Inductance

Q: What is self-inductance in simple words?

Self-inductance is a coil's unwillingness to let its own current change. Changing the current changes the coil's own flux, which induces an emf in the coil that fights the change. The number measuring this is L, defined by Nφ = Li and ε = -L di/dt.

Q: Why is inductance called electrical inertia?

Because the equations match. Force equals mass times acceleration becomes emf equals inductance times rate of change of current, and kinetic energy 1/2mv² becomes stored energy 1/2Li². Current plays the part of velocity and inductance the part of mass.

Q: How does an inductor behave just after and long after a switch is closed?

Just after closing, the current cannot jump, so the inductor acts like an open circuit and carries no current. Long afterwards the current is steady, di/dt = 0, so it acts like a plain wire. A capacitor behaves in exactly the opposite way.

Q: What does the time constant of an L-R circuit mean?

It is τ = L/R, the time in which a growing current reaches 63% of its final value or a decaying current falls to 37% of its starting value. A large inductance or a small resistance makes the circuit slow to respond.

Q: Where is the energy of an inductor stored?

In the magnetic field itself, not in the wire. The energy density is u = B²/2μ₀ joule per cubic metre, and integrating that over the volume of a solenoid gives back exactly 1/2Li², which is a good check in any energy problem.

Q: What is the reciprocity theorem for mutual inductance?

It says M₁₂ = M₂₁: the mutual inductance is the same whichever coil you treat as the primary. This is very useful, because one direction is often easy to compute and the other needs a difficult integral.

Q: How is inductance tested in NEET?

NEET asks direct substitutions: the solenoid formula L = μ₀n²Al, the energy 1/2Li², the emf L di/dt for a given rate of change, the time constant L/R and simple mutual-inductance numericals such as a coil wound on a solenoid or a transformer primary switched off.

Q: What inductance questions come in JEE Main and JEE Advanced?

JEE Main uses L-R circuits solved with the open-circuit and plain-wire substitutions, energy and time constant. JEE Advanced adds sudden changes where the flux linkage Li is conserved, coupled coils with ± 2M, and magnetic field energy integrals inside wires and solenoids.

PhysicsElectromagnetic InductionFor NEET aspirants

Inductance measures how strongly a coil resists a change in the current through it. Self-inductance $L$ links a coil's own flux to its own current, $N ϕ = L i$ , and produces a back emf $ε = - L \frac{d i}{d t}$ ; mutual inductance $M$ does the same between two coils. Inductance depends only on geometry and the core material, never on the current. This page covers the solenoid, the L-R circuit, stored energy and mutual inductance for JEE Main, JEE Advanced and NEET.

On this page1Self-induction2Solenoid3Inductor in a circuit4Energy stored5L-R circuit6Mutual inductance7Combinations

Key Formulas - Quick Reference

★ Must learn Self-inductance: $N ϕ = L i$ and $ε = - L \frac{d i}{d t}$ ; the SI unit is the henry (H)
★ Must learn Long solenoid: $L = μ_{0} n^{2} A l = \frac{μ _{0} N ^{2} A}{l}$ (multiply by $μ_{r}$ for a core)
Terminal voltage: walking along the current, the potential falls by $L \frac{d i}{d t}$
★ Must learn Energy stored: $U = \frac{1}{2} L i^{2}$ ; energy density of the field $u = \frac{B ^{2}}{2 μ _{0}}$
Inductor at $t = 0$ behaves as an open circuit; at $t \to \infty$ as a plain wire
★ Must learn L-R growth: $i = i_{0} (1 - e^{- t / τ})$ with $i_{0} = \frac{ε}{R}$ and $τ = \frac{L}{R}$
L-R decay: $i = i_{0} e^{- t / τ}$ ; the total heat produced equals $\frac{1}{2} L i_{0}^{2}$
★ Must learn Mutual inductance: $ϕ_{2} = M i_{1}$ and $ε_{2} = - M \frac{d i _{1}}{d t}$ ; reciprocity gives $M_{12} = M_{21}$
Coil of $N_{2}$ turns on a solenoid: $M = \frac{μ _{0} N _{1} N _{2} A}{l}$ ; in general $M = k L_{1} L_{2}$ with $k \leq 1$
Combinations: series $L = L_{1} + L_{2} \pm 2 M$ ; parallel (uncoupled) $\frac{1}{L} = \frac{1}{L _{1}} + \frac{1}{L _{2}}$

1. Self-Induction and the Coefficient $L$

A coil carrying a current sits in its own magnetic field. Change that current and the coil's own flux changes, so the coil induces an emf in itself. This is self-induction, and by Lenz's law it always fights the change.

Figure 1: A coil opposes the change in its own current, never the current itself. This is Lenz's law for a coil's own flux,

ε = - L \frac{d i}{d t}

, and it is why inductance behaves like inertia.

The total flux linked with a coil is proportional to its own current:

N ϕ = L i

where $L$ is the self-inductance of the coil. Differentiating and using Faraday's law,

ε = - \frac{d ( N ϕ )}{d t} = - L \frac{d i}{d t}

The SI unit is the henry (H): a coil has an inductance of $1 H$ if a current changing at $1 A s^{- 1}$ induces $1 V$ across it.

$L$ is a positive scalar and is purely geometrical: it depends on the shape and size of the coil, the number of turns and the core material.
$L$ does not depend on the current, the emf or how fast the current is changing. Doubling the current does not change $L$ .
Doubling the number of turns quadruples $L$ , because both the field and the number of turns linking it double.

