An inductor in a circuit: potential, power and energy
An inductor is a wire for steady current and a temporary battery for changing current. This file sets the sign rule for the potential across it, then follows the energy it stores.
Inductor as a battery. A changing current turns the inductor into a temporary cell. Rising current: it opposes, so potential drops from A to B. Falling current: it pushes, so potential rises from A to B. Steady current: nothing.
Picture it
Think of a heavy revolving door. When you push it to speed it up, it pushes back on you: you have to spend effort. When it is spinning and you try to slow it down, it pushes you forward. When it turns at a steady speed, it neither helps nor resists. An inductor is that door for current.
In exam language
Walking through an inductor in the direction of the current, from A to B:
V_A − V_B = L dI/dt
If dI/dt is positive the potential drops from A to B. If negative, it drops from B to A. If zero, there is no potential difference across an ideal inductor. A real coil also has resistance: treat it as an ideal inductor L in series with a resistor R.
Kirchhoff's loop rule with inductors
Walk from A to B along the direction of current and add up the changes in potential:
Element (walking along the current)
Change in potential
Resistor R
−IR
Inductor L
−L dI/dt
Cell from − terminal to + terminal
+E
Cell from + terminal to − terminal
−E
V_A − IR − E − L dI/dt = V_B (Allen Illustrations 15–17, cell giving a 15 V drop from A to B)
Potential staircase. Only the inductor step changes between the three illustrations. Increasing current adds a 5 V drop, steady current adds nothing, decreasing current gives a 5 V rise.
Illustration
dI/dt
L dI/dt
V_A − V_B
15
−10³ A/s
−5 V
5 + 15 − 5 = 15 V
16
+10³ A/s
+5 V
5 + 15 + 5 = 25 V
17
0
0
5 + 15 = 20 V
18: 4 Ω, 0.5 H, 3 Ω, I = t² + 2 at t = 2 s
2t = 4 A/s
2 V
6 × 4 + 2 + 6 × 3 = 44 V
Trap
The scan does not show the cell's polarity clearly. The answers above follow the module's working, which takes a 15 V drop across the cell going from A to B. If a question draws the cell the other way round, the cell term becomes +15 V.
Power: where the battery's energy goes
For a cell E driving current through L and R in series, KVL gives E = L dI/dt + IR. Multiply by I:
EI = LI dI/dt + I²R
power from cell = power stored in the inductor + power turned into heat
While the current is still rising, part of the cell's power builds up the inductor's magnetic field. Once the current is steady, dI/dt = 0 and all the power goes into heat in R.
Energy stored in an inductor
Energy grows as I². The current bar rises steadily; the energy bar lags behind at first and then races up. Half the current stores only a quarter of the energy.
Picture it
A spinning flywheel stores energy ½mv² because it took work to spin it up. An inductor stores energy ½LI² because it took work to push the current up against its back emf. And just as the flywheel keeps turning when you let go, the inductor keeps the current flowing for a moment when the cell is removed, paying back the stored energy.
In exam language
dU/dt = LI dI/dt → dU = LI dI → U = ∫₀ᴵ LI dI = ½LI²
The energy is stored in the magnetic field of the inductor.
Energy density of a magnetic field
For a solenoid, L = μ₀n²V and B = μ₀nI, so I = B/μ₀n. Substitute:
U = ½(μ₀n²V)(B/μ₀n)² = (B²/2μ₀) V → u = U/V = B²/2μ₀
Allen notes that this result, derived for a solenoid, holds for any magnetic field in free space. Compare the electric version:
Field
Energy per unit volume
SI unit
Magnetic field B
u = B²/2μ₀
J m⁻³
Electric field E
u = ½ε₀E²
J m⁻³
Trap
B²/2μ₀ is not only for solenoids. It is true for every magnetic field in vacuum; the solenoid was just a convenient case in which to derive it.
