Self-inductance of a current-carrying coil is the property by which it opposes the change in the current flowing through it. It is mainly due to the self-induced emf within the coil itself. In simple terms, we can tell that self-inductance is a phenomenon where there is the voltage(e.m.f) induced in a wire carrying an electric current.

When a charged capacitor is connected across a coil, an alternating potential difference is observed across the coil. This combination is called an oscillator. The alternating potential is observed due to the principle of electromagnetic self-induction. When a time-varying potential difference is applied across a single-coil, an emf is induced in the coil due to self-induction. When a second coil is brought near the first coil, the other coil also gets an induced emf. This process is called mutual induction.

We all are very fond of the idea of wirelessly charging our phones. The foundation of wireless charging is based on the concept of mutual induction between two coils. There is a coil inside the phone and a coil inside the charger. The magnetic flux due to the coil in the charger continuously changes, due to which an emf is induced in the coil inside the phone.

In this article, we will study the concept of self-induction, the factors on which it depends, its applications, and its limitations.

Learn Everything About Electromagnetic Induction

Magnetic Flux

The total number of magnetic field lines(magnetic lines of force) passing normally through an area placed in a magnetic field is called the magnetic flux linked with that area.

For elementary area $dA$ of a surface flux linked $d\varphi =BdA\cos \theta$ or $d\varphi =\overrightarrow{B}\overrightarrow{dA}$

So, Net flux through the surface $\varphi =\oint \overrightarrow{B}.\overrightarrow{dA}=BA\cos \theta$

For – $N$ turns coil $\varphi =NBA\cos \theta$

(1) Unit and Dimension

Magnetic flux is a scalar quantity.

 Its S.I. unit is weber (wb),

CGS unit is Maxwell or

$\mathrm{G}\mathrm{a}\mathrm{u}\mathrm{s}\mathrm{s}\times \mathrm{c}{\mathrm{m}}^2;1\mathrm{w}\mathrm{b}={10}^8\mathrm{M}\mathrm{a}\mathrm{x}\mathrm{w}\mathrm{e}\mathrm{l}$

Other units:

$Tesla\times m^2=\frac{N\times m}{Amp}=\frac{Joule}{Amp}=\frac{volt\times Coulomb}{Amp}=Volt\times \sec =ohm\times Coulomb$
$=Henry\times Amp$

Its dimensional formula

$[\varphi ]=\left[M{L}^2{T}^{–2}{A}^{–1}\right]$

(2) Maximum and Zero flux

If $\theta =0^{\circ },$ i.e. plane of the area is held perpendicular to the direction of the magnetic field, then flux from the surface is maximum, and if $\theta ={90}^{\circ }$ i.e. plane is held parallel to the direction of the magnetic field, then the flux linked with the surface is zero.

${\varphi }_{max}=BA;{\varphi }_{min}=0$

 

Note:

1. In case of a body present in a field, either a uniform or non-uniform, outward flux is taken to be positive while inward negative and Net flux linked with a closed surface is zero i.e. $\varphi =\oint \overrightarrow{B}.\overrightarrow{ds}$

Faraday’s Experiment and Laws

(1) First experiment 

A coil links some of the magnetic flux from a source $S$. Suppose relative motion occurs between source $S$ and coil such that the flux linked with the coil changes, a current is induced in it. 

(2) Second experiment 

Consider two coils arranged as shown in the figure. Pass a steady current in one coil. The magnetic flux of the first coil links the other. If the current in the first coil changes, an electric current is induced in the second.

(3) Faraday’s first law 

Whenever magnetic flux linked with a circuit changes (or a moving conductor cuts the magnetic flux), an emf is induced in the circuit (or emf induces across the ends of the conductor) called induced emf. The induced emf persists only if there is a change or cutting of flux.  This law is called as Faraday’s first law.