1.1 Inductance is electrical inertia

The single most useful way to think about inductance is as the electrical version of mass.

Mechanics	Electricity	Why they match
Mass $m$	Inductance $L$	both resist a change in the thing that flows
Velocity $v$	Current $i$	the quantity that cannot jump suddenly
Force $F$	emf $ε$	what drives the change
$F = m \frac{d v}{d t}$	$ε = L \frac{d i}{d t}$	the same equation
$\frac{1}{2} m v^{2}$	$\frac{1}{2} L i^{2}$	energy stored while building up

Exam Trick

Whenever a circuit question about inductors confuses you, translate it into mechanics. A current cannot change instantly for the same reason a heavy trolley cannot change speed instantly, and an inductor stores $\frac{1}{2} L i^{2}$ for the same reason a moving trolley stores $\frac{1}{2} m v^{2}$ .

Resistor

R

Opposes the current: $V = i R$ even for a steady current. Turns electrical energy into heat.

Inductor

L

Opposes the change of current: $V = L \frac{d i}{d t}$ , zero for a steady current. Stores energy $\frac{1}{2} L i^{2}$ and gives it back.

Key idea

Inductance is electrical inertia:

L

resists changes in current the way mass resists changes in velocity, and it depends only on geometry and core.

2. Inductance of a Long Solenoid

The solenoid is the one shape whose inductance can be worked out in three lines, and almost every numerical in this chapter uses it.

Figure 2: For a long solenoid

B = μ_{0} ni

inside and almost zero outside, so

N ϕ = μ_{0} n^{2} A l i

and

L = μ_{0} n^{2} A l

. Inductance depends only on the geometry and the core, never on the current.

Field inside a long solenoid with $n$ turns per unit length: $B = μ_{0} ni$ , and almost zero outside.
Flux through one turn: $ϕ = B A = μ_{0} ni A$ . The number of turns in a length $l$ is $N = n l$ , so the total flux linkage is
$N ϕ = (n l) (μ_{0} ni A) = μ_{0} n^{2} A l i$
Comparing with $N ϕ = L i$ ,
$L = μ_{0} n^{2} A l = \frac{μ _{0} N ^{2} A}{l}$

Since $A l$ is the volume $V$ of the solenoid, this is also $L = μ_{0} n^{2} V$ , so the inductance per unit volume is just $μ_{0} n^{2}$ . Filling the core with a material of relative permeability $μ_{r}$ multiplies everything by $μ_{r}$ , which is why iron cores are used to get large inductance from a small coil.

Watch the difference between $n$ (turns per metre) and $N$ (total turns). Using $N$ where the formula wants $n$ is the most common slip in solenoid numericals, and it changes the answer by a factor of $l^{2}$ .

3. The Inductor in a Circuit

3.1 Which way does the voltage go?

An inductor is not a resistor: its voltage has nothing to do with $i R$ and everything to do with how fast the current is changing.

Figure 3: Walking through an inductor in the direction of the current, the potential falls by

L \frac{d i}{d t}

. When

\frac{d i}{d t}

is negative that fall becomes a rise, which is how an inductor keeps a dying current alive.

Travelling through an inductor in the direction of the current, the potential changes by

V_{A} - V_{B} = L \frac{d i}{d t}

If the current is rising, $\frac{d i}{d t} > 0$ and the potential drops, so the inductor is absorbing energy. If the current is falling, $\frac{d i}{d t} < 0$ and the potential rises, so the inductor is returning energy to the circuit.

3.2 The two limits that solve most problems

You almost never need to solve a differential equation to answer "find the current just after the switch is closed" or "find the current long afterwards". Two substitutions do the whole job.

Figure 4: Two substitutions solve most L-R questions without calculus: an inductor is an open circuit at

t = 0

(if it carried no current before) and a plain wire at

t \to \infty

. A capacitor does exactly the opposite.

At $t = 0$ : the current through an inductor cannot change instantly, so it is still whatever it was an instant earlier, usually zero. Replace the inductor by a break in the wire and solve the remaining circuit.
At $t \to \infty$ : the current has settled, so $\frac{d i}{d t} = 0$ and the voltage across the inductor is zero. Replace the inductor by a plain wire.
A capacitor does exactly the opposite: a wire at $t = 0$ and a break at $t \to \infty$ .

Exam Trick

In any L-R network, redraw the circuit twice, once with the inductor as a gap and once as a wire. Those two pictures give the initial and final currents in every branch, and a JEE Main question usually asks for nothing else.

Figure 5: Every single-inductor circuit follows

i (t) = i_{\infty} + (i_{0} - i_{\infty}) e^{- t / τ}

: find the start value (

t = 0

), the end value (

t \to \infty

) and

τ = \frac{L}{R _{e q}}

Quick Recall: tap to check

What does an uncharged inductor look like just after a switch is closed?

An open circuit: its current starts at zero.

What does it look like long afterwards?

A plain wire:

\frac{d i}{d t} = 0

, so

V_{L} = 0

Current of

2 A

flows from

A

B

through a

1 H

coil and is rising at

1 A s^{- 1}

V_{A} - V_{B}

+ 1 V

: walking along the current, the potential falls by

L \frac{d i}{d t}

Does doubling the current in a coil change its

L

No.