Formula sheet
Formula
Meaning
When to use
Watch out
V_A − V_B = L dI/dt
Potential across an inductor
Current flows from A to B
Sign follows dI/dt
V_A − IR − L dI/dt ± E = V_B
KVL along a branch
Mixed R, L, cell branches
Cell sign depends on polarity
EI = LI dI/dt + I²R
Power balance
Growing current in LR circuit
Steady current: LI dI/dt = 0
U = ½LI²
Energy stored
Any inductor
∝ I²
ΔU = ½L(I₂² − I₁²)
Change in stored energy
Current changes between values
Not ½L(I₂ − I₁)²
u = B²/2μ₀
Magnetic energy density
Any field in vacuum
Unit J m⁻³
u = ½ε₀E²
Electric energy density
Any field in vacuum
Compare with B²/2μ₀
Allen pages 108–110 checked
Item
Check
V_A − V_B = L dI/dt and the three sign cases
Correct
Power: EI = LI dI/dt + I²R
Correct
U = ½LI², u = B²/2μ₀, u = ½ε₀E²
Correct
Inductor as a battery (three cases)
Correct
Illustration 15: 15 V
Correct, with the cell polarity used in the module
Illustration 16: 25 V
Correct
Illustration 17: 20 V
Correct
Illustration 18: 44 V
Correct
There is no Beginner's Box on these three pages.
NEET practice: 32 questions
Sign errors across inductors are the main risk here. Every number was recalculated in Python.
Q1Numerical
An inductor of 50 mH carries a current of 2 A. The energy stored in it is
(A)100 J
(B)0.05 J
(C)0.2 J
(D)0.1 J
Show the solution
Given
L = 0.05 H, I = 2 A
Asked
U
Concept
Energy stored in an inductor.
Formula
U = ½LI²
Baby steps
I² = 4.
U = ½ × 0.05 × 4 = 0.1 J.
Answer
(D) 0.1 J
Why not the others
0.05 J uses I instead of I². 0.2 J forgets the ½. 100 J uses 50 H.
Shortcut
½ × 0.05 × 4.
Where it went wrong
Forgetting to square the current.
Q2Concept
The current through an inductor is doubled. The energy stored becomes
(A)unchanged
(B)twice as large
(C)half as large
(D)four times as large
Show the solution
Given
I → 2I, L fixed
Asked
New U / old U
Concept
U ∝ I².
Formula
U = ½LI²
Baby steps
(2)² = 4.
Answer
(D) four times as large
Why not the others
Twice treats U ∝ I. The others ignore the current.
Shortcut
Square the factor.
Where it went wrong
Treating U as linear in I.
Q3Numerical
A 0.5 H inductor stores 0.25 J. The current through it is
(A)0.5 A
(B)1 A
(C)0.71 A
(D)2 A
Show the solution
Given
L = 0.5 H, U = 0.25 J
Asked
I
Concept
Rearrange the energy formula.
Formula
I = √(2U/L)
Baby steps
2U/L = 0.5/0.5 = 1.
I = √1 = 1 A.
Answer
(B) 1 A
Why not the others
0.5 A forgets the 2. 0.71 A forgets the 2 and takes the root of 0.5. 2 A forgets the square root.
Shortcut
√(2U/L).
Where it went wrong
Forgetting the square root.
Q4Numerical
The energy density of a magnetic field of 0.5 T in vacuum is about
(A)1.99 × 10⁵ J m⁻³
(B)9.9 × 10⁴ J m⁻³
(C)3.1 × 10⁻⁷ J m⁻³
(D)5.0 × 10⁴ J m⁻³
Show the solution
Given
B = 0.5 T
Asked
u
Concept
Magnetic energy density.
Formula
u = B²/2μ₀
Baby steps
B² = 0.25.
2μ₀ = 8π × 10⁻⁷ = 2.513 × 10⁻⁶.
u = 0.25/2.513 × 10⁻⁶ = 9.95 × 10⁴ J m⁻³.
Answer
(B) 9.9 × 10⁴ J m⁻³
Why not the others
1.99 × 10⁵ forgets the 2. 3.1 × 10⁻⁷ multiplies by μ₀ instead of dividing. 5.0 × 10⁴ puts in an extra ½.
Shortcut
B²/2μ₀ ≈ 0.25 × 4 × 10⁵.
Where it went wrong
Multiplying by μ₀.
