A changing magnetic field pushes electrons that are standing still, so it must be making an electric field. This file shows what that field looks like, how it differs from an ordinary electric field, and how strong it is.
Induced electric field. A changing magnetic field creates circles of electric field around it. They exist whether or not a wire is there. If a ring of wire is there, these circles are what push its electrons round.
Picture it
A loop of wire sits still in a magnetic field that is getting stronger. Its electrons start to flow. But a magnetic force only acts on moving charges (F = qvB sin θ), and these electrons were sitting still. So something else must be pushing them. The only thing that pushes a charge at rest is an electric field. Conclusion: a changing magnetic field makes an electric field.
In exam language
A time-varying magnetic field produces an induced electric field E. The induced emf in any closed path equals the work done by this field on a unit charge taken once round it:
e = ∮ E · dl = −dφ/dt
How it differs from an electrostatic field
Closed lines mean real work. Round a closed path, an electrostatic field gives back all the work it took. An induced field keeps doing work lap after lap, which is why it can drive a current.
Property
Electrostatic field
Induced electric field
Produced by
Charges at rest
A changing magnetic field
Field lines
Start on + charges, end on − charges
Closed loops
∮ E · dl
0
−dφ/dt (not zero)
Nature
Conservative
Non-conservative
Potential V
Can be defined
Cannot be defined
Force on a charge
F = qE
F = qE (same rule)
The induced field points the same way as the induced current would flow in a ring placed along its lines. So Lenz's law gives its direction: for a field into the page that is growing, the induced E circulates anticlockwise as seen from the front.
How strong it is around a cylinder of changing field
E against r. Inside, a bigger circle encloses more changing flux in proportion to its length, so E grows with r. Outside, the enclosed flux stops growing but the circle keeps getting longer, so E falls as 1/r.
By symmetry, E has the same size everywhere on a circle of radius r centred on the region, and points along the circle. So ∮ E · dl = E(2πr). Set that equal to the rate of change of flux enclosed:
Where
Flux enclosed
Result
Inside (r < R)
πr² B
E = (r/2)(dB/dt)
On the surface (r = R)
πR² B
E = (R/2)(dB/dt), the maximum
Outside (r > R)
πR² B (only the region has field)
E = (R²/2r)(dB/dt)
Allen Illustration 32. A loop of radius 2 cm in a field changing at 2 T/s: E = (r/2)(dB/dt) = 0.01 × 2 = 0.02 V/m. Illustration 33. For dB/dt = α: E = αr/2 inside (a straight line through the origin) and αR²/2r outside. Both checked: correct. Beginner's Box 3 Q9 is the same calculation.
Trap
The field outside the region is not zero, even though B is zero there. What matters is the changing flux enclosed by the circle, not the field at the point itself.
Formula sheet
Formula
Meaning
When to use
Watch out
∮ E · dl = −dφ/dt
Faraday's law in field form
Emf around any closed path
Only the enclosed changing flux matters
E = (r/2) dB/dt
Inside a cylindrical region
r < R
E ∝ r
E = (R/2) dB/dt
At the surface
r = R
Largest value
E = (R²/2r) dB/dt
Outside the region
r > R
E ∝ 1/r, not 1/r²
e = πr² dB/dt or πR² dB/dt
Emf round a circle
Inside or outside
Outside: use the region's area
W = qe
Work on a charge per lap
Charge carried round once
Non-zero: non-conservative
F = qE
Force on a charge
Force on a stationary charge
Same as electrostatics
Allen pages 128–130 checked
Item
Check
Argument that changing B must create an electric field
Correct
Properties: closed loops, non-conservative, F = qE, same direction as induced current
Correct
∮ E · dl = −dφ/dt
Correct
E inside, on the surface and outside a cylindrical region
Correct
Illustration 32: 0.02 V/m
Correct
Illustration 33: E ∝ r inside, E ∝ 1/r outside
Correct
NEET practice: 29 questions
This is Beyond NCERT material, so NEET touches it mainly through conceptual questions and the E-against-r graph. Every number was recalculated in Python.