L

depends only on geometry and core.

4. Energy Stored in an Inductor

Building up a current costs work, because the back emf fights you every step of the way. That work is not lost; it is stored, and you get it back when the current dies.

Figure 6: Building the current from

0

I

costs

U = \int_{0}^{I} L i d i = \frac{1}{2} L I^{2}

, the shaded triangle. That energy sits in the magnetic field, at

u = \frac{B ^{2}}{2 μ _{0}}

per unit volume.

To keep the current growing, the source must supply power $P = ε i = (L \frac{d i}{d t}) i$ .
Work done in time $d t$ : $d W = L i d i$ .
Total work in taking the current from $0$ to $I$ :
$U = \int_{0}^{I} L i d i = \frac{1}{2} L I^{2}$

The energy is stored in the magnetic field itself, with energy density

u = \frac{B ^{2}}{2 μ _{0}}

joule per cubic metre (replace $μ_{0}$ by $μ_{0} μ_{r}$ inside a magnetic material).

The two expressions must agree. For a solenoid, $B = μ_{0} ni$ and the volume is $A l$ , so

U = \frac{B ^{2}}{2 μ _{0}} (A l) = \frac{μ _{0}^{2} n ^{2} i ^{2}}{2 μ _{0}} A l = \frac{1}{2} (μ_{0} n^{2} A l) i^{2} = \frac{1}{2} L i^{2}

which is exactly the same thing, a useful check in any energy question.

JEE Advanced

The field energy formula works even where there is no coil at all. Inside a straight wire of radius $R$ carrying a uniform current density $J$ , Ampere's law gives $B = \frac{μ _{0} J r}{2}$ at radius $r$ , so the energy stored per unit length is

\frac{U}{l} = \int_{0}^{R} \frac{B ^{2}}{2 μ _{0}} (2 π r d r) = \frac{μ _{0} J ^{2} π}{4} \int_{0}^{R} r^{3} d r = \frac{μ _{0} J ^{2} π R ^{4}}{16}

This is the standard way JEE Advanced turns an inductance question into a calculus question.

Key idea

An inductor carrying current

i

stores

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

in its magnetic field, at

u = \frac{B ^{2}}{2 μ _{0}}

per unit volume; the flux linkage

L i

can never jump.

5. The L-R Circuit

Put a coil and a resistor in series with a cell and the current does not jump to its final value; it eases into it. Take the cell away and it eases back down.

Figure 7: Key at 1:

ε - i R - L \frac{d i}{d t} = 0

(growth). Key at 2:

i R + L \frac{d i}{d t} = 0

(decay, same direction of current). Both are first-order equations with

τ = \frac{L}{R}

5.1 Growth of current

Loop rule with the key closed: $ε - i R - L \frac{d i}{d t} = 0$ .
Separating the variables and integrating from $i = 0$ at $t = 0$ :
$\int_{0}^{i} \frac{d i}{ε - i R} = \frac{1}{L} \int_{0}^{t} d t$
Rearranging,
$i = \frac{ε}{R} (1 - e^{- R t / L}) = i_{0} (1 - e^{- t / τ})$

The voltage across the inductor falls the other way, $V_{L} = ε e^{- t / τ}$ : it takes the whole emf at the first instant and nothing at all in the end.

Figure 8: The inductor takes the whole emf at first (

V_{L} = ε e^{- t / τ}

) and none at the end; the resistor does the opposite. The two are equal at

t = τ ln 2 \approx 0.69 τ

5.2 Decay of current

With the cell removed and the loop closed on itself, $i R + L \frac{d i}{d t} = 0$ , which integrates to

i = i_{0} e^{- R t / L} = i_{0} e^{- t / τ}

All the energy that was stored in the field, $\frac{1}{2} L i_{0}^{2}$ , eventually comes out as heat in $R$ .

Figure 9: Growth

i = i_{0} (1 - e^{- t / τ})

and decay

i = i_{0} e^{- t / τ}

, both with

τ = \frac{L}{R}

. After one

τ

the current has reached

63%

of its final value, or fallen to

37%

5.3 The time constant

τ = \frac{L}{R}

After one time constant the growing current has reached $63%$ of its final value, and the decaying current has fallen to $37%$ of its initial value. A large $L$ or a small $R$ makes the circuit sluggish.

Do not mix this up with the capacitor circuit, where $τ = R C$ . In an L-R circuit the resistance is in the denominator, so increasing $R$ makes the current settle faster, which is the opposite of what happens with a capacitor.

Quick Recall: tap to check

After one time constant, what fraction of the final current has a growing current reached?

63%

(

1 - e^{- 1}

How long does a growing current take to reach half its final value?

t = τ ln 2 \approx 0.69 τ

Increasing

R

in an L-R circuit makes the current settle faster or slower?

Faster, because

τ = \frac{L}{R}

gets smaller.

When the cell is removed, where does the stored energy go?

All of

\frac{1}{2} L i_{0}^{2}

becomes heat in

R

Key idea

Every L-R transient is one exponential with

τ = \frac{L}{R}

i = i_{\infty} + (i_{0} - i_{\infty}) e^{- t / τ}

6. Mutual Inductance

Put two coils near each other and the flux of one threads the other. Change the current in the first and an emf appears in the second, even though nothing is connected between them. That is the whole principle of the transformer.