Q5Numerical
A solenoid has a uniform field of 0.2 T inside a volume of 1 × 10⁻³ m³. (Take the field outside as zero.) The magnetic energy stored is about
(A)0.016 J
(B)32 J
(C)1.6 × 10⁴ J
(D)16 J
Show the solution
Given
B = 0.2 T, V = 10⁻³ m³
Asked
U
Concept
Energy = density × volume.
Formula
U = (B²/2μ₀)V
Baby steps
u = 0.04/(8π × 10⁻⁷) = 1.59 × 10⁴ J m⁻³.
U = 1.59 × 10⁴ × 10⁻³ ≈ 16 J.
Answer
(D) 16 J
Why not the others
32 J forgets the 2. 1.6 × 10⁴ J is the density, not the energy. 0.016 J shifts by 10³.
Shortcut
Density first, then × volume.
Where it went wrong
Stopping at the energy density.
Q6Numerical
An electric field of 3 × 10⁶ V m⁻¹ exists in air (ε₀ = 8.85 × 10⁻¹² F m⁻¹). The electric energy density is about
(A)40 J m⁻³
(B)80 J m⁻³
(C)13 J m⁻³
(D)2.7 × 10⁻⁵ J m⁻³
Show the solution
Given
E = 3 × 10⁶ V/m
Asked
u
Concept
Electric energy density.
Formula
u = ½ε₀E²
Baby steps
E² = 9 × 10¹².
ε₀E² = 8.85 × 10⁻¹² × 9 × 10¹² = 79.7.
u = 39.8 ≈ 40 J m⁻³.
Answer
(A) 40 J m⁻³
Why not the others
80 J m⁻³ forgets the ½. 13 J m⁻³ uses E instead of E² somewhere. 2.7 × 10⁻⁵ uses E without squaring.
Shortcut
½ × 8.85 × 9.
Where it went wrong
Forgetting to square E.
Q7Numerical
Current flows from A to B through a 0.2 H ideal inductor and is increasing at 5 A s⁻¹. V_A − V_B is
(A)−1 V
(B)+1 V
(C)zero
(D)+0.04 V
Show the solution
Given
L = 0.2 H, dI/dt = +5 A/s, current A → B
Asked
V_A − V_B
Concept
Walking along the current through an inductor, the potential drops by L dI/dt.
Formula
V_A − V_B = L dI/dt
Baby steps
L dI/dt = 0.2 × 5 = 1 V.
Positive, so A is higher: +1 V.
Answer
(B) +1 V
Why not the others
−1 V is the decreasing-current answer. Zero is the steady-current answer. 0.04 V divides by 5.
Shortcut
Rising current: higher potential where current enters.
Where it went wrong
Direction reversal of the sign.
Q8Numerical
Current flows from A to B through a 0.2 H ideal inductor and is decreasing at 5 A s⁻¹. V_A − V_B is
(A)−0.04 V
(B)+1 V
(C)zero
(D)−1 V
Show the solution
Given
L = 0.2 H, dI/dt = −5 A/s
Asked
V_A − V_B
Concept
Falling current: the inductor pushes the current on, so B is higher.
Formula
V_A − V_B = L dI/dt
Baby steps
L dI/dt = 0.2 × (−5) = −1 V.
So V_A − V_B = −1 V.
Answer
(D) −1 V
Why not the others
+1 V is the rising-current answer. Zero is the steady case. −0.04 V divides by 5.
Shortcut
Put the sign of dI/dt in.
Where it went wrong
Using the size of dI/dt and forgetting its sign.
Q9Numerical
In a branch A → 2 Ω → 10 mH → B, a current of 3 A flows from A to B and is increasing at 200 A s⁻¹. V_A − V_B is
(A)6 V
(B)4 V
(C)8 V
(D)2 V
Show the solution
Given
R = 2 Ω, L = 0.01 H, I = 3 A, dI/dt = +200 A/s
Asked
V_A − V_B
Concept
KVL along the current.
Formula
V_A − V_B = IR + L dI/dt
Baby steps
IR = 6 V.
L dI/dt = 0.01 × 200 = 2 V.
V_A − V_B = 6 + 2 = 8 V.