Q1Numerical
A uniform magnetic field inside a long cylindrical region is changing at 0.4 T/s. The induced electric field at 5 cm from the axis (inside the region) is
(A)0.02 V/m
(B)0.01 V/m
(C)0.1 V/m
(D)0.005 V/m
Show the solution
Given
dB/dt = 0.4 T/s, r = 0.05 m, inside
Asked
E
Concept
Inside a cylindrical region.
Formula
E = (r/2) dB/dt
Baby steps
E = 0.025 × 0.4 = 0.01 V/m.
Answer
(B) 0.01 V/m
Why not the others
0.02 V/m forgets the ½. 0.1 V/m drops a factor of 10. 0.005 V/m halves again.
Shortcut
r/2 × rate.
Where it went wrong
Forgetting the ½.
Q2Numerical
A cylindrical region of radius 10 cm has a uniform field changing at 0.2 T/s. The induced electric field 20 cm from the axis is
(A)2.5 × 10⁻³ V/m
(B)0.02 V/m
(C)0.01 V/m
(D)5 × 10⁻³ V/m
Show the solution
Given
R = 0.1 m, r = 0.2 m (outside), dB/dt = 0.2 T/s
Asked
E
Concept
Outside: flux enclosed is that of the region only.
Formula
E = (R²/2r) dB/dt
Baby steps
R² = 0.01.
E = 0.01/(0.4) × 0.2 = 5 × 10⁻³ V/m.
Answer
(D) 5 × 10⁻³ V/m
Why not the others
0.02 V/m uses the inside formula with r = 0.2. 0.01 V/m is the surface value. 2.5 × 10⁻³ V/m uses 1/r².
Shortcut
R²/2r.
Where it went wrong
Using the inside formula outside.
Q3Numerical
For the same region (R = 10 cm, dB/dt = 0.2 T/s), the induced electric field at its surface is
(A)0.02 V/m
(B)0.01 V/m
(C)5 × 10⁻³ V/m
(D)0.1 V/m
Show the solution
Given
r = R = 0.1 m
Asked
E
Concept
Surface: both formulas agree.
Formula
E = (R/2) dB/dt
Baby steps
0.05 × 0.2 = 0.01 V/m.
Answer
(B) 0.01 V/m
Why not the others
0.02 V/m forgets the ½. 5 × 10⁻³ V/m is the value at 2R. 0.1 V/m drops the R.
Shortcut
Maximum value.
Where it went wrong
Forgetting the ½.
Q4Numerical
A uniform field changing at 0.4 T/s fills a large region. The emf around a circular path of radius 5 cm inside it is about
(A)0.13 V
(B)6.3 × 10⁻³ V
(C)0.01 V
(D)3.1 × 10⁻³ V
Show the solution
Given
r = 0.05 m, dB/dt = 0.4 T/s
Asked
e
Concept
Emf = rate of change of enclosed flux.
Formula
e = πr² dB/dt
Baby steps
πr² = 7.85 × 10⁻³ m².
× 0.4 = 3.1 × 10⁻³ V.
Answer
(D) 3.1 × 10⁻³ V
Why not the others
6.3 × 10⁻³ V doubles the area. 0.01 V is the field E in V/m, not an emf. 0.13 V multiplies the circumference, not the area, by the rate.
Shortcut
Area × rate.
Where it went wrong
Using circumference instead of area.
Q5Numerical
A cylindrical region of radius 10 cm has a field changing at 0.2 T/s. The emf around a circle of radius 20 cm centred on the region's axis is about
(A)6.3 × 10⁻³ V
(B)2.5 × 10⁻² V
(C)3.1 × 10⁻³ V
(D)zero
Show the solution
Given
R = 0.1 m, circle r = 0.2 m, dB/dt = 0.2 T/s
Asked
e
Concept
Only the region itself holds changing flux.
Formula
e = πR² dB/dt
Baby steps
Flux area = πR² = 0.0314 m².
e = 0.0314 × 0.2 = 6.3 × 10⁻³ V.
Answer
(A) 6.3 × 10⁻³ V
Why not the others
2.5 × 10⁻² V uses πr² for the bigger circle. 3.1 × 10⁻³ V halves. Zero forgets the flux inside the region.
Shortcut
Enclosed changing flux only.
Where it went wrong
Using the area of the bigger circle.