Figure 10: A changing current in one circuit induces an emf in the other:

N_{2} ϕ_{2} = M i_{1}

and

ε_{2} = - M \frac{d i _{1}}{d t}

. By reciprocity the same

M

works both ways, and

M = k L_{1} L_{2}

with

k \leq 1

If a current $i_{1}$ in coil 1 produces a flux linkage $N_{2} ϕ_{2}$ in coil 2, then

N_{2} ϕ_{2} = M i_{1} and ε_{2} = - M \frac{d i _{1}}{d t}

where $M$ is the mutual inductance of the pair, measured in henry.

Reciprocity theorem: $M_{12} = M_{21} = M$ . Whichever coil you drive, the same number comes out, which is often the quickest route to an answer.
$M$ depends on how close the coils are, their sizes and turns, and above all their orientation. It is largest when they share an axis and zero when their axes are at right angles.
$M = k L_{1} L_{2}$ , where the coupling coefficient satisfies $0 \leq k \leq 1$ . So $M$ can never exceed $L_{1} L_{2}$ .

6.1 The standard calculation

Wind a small coil of $N_{2}$ turns round the middle of a long solenoid of $N_{1}$ turns, length $l$ and cross-section $A$ . Driving the solenoid gives $B = \frac{μ _{0} N _{1} i}{l}$ inside, so

N_{2} ϕ_{2} = N_{2} (\frac{μ _{0} N _{1} i}{l}) A \Rightarrow M = \frac{μ _{0} N _{1} N _{2} A}{l}

Notice that $M$ does not depend on the radius of the outer coil, because the solenoid keeps all its field inside.

Exam Trick

Use reciprocity to pick the easy direction. Two concentric coils of radii $a_{1} ≪ a_{2}$ look hard if you drive the small one, because its field is awkward far away. Drive the large one instead: its field at the centre is uniform over the small coil, and $M = \frac{μ _{0} π a _{1}^{2}}{2 a _{2}}$ falls out in one line.

Figure 11: Use the easy coil for the flux. Coaxial solenoids: drive the outer one, its field

μ_{0} n_{2} i

fills the inner area

π r_{1}^{2}

, so

M = μ_{0} n_{1} n_{2} π r_{1}^{2} l

. Concentric loops: drive the big one,

M = \frac{μ _{0} π a _{1}^{2}}{2 a _{2}}

7. Combinations of Inductors

Well separated inductors combine exactly like resistors, because the same equations govern them.

Figure 12: Well-separated inductors combine like resistors:

L = L_{1} + L_{2}

in series and

\frac{1}{L} = \frac{1}{L _{1}} + \frac{1}{L _{2}}

in parallel. If they share flux, the series value becomes

L_{1} + L_{2} \pm 2 M

Series: the same current flows through both, so the emfs add and $L = L_{1} + L_{2}$ .
Parallel: the same voltage sits across both, so the currents add and $\frac{1}{L} = \frac{1}{L _{1}} + \frac{1}{L _{2}}$ .
Coupled in series: if the two coils share flux, the series result becomes $L = L_{1} + L_{2} \pm 2 M$ , with a plus sign when the windings help each other and a minus sign when they oppose.

The $\pm 2 M$ term is easy to check: with maximum coupling ( $k = 1$ ) and $L_{1} = L_{2} = L$ , the aiding combination gives $4 L$ , which is correct because the two coils together act as one coil of twice the turns, and inductance goes as the square of the turns.

Figure 13: Mind map of inductance.

L

and

M

depend only on geometry and the core; everything else follows from

ε = - L \frac{d i}{d t}

8. Solved Examples

Solved Example 1

A coil of inductance

1.0 H

and resistance

100 Ω

is connected to a battery of emf

12 V

. Find the energy stored in the magnetic field of the coil

10 ms

after the circuit is switched on.

Solution:

Given: $L = 1.0 H$ , $R = 100 Ω$ , $ε = 12 V$ , $t = 10 ms$ .

Time constant: $\tau = \dfrac{L}{R} = \dfrac{1.0}{100} = 0.010\,\text{s} = 10\,\text{ms}$, so we are asked for the state after exactly one time constant.

Current:

i = \frac{ε}{R} (1 - e^{- t / τ}) = \frac{12}{100} (1 - e^{- 1}) = (0.12) (0.632) = 0.0759 A

Energy:

U = \frac{1}{2} L i^{2} = \frac{1}{2} (1.0) (0.0759)^{2} = 2.88 \times 1 0^{- 3} J

Answer: $U \approx 2.88 mJ$ . Note that this is only about $40%$ of the final stored energy $\frac{1}{2} L i_{0}^{2} = 7.2 mJ$ , because the energy goes as the square of the current.

Solved Example 2

A branch

A B

of a circuit contains, in order from

A

B

, an inductor of

1 H

, a cell of

5 V

(positive terminal towards

A

) and a resistor of

2 Ω

. The current of

2 A

flows from

A

B

. Find

V_{A} - V_{B}

when the current is (i) steady, (ii) increasing at

1 A s^{- 1}

, (iii) decreasing at

1 A s^{- 1}

Solution:

Write Kirchhoff's rule walking from $A$ to $B$ :

V_{A} - L \frac{d i}{d t} - 5 - 2 i = V_{B}

(i) $i = 2 A$ , $\frac{d i}{d t} = 0$ : $V_{A} - V_{B} = 0 + 5 + 4 = 9 V$ .

