Home › Physics › Electricity and Magnetism › Self Induction
Electricity and Magnetism
Self Induction
Self induction is the phenomenon by which a changing current in a single electric circuit induces an opposing electromotive force (emf) within the same circuit. The magnitude of this induced emf is proportional to the time rate of change of the current through the inductor.
UP TGT ScienceUP PGT PhysicsStudy Notes + Practice
Scope and prerequisites
TGT core
Definition of self-inductance, factors affecting self-inductance, self-inductance of a long solenoid, and growth/decay of current in an L-R circuit.
PGT extension
Detailed field-theoretic derivation of magnetic energy density stored in an inductor, coefficient of self-induction for complex geometries like toroidal coils, and transient response analysis in coupled inductive circuits.
Prerequisites
Faraday's laws of electromagnetic induction
Lenz's law and magnetic flux definition
Ampere's circuital law and magnetic field of a solenoid
Origin and Nature of Self-Induction
When a varying electric current passes through a coil, the magnetic flux linked with each turn of the coil changes in proportion to the current.
By Faraday's law, this change in flux induces an emf in the coil itself, termed as the back emf.
According to Lenz's law, this induced emf always opposes the change in current that produces it, acting as electrical inertia.
Coefficient of Self-Induction (Self-Inductance)
Magnetic flux linkage NΦ is directly proportional to the current I flowing through the circuit, given by N$\Phi = L I$.
The constant of proportionality L is defined as the coefficient of self-induction or self-inductance of the coil.
Self-inductance depends strictly on the geometrical parameters of the conductor, number of turns, and the magnetic permeability of the core medium.
Stored Magnetic Energy in an Inductor
Work must be done by an external source to establish a current against the opposing self-induced back emf.
This electrical work is stored entirely within the magnetic field enclosed by the inductor.
The total magnetic potential energy stored when current reaches steady value I is given by $U = (1/2) L I^{2}$.
Growth and Decay of Current in L-R Circuit
When a DC voltage source is connected to a series combination of resistor R and inductor L, the current grows exponentially according to $I = (V/R)(1 - exp(-t/\tau))$.
The time constant $\tau = L/R characterizes the rate at$ which the current approaches its maximum steady-state value.
During decay, when the voltage source is shorted, the current decays exponentially as $I = I_0 exp(-t/\tau)$.
Derivations
Self-Inductance of a Long Solenoid
Conditions: Air-core long solenoid of length l, cross-sectional area A, and total turns N, where l is much greater than its radius.
Write the magnetic field inside an ideal long solenoid carrying current I: $B = \mu_0 n I$, where $n = N/l$ is the number of turns per unit length.
Calculate the magnetic flux Φ through a single turn of the solenoid: Ph$i = B A = \mu_0 \left(\frac{N}{l}\right) I A$.
Determine the total magnetic flux linkage for N turns: NPh$i = N \left[ \mu_0 \left(\frac{N}{l}\right) I A \right] = \frac{\mu_0 N^2 A}{l} I$.
Equate total flux linkage to L I to extract the self-inductance L: $L = \frac{NPhi}{I} = \frac{\mu_0 N^2 A}{l}$.
Result: $L = \frac{\mu_0 N^2 A}{l} or in terms of magnetic permeability \mu$ and small n: $L = \mu n^2 A l$
Energy Stored in an Inductor
Conditions: An inductor of self-inductance L carrying a growing current I at any instant t.
The instantaneous power supplied to overcome the back emf e is P = e I, where magnitude of back emf is e = L \frac{dI}{dt}.
Express the incremental work done dW in time dt as dW = P dt = left( L \frac{dI}{dt} \right) I dt = L I dI.
Integrate the incremental work from initial current 0 to final steady current I_0: $U = \int_{0}^{I_0} L I dI$.
Evaluate the integral to obtain the total magnetic energy: $U = L \left[ \frac{I^2}{2} \right]_{0}^{I_0} = \frac{1}{2} L I_0^2$.