Q6Numerical
A charge of 2 μC is carried once round a circle of radius 10 cm inside a region where the field changes at 1 T/s. The work done on it by the induced electric field is about
(A)6.3 × 10⁻⁸ J
(B)1.3 × 10⁻⁷ J
(C)3.1 × 10⁻² J
(D)zero
Show the solution
Given
q = 2 × 10⁻⁶ C, r = 0.1 m, dB/dt = 1 T/s
Asked
W
Concept
Work per lap = q × emf.
Formula
W = q πr² dB/dt
Baby steps
emf = π × 0.01 × 1 = 0.0314 V.
W = 2 × 10⁻⁶ × 0.0314 = 6.3 × 10⁻⁸ J.
Answer
(A) 6.3 × 10⁻⁸ J
Why not the others
1.3 × 10⁻⁷ J doubles. 3.1 × 10⁻² J is the emf, not the work. Zero would be true only for an electrostatic field.
Shortcut
q × emf.
Where it went wrong
Saying zero because 'the path is closed'.
Q7Numerical
At 1 cm from the axis of a cylindrical region where B changes at 100 T/s, the force on an electron due to the induced electric field is about
(A)0.5 N
(B)1.6 × 10⁻¹⁹ N
(C)8 × 10⁻²⁰ N
(D)3.2 × 10⁻¹⁹ N
Show the solution
Given
r = 0.01 m, dB/dt = 100 T/s, e = 1.6 × 10⁻¹⁹ C
Asked
F
Concept
Find E, then F = eE.
Formula
E = (r/2) dB/dt
Baby steps
E = 0.005 × 100 = 0.5 V/m.
F = 1.6 × 10⁻¹⁹ × 0.5 = 8 × 10⁻²⁰ N.
Answer
(C) 8 × 10⁻²⁰ N
Why not the others
1.6 × 10⁻¹⁹ N takes E = 1 V/m. 0.5 N is E, not a force. 3.2 × 10⁻¹⁹ N uses E = 2 V/m.
Shortcut
E first.
Where it went wrong
Forgetting to multiply by the charge.
Q8Concept
A cylindrical region of radius R has a uniformly changing field. The ratio of the induced electric field at r = R/2 to that at r = 2R is
(A)4 : 1
(B)1 : 4
(C)1 : 1
(D)1 : 2
Show the solution
Given
r₁ = R/2 (inside), r₂ = 2R (outside)
Asked
E₁ : E₂
Concept
Inside E ∝ r, outside E ∝ 1/r.
Formula
E₁ = (R/4) dB/dt, E₂ = (R²/4R) dB/dt
Baby steps
E₁ = (R/2)/2 × dB/dt = (R/4) dB/dt.
E₂ = R²/(2 × 2R) × dB/dt = (R/4) dB/dt.
Ratio 1 : 1.
Answer
(C) 1 : 1
Why not the others
1 : 4 and 4 : 1 use one formula for both points. 1 : 2 uses 1/r² outside.
Shortcut
Half-way in equals twice out.
Where it went wrong
Using the inside formula for both.
Q9Concept
In the same region, the ratio of the induced electric field at r = R/2 to that at r = R is
(A)1 : 1
(B)2 : 1
(C)1 : 4
(D)1 : 2
Show the solution
Given
Both points inside or on the surface
Asked
E₁ : E₂
Concept
Inside, E ∝ r.
Formula
E = (r/2) dB/dt
Baby steps
R/2 : R = 1 : 2.
Answer
(D) 1 : 2
Why not the others
2 : 1 inverts. 1 : 4 squares. 1 : 1 ignores r.
Shortcut
Linear inside.
Where it went wrong
Ratio reversal.
Q10Concept
For a cylindrical region of uniformly changing magnetic field, the induced electric field is largest
(A)on the axis
(B)at the surface of the region
(C)far outside the region
(D)everywhere equally
Show the solution
Given
E against r
Asked
Location of maximum
Concept
E rises as r inside and falls as 1/r outside.
Formula
Emax = (R/2) dB/dt
Baby steps
Inside: increasing with r.
Outside: decreasing with r.
Peak at r = R.