(ii) $i = 2 A$ , $\frac{d i}{d t} = + 1$ : $V_{A} - V_{B} = 1 + 5 + 4 = 10 V$ .

(iii) $i = 2 A$ , $\frac{d i}{d t} = - 1$ : $V_{A} - V_{B} = - 1 + 5 + 4 = 8 V$ .

Answer: $9 V$ , $10 V$ , $8 V$ . The inductor adds a volt when the current is rising and gives one back when it is falling; the resistor and the cell do not care.

Solved Example 3

An ideal cell of emf

ε

is connected through a switch to a pure inductor

L

with no resistance anywhere. The switch is closed at

t = 0

. Find the current as a function of time.

Solution:

Loop rule with $R = 0$ :

ε = L \frac{d i}{d t}

Integrate from $i = 0$ at $t = 0$ :

\int_{0}^{i} L d i = \int_{0}^{t} ε d t \Rightarrow L i = εt

i = \frac{εt}{L}

Answer: $i = \frac{εt}{L}$ , rising without limit. With no resistance there is no final current: the time constant $τ = L / R$ is infinite, and $1 - e^{- t / τ} \to t / τ$ for small $t / τ$ , which is exactly this answer.

Solved Example 4

The current in a coil of self-inductance

L = 2 H

is increased slowly from zero. Find the energy supplied by the source by the time the current reaches

2 A

Solution:

Key idea. The work done against the back emf depends only on the initial and final currents, not on how the current got there. So the detailed time dependence in the question is irrelevant.

U = \frac{1}{2} L i^{2} = \frac{1}{2} (2) (2)^{2} = 4 J

Answer: $4 J$ . If you are asked to prove it, start from $d W = ε i d t = L i d i$ and integrate from $0$ to $2 A$ : the time variable cancels, exactly as it does for induced charge in Faraday's law.

Solved Example 5

A cell of emf

12 V

is connected in series with a resistance

6 Ω

and an inductance

3 H

. Find (a) the final current, (b) the time constant, (c) the current

0.5 s

after the key is closed, and (d) the time taken to reach half the final current.

Solution:

(a) Final current: at $t \to \infty$ the inductor is a plain wire, so

i_{0} = \frac{ε}{R} = \frac{12}{6} = 2.0 A

(b) Time constant: $τ = \frac{L}{R} = \frac{3}{6} = 0.50 s$ .

(c) At $t = 0.5 s = τ$ :

i = i_{0} (1 - e^{- 1}) = 2.0 (0.632) = 1.26 A

(d) Half the final current: put $i = 1.0 A$ :

1.0 = 2.0 (1 - e^{- t / τ}) \Rightarrow e^{- t / τ} = 0.5 \Rightarrow t = τ ln 2 = (0.50) (0.693) = 0.35 s

Answer: $2.0 A$ , $0.50 s$ , $1.26 A$ and $0.35 s$ . The half-value time $τ ln 2$ is worth remembering; it is the same expression as a radioactive half-life.

Solved Example 6

In the same circuit the current has reached its steady value. The cell is now removed and the coil is left short-circuited through the same

6 Ω

. Find the current

1.0 s

later, and the total heat produced in the resistance.

Solution:

Decay law: $i = i_{0} e^{- t / τ}$ with $i_{0} = 2.0 A$ and $τ = 0.50 s$ .

At $t = 1.0 s = 2 τ$ :

i = 2.0 e^{- 2} = 2.0 (0.135) = 0.27 A

Total heat. Once the cell is gone, the only energy available is what was stored in the field, and all of it ends up in $R$ :

Q = \frac{1}{2} L i_{0}^{2} = \frac{1}{2} (3) (2.0)^{2} = 6.0 J

Answer: $0.27 A$ and $6.0 J$ . You could get the heat by integrating $i^{2} R$ from $0$ to infinity, but energy conservation gives it in one step.

Solved Example 7

A solenoid of length

0.50 m

and cross-sectional area

4.0 cm^{2}

has

1000

turns. Find its self-inductance, and the energy stored when it carries

2.0 A

Solution:

Given: $l = 0.50 m$ , $A = 4.0 \times 1 0^{- 4} m^{2}$ , $N = 1000$ .

Inductance:

L = \frac{μ _{0} N ^{2} A}{l} = \frac{( 4 π \times 1 0 ^{- 7} ) ( 1000 ) ^{2} ( 4.0 \times 1 0 ^{- 4} )}{0.50}

L = 1.0 \times 1 0^{- 3} H = 1.0 mH

Energy at $2.0 A$ :

U = \frac{1}{2} L i^{2} = \frac{1}{2} (1.0 \times 1 0^{- 3}) (2.0)^{2} = 2.0 \times 1 0^{- 3} J

Answer: $L = 1.0 mH$ and $U = 2.0 mJ$ . Convert $cm^{2}$ to $m^{2}$ first; forgetting the factor of $1 0^{- 4}$ is the usual way this question goes wrong.

Solved Example 8

A circuit contains a cell of emf

ε

, a resistance

R

and an inductor

L

in series, and has been running long enough for the current to be steady. The inductance is now suddenly reduced to

\frac{L}{λ}

. Find the current immediately afterwards.

Solution:

Key idea. The flux linkage of an inductor cannot change instantly, because an instant jump in $N ϕ$ would need an infinite emf. So $L i$ is conserved across the sudden change, even though $i$ itself is not.

