
Answer
(D) the graph in option D
Why not the others
The parabola's slope keeps growing. The flat line gives zero emf. The sine curve's slope keeps changing.
Shortcut
Constant emf ↔ straight sloping flux line.
Where it went wrong
Choosing the flat line because 'constant' appears in the question.
Q26Graph
The graph shows the flux through a coil in four stretches P, Q, R and S. In which stretch is the size of the induced emf largest?
(A)P
(B)R
(C)Q
(D)S
Show the solution
Given
P: 1 → 3 Wb in 2 s. Q: flat. R: 3 → 0 Wb in 1 s. S: 0 → 1 Wb in 2 s
Asked
Stretch with the largest |e|
Concept
|e| = size of the slope.
Formula
|e| = |Δφ|/Δt
Baby steps
P: 2/2 = 1 V.
Q: 0.
R: 3/1 = 3 V.
S: 1/2 = 0.5 V.
Largest is R.
Answer
(B) R
Why not the others
P has a large flux but a gentler slope. Q is flat (zero emf) even though its flux is the highest. S is the gentlest slope.
Shortcut
Pick the steepest stretch, whatever its height.
Where it went wrong
Picking Q because the flux is largest there.
Q27Assertion–reason
Assertion (A): A bar magnet held at rest inside a coil induces no current in the coil. Reason (R): An emf is induced only when the magnetic flux linked with the coil 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: still magnet gives no current. R: emf needs changing flux.
Asked
Truth and link
Concept
Faraday's law.
Formula
e = −N dφ/dt
Baby steps
A is true.
R is true.
The flux is constant, so by R there is no emf, so no current. 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 link that is direct. (C) and (D) need a false statement.
Shortcut
'No current because flux is not changing' reads correctly.
Where it went wrong
Choosing (B) whenever the reason sounds like a definition.
Q28Assertion–reason
Assertion (A): An emf is induced in a plastic ring when the flux through it changes. Reason (R): A plastic ring has no free electrons to carry a current.
(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: emf in plastic. R: plastic cannot carry current.
Asked
Truth and link
Concept
Emf depends on changing flux; current depends on free charges.
Formula
e = −dφ/dt; I = e/R
Baby steps
A is true: emf does not need a conductor.
R is true.
But R is about current, not emf, so it 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) would need R to explain why emf appears; it does not. (C) and (D) need a false statement.
Shortcut
Test the 'because': 'emf is induced because there are no free electrons' makes no sense.
Where it went wrong
Linking two true statements just because they share a topic.
Q29Assertion–reason
Assertion (A): A magnet pushed quickly into a coil gives a larger deflection than when pushed slowly through the same distance. Reason (R): The induced emf depends only on the total change in flux, not on how quickly it happens.
(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 → larger. R: only total change matters.
Asked
Truth and link
Concept
Emf depends on the rate of change of flux.
Formula
|e| = N Δφ/Δt
Baby steps
A is true (Allen p. 99).
R is false: the same Δφ in a shorter Δt gives a larger emf.
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
Emf is a rate; the time is in the formula.
Where it went wrong
Confusing emf with induced charge (which does depend only on Δφ, Topic 04).
Q30Assertion–reason
Assertion (A): An emf can be induced in a coil only if the coil or a magnet near it moves. Reason (R): A changing current in a nearby stationary coil can induce an emf in a stationary coil.
(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: motion is necessary. R: switching a current works without motion.
Asked
Truth and link
Concept
Any change in flux induces an emf, with or without motion.
Formula
e = −N dφ/dt
Baby steps
R is true: Experiment 3 has no motion at all.
So A is false: motion is not necessary.
A false, R true.
Answer
(D) A is false, but R is true.
Why not the others
(A), (B) and (C) all need A to be true.
Shortcut
Experiment 3 is the counter-example to A.
Where it went wrong
Remembering only the magnet experiment.
Q31Two statements
Statement I: Faraday's second law gives the magnitude of the induced emf as the rate of change of flux linkage. Statement II: The negative sign in e = −dφ/dt comes from Faraday's first law.
(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
Two claims about Faraday's laws
Asked
Which are true
Concept
The size comes from Faraday; the sign comes from Lenz.
Formula
e = −N dφ/dt
Baby steps
Statement I is true.
The minus sign expresses Lenz's law, not Faraday's first law. Statement II is false.
Answer
(C) Statement I is true, but Statement II is false.
Why not the others
Both-true and I-false fail because I is correct and II is not. Both-false fails because I is correct.
Shortcut
Faraday: how much. Lenz: which way.
Where it went wrong
Attribution error: giving Lenz's contribution to Faraday.
Q32Two statements
Statement I: An emf is induced in a loop only when the loop forms a closed circuit. Statement II: An induced current flows only when the loop forms a closed conducting circuit.
(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
Emf versus current in open and closed loops
Asked
Which are true
Concept
Emf needs changing flux; current also needs a closed conducting path.
Formula
I = e/R
Baby steps
Statement I is false: an open loop still gets an emf across its ends.
Statement II is true.
Answer
(D) Statement I is false, but Statement II is true.
Why not the others
Both-true would make I correct. Both-false would make II wrong. I-true-II-false has them the wrong way round.
Shortcut
Open loop: emf yes, current no.
Where it went wrong
Treating emf and current as the same thing.
Answer key
1 C
2 A
3 D
4 C
5 A
6 A
7 B
8 B
9 D
10 D
11 A
12 A
13 D
14 A
15 C
16 B
17 C
18 B
19 D
20 C
21 C
22 B
23 A
24 B
25 D
26 B
27 A
28 B
29 C
30 D
31 C
32 D
Spread across letters: A 8, B 8, C 8, D 8. No letter repeats more than twice in a row. Question mix: Concept 14, Numerical 8, Graph 4, Assertion–reason 4, Two statements 2. Balancing seed 0.