Answer
(B) at the surface of the region
Why not the others
On the axis E = 0. Far away E → 0. It is not uniform.
Shortcut
Peak at the boundary.
Where it went wrong
Guessing the axis because the field is 'strongest' there.
Q11Numerical
Inside a region where B = 0.5t² tesla, the induced electric field at t = 2 s, 4 cm from the axis, is
A copper ring of radius 10 cm and resistance 0.1 Ω lies inside a region where the perpendicular field changes at 0.5 T/s. The induced current is about
(A)0.16 A
(B)0.31 A
(C)1.6 × 10⁻² A
(D)0.05 A
Show the solution
Given
r = 0.1 m, R = 0.1 Ω, dB/dt = 0.5 T/s
Asked
I
Concept
Emf from enclosed changing flux, then Ohm's law.
Formula
I = πr²(dB/dt)/R
Baby steps
emf = π × 0.01 × 0.5 = 0.0157 V.
I = 0.0157/0.1 = 0.157 A.
Answer
(A) 0.16 A
Why not the others
0.31 A doubles. 1.6 × 10⁻² A is the emf in volts. 0.05 A forgets π.
Shortcut
Emf ÷ R.
Where it went wrong
Stopping at the emf.
Q13Concept
The lines of an induced electric field
(A)start on positive charges and end on negative charges
(B)form closed loops
(C)are always straight
(D)are parallel to the changing magnetic field
Show the solution
Given
Induced E
Asked
Shape of field lines
Concept
There are no charges for the lines to start or end on.
Formula
∮ E · dl ≠ 0
Baby steps
Induced E is created by changing flux, not charges.
Its lines circle the region of changing flux: closed loops.
Answer
(B) form closed loops
Why not the others
Starting and ending on charges describes an electrostatic field. They are circles, not straight. They lie in planes perpendicular to B.
Shortcut
Changing B → circling E.
Where it went wrong
Carrying over electrostatic pictures.
Q14Concept
The induced electric field is
(A)zero wherever B is zero
(B)conservative
(C)conservative only outside the region
(D)non-conservative
Show the solution
Given
Induced E
Asked
Nature
Concept
Its line integral round a closed path is not zero.
Formula
∮ E · dl = −dφ/dt
Baby steps
Work round a closed path ≠ 0.
So it is non-conservative.
Answer
(D) non-conservative
Why not the others
Conservative fields have zero closed-loop work. Outside the region it is still non-conservative. It exists outside where B = 0.
Shortcut
Closed-loop work ≠ 0.
Where it went wrong
Assuming all electric fields are conservative.
Q15Concept
For the induced electric field, the line integral ∮ E · dl around a closed path equals
(A)the enclosed charge divided by ε₀
(B)zero
(C)the negative rate of change of magnetic flux through the path
(D)the magnetic flux through the path
Show the solution
Given
Faraday's law in field form
Asked
Value
Concept
Emf is the work per unit charge round the loop.
Formula
∮ E · dl = −dφ/dt
Baby steps
Emf = ∮ E · dl.
Faraday: emf = −dφ/dt.
Answer
(C) the negative rate of change of magnetic flux through the path
Why not the others
Zero is electrostatics. q/ε₀ is Gauss's law for a closed surface. φ itself is not a rate.
Shortcut
Line integral = emf.
Where it went wrong
Mixing up Gauss's law and Faraday's law.
Q16Concept
A changing magnetic field fills a region, with no conductors anywhere. The induced electric field
(A)exists anyway
(B)exists only if a conducting ring is placed there
(C)exists only inside a conductor
(D)is cancelled by the magnetic field
Show the solution
Given
No conductors present
Asked
Does E exist?
Concept
The field comes from the changing flux, not from the conductor.
Formula
∮ E · dl = −dφ/dt
Baby steps
The changing flux alone creates E.
A ring would only reveal it by carrying a current.
Answer
(A) exists anyway
Why not the others
The other options tie E to a conductor, but the conductor only responds to it.
Shortcut
Field first, current second.
Where it went wrong
Thinking induced effects need wires.