Before the change: the steady current is $i_{1} = \frac{ε}{R}$ with inductance $L$ .

Conserve the flux linkage:

L i_{1} = \frac{L}{λ} i_{2} \Rightarrow i_{2} = λ i_{1}

i_{2} = \frac{λ ε}{R}

Answer: $i_{2} = \frac{λ ε}{R}$ . The current jumps up by a factor $λ$ and then relaxes back to $\frac{ε}{R}$ with the new, shorter time constant. This is the inductor version of "momentum is conserved in a sudden change".

Solved Example 9

A long solenoid of length

1 m

and cross-sectional area

10 cm^{2}

has

1000

turns. A small coil of

20

turns is wound about its centre. Find the mutual inductance, and the emf induced in the coil when the current in the solenoid changes at

10 A s^{- 1}

Solution:

Drive the solenoid. Its field inside is $B_{1} = \frac{μ _{0} N _{1} i _{1}}{l}$ , uniform over the small coil, so

N_{2} ϕ_{2} = N_{2} (\frac{μ _{0} N _{1} i _{1}}{l}) A \Rightarrow M = \frac{μ _{0} N _{1} N _{2} A}{l}

Substitute $N_{1} = 1000$ , $N_{2} = 20$ , $A = 10 \times 1 0^{- 4} m^{2}$ , $l = 1 m$ :

M = \frac{( 4 π \times 1 0 ^{- 7} ) ( 1000 ) ( 20 ) ( 10 \times 1 0 ^{- 4} )}{1} = 2.51 \times 1 0^{- 5} H

Induced emf:

∣ ε_{2} ∣ = M \frac{d i _{1}}{d t} = (2.51 \times 1 0^{- 5}) (10) = 2.51 \times 1 0^{- 4} V

Answer: $M = 2.51 \times 1 0^{- 5} H$ and $∣ ε_{2} ∣ = 2.51 \times 1 0^{- 4} V$ . The radius of the outer coil never appears, because a long solenoid keeps its field entirely inside.

Solved Example 10

The coefficient of mutual induction between the primary and the secondary of a transformer is

5 H

. Find the emf induced in the secondary when a current of

3 A

in the primary is cut off in

2.5 \times 1 0^{- 4} s

Solution:

Given: $M = 5 H$ , $∣Δ i ∣ = 3 A$ , $Δ t = 2.5 \times 1 0^{- 4} s$ .

∣ ε_{2} ∣ = M \frac{Δ i}{Δ t} = 5 \times \frac{3}{2.5 \times 1 0 ^{- 4}} = 6 \times 1 0^{4} V

Answer: $6 \times 1 0^{4} V$ , that is $60 kV$ from a supply of a few volts. This is exactly how the spark coil of a petrol engine works, and it is also why switches in inductive circuits arc.

Solved Example 11

Two circular coils of radii

a_{1}

and

a_{2}

with

a_{1} ≪ a_{2}

are placed with a common centre. Find their mutual inductance when (a) the coils are coplanar, (b) their planes are perpendicular, (c) their planes make an angle

θ

Solution:

Choose the easy direction. By reciprocity it does not matter which coil we drive, so drive the large one: its field near the centre, $B = \frac{μ _{0} i}{2 a _{2}}$ , is uniform over the whole of the small coil.

(a) Coplanar. The field is perpendicular to the plane of both coils, so

ϕ_{1} = \frac{μ _{0} i}{2 a _{2}} (π a_{1}^{2}) \Rightarrow M = \frac{μ _{0} π a _{1}^{2}}{2 a _{2}}

(b) Planes perpendicular. The field of the large coil now lies in the plane of the small coil, so no flux passes through it and $M = 0$ .

(c) Planes at angle $θ$ . Only the component along the small coil's normal counts:

M = \frac{μ _{0} π a _{1}^{2} cos θ}{2 a _{2}}

Answer: $\frac{μ _{0} π a _{1}^{2}}{2 a _{2}}$ , zero, and $\frac{μ _{0} π a _{1}^{2} cos θ}{2 a _{2}}$ . Driving the small coil instead would need a messy integral over the large one, and would give the same answer.

Solved Example 12

A long straight wire of radius

R

carries a uniform current density

J

. Find the magnetic energy stored per unit length inside the wire.

Solution:

Field inside at radius $r$ , from Ampere's law applied to a circle of radius $r < R$ :

B (2 π r) = μ_{0} J (π r^{2}) \Rightarrow B = \frac{μ _{0} J r}{2}

Energy in a shell of radius $r$ , thickness $d r$ and unit length, whose volume is $2 π r d r$ :

d U = \frac{B ^{2}}{2 μ _{0}} (2 π r d r) = \frac{1}{2 μ _{0}} \cdot \frac{μ _{0}^{2} J ^{2} r ^{2}}{4} \cdot 2 π r d r = \frac{μ _{0} J ^{2} π}{4} r^{3} d r

Integrate from $0$ to $R$ :

\frac{U}{l} = \frac{μ _{0} J ^{2} π}{4} \cdot \frac{R ^{4}}{4} = \frac{μ _{0} J ^{2} π R ^{4}}{16}

Answer: $\frac{U}{l} = \frac{μ _{0} J ^{2} π R ^{4}}{16}$ . Writing it in terms of the total current $I = J π R^{2}$ gives $\frac{μ _{0} I ^{2}}{16 π}$ , which is the famous result that the internal self-inductance of any straight wire is $\frac{μ _{0}}{8 π}$ per unit length, whatever its radius.