(4) Faraday’s second law 

The induced emf is defined as the rate of change of magnetic flux linked with the circuit, i.e. $e=–\frac{d\varphi }{dt}$ For $N$ turns, $\mathrm{e}=–\frac{Nd\varphi }{dt}$

The negative sign indicates that induced emf $(e)$ opposes the change of flux. This law is called Faraday’s second law.

Static EMI

Inductance is the property of electrical circuits which opposes any change in the current in the circuit. 

Inductance is the inbuilt property of electrical circuits. It will always be present in an electrical circuit whether we want it or not; if a large emf is induced when the current in the circuit changes, it is said to have greater inductance. A straight current-carrying wire with no iron part in the circuit will have a lesser inductance value. If the circuit contains a circular coil having numerous turns, the induced emf which opposes the cause will be greater, and the circuit is said to have a higher value of self-inductance.

Inductance is called electrical inertia: Inductance is analogous to inertia in Newton’s laws of motion(mechanics). A body at rest opposes any attempt which tries to bring it in motion, and a body in motion opposes any attempt which tries to bring it to rest due to inertia. The inductance of an electrical circuit opposes any change of current in the circuit; thus, it is also called electrical inertia. 

Self-Induction

Whenever the electric current is passing through a coil or circuit changes, the magnetic flux linked with it also changes. As a result, an emf is induced in the coil or the circuit according to Faraday’s laws of electromagnetic induction. The induced e.m.f opposes the change that causes it. This phenomenon is called ‘self-induction’, and the resultant emf induced is called back emf; the current produced in the coil is called induced current.

i) Coefficient of self-induction: If no magnetic substances are present near the coil, the number of flux linkages with the coil is proportional to the current $i$. i.e. $N\varphi \propto i$ or  $N\varphi =Li$ ($N$ is the number of turns in the coil and $N\varphi –$ total flux linkage) where $L=\frac{N\varphi }{i}=$ coefficient of self-induction. If , $i=1\mathrm{a}\mathrm{m}\mathrm{p}$ $N=1$ then,$L=\varphi$ i.e. the coefficient of self-induction of a coil is equal to the flux linked with the coil when the current in it is $1$ amp.

By Faraday’s second law induced emf . $e=–N\frac{d\varphi }{dt}$ Which gives $e=–L\frac{di}{dt}$ ; If

$e=–L\frac{di}{dt}$ then $|e|=L$.

Hence, the coefficient of self-induction is equal to the emf induced in the coil when the rate of change of current is unity.

Note:

1. If we want to calculate the induced emf in an inductor, then we use the formula , and when we are asked to calculate the voltage across the inductor, then we use the formula $e=–L\frac{di}{dt}$ and when we are asked to calculate the voltage $V$  across the inductor, then we use the formula $V=|e|=\frac{di}{dt}\times L$

(ii) Units and dimensional formula of ‘L’S.I. unit: $\frac{\mathrm{w}\mathrm{e}\mathrm{b}\mathrm{e}\mathrm{r}}{Amp}=\frac{\mathrm{T}\mathrm{e}\mathrm{s}\mathrm{l}\mathrm{a}\mathrm{x}{m}^2}{Amp}=\frac{N\times m}{Am{p}^2}=\frac{\mathrm{J}\mathrm{o}\mathrm{u}\mathrm{l}\mathrm{e}}{Am{p}^2}=\frac{\mathrm{C}\mathrm{o}\mathrm{u}\mathrm{l}\mathrm{o}\mathrm{m}\mathrm{b}\mathrm{v}\mathrm{o}\mathrm{l}\mathrm{t}}{Am{p}^2}=\frac{\mathrm{v}\mathrm{o}\mathrm{l}\mathrm{t}\times \mathrm{s}\mathrm{e}\mathrm{c}}{amp}=ohm\times \sec$

But the practical unit is henry $(\mathrm{H})$. Its dimensional formula $[\mathrm{L}]=\left[\mathrm{M}{\mathrm{L}}^2{\mathrm{T}}^{–2}{\mathrm{A}}^{–2}\right]$

Note:  

1. One henry $={10}^9\mathrm{e}.\mathrm{m}.\mathrm{u}$ of inductance or${10}^9$
    ab-henry.

(iii) Dependence of self-inductance $(L)$:$‘L^{\prime }$ does not depend upon current flowing or change in current flowing. Still, it depends upon the number of turns$(N)$, area of cross-section $(A)$ and permeability of medium $(\mu )$. (Soft iron has greater permeability—hence greater self-inductance),” does not play any role till there is a constant current flowing in the circuit. ” comes into play only when there is a change in current $L$.