Q17Concept
A charge at rest in a region where the magnetic field is changing experiences a force because
(A)the magnetic field acts on stationary charges when it changes
(B)the changing magnetic field produces an electric field that acts on it
(C)gravity is altered by the changing field
(D)the charge begins to move by itself
Show the solution
Given
Charge at rest, changing B
Asked
Origin of the force
Concept
Magnetic force needs motion; electric force does not.
Formula
F = qvB sin θ = 0 at rest; F = qE
Baby steps
v = 0, so qvB = 0.
Only an electric field can push a charge at rest.
So the changing B has produced E.
Answer
(B) the changing magnetic field produces an electric field that acts on it
Why not the others
Magnetic force on a charge at rest is always zero. Gravity is unaffected. Charges do not start moving without a force.
Shortcut
At rest → electric force.
Where it went wrong
Saying the magnetic field pushes it.
Q18Direction
A magnetic field into the page, confined to a circular region, is increasing. Seen from the front, the induced electric field lines circulate
(A)anticlockwise
(B)clockwise
(C)radially outward
(D)radially inward
Show the solution
Given
B into page, growing
Asked
Sense of E
Concept
E follows the induced current (Lenz's law).
Formula
Growing into → induced current anticlockwise
Baby steps
A ring here would carry anticlockwise current.
E points along that current: anticlockwise.
Answer
(A) anticlockwise
Why not the others
Clockwise would add to the growth. Radial lines belong to electrostatic fields.
Shortcut
Same as the Lenz direction.
Where it went wrong
Direction reversal.
Q19Direction
A magnetic field out of the page, confined to a circular region, is increasing. Seen from the front, the induced electric field lines circulate
(A)they do not circulate
(B)anticlockwise
(C)radially outward
(D)clockwise
Show the solution
Given
B out of page, growing
Asked
Sense of E
Concept
Lenz's law.
Formula
Growing out → induced current clockwise
Baby steps
Induced field must point into page.
Clockwise circulation.
Answer
(D) clockwise
Why not the others
Anticlockwise is the into-page answer. Radial and non-circulating options describe other fields.
Shortcut
Out and growing → clockwise.
Where it went wrong
Using the into-page answer.
Q20Concept
An electric potential V cannot be defined for an induced electric field because
(A)the work done round a closed path is not zero
(B)the field is too weak
(C)it has no direction
(D)it exists only inside conductors
Show the solution
Given
Potential for induced E
Asked
Reason
Concept
A potential needs work to be independent of path.
Formula
∮ E · dl ≠ 0
Baby steps
If V existed, returning to the start would give zero work.
Here it gives non-zero work.
So no potential can be defined.
Answer
(A) the work done round a closed path is not zero
Why not the others
Strength, direction and conductors are irrelevant to whether a potential exists.
Shortcut
Path dependence kills potential.
Where it went wrong
Treating induced E like electrostatic E.
Q21Graph
A cylindrical region of radius R has a uniform magnetic field changing at a constant rate. Which graph shows the induced electric field against distance r from the axis?
(A)
(B)
(C)
(D)
Show the solution
Given
Cylindrical region, constant dB/dt
Asked
E against r
Concept
E ∝ r inside, E ∝ 1/r outside.
Formula
E = (r/2)B′ ; (R²/2r)B′
Baby steps
Straight rise from zero at the axis to a peak at R.
Then a 1/r fall outside.
Answer
(B) the graph in option B
Why not the others
The flat top ignores the fall outside. The drop to zero at R forgets that E exists outside. The steeper 1/r² tail is too fast.
Shortcut
Rise, peak at R, 1/r tail.
Where it went wrong
Dropping E to zero outside the region.
Q22Graph
In the same region, which graph shows the emf round a circle of radius r (centred on the axis) against r?
(A)
(B)
(C)
(D)
Show the solution
Given
Emf round a circle
Asked
e against r
Concept
Emf = rate of change of enclosed flux.
Formula
e = πr²B′ inside, πR²B′ outside
Baby steps
Inside: e ∝ r² (parabola).
Outside: enclosed flux is fixed, so e is constant.
Answer
(C) the graph in option C
Why not the others
The straight rise treats e ∝ r. The parabola forever ignores the region's edge. The falling tail confuses emf with E.
Shortcut
Parabola, then flat.
Where it went wrong
Plotting E instead of the emf.
