Self-inductance L, mutual inductance M: Question bank

True or false — justify

A large steady current in a coil produces a large self-induced EMF.

False. Self-EMF is

ε = - L d I / d t

; a steady current has

d I / d t = 0

, so the EMF is exactly zero no matter how large the current. Flux must change for Faraday's law to bite.

Doubling the number of turns on a solenoid doubles its self-inductance.

False.

L = μ_{0} N^{2} A / ℓ \propto N^{2}

, so doubling

N

gives

4 \times L

. Each extra turn both makes more flux and links the flux, so the effect compounds.

Inductance

L

depends on the current flowing through the coil.

False.

L = N Φ/ I

is purely geometric (turns, area, length, core material). The

I

cancels because

Φ \propto I

;

L

is a fixed property of the coil's shape, like a container's volume.

For two coils of very different sizes,

M_{12}

and

M_{21}

must differ.

False. The reciprocity theorem proves

M_{12} = M_{21} = M

always, regardless of size asymmetry — it follows from the symmetry of the Neumann formula / energy argument.

The coupling coefficient

k

M = k L_{1} L_{2}

can exceed 1 if the coils are wound very tightly.

False.

0 \leq k \leq 1

always.

k = 1

already means all of one coil's flux links the other; you cannot link more flux than exists, so

k > 1

is physically impossible.

Energy

U = \frac{1}{2} L I^{2}

is stored in the wire's resistance.

False. The energy lives in the magnetic field filling the coil's volume, not in resistive heating. An ideal (zero-resistance) inductor still stores

\frac{1}{2} L I^{2}

; see Energy stored in magnetic field.

The minus sign in

ε = - L d I / d t

means the EMF is always negative.

False. The sign tracks direction, not value. If

I

is decreasing (

d I / d t < 0

), the EMF is positive — it acts to keep the current flowing. The sign encodes Lenz's "oppose the change."

If a solenoid's current is momentarily constant (at a peak of a sine wave), the self-EMF at that instant is zero.

True. At the peak of

I = I_{0} sin ω t

, the slope

d I / d t = I_{0} ω cos ω t = 0

, so

ε = 0

right there — even though

I

is at its maximum.

Spot the error

ε = - L I

, so a coil with 5 A has five times the back-EMF of one with 1 A."

The formula is wrong: it's

ε = - L d I / d t

, involving the rate of change, not the current itself. A constant 5 A gives zero back-EMF just like a constant 1 A.

L = Φ/ I

for a solenoid, so

L = μ_{0} N A / ℓ

The definition uses the total flux linkage

N Φ

, not

Φ

alone:

L = N Φ/ I = μ_{0} N^{2} A / ℓ

. Dropping the extra

N

loses the whole

N^{2}

dependence.

"Since

B = μ_{0} n I

depends only on turns-per-metre

n

, two solenoids with the same

n

have the same

L

B

depends on

n

, but

L = μ_{0} n^{2} A ℓ

also depends on cross-section

A

and length

ℓ

. Same

n

but different

A

ℓ

gives different

L

"Mutual inductance needs the two coils to touch electrically."

No electrical connection is needed — coupling is through shared magnetic flux (Magnetic flux), not shared current. That is precisely how a transformer works: two isolated windings.

"To find

L

we differentiate

N Φ = L I

and get

ε = - L d I / d t - I d L / d t

For a rigid coil the geometry is fixed, so

d L / d t = 0

and the second term vanishes. The extra term only appears if the coil's shape or core is physically changing.

"A single straight wire has no inductance because it isn't a loop."

Even a straight wire carries flux around itself and has a (small) self-inductance. Any current-carrying conductor stores magnetic energy; loops just make it much larger and easier to compute.

Why questions

Why is

L

often called "flux per amp"?

Because

L = N Φ/ I

literally measures how much flux linkage each ampere of current threads through the coil — a bookkeeping constant so you never re-integrate the field each time.

Why does the induced EMF oppose the change rather than aid it?

Conservation of energy via Lenz's law: if it aided the change, current would grow without an energy source — a perpetual motion machine. Opposition means work must be done to change the current.

Why is an inductor described as a "flywheel for current"?

Just as a spinning flywheel resists changes in its rotation because it stores kinetic energy, an inductor resists changes in current because it stores magnetic energy

\frac{1}{2} L I^{2}

; both smooth out sudden changes.

Why does

L \propto N^{2}

but

M \propto N_{1} N_{2}

Self-inductance counts one coil doing "double duty" (its own turns both make and link flux), giving

N^{2}

. Mutual inductance pairs coil 1's flux-making (

N_{1}

) with coil 2's flux-linking (

N_{2}

), giving the product.

Why does inserting an iron core dramatically raise

L

Iron's high permeability

μ ≫ μ_{0}

multiplies

B

for the same current, so

Φ

and hence

L = μ n^{2} A ℓ

rise by the relative permeability factor — often thousands.

Why can we treat the flux through every turn of a long solenoid as equal?

Inside a long ideal solenoid the field is uniform and axial, so each turn (same area, same location) is threaded by the identical

Φ = B A

, letting us write total linkage as simply

N Φ

Why does the RL circuit current rise gradually rather than jump to its final value?

Any sudden jump means infinite

d I / d t

, hence infinite back-EMF opposing it — impossible with finite battery voltage. So current in an RL circuit rises smoothly over the timescale

τ = L / R

Edge cases

What is the self-EMF the instant a switch is opened on an inductor carrying steady current?

Very large (ideally infinite): the current is forced toward zero almost instantly, making

d I / d t

huge and negative, so a big positive EMF appears — this is why switches spark.

What happens to

M

if one coil is rotated 90° so their axes are perpendicular?

M \to 0

(and

k \to 0

): the flux from coil 1 no longer threads coil 2's area, so no linkage and no induced EMF, even though both coils still exist.

What is

L

for a coil in the limit

N \to 0

(no turns)?

L = μ_{0} N^{2} A / ℓ \to 0

. With no turns there is no loop to carry current, no flux, no linkage — inductance vanishes, consistent with the formula.

What is the self-EMF for a perfectly DC circuit that has been running for a long time?

Zero. Once current is steady,

d I / d t = 0

, so the inductor behaves like an ordinary wire (a short) — it only "reacts" while current changes.

If two coils are identical (

L_{1} = L_{2} = L

) and perfectly coupled, what is

M

M = k L_{1} L_{2} = 1 \cdot L \cdot L = L

. Perfect coupling of identical coils makes the mutual inductance equal to each coil's self-inductance.

What self-EMF arises if the current increases linearly with time,

I = a t

A constant EMF:

ε = - L d I / d t = - L a

. A steady ramp of current produces a steady (unchanging) back-EMF, not a growing one.

Connections

Faraday's law of induction — the source of every induced EMF here.
Lenz's law — the reason for the minus sign and all "opposition" answers.
Ampère's law — gives the field these traps assume.
Magnetic flux — the quantity that must change.
Energy stored in magnetic field — where $\frac{1}{2} L I^{2}$ actually lives.
RL circuits — the gradual-rise edge case.
Transformers — no-electrical-contact coupling.