Result: $U = \frac{1}{2} L I^2$
Formula sheet
Formula / law
Equation
Variables and units
Conditions
Definition of Self-Inductance
$$NPhi = L I$$
N = number of turns, Phi = magnetic flux per turn, $L = self-inductance$, I = current Weber (Wb), Henry (H), Ampere (A)
Linear magnetic medium where permeability is constant
Induced EMF in an Inductor
$$e = -L \frac{dI}{dt}$$
e = induced electromotive force, $L = self-inductance$, dI/dt = rate of change of current Volt (V), Henry (H), Ampere per second (A/s)
L is constant with respect to time
Self-Inductance of Long Solenoid
$$L = \frac{\mu_0 N^2 A}{l}$$
m$u_0 = permeability of free space$, N = total turns, $A = cross-sectional area$, l = length H, H/m, dimensionless, m^2, m
End effects neglected (l >> radius)
Magnetic Energy Stored
$$U = \frac{1}{2} L I^2$$
U = stored magnetic potential energy, $L = self-inductance$, I = current Joule (J), Henry (H), Ampere (A)
Current established from zero to I
Magnetic Energy Density
$$u = \frac{B^2}{2 \mu_0}$$
u = magnetic energy per unit volume, B = magnetic field magnitude, m$u_0 = permeability$ J/m^3, Tesla (T), T m/A
Uniform magnetic field
Time Constant of L-R Circuit
$$\tau = \frac{L}{R}$$
\tau = time constant, $L = self-inductance$, R = resistance Second (s), Henry (H), Ohm (Omega)
Series connection of pure L and R
Growth of Current in L-R Circuit
$$I(t) = \frac{V}{R} (1 - e^{-\frac{Rt}{L}})$$
I(t) = current at time t, V = applied DC voltage, R = resistance, L = inductance A, V, Omega, H, s
Switch closed at t = 0
Dimensions of Self-Inductance
$$[L] = [M L^2 T^{-2} A^{-2}]$$
M = mass, L = length, T = time, A = electric current kg m^2 s^{-2} A^{-2}
Dimensional homogeneity
Graphs and experiments
Growth of Current in an L-R Circuit
Draw: A plot of current I on the y-axis versus time t on the x-axis, showing an asymptotic exponential curve approaching maximum value $I_0 = V/R$, with a tangent at t=0 intersecting the asymptote at t = \tau.
Interpret: Indicates the inertia of the inductor delaying the instantaneous rise of current due to back emf.
Exam use: Questions testing the definition of time constant as the time required for current to reach (1 - 1/e) or about 63.2% of its maximum value.
Decay of Current in an L-R Circuit
Draw: A downward sloping exponential decay curve of current I versus time t starting from initial value I_0 and asymptotically approaching zero.
Interpret: Represents the dissipation of stored magnetic energy through the resistor when the driving voltage source is removed.
Exam use: Evaluating current magnitude at t = \tau, where current falls to 1/e (approx 36.8%) of its initial value.
Magnetic Flux Linkage versus Current
Draw: A straight line passing through the origin on a NPhi vs I coordinate system.
Interpret: The slope of this linear graph represents the constant coefficient of self-induction L for linear ferromagnetic or air-core systems.
Exam use: Numerical interpretation where slope calculation yields self-inductance.
Solved numericals
TGT core
A coil of 500 turns has a magnetic flux of 4 \times 10^{-3} Wb linked with it when a current of 2 A flows through it. Calculate the self-inductance of the coil.
Identify given values: N = 500, Ph$i = 4 \times 10^{-3} Wb$, I = 2 A.
Use the formula for flux linkage: NPhi = L I.
Substitute the values: 500 \times 4 \times 10^{-3} = L \times 2.
Calculate L: 2 = 2 L implies L = 1 H.
Answer: 1 H
PGT extension
A long solenoid has 1000 turns per meter and a cross-sectional area of 10 cm^2. If a current of 5 A flows through it, calculate the total magnetic energy stored in the solenoid within a length of 0.5 meters.
Find self-inductance per unit length or total inductance for length l = 0.5 m. Using $n = 1000 m^{-1}$, N = n \times l = 1000 \times 0.5 = 500 turns.
Convert area to SI units: $A = 10 cm^2 = 10 \times 10^{-4} m^2 = 10^{-3} m^2$.
Calculate self-inductance L using formula $L = \mu_0 n^2 A l: L = (4pi \times 10^{-7}) \times (1000)^2 \times (10^{-3}) \times 0.5$.
An inductor of inductance 2 H and negligible resistance is connected across a source of emf. If the current changes at a constant rate of 10 A/s, find the magnitude of the induced back emf and the power absorbed by the inductor at t = 2 seconds.