Q23Graph
The magnetic field in a region grows as B = kt. Which graph shows the induced electric field at a fixed point against time?
(A)
(B)
(C)
(D)
Show the solution
Given
B = kt
Asked
E against t
Concept
E depends on dB/dt, which is constant.
Formula
E = (r/2)k
Baby steps
dB/dt = k.
E is constant in time.
Answer
(A) the graph in option A
Why not the others
The rising line copies B. The parabola has no basis. Zero would need B steady.
Shortcut
Constant rate → constant E.
Where it went wrong
Plotting B.
Q24Graph
The magnetic field in a region varies as B = B₀ sin ωt. Which graph shows the induced electric field (with sign) at a fixed point inside, against time?
(A)
(B)
(C)
(D)
Show the solution
Given
B = B₀ sin ωt
Asked
E(t)
Concept
E ∝ dB/dt ∝ cos ωt.
Formula
E = (r/2)B₀ω cos ωt
Baby steps
Differentiate sin → cos.
E is largest when B passes through zero.
Answer
(C) the graph in option C
Why not the others
The sine copies B. The flat line needs a constant rate. The rectified curve never goes negative.
Shortcut
Field follows the slope of B.
Where it went wrong
Plotting B instead of its rate.
Q25Assertion–reason
Assertion (A): A time-varying magnetic field can set charges that are at rest into motion. Reason (R): A time-varying magnetic field produces an induced electric field.
(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: charges at rest move. R: changing B → E.
Asked
Truth and link
Concept
Only an electric field acts on charges at rest.
Formula
F = qE
Baby steps
A is true.
R is true.
The induced E is what pushes them: 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
Mechanism explains effect.
Where it went wrong
Choosing (B) by habit.
Q26Assertion–reason
Assertion (A): The line integral of the induced electric field around any closed path is zero. Reason (R): The induced electric field is non-conservative.
(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: ∮ = 0. R: non-conservative.
Asked
Truth and link
Concept
Non-conservative means ∮ ≠ 0.
Formula
∮ E · dl = −dφ/dt
Baby steps
R is true.
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
R contradicts A.
Where it went wrong
Carrying over the electrostatic result.
Q27Assertion–reason
Assertion (A): Induced electric field lines form closed loops. Reason (R): The induced electric field is produced by charges at rest.
(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: closed loops. R: produced by static charges.
Asked
Truth and link
Concept
Induced E comes from changing B, not charges.
Formula
—
Baby steps
A is true.
R is false.
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
Static charges make electrostatic fields.
Where it went wrong
Mixing up the two kinds of electric field.
Q28Assertion–reason
Assertion (A): Outside a cylindrical region of changing magnetic field, the induced electric field is not zero. Reason (R): The magnetic field outside that region is zero.
(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: E ≠ 0 outside. R: B = 0 outside.
Asked
Truth and link
Concept
E outside comes from the changing flux enclosed, not from B at the point.
Formula
E = (R²/2r) dB/dt
Baby steps
A is true.
R is true.
R does not explain A; if anything, it makes A surprising.
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
Enclosed flux, not local B.
Where it went wrong
Linking two true facts that point in opposite directions.
Q29Two statements
Statement I: Inside a cylindrical region of uniformly changing field, the induced electric field is proportional to the distance from the axis. Statement II: Outside the region, it is inversely proportional to the square of the distance.
(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
E inside and outside
Asked
Which are true
Concept
E ∝ r inside, E ∝ 1/r outside.
Formula
—
Baby steps
Statement I is true.
Statement II is false: it is 1/r, not 1/r².
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
1/r, not 1/r².
Where it went wrong
Borrowing the point-charge law.
Answer key
1 B
2 D
3 B
4 D
5 A
6 A
7 C
8 C
9 D
10 B
11 D
12 A
13 B
14 D
15 C
16 A
17 B
18 A
19 D
20 A
21 B
22 C
23 A
24 C
25 A
26 D
27 C
28 B
29 C
Spread across letters: A 8, B 7, C 7, D 7. No letter repeats more than twice in a row. Question mix: Numerical 9, Concept 9, Direction 2, Graph 4, Assertion–reason 4, Two statements 1. Balancing seed 2000.