Inductance

SELF INDUCTION

We have already discussed capacitors – devices that store energy using electric fields. Like a capacitor, an inductor is also quite a commonly used element in electric circuits. It stores magnetic energy. As we know that when current flows through a conductor a magnetic field is set-up in surrounding of it, and hence it is associated with magnetic flux. If magnetic flux associated with a coil is $\varphi$ and current flowing through it is I, then its inductance is given by the expression$L=\frac{\phi }{I}$. The quantity 'L' is called self-inductance of the coil. It does not depend on the current, but it depends on the permeability of the core and the dimensions of the coil.

S.I. unit of inductance is Henry.

Consider the circuit, in which a solenoid is connected across a cell through a resistor. When the switch is open, the current in the circuit is zero. When the switch is closed, a current flow in it. Since current in the circuit increases from zero to a certain value, magnetic field associated with it changes that causes induction of an emf across the solenoid.

Induction of an emf due to variation in current flowing through the coil itself is known as self induction.

Since $\varphi$_B = LI, and

$\omega$ = $-\frac{d{\phi }_B}{dt}$ $\Rightarrow \epsilon =-L\frac{dI}{dt}$.

Inductance of an ideal solenoid:

Let a current I flow through a solenoid. The magnetic field due to the current flowing within the solenoid is, B = $\mu$₀nI, where n is the number of turns per unit length.

If area of cross section of the solenoid is A then flux associated with length $\ell$ is equal to

$\varphi$ = n$\ell$BA. (Assuming that the solenoid is ideal and long)

where $\ell$ is the length of the solenoid.

Now B = $\mu$nI

$\Rightarrow$ $\frac{(n\ell )({\mu }_{0}n\ell A)}{\ell }$ = $\mu$₀ n^2$\ell$A

Self Inductance of a Coil

Consider a coil of N turns and area of crossection A carrying a current i. The length of the coil is $\ell (\ell \ge$ÖA).

$\varphi =NBA=N\left(\frac{{\mu }_{0}iN}{\ell }\right)A$

$\varphi =\frac{{\mu }_0{N}^{2}A}{\ell }i$

Comparing with $\varphi$ = L i, we get: L = $\frac{{\mu }_0{N}^{2}A}{\ell }$

R - L CIRCUIT

GROWTH OF CURRENT

When switch is on at t = 0. Then current starts from 0 to I. And self inductor opposes the grow of current

$\frac{-LdI}{dt}-IR+\epsilon =0$

$\frac{LdI}{dt}=\epsilon -IR\Rightarrow$ $\underset{0}{\overset{I}{\int }}\frac{LdI}{\epsilon -IR}={\int }_0^{t}dt$$\Rightarrow$

$-\frac{L}{R}{\int }_0^I\frac{1}{I-\frac{\epsilon }{R}}dI=t$

$-\frac{L}{R}{\left[In\left(I-\frac{\epsilon }{R}\right)\right]}_0^I=t$ $\Rightarrow$ In $\left(\frac{I-\epsilon /R}{-\epsilon /R}\right)=\frac{-Rt}{L}$

$e^{-\frac{Rt}{L}}=\frac{\frac{\epsilon }{R}-I}{\epsilon /R}$ $\Rightarrow$ I = $\frac{\epsilon }{R}\left[1-e^{-\frac{Rt}{L}}\right]$

$\frac{dI}{dt}=\frac{\epsilon }{L}{e}^{-\frac{Rt}{L}}$

$\frac{LdI}{dt}=$ potential difference across inductor = $\epsilon e^{-\frac{Rt}{L}}$

I = $\frac{\epsilon }{R}\left[I-e^{-\frac{Rt}{L}}\right]$

At t = 0, I = 0

At t $\to \infty$ $\Rightarrow$ I $\to$ $\frac{\epsilon }{R}$

V = $\epsilon e^{-\frac{Rt}{L}}$.

At t = 0 $\Rightarrow$ potential difference across inductor = $\epsilon$

At t $\to \infty$ $\Rightarrow$ potential difference across inductor = 0.

So inductor will behave like plane wire.

Consider graph betⁿ I and time.

Consider graph betⁿ potential difference across

Inductor and time

Illustration 1: A coil of inductance 1.0 H and resistance 100 $\Omega$ 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: L = 1.0 H R = 100 $\Omega$

E = 12 V

i(t) = $\frac{E}{R}\left(1-e^{-\frac{Rt}{L}}\right)$

Energy stored in the magnetic field is 1/2 L i².

= $\frac{1}{2}\times 1.0\times \frac{{(12)}^2}{{10}^4}{\left(1-e^{-\frac{100\times 10\times {10}^{-3}}{1}}\right)}^2$

= $\frac{144}{2\times {10}^4}{\left(1-\frac{1}{e}\right)}^2=28.8mJ$

Illustration 2: A coil of metal wire is kept stationary in a non-uniform magnetic field. An e.m.f. is induced in the coil.

Solution: For induced emf to develop in a coil the magnetic flux through it must change. But in this case the number of magnetic lines of force through the coil is not changing.

Therefore the statement is false.

MUTUAL INDUCTION

Consider two coils C₁ and C₂ placed as shown. By varying current i₁ in coil C₁, we change the flux not only through C₁ but also through coil C₂. The change in flux $\varphi$₂ through C₂ (due to change in current i₁) induces an emf in the coil C₂. This emf is known as mutually induced emf and the process is known as mutual induction.