Answer
(C) 8 V
Why not the others
4 V subtracts the inductor term (decreasing case). 6 V ignores the inductor. 2 V ignores the resistor.
Shortcut
Both drops add when current is rising.
Where it went wrong
Treating the inductor's term as negative.
Q10Numerical
In the same branch (A → 2 Ω → 10 mH → B, 3 A from A to B), the current is now decreasing at 200 A s⁻¹. V_A − V_B is
(A)6 V
(B)8 V
(C)4 V
(D)−2 V
Show the solution
Given
R = 2 Ω, L = 0.01 H, I = 3 A, dI/dt = −200 A/s
Asked
V_A − V_B
Concept
The inductor term changes sign with dI/dt.
Formula
V_A − V_B = IR + L dI/dt
Baby steps
IR = 6 V.
L dI/dt = −2 V.
V_A − V_B = 6 − 2 = 4 V.
Answer
(C) 4 V
Why not the others
8 V is the increasing case. 6 V ignores the inductor. −2 V ignores the resistor.
Shortcut
Falling current: inductor term subtracts.
Where it went wrong
Keeping the + sign from the rising case.
Q11Numerical
A branch runs A → 1 Ω → cell of 10 V with its positive terminal towards A → 5 mH → B. A current of 2 A flows from A to B and is decreasing at 400 A s⁻¹. V_A − V_B is
(A)14 V
(B)10 V
(C)12 V
(D)6 V
Show the solution
Given
R = 1 Ω, E = 10 V (+ towards A), L = 5 mH, I = 2 A, dI/dt = −400 A/s
Asked
V_A − V_B
Concept
KVL: going through the cell from + to − the potential drops by E.
Formula
V_A − IR − E − L dI/dt = V_B
Baby steps
IR = 2 V.
Cell, + to −: drop 10 V.
L dI/dt = 5 × 10⁻³ × (−400) = −2 V, so −L dI/dt = +2 V.
V_A − V_B = 2 + 10 − 2 = 10 V.
Answer
(B) 10 V
Why not the others
14 V uses +2 V for the inductor term (rising current). 12 V ignores the inductor. 6 V takes the cell as a rise.
Shortcut
Add the three terms with signs: +2, +10, −2.
Where it went wrong
Getting the inductor's sign wrong for a falling current.
Q12Numerical
A branch A → 4 Ω → 0.5 H → B carries I = (2t + 1) A from A to B. At t = 2 s, V_A − V_B is
(A)22 V
(B)20 V
(C)21 V
(D)10 V
Show the solution
Given
R = 4 Ω, L = 0.5 H, I = 2t + 1, t = 2 s
Asked
V_A − V_B
Concept
KVL with a time-varying current.
Formula
V_A − V_B = IR + L dI/dt
Baby steps
I = 5 A, IR = 20 V.
dI/dt = 2 A/s, L dI/dt = 1 V.
Total = 21 V.
Answer
(C) 21 V
Why not the others
20 V ignores the inductor. 22 V doubles the inductor term. 10 V uses I = 2.5 A.
Shortcut
Find I and dI/dt separately.
Where it went wrong
Using I instead of dI/dt in the inductor term.
Q13Numerical
At some instant, a 0.2 H inductor carries 3 A and the current is increasing at 5 A s⁻¹. The rate at which energy is being stored in it is
(A)15 W
(B)0.9 W
(C)3 W
(D)1.5 W
Show the solution
Given
L = 0.2 H, I = 3 A, dI/dt = 5 A/s
Asked
dU/dt
Concept
Power into an inductor.
Formula
dU/dt = LI dI/dt
Baby steps
LI dI/dt = 0.2 × 3 × 5 = 3 W.
Answer
(C) 3 W
Why not the others
0.9 W is ½LI² (energy, not power). 15 W forgets L. 1.5 W includes a stray ½.
Shortcut
Differentiate ½LI².
Where it went wrong
Giving the stored energy instead of its rate.
Q14Numerical
A 12 V cell drives a current through a 4 Ω resistor and an inductor in series. At the instant the current is 2 A, the power going into the inductor is
(A)8 W
(B)16 W
(C)24 W
(D)32 W
Show the solution
Given
E = 12 V, R = 4 Ω, I = 2 A
Asked
Power into the inductor
Concept
Power balance: EI = I²R + (power into L).