Recall the formula for induced emf: e = L \frac{dI}{dt}.
Substitute given values: L = 2 H and \frac{dI}{dt} = 10 A/s.
Calculate emf: e = 2 \times 10 = 20 V.
Power absorbed by the inductor against back emf is given by P = e I. Since current increases at 10 A/s, at t = 2 s, current $I = (dI/dt) \times t = 10 \times 2 = 20 A$.
Calculate power: P = 20 V \times 20 A = 400 W.
Answer: Emf = 20 V, Power = 400 W
Assumptions and limitations
The formulas assume a linear magnetic core where magnetic permeability mu remains constant and independent of the applied magnetic field strength (absence of magnetic saturation and hysteresis losses).
Common misconceptions
Common error: An inductor opposes the flow of electric current through it.
Correct: An inductor opposes only the change in electric current, not the steady current itself.
When current is steady (dI/dt = 0), the induced emf is zero, and a pure inductor acts as a simple conducting wire with zero resistance.
Common error: Self-inductance depends on the magnitude of current flowing through the coil.
Correct: Self-inductance is purely a geometric and material property of the inductor.
Although flux linkage NPhi is proportional to current I, their ratio $L = NPhi/I remains constant$ for a rigid geometrical configuration.
Quick revision
Self-inductance L represents electrical inertia, measured in Henry (H) or Wb/A.
Stored magnetic energy in an inductor is given by $U = (1/2) L I^2$.
Time constant of an L-R circuit is $\tau = L/R$, representing the inertia of transient response.
Must remember
Self-induction is the phenomenon where a changing current in a coil induces a back emf in the same coil, opposing the change.
The self-inductance $L$ is defined by the flux linkage relation $N\Phi = L I$, where $N\Phi$ is total flux linkage and $I$ is current.
Self-inductance $L$ depends entirely on geometric parameters, number of turns, and core material permeability, not on current or voltage.
The magnetic energy stored in an inductor carrying a steady current $I$ is given by $U = \frac{1}{2} L I^2$.
The time constant of an $L-R$ circuit is given by $\tau = \frac{L}{R}$, representing the time taken for current to reach about $63.2\%$ of its maximum value during growth.
Glossary
Self-Inductance ($L$)The property of a coil by which it opposes any change in the strength of electric current flowing through it; measured in Henry ($H$).
Back EMF ($e$)The induced electromotive force in a circuit that opposes the primary change in current, given by $e = -L \frac{dI}{dt}$.
Time Constant ($\tau$)In an $L-R$ circuit, the ratio $\frac{L}{R}$ that characterizes the rate of growth or decay of current.
Magnetic Energy Density ($u$)The magnetic potential energy stored per unit volume in a magnetic field, expressed as $u = \frac{B^2}{2\mu_0}$.
Growth vs Decay of Current in $L-R$ Circuit
Parameter
Growth of Current
Decay of Current
Equation
$I(t) = \frac{V}{R} (1 - e^{-\frac{Rt}{L}})$
$I(t) = I_0 e^{-\frac{Rt}{L}}$
Initial Current ($t=0$)
$0$
$I_0 = \frac{V}{R}$
Final Steady Current ($t=\infty$)
$\frac{V}{R}$
$0$
Role of Inductor
Opposes current growth (electrical inertia)
Opposes current decay by sustaining current
Practice Quiz
Choose an option to reveal the explanation.
1. What is the physical significance of self-induction acting as 'electrical inertia'?
Explanation: According to Lenz's law, the induced back emf opposes the change in current that produces it, analogous to mechanical mass opposing changes in velocity.
2. The self-inductance $L$ of a coil depends upon:
Explanation: Self-inductance $L$ is strictly a structural property determined by geometry, turns, and core material, independent of current or voltage.
3. What is the dimensional formula of self-inductance $L$?
Explanation: From $e = -L \frac{dI}{dt}$, $L = \frac{e dt}{dI}$. Dimensions of $e$ are $[M L^2 T^{-3} A^{-1}]$, time is $[T]$, and current is $[A]$, yielding $[M L^2 T^{-2} A^{-2}]$. বাঙ্গा
4. If the number of turns in a long solenoid is doubled while keeping its length and radius constant, its self-inductance will:
Explanation: The self-inductance of a solenoid is $L = \frac{\mu_0 N^2 A}{l}$, meaning $L \propto N^2$. Doubling $N$ increases $L$ by a factor of $4$.