Solved Example 13

12 V

ideal cell is connected through a switch to a

4 Ω

resistor in series with a parallel combination of a

6 Ω

resistor and an ideal inductor (no resistance). The current drawn from the cell just after the switch is closed, and a long time later, are
(A)

1.2 A

and

3.0 A

(B)

3.0 A

and

1.2 A

(C)

0

and

3.0 A

(D)

1.2 A

and

1.2 A

Solution:

Just after closing: the inductor is an open circuit, so the current goes through $4 Ω$ and $6 Ω$ in series: $i = \frac{12}{10} = 1.2 A$ .

Long after: the inductor is a plain wire and shorts out the $6 Ω$ : $i = \frac{12}{4} = 3.0 A$ .

Answer: (A). (C) is the trap: only the inductor's own branch starts at zero, not the whole circuit.

Solved Example 14

The time constant of an L-R circuit is doubled when
(A) both

L

and

R

are doubled
(B)

L

is doubled, or

R

is halved
(C)

L

is halved
(D)

R

is doubled

Solution:

$τ = \frac{L}{R}$ doubles if the numerator doubles or the denominator halves.

Answer: (B). (A) leaves $τ$ unchanged; (C) and (D) halve it. Compare $R C$ , where doubling $R$ doubles $τ$ .

Solved Example 15

The current in a coil falls uniformly from

5.0 A

2.0 A

0.10 s

, and an emf of

50 V

is induced in it. Find its self-inductance and the energy it gives up.

Solution:

Inductance: $L = \frac{∣ ε ∣}{∣ d i / d t ∣} = \frac{50}{3.0/0.10} = 1.67 H$ .

Energy released: $Δ U = \frac{1}{2} L (i_{1}^{2} - i_{2}^{2}) = \frac{1}{2} (1.67) (25 - 4) = 17.5 J$ .

Answer: $L \approx 1.7 H$ , $Δ U \approx 17.5 J$ . Energy goes as $i^{2}$ , so subtract the squares, not the currents.

Solved Example 16

Two long coaxial solenoids share a length of

0.50 m

. The inner one has radius

2.0 cm

and

1000

turns per metre; the outer one has radius

3.0 cm

and

500

turns per metre. Find their mutual inductance, their self-inductances and the coupling coefficient

k

Solution:

Mutual inductance: drive the outer solenoid; its field $μ_{0} n_{2} i$ passes through the inner area $π r_{1}^{2}$ and links $n_{1} l$ turns:

M = μ_{0} n_{1} n_{2} π r_{1}^{2} l = (4 π \times 1 0^{- 7}) (1000) (500) π (0.020)^{2} (0.50) = 3.9 \times 1 0^{- 4} H

Self-inductances: $L_{1} = μ_{0} n_{1}^{2} π r_{1}^{2} l = 7.9 \times 1 0^{- 4} H$ and $L_{2} = μ_{0} n_{2}^{2} π r_{2}^{2} l = 4.4 \times 1 0^{- 4} H$ .

Coupling: $k = \frac{M}{L _{1} L _{2}} = \frac{3.9 \times 1 0 ^{- 4}}{5.9 \times 1 0 ^{- 4}} = 0.67$ , which is exactly $\frac{r _{1}}{r _{2}}$ .

Answer: $M \approx 0.39 mH$ , $L_{1} \approx 0.79 mH$ , $L_{2} \approx 0.44 mH$ , $k \approx 0.67$ . $k < 1$ because part of the outer solenoid's flux passes outside the inner one.

Practice Questions

Two concentric coplanar circular loops have radii $a$ and $b$ with $a ≪ b$ . A current $i = 2 t$ flows in the smaller loop, and the larger loop has resistance $R$ . Find the mutual inductance, the emf induced in the larger loop and the current in it.Answer: $M = \frac{μ _{0} π a ^{2}}{2 b}$ (use reciprocity), $∣ ε ∣ = M \frac{d i}{d t} = \frac{μ _{0} π a ^{2}}{b}$ and $i = \frac{μ _{0} π a ^{2}}{b R}$ .
Two coils of self-inductance $4 H$ and $6 H$ have a mutual inductance of $2 H$ . Find the equivalent inductance when they are joined in series, first aiding and then opposing. What would the parallel value be if they were far apart?Answer: Aiding $4 + 6 + 4 = 14 H$ ; opposing $4 + 6 - 4 = 6 H$ ; parallel with no coupling $\frac{4 \times 6}{4 + 6} = 2.4 H$ .
An inductor of $200 mH$ carries a steady current of $3.0 A$ . Find the energy stored, and the average emf induced if the current is switched off in $20 ms$ .Answer: $U = \frac{1}{2} L i^{2} = 0.90 J$ and $∣ ε ∣ = L \frac{Δ i}{Δ t} = 30 V$ .
Two current-time curves for different L-R circuits are drawn on the same axes, one rising much more steeply than the other. Which has the smaller time constant, and what does that mean physically?Answer: The steeper curve. A smaller $τ = L / R$ means a smaller inductance or a larger resistance, so the current settles sooner.
Prove that in an L-R circuit the growing current reaches $63%$ of its final value after one time constant.Answer: Put $t = τ$ in $i = i_{0} (1 - e^{- t / τ})$ : $i = i_{0} (1 - e^{- 1}) = 0.632 i_{0}$ .
In an L-R circuit the current is already $i_{0}$ when a cell of emf $ε$ is switched in. Find the current as a function of time.Answer: Solving $ε - i R - L d i / d t = 0$ with $i = i_{0}$ at $t = 0$ gives $i = \frac{ε - ( ε - i _{0} R ) e ^{- R t / L}}{R}$ .
Find the self-inductance of a solenoid of $500$ turns, length $25 cm$ and cross-section $5.0 cm^{2}$ wound on an iron core of relative permeability $800$ .Answer: $L = \frac{μ _{r} μ _{0} N ^{2} A}{l} \approx 0.50 H$ .