Formula
PL = EI − I²R
Baby steps
EI = 24 W.
I²R = 16 W.
PL = 24 − 16 = 8 W.
Answer
(A) 8 W
Why not the others
16 W is the heat in R. 24 W is the cell's total power. 32 W adds instead of subtracting.
Shortcut
Cell power minus heating power.
Where it went wrong
Giving the resistor's share.
Q15Numerical
The current in a 3 H inductor rises from 2 A to 4 A. The increase in stored energy is
(A)6 J
(B)18 J
(C)24 J
(D)12 J
Show the solution
Given
L = 3 H, I₁ = 2 A, I₂ = 4 A
Asked
ΔU
Concept
Energy depends on I², so subtract the two energies.
Formula
ΔU = ½L(I₂² − I₁²)
Baby steps
U₂ = ½ × 3 × 16 = 24 J.
U₁ = ½ × 3 × 4 = 6 J.
ΔU = 18 J.
Answer
(B) 18 J
Why not the others
6 J uses ½L(I₂ − I₁)², which is also the starting energy. 24 J is the final energy. 12 J is half the true increase.
Shortcut
Energy at the end minus energy at the start.
Where it went wrong
Squaring the difference instead of taking the difference of squares.
Q16Concept
Inductor P has twice the inductance of Q but carries half the current. The ratio of their stored energies UP : UQ is
(A)1 : 4
(B)2 : 1
(C)1 : 1
(D)1 : 2
Show the solution
Given
LP = 2LQ, IP = IQ/2
Asked
UP : UQ
Concept
U ∝ LI².
Formula
U = ½LI²
Baby steps
L factor: 2.
I² factor: ¼.
UP/UQ = 2 × ¼ = ½.
Answer
(D) 1 : 2
Why not the others
2 : 1 is upside down. 1 : 1 treats U ∝ LI. 1 : 4 ignores the L factor.
Shortcut
Square the current factor, not the inductance factor.
Where it went wrong
Ratio reversal.
Q17Concept
The number of turns of an air-cored solenoid is doubled (same length and area) and the current is kept the same. The stored energy becomes
(A)twice as large
(B)four times as large
(C)eight times as large
(D)unchanged
Show the solution
Given
N → 2N, I fixed
Asked
New U
Concept
L ∝ N², U ∝ L at fixed I.
Formula
U = ½LI², L = μ₀N²A/ℓ
Baby steps
L × 4.
U × 4.
Answer
(B) four times as large
Why not the others
Twice treats L ∝ N. Eight times cubes the factor. Unchanged ignores L.
Shortcut
Follow L.
Where it went wrong
Forgetting L ∝ N².
Q18Concept
The magnetic field inside an air-cored solenoid is doubled. The energy stored in that volume becomes
(A)twice as large
(B)four times as large
(C)half as large
(D)eight times as large
Show the solution
Given
B → 2B, same volume
Asked
New U
Concept
u = B²/2μ₀.
Formula
U ∝ B²
Baby steps
(2)² = 4.
Answer
(B) four times as large
Why not the others
Twice treats u ∝ B. Half and eight times have no basis.
Shortcut
Energy goes with the square of the field.
Where it went wrong
Treating energy density as linear in B.
Q19Concept
The energy stored in a current-carrying inductor resides in
(A)the resistance of the wire
(B)its electric field
(C)the kinetic energy of the electrons in the wire
(D)its magnetic field
Show the solution
Given
Where U = ½LI² is stored
Asked
Location
Concept
The derivation u = B²/2μ₀ ties the energy to the magnetic field.
Formula
u = B²/2μ₀
Baby steps
The field inside the coil holds energy B²/2μ₀ per unit volume.
Integrating over the volume gives ½LI².
Answer
(D) its magnetic field
Why not the others
An electric field stores energy in a capacitor. The electrons' kinetic energy is tiny. Resistance turns energy into heat; it does not store it.
Shortcut
Inductor ↔ magnetic field; capacitor ↔ electric field.
Where it went wrong
Mixing up inductors and capacitors.