(iv) Magnetic potential energy of inductor: The source emf has to do some work against the self-inductance of the coil in building a steady current in the circuit. Whatever energy is consumed for this work is stored in the magnetic field of the coil. This energy is called magnetic potential energy $(U)$ of the coil.

Note: 

1. Energy density is given as $U=\frac{1}{2}\frac{B^2}{{\mu }_0}$

$(V)$Calculation of self-inductance for a current-carrying coil: If a coil of any shape having $N$ turns carries current , then total flux linked with the coil $N\varphi =Li$ Also $\varphi =BA\cos \theta$, where $B=$ magnetic field produced at the centre of the coil due to its current; = Area of each turn;   = Angle between normal to the plane of coil and direction of the magnetic field.

Self-Induction of Circular Coil

Consider a circular coil of $N$ turns carrying current $i$ , and it’s every turn is of the radius $r$. The self-inductance of the coil can be calculated as follows-

The magnetic field at the centre of the coil due to its own current$B=\frac{{\mu }_0}{4\pi }\cdot \frac{2\pi Ni}{r}\Rightarrow$$L=\frac{N\left(\frac{{\mu }_0}{4\pi }\cdot \frac{2\pi Ni}{r}\right)\left(\pi r^2\right)}{i}=\frac{{\mu }_0\pi N^{2}r}{2}$

Note:
1. If the radius is doubled so self-inductance will also double $\left(L\alpha r\right)$
(If $N=$ constant)

2. If area across section is doubled ($N=$ constant) i.e. $A^{\prime }=2A\Rightarrow \pi r{‘}^2=2\times \pi r^2\to r^{\prime }=\sqrt{2}r$
So $L^{\prime }=\sqrt{2}L$ i.e., increase in self-induction is $41.4\mathrm{\%}.$

3. If a current-carrying wire of constant length is bent into a circular coil of $N$ turns, then $N\left(2\pi r\right)=l;N\alpha \frac{l}{r}$
Now, as $L\alpha N^{2}r;L\alpha N^2\left(\frac{1}{N}\right)\Rightarrow L\alpha N;L\alpha \frac{l}{r^2}\left(r\right)\Rightarrow L\alpha \frac{l}{r}$

For example, if a wire of length $l$ is first bent in single turn circular coil then in double turn (concentric coplanar) coil, by using $L\alpha N$ we can say that $L$ in the second case twice that in the first case.

Self-Induction of Other Bodies

Square coil	Triangular coil	Solenoid	Toroid
$B=\frac{{\mu }_0}{4\pi }.\frac{8\sqrt{2}i}{a}N$   $L=\frac{N\left(\frac{{\mu }_0}{4\pi }.\frac{8\sqrt{2}Ni}{a}\right)a^2}{i}$ $L=\frac{2\sqrt{2}{\mu }_0{N}^{2}a}{\pi }\Rightarrow L\alpha N^2$	$B=\frac{{\mu }_0}{4\pi }.\frac{18Ni}{l}$   $L=\frac{N\left(\frac{{\mu }_0}{4\pi }\frac{18Ni}{l}\right)\times \left(\frac{\sqrt{3}}{4}{l}^2\right)}{i}$ $L=\frac{9\sqrt{3}{\mu }_0{N}^{2}l}{8\pi }\Rightarrow L\alpha N^2$	$B={\mu }_{0}ni=\frac{{\mu }_{0}Ni}{l}$   $L=\frac{N\left(\frac{{\mu }_{0}Ni}{l}\right).A}{i}$ $L=\frac{{\mu }_0{N}^{2}A}{l}\Rightarrow L\alpha N^2$ For iron cored solenoid $L=\frac{{\mu }_0{\mu }_r{N}^{2}A}{l}=\frac{\mu N^{2}A}{l}\left(\mu ={\mu }_0{\mu }_r\right)$	$B=\frac{{\mu }_{0}Ni}{2\pi r}$   $L=\frac{N\left(\frac{{\mu }_{0}Ni}{2\pi r}\right)\pi r^2}{i}=\frac{{\mu }_0{N}^{2}r}{2}$