Common Mistakes to Avoid

Watch out

Saying an inductor opposes current. It opposes only the change in current; a steady current passes through it with no voltage drop at all.
Treating the inductor as a wire at $t = 0$ . At the first instant it is an open circuit; it becomes a wire only after a long time.
Using $τ = R C$ for an L-R circuit. Here $τ = \frac{L}{R}$ , so a larger resistance makes the circuit settle faster, not slower.
Thinking $L$ depends on the current or the emf. It depends only on the geometry, the number of turns and the core material.
Using the final steady current in $\frac{1}{2} L i^{2}$ when the question asks for the energy at some intermediate time. Find $i$ at that instant first.
Forgetting the $\pm 2 M$ term for coupled coils in series, or thinking $M$ can exceed $L_{1} L_{2}$ . The coupling coefficient satisfies $k \leq 1$ .
Assuming the current through an inductor can jump when something in the circuit is changed suddenly. It cannot; the flux linkage $L i$ is what stays continuous.
Writing the voltage across an inductor as $i R$ or as $\frac{q}{C}$ . It is $L \frac{d i}{d t}$ , and nothing else.

Frequently Asked Questions

What is self-inductance in simple words?

Self-inductance is a coil's unwillingness to let its own current change. Changing the current changes the coil's own flux, which induces an emf in the coil that fights the change. The number measuring this is $L$ , defined by $N ϕ = L i$ and $ε = - L d i / d t$ .

Why is inductance called electrical inertia?

Because the equations match. Force equals mass times acceleration becomes emf equals inductance times rate of change of current, and kinetic energy $\frac{1}{2} m v^{2}$ becomes stored energy $\frac{1}{2} L i^{2}$ . Current plays the part of velocity and inductance the part of mass.

How does an inductor behave just after and long after a switch is closed?

Just after closing, the current cannot jump, so the inductor acts like an open circuit and carries no current. Long afterwards the current is steady, $d i / d t = 0$ , so it acts like a plain wire. A capacitor behaves in exactly the opposite way.

What does the time constant of an L-R circuit mean?

It is $τ = L / R$ , the time in which a growing current reaches $63%$ of its final value or a decaying current falls to $37%$ of its starting value. A large inductance or a small resistance makes the circuit slow to respond.

Where is the energy of an inductor stored?

In the magnetic field itself, not in the wire. The energy density is $u = B^{2} /2 μ_{0}$ joule per cubic metre, and integrating that over the volume of a solenoid gives back exactly $\frac{1}{2} L i^{2}$ , which is a good check in any energy problem.

What is the reciprocity theorem for mutual inductance?

It says $M_{12} = M_{21}$ : the mutual inductance is the same whichever coil you treat as the primary. This is very useful, because one direction is often easy to compute and the other needs a difficult integral.

How is inductance tested in NEET?

NEET asks direct substitutions: the solenoid formula $L = μ_{0} n^{2} A l$ , the energy $\frac{1}{2} L i^{2}$ , the emf $L d i / d t$ for a given rate of change, the time constant $L / R$ and simple mutual-inductance numericals such as a coil wound on a solenoid or a transformer primary switched off.

What inductance questions come in JEE Main and JEE Advanced?

JEE Main uses L-R circuits solved with the open-circuit and plain-wire substitutions, energy and time constant. JEE Advanced adds sudden changes where the flux linkage $L i$ is conserved, coupled coils with $\pm 2 M$ , and magnetic field energy integrals inside wires and solenoids.

Previous year questions on Inductance

15 questions from past papers, each with a step-by-step solution.

Show all 15 questions

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Practice questions →Take a mock test

1. Self-Induction and the Coefficient L

1.1 Inductance is electrical inertia

2. Inductance of a Long Solenoid

3. The Inductor in a Circuit

3.1 Which way does the voltage go?

3.2 The two limits that solve most problems

4. Energy Stored in an Inductor

5. The L-R Circuit

5.1 Growth of current

5.2 Decay of current

5.3 The time constant

6. Mutual Inductance

6.1 The standard calculation

7. Combinations of Inductors

8. Solved Examples

Common Mistakes to Avoid

Frequently Asked Questions

What is self-inductance in simple words?

Why is inductance called electrical inertia?

How does an inductor behave just after and long after a switch is closed?

What does the time constant of an L-R circuit mean?

Where is the energy of an inductor stored?

What is the reciprocity theorem for mutual inductance?

How is inductance tested in NEET?

What inductance questions come in JEE Main and JEE Advanced?

Previous year questions on Inductance

Ready to master Electromagnetic Induction?

1. Self-Induction and the Coefficient $L$