5. The magnetic energy stored in an inductor of inductance $L$ carrying current $I$ is given by:
Explanation: The work done against the back emf to establish current $I$ is integrated as $U = \int_0^I L I' dI' = \frac{1}{2} L I^2$.
6. What is the time constant $\tau$ of an $L-R$ circuit containing an inductance $L$ and resistance $R$ in series?
Explanation: The time constant characterizing the exponential growth or decay of current in an $L-R$ circuit is defined as $\tau = \frac{L}{R}$.
7. In a pure inductor connected to a steady DC source after a long time ($t \to \infty$), what is the induced emf?
Explanation: When current reaches a steady state, $\frac{dI}{dt} = 0$, so the induced back emf $e = -L \frac{dI}{dt}$ becomes zero.
8. What is the magnetic energy density $u$ inside a uniform magnetic field $B$?
Explanation: Magnetic energy per unit volume is derived as $u = \frac{B^2}{2\mu_0}$ for a linear magnetic medium.
9. Which of the following correctly describes the behavior of a pure inductor in a DC circuit at $t = 0$ right after the switch is closed?
Explanation: At $t = 0$, the rate of change of current is maximum, producing a back emf equal and opposite to the applied voltage, making the inductor act initially as an open circuit.
10. If the current in a coil changes at the rate of $2 \text{ A/s}$ and induces an emf of $4 \text{ V}$, what is the self-inductance of the coil?
Explanation: Using $|e| = L \frac{dI}{dt}$, we get $4 = L \times 2$, which gives $L = 2 \text{ H}$.
11. What happens to the time constant of an $L-R$ circuit if both the inductance $L$ and resistance $R$ are doubled?
Explanation: The time constant is $\tau = \frac{L}{R}$. If both $L$ and $R$ are doubled, the ratio $\frac{2L}{2R}$ remains $\frac{L}{R}$, so $\tau$ is unchanged.
12. Which misconception about self-induction is scientifically false?
Explanation: An inductor does not oppose steady DC current ($\frac{dI}{dt} = 0$); it only opposes changes in current.
13. During the decay of current in an $L-R$ circuit, the current drops to what fraction of its initial value $I_0$ in one time constant ($\tau = L/R$)?
Explanation: For decay, $I(t) = I_0 e^{-t/\tau}$. At $t = \tau$, $I(\tau) = I_0 e^{-1} \approx 0.368 I_0$.
14. A toroidal coil has a mean radius $r$ and cross-sectional area $A$ with total turns $N$. Its self-inductance is primarily determined by:
Explanation: For a toroid, $L = \frac{\mu_0 N^2 A}{2 \pi r}$, depending on core permeability, turns, area, and magnetic path length ($2 \pi r$).
15. Why is self-induction sometimes referred to as electromagnetic inertia?
Explanation: In mechanics, mass is inertia against motion change; in electromagnetism, inductance is electrical inertia against current change.
PYQ Patterns
These are pattern-based practice questions. Authentic year labels are only used when verified from a source.
1. The self-inductance of a coil of $500$ turns is $5 \text{ mH}$. What is the magnetic flux linkage through the coil when a current of $2 \text{ A}$ flows through it?
Explanation: Using $N\Phi = L I$, we find $N\Phi = (5 \times 10^{-3} \text{ H}) \times (2 \text{ A}) = 10 \times 10^{-3} \text{ Wb} = 0.01 \text{ Wb}$.
2. A solenoid of length $1.0 \text{ m}$ has a cross-sectional area of $10 \text{ cm}^2$ and $1000$ turns. What is its self-inductance? (Take $\mu_0 = 4\pi \times 10^{-7} \text{ T m/A}$)
3. An inductor with $L = 0.5 \text{ H}$ and $R = 10 \Omega$ is connected to a $100 \text{ V}$ DC supply. What is the final steady current in the circuit?
Explanation: At steady state, the inductor offers zero DC resistance, so $I_{\max} = \frac{V}{R} = \frac{100 \text{ V}}{10 \text{ \Omega}} = 10 \text{ A}$.
4. If current in a coil drops from $5 \text{ A}$ to $0 \text{ A}$ in $0.1 \text{ s}$ and induces an average emf of $50 \text{ V}$, what is the self-inductance of the coil?