Q20Concept
An ideal inductor carries a large steady current. The potential difference across it is
(A)½LI²
(B)LI
(C)zero
(D)infinite
Show the solution
Given
Ideal inductor, dI/dt = 0
Asked
p.d.
Concept
p.d. = L dI/dt.
Formula
V = L dI/dt
Baby steps
Steady current → dI/dt = 0.
No resistance in an ideal inductor.
V = 0.
Answer
(C) zero
Why not the others
LI is flux linkage. ½LI² is energy. Infinite would need a sudden jump in current.
Shortcut
Steady → no voltage across an ideal inductor.
Where it went wrong
Thinking a large current needs a large voltage across L.
Q21Concept
The quantity ½LI² has the same dimensions as
(A)force
(B)power
(C)work
(D)momentum
Show the solution
Given
½LI²
Asked
Dimensions
Concept
[L][A²] = [M L² T⁻² A⁻²][A²].
Formula
—
Baby steps
The A² cancels.
Left with [M L² T⁻²], the dimensions of energy or work.
Answer
(C) work
Why not the others
Power has T⁻³. Force has L¹. Momentum is [M L T⁻¹].
Shortcut
It is an energy.
Where it went wrong
Stopping at the dimensions of L.
Q22Concept
The SI unit of B²/2μ₀ is
(A)J m⁻³
(B)J m⁻²
(C)W m⁻³
(D)J
Show the solution
Given
B²/2μ₀
Asked
Unit
Concept
It is energy per unit volume.
Formula
u = U/V
Baby steps
Energy in joules.
Per cubic metre.
J m⁻³.
Answer
(A) J m⁻³
Why not the others
J m⁻² is per area. W m⁻³ is power per volume. J alone is the total energy.
Shortcut
Density → per volume.
Where it went wrong
Confusing energy density with energy.
Q23Graph
Which graph shows the energy stored in an inductor against the current through it?
(A)
(B)
(C)
(D)
Show the solution
Given
L fixed
Asked
U against I
Concept
U ∝ I².
Formula
U = ½LI²
Baby steps
A square law gives a parabola through the origin.
Answer
(B) the graph in option B
Why not the others
The straight line treats U ∝ I. The square-root curve treats I ∝ U². The flat line ignores the current.
Shortcut
I² → parabola.
Where it went wrong
Choosing the straight line.
Q24Graph
Inductors of different inductance all carry the same current. Which graph shows the stored energy against L?
(A)
(B)
(C)
(D)
Show the solution
Given
I fixed
Asked
U against L
Concept
U ∝ L at fixed I.
Formula
U = ½LI²
Baby steps
½I² is a constant.
U = (constant) × L: straight line through the origin.
Answer
(A) the graph in option A
Why not the others
The parabola squares L. The hyperbola inverts it. The flat line ignores L.
Shortcut
Only I is squared.
Where it went wrong
Squaring L as well.
Q25Graph
The current through an inductor increases in direct proportion to time, I = kt. Which graph shows the potential difference across the ideal inductor against time?
(A)
(B)
(C)
(D)
Show the solution
Given
I = kt
Asked
V against t
Concept
V = L dI/dt.
Formula
V = Lk
Baby steps
dI/dt = k, a constant.
V = Lk, the same at all times.
Answer
(C) the graph in option C
Why not the others
The rising line copies the current. The parabola is the energy. Zero would need a steady current.
Shortcut
Constant slope → constant voltage.
Where it went wrong
Drawing the current instead of its rate.
Q26Graph
The current in an inductor rises quickly, stays steady, then falls slowly back to zero, as shown. Which graph shows V_A − V_B (current flowing from A to B)?
(A)
(B)
(C)
(D)
Show the solution
Given
Fast rise, flat, slow fall
Asked
V_A − V_B against t
Concept
V_A − V_B = L dI/dt.
Formula
V ∝ slope
Baby steps
Fast rise: large positive slope → large positive V.
Flat: zero.
Slow fall: small negative slope → small negative V.
Answer
(A) the graph in option A
Why not the others
The mirrored graph has the wrong signs (that would be −L dI/dt, the emf). The equal-height graph ignores the different slopes. The trapezoid copies the current.