$\varphi$₂ $\propto$i₁ $\Rightarrow$ $\varphi$₂ = M i₁

Where M is called as mutual inductance of the pair of coils. The coil C₁ in which i varies is often called primary coil and the coil C₂ in which the emf is induced is called secondary coil.

induced emf in coil C₂ = E₂

$E_2=\left|\frac{d{\varphi }_2}{dt}\right|=M\left|\frac{di}{dt}\right|$

The mutual inductance is maximum when the coils arc wound up on the same axis. It is minimum when the axes of coils are normal to each other.

Note down the following points regarding the mutual inductance:

1. The SI unit of mutual inductance is henry (1H).

2. M depends upon closeness of the two circuits, their orientations and sizes and the number of turns etc.

3. Reciprocity theorem:

M₂₁ = M₁₂ = M

e₂ = - M(di₁/dt) and e₁ = - M(di₂/dt)

$M_{12}=\frac{N_2{\varphi }_{B_2}}{i_1}and{M}_{21}=\frac{N_1{\varphi }_{B_1}}{i_2}$

4. A good approach for calculating the mutual inductance of two circuits consists of the following steps:

(a) Assume any one of the circuits as primary (first) and the other as secondary (second).

(b) Suppose a current i₁ flows through the primary circuit.

(c) Determine the magnetic field $\overrightarrow{B}$ produced by the current i₁.

(d) Obtain the magnetic flux ${\varphi }_{B_2}$.

(e) With the flux known, the mutual inductance can be found from,

$M=\frac{N_2{\varphi }_{B_2}}{i_1}$

Illustration 3: A long solenoid of length 1 m, cross sectional area 10 cm², having 1000 turns has wound about its centre a small coil of 20 turns. Compute the mutual inductance of the two circuits. What is the emf in the coil when the current in the solenoid changes at the rate of 10 Amp/s?

Solution: Let N₁ = number of turns in solenoid;

N₂ = number of turns in coil

A₁ and A₂ be their respective areas of crossection. (A₁ = A₂ in this problem)

Flux $\varphi$₂ through coil crated by current i₁ in solenoid is $\varphi$₂ = N₂(B₁A₂)

${\varphi }_2=N_2\left(\frac{{\mu }_0{i}_1{N}_1}{\ell }\right)A_2$ $\Rightarrow$

${\varphi }_2=\left(\frac{{\mu }_0{N}_1{N}_2{A}_2}{\ell }\right)i_2$

Comparing with $\varphi$₂ = M i₁; we get

Mutual inductance = M = $\frac{{\mu }_0{N}_1{N}_2{A}_2}{\ell }$

$M=\frac{4\pi \times {10}^{-7}\times 1000\times 20\times 10\times {10}^{-4}}{1}=2.51\times {10}^{-5}H$

Magnitude of induced emf = E₂ = M $\left|\frac{d{i}_1}{dt}\right|$

L–C CIRCUIT

L.C. OSCILLATIONS

A capacitor is charged to a potential difference of V_o by connecting it across a battery and then is allowed to discharge through an inductor of inductance L.

Initial charge on the plates of the capacitor q_o = CV_o

At any instant, let the charge flown through the circuit be q and the current in the circuit be i. Applying Kirchhoff's law

$\frac{q_0-q}{C}$ - $L\frac{di}{dt}$ = 0

Differentiating w.r.t. time we get

$-\frac{dq}{dt}-LC\frac{d^{2}i}{d{t}^2}$ = 0

$\frac{d^{2}i}{d{t}^2}$=$\frac{-1}{LC}i$ = -$\omega$²i, f = $\frac{1}{2\pi \sqrt{LC}}$

$\therefore$ The charge q on the plates of the capacitor and current I in the circuit vary sinusoidally as q = q₀ sin ($\omega$t + $\varphi$) and I = q₀ $\omega$ cos ($\omega$t + $\varphi$). where $\varphi$ is the initial phase and it depends on initial conditions of the circuit and $\omega$ = $\frac{1}{\sqrt{LC}}$.

The total energy of the system remains conserved

$\frac{1}{2}C{V}^2+\frac{1}{2}L{i}^2$ = constant = $\frac{1}{2}C{V}_0^2=\frac{1}{2}L{i}_0^2$

Illustration 4: A–2–$\mu$F capacitor is initially charged to 20 V and then shorted across a 6–$\mu$H inductor. What are the frequency of oscillation and the maximum value of the current?

Solution: The frequency of oscillation is independent of the initial charge and depends only on the values of the capacitance and inductance. The frequency is

f = $\frac{\omega }{2\pi }=\frac{1}{2\pi }\sqrt{\frac{1}{LC}}=\frac{1}{2\pi \sqrt{LC}}$

= $\frac{1}{2\pi \sqrt{(6\times {10}^{-6})(2\times {10}^{-6})}}$ = 4.59 x 10⁴ Hz

According to equation the maximum value of the current is related to the maximum value of the charge by

I_m = $\omega$Q₀ = .

The initial charge on the capacitor is

Q₀ = CV₀ = (2$\mu$F)(20 V) = 40$\mu$C

Thus

I_m = $\frac{40\mu C}{\sqrt{(6\mu H)(2\mu F)}}$ = 11.5 A

MASS–SPRING SYSTEM VS INDUCTOR–CAPACITOR CIRCUIT

A comparison of oscillations of a mass spring system and an L - C circuit.