Explanation: Magnitude of induced emf is $|e| = L \frac{\Delta I}{\Delta t}$. Thus, $50 = L \left(\frac{5}{0.1}\right) = L \times 50$, which yields $L = 1.0 \text{ H}$.
5. What is the time constant of an $L-R$ circuit where $L = 2 \text{ H}$ and $R = 40 \text{ \Omega}$?
6. What is the energy stored in a $2 \text{ H}$ inductor when a steady current of $3 \text{ A}$ flows through it?
Explanation: Using $U = \frac{1}{2} L I^2$, we get $U = \frac{1}{2} (2 \text{ H}) (3 \text{ A})^2 = 9 \text{ J}$.
Timed Topic Test
Attempt these questions under a ten-minute limit using a separate timer.
1. Which of the following expressions correctly represents the growth of current in an $L-R$ circuit connected to a DC voltage $V$?
Explanation: The standard solution for current growth in a series $L-R$ circuit is $I(t) = \frac{V}{R} \left(1 - e^{-\frac{Rt}{L}}\right)$.
2. If the total number of turns of a solenoid is doubled while its length is halved, keeping area constant, the self-inductance becomes:
Explanation: Since $L = \frac{\mu_0 N^2 A}{l}$, if $N \to 2N$ and $l \to l/2$, then $L' = \frac{\mu_0 (2N)^2 A}{l/2} = 8 \left(\frac{\mu_0 N^2 A}{l}\right) = 8L$.
3. A coil of self-inductance $10 \text{ H}$ carries a current of $2 \text{ A}$. If the current is reduced to zero in $1 \text{ ms}$, the average induced emf is:
Explanation: $|e| = L \frac{\Delta I}{\Delta t} = 10 \times \frac{2 - 0}{1 \times 10^{-3}} = \frac{20}{10^{-3}} = 20,000 \text{ V}$? Wait: $10 \times 2 = 20$, $20 / 10^{-3} = 20,000 \text{ V}$. Let's verify options: $20000$ is option 3 index? Wait, options are $20$, $200$, $2000$, $20000$. Let's check: index 3 is $20000$. Correct index is 3.
4. The magnetic energy density inside an inductor carrying a magnetic field of magnitude $0.2 \text{ T}$ in free space is:
Explanation: Energy density is $u = \frac{B^2}{2\mu_0} = \frac{(0.2)^2}{2\mu_0} = \frac{0.04}{2\mu_0}$.
5. What is the unit of self-inductance in fundamental SI base units?
Explanation: From dimensional formula $[M L^2 T^{-2} A^{-2}]$, the SI base units are $\text{kg m}^2 \text{s}^{-2} \text{A}^{-2}$ (or $\text{V s A}^{-1}$).
6. An inductor and a resistor are connected in series with a DC source. At what time $t$ will the current reach $63.2\%$ of its maximum value?
7. Two inductors of self-inductances $L_1$ and $L_2$ are connected in series opposition (with negligible mutual inductance). The equivalent self-inductance is:
Explanation: Without mutual coupling, inductances in series always add up algebraically: $L_{eq} = L_1 + L_2$ regardless of orientation terminology unless mutual inductance $M$ is specified.
8. Which of the following statements about self-induction is correct?
Explanation: Faraday-Lenz law states $e = -L \frac{dI}{dt}$, meaning induced emf is directly proportional to the rate of change of current $\frac{dI}{dt}$.
9. If a soft iron core is inserted into a current-carrying solenoid, its self-inductance:
Explanation: Inserting a ferromagnetic core like soft iron greatly increases the magnetic permeability $\mu$ of the core, thereby increasing self-inductance $L$ ($L = \frac{\mu N^2 A}{l}$).
10. A current of $4 \text{ A}$ is flowing in a coil of self-inductance $2 \text{ H}$. The magnetic energy stored per unit volume in a uniform magnetic field region of volume $0.5 \text{ m}^3$ is:
Explanation: Total energy $U = \frac{1}{2} L I^2 = \frac{1}{2} (2) (4)^2 = 16 \text{ J}$. Energy density $u = \frac{U}{\text{Volume}} = \frac{16 \text{ J}}{0.5 \text{ m}^3} = 32 \text{ J/m}^3$.