Note:
1. Inductance at the ends of a solenoid is half of its inductance at the centre, i.e.

Applications of Self Inductance

The main function of an inductor is to store electrical energy in the form of a magnetic field. Inductors are used in the following:

1. Tuning circuits
2. Sensors
3. Store energy in a device
4. Induction motors
5. Transformers
6. Filters
7. Chokes
8. Ferrite beads
9. Inductors used as a relay

Limitations of Inductors

1. An inductor is limited in its current carrying capacity by its resistance and dissipates heat as per Joule’s law of heating.
2.  Inductors in pure form aren’t easy to manufacture due to size and stray effects, whereas capacitors are relatively easy to manufacture because of negligible stray effects.
3. Inductors in the circuit may affect the nearby components with their magnetic fields.

Solved Examples on Self Induction

Q.1. A circular coil of radius $5\text{cm}$ has $500$ turns of a wire. The approximate value of the coefficient of self-induction of the coil will be
(a) $25$ millihenry
(b) $25\times {10}^{–3}$ millihenry
(c) $50\times {10}^{–34}$ millihenry
(d) $50\times {10}^{–3}$ henry
Solution: (a) By using
$L=\frac{\pi {\mu }_0{N}^{2}r}{2};L=\frac{\left(3.14\right)\times 4\times \left(3.14\right)\times {10}^{–7}\times {\left(500\right)}^2\times 5\times {10}^{–2}}{2}\approx 25\times {10}^{–3}H\approx 25mH$

Q.2. The coefficient of self-inductance of a solenoid is $0.18mH,$ If a core of soft iron of relative permeability ${\mu }_{r}900$ is inserted, then the coefficient of self-inductance will become nearly
(a) $5.4mH$
(b) $162mH$
(c) $0.006mH$
(d) $0.0002mH$
Solution: (b) We know for air-cored solenoid $L=\frac{{\mu }_0{N}^{2}A}{I}$
In case of soft of iron core it’s self-inductance $L^{\prime }=\frac{{\mu }_0{\mu }_r{N}^{2}A}{I}$
$L^{\prime }={\mu }_{r}L.$ So here $L^{\prime }=900\times 0.18=162mH.$

Q.3. The current in an inductor is given by $i=2+3t\text{amp}$ where t is in second. The self-induced emf in it is $9\text{mV}$ the energy stored in the inductor at $t=1$ second is
(a) $10\text{mJ}$
(b) $37.5\text{mJ}$
(c) $75\text{mJ}$
(d) Zero
Solution: (b) At $t=1$ sec, $i=2+3\times 1=5A$ and $\left|e\right|=L\frac{di}{dt}\Rightarrow 9\times {10}^{–6}=L\times \frac{d}{dt}\left(2+3t\right)\Rightarrow L=3\times {10}^{–3}H$
So energy $U=\frac{1}{2}L{i}^2=\frac{1}{2}\left(3\times {10}^{–3}\right)\times {\left(5\right)}^2=37.5\text{mJ}.$