Shortcut
V_A − V_B has the same sign as dI/dt.
Where it went wrong
Mixing up p.d. (L dI/dt) with emf (−L dI/dt).
Q27Assertion–reason
Assertion (A): The energy stored in an inductor carrying current I is ½LI². Reason (R): Work has to be done against the self-induced emf while the current is being built up.
(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: U = ½LI². R: work against back emf.
Asked
Truth and link
Concept
The stored energy is the work done against the back emf.
Formula
U = ∫LI dI
Baby steps
A is true.
R is true.
Integrating the power LI dI/dt that the source spends against the back emf gives ½LI²: 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
The derivation of A starts from R.
Where it went wrong
Choosing (B) by habit.
Q28Assertion–reason
Assertion (A): The magnetic energy density B²/2μ₀ applies only inside a solenoid. Reason (R): The formula B²/2μ₀ is usually derived using a solenoid.
(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: only in solenoids. R: derived with a solenoid.
Asked
Truth and link
Concept
The result holds for any magnetic field in vacuum.
Formula
u = B²/2μ₀
Baby steps
R is true.
A is false: the derivation uses a solenoid for convenience, but the result is general.
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
How a formula is derived does not limit where it applies.
Where it went wrong
Treating a derivation as a restriction.
Q29Assertion–reason
Assertion (A): The potential difference across an ideal inductor carrying a steady current is zero. Reason (R): The energy stored in an inductor carrying a steady current 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.
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Given
A: no p.d. R: no energy.
Asked
Truth and link
Concept
p.d. depends on dI/dt; energy depends on I.
Formula
V = L dI/dt; U = ½LI²
Baby steps
A is true: dI/dt = 0.
R is false: U = ½LI² is not zero for a steady non-zero current.
A true, R false.
Answer
(C) A is true, but R is false.
Why not the others
(A) and (B) need R true. (D) needs A false.
Shortcut
Steady current: V = 0 but U ≠ 0.
Where it went wrong
Thinking zero voltage means zero energy.
Q30Assertion–reason
Assertion (A): When the current through an inductor decreases, the inductor gives energy back to the circuit. Reason (R): The stored energy ½LI² decreases as the current decreases.
(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: inductor supplies energy. R: U falls with I.
Asked
Truth and link
Concept
Energy conservation.
Formula
U = ½LI²
Baby steps
A is true.
R is true.
The energy given back is exactly the fall in ½LI²: 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 the direct link. (C) and (D) need a false statement.
Shortcut
Falling U has to go somewhere: into the circuit.
Where it went wrong
Choosing (B) out of caution.
Q31Two statements
Statement I: The SI unit of magnetic energy density is J m⁻³. Statement II: The electric energy density in free space is ½ε₀E².
(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
Units and electric energy density
Asked
Which are true
Concept
Both are standard results.
Formula
u = B²/2μ₀; u = ½ε₀E²
Baby steps
Statement I is true.
Statement II is true.
Answer
(A) Both Statement I and Statement II are true.
Why not the others
Every other option calls a correct statement false.
Shortcut
Both are formula-sheet facts.
Where it went wrong
Doubting a true statement because it looks too simple.
Q32Two statements
Statement I: If the current in an inductor is doubled, the energy stored doubles. Statement II: At a fixed current, the energy stored in an inductor is proportional to its inductance.
(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
Energy scaling with I and with L
Asked
Which are true
Concept
U = ½LI².
Formula
U ∝ I², U ∝ L
Baby steps
Statement I is false: doubling I gives four times U.
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
I is squared, L is not.
Where it went wrong
Treating both dependences the same way.
Answer key
1 D
2 D
3 B
4 B
5 D
6 A
7 B
8 D
9 C
10 C
11 B
12 C
13 C
14 A
15 B
16 D
17 B
18 B
19 D
20 C
21 C
22 A
23 B
24 A
25 C
26 A
27 A
28 D
29 C
30 A
31 A
32 D
Spread across letters: A 8, B 8, C 8, D 8. No letter repeats more than twice in a row. Question mix: Numerical 14, Concept 8, Graph 4, Assertion–reason 4, Two statements 2. Balancing seed 1.