Q.4. A coil of wire of a certain radius has $600$ turns and a self-inductance of $108\text{mH}.$ The self-inductance of another similar coil of $500$ turns will be
(a) $74\text{mH}$
(b) $75\text{mH}$
(c) $76\text{mH}$
(d) $77\text{mH}$
Solution: (b) $\because L\alpha N^2\Rightarrow \frac{L_1}{L_2}={\left(\frac{N_1}{N_2}\right)}^2\Rightarrow \frac{108}{L_2}={\left(\frac{600}{500}\right)}^2;L_2=75\text{mH}.$

Q.5. A current increases uniformly from zero to one ampere in $0.01$ seconds, in a coil of inductance $10\text{mH}.$ The induced emf will be
(a) $1\text{V}$
(b) $2\text{V}$
(c) $3\text{V}$
(d) $4\text{V}$
Solution: (a) $e=–L\frac{dl}{dt}=–10\times {10}^{–3}\frac{1.0}{0.01}=–1\text{volt}\therefore \left|e\right|=1\text{volt}.$

Q.6. What inductance is required to store $1\text{kWh}$ of energy in a coil carrying a $200\text{A}$ current
(a) $100\text{H}$
(b) $180\text{H}$
(c) $200\text{H}$
(d) $450\text{H}$
Solution: (b) $\text{U}=1\text{KWH = 3}\text{.6}\times \text{1}{\text{0}}^6\text{J}.$ By using $\text{U}=\frac{1}{2}L{i}^2\Rightarrow 3.6\times {10}^6=\frac{1}{2}\times L\times {\left(200\right)}^2\Rightarrow L=180H.$

Q.7. The self-inductance of a coil is $L.$ Keeping the length and area the same, the number of turns in the coil is increased to four times. The self-inductance of the coil will now be
(a) $\frac{1}{4}\text{L}$
(b) $\text{L}$
(c) $\text{4}\text{L}$
(d) $\text{16}\text{L}$
Solution: (d) $\text{L}\alpha {\text{N}}^2.$

Summary

1. If a bar magnet moves towards a fixed conducting coil, then an emf, current, and charge are induced in the coil due to the flux changes. If the speed of the magnet increases, then induced emf and induced current increase, but the induced charge remains the same.
2. A thin long wire made up of a material of high resistivity behaves predominantly as a resistor. It also has some amount of inductance as well as capacitance. It isn’t easy to obtain a pure resistor. Similarly, it isn’t easy to obtain a pure capacitor as well as a pure inductor.
3. The effect of self-inductance can be eliminated just like the coils of a resistance box by doubling the coil on itself. 
4. It is impossible to have mutual inductance without self-inductance, but it may or may not be possible without mutual inductance.
5. The circuit behaviour of an inductor is quite different from that of a resistor. While a resistor opposes the current , an inductor opposes the change $\frac{di}{dt}$ in the circuit.
6. The self-inductance of a solenoid can be increased by inserting a soft-iron core. The function of the core is to improve the flux linkage between the turns of the coil.

FAQ’s on Self-induction

Q.1. On what factors does the self-induction depend?
Ans: Self-inductance of the coil depends on the area of cross-section of the coil(A), the number of turns per unit length in the coil(n), the length of the solenoid(l) and the permeability of the core material (μ).

Q.2. What devices use self-induction?
Ans: Induction motors, transformers, the potentiometer is a device that works on the self-inductance principle.

Q.3. What is the self-induction of a coil?
Ans: Self-inductance is the tendency of a coil to resist changes in current in itself. Whenever current changes through a coil, they induce an EMF, which is proportional to the rate of change of current through the coil.

Q.4. How does self-induction occur?
Ans: Self-inductance is defined as the induction of a voltage in a current-carrying wire when the current in the wire itself is changing.

Q.5. Why is self-induction called inertia?
Ans: The self-induction of the inductor resists the change of current in the circuit. This property is also called the inertia of electricity.

Q.6.What do you mean by self-induction?
Ans: Induction of an emf in a circuit by a varying current in the same circuit.

Q.7. What is the unit of self-induction?
Ans: The unit of self-induction is Henry $(H)$.

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