**CBSE NCERT NOTES CLASS 12 PHYSICS CHAPTER 6**

**ELECTROMAGNETIC INDUCTION**

Whenever the magnetic flux linked with an electric circuit changes, an emf is induced in the circuit. This phenomenon is called **electromagnetic induction**.

**Magnetic flux**

Magnetic flux through a plane of area *A *placed in a uniform magnetic field *B*** **can be written as

$${\mathrm{\Phi}}_{\mathrm{B}}\mathrm{}=\mathrm{}\mathrm{}\overrightarrow{\mathrm{B}}.\overrightarrow{\mathrm{A}}=\mathrm{}\mathrm{B}\mathrm{A}\mathrm{cos\; \theta}$$

SI unit of magnetic flux is** **weber,

CGS unit of magnetic flux is maxwell,

1 weber = 10^{8} maxwell,

Dimensional formula of magnetic flux [Φ_{B}] = [ML^{2}T^{-2}A^{-2}]

The magnetic flux can be changed by

- varying
*B*, - changing the shape of a coil (that is, by shrinking it or stretching it, we can change the area of the coil) in a magnetic field,
- rotating a coil in a magnetic field such that the angle
*θ*between $\overrightarrow{\mathrm{B}}$ - when strength of current flowing in a coil is increased or decreased, induced current is developed in the coil in same or opposite direction.

**Faraday’s laws of electromagnetic induction**

- Whenever the magnetic flux linked with a circuit changes, an induced emf is produced in it.
- The induced emf lasts so long as the change in magnetic flux continues.
- The magnitude of the induced emf in a circuit is equal to the time rate of change of magnetic flux through the circuit. In the case of a closely wound coil of
*N*$$\mathrm{\varepsilon}\mathrm{}=\mathrm{}\u2013\mathrm{N}\frac{\mathrm{d}{\mathrm{\Phi}}_{\mathrm{B}}}{\mathrm{d}\mathrm{t}}$$

Here constant of proportionality is one and negative sign is as per Lenz’s law.

**Lenz’s law**

The polarity of induced **emf** is such that it tends to produce a current which opposes the change in magnetic flux that produced it.

**Justification of Lenz’s law**

Lenz’s law conforms to conservation of energy. Suppose the direction were opposite to that depicted in the figure. The south pole due to the induced current will face the approaching north pole of the magnet. The bar magnet will then be attracted towards the coil at an ever increasing acceleration. A gentle push on the magnet will initiate the process and its velocity and kinetic energy will continuously increase without expending any energy. If this can happen, one could construct a perpetual-motion machine by a suitable arrangement. This violates the law of conservation of energy and hence cannot happen.

Now consider the correct case shown. In this situation, the bar magnet experiences a repulsive force due to the induced current. Therefore, we have to do work in moving the magnet. Where does the energy spent by the person go? This energy is dissipated by Joule heating produced by the induced current.

**MOTIONAL EMF**

If a rod of length *l* moves perpendicular to a magnetic field *B*, with a velocity* v*, then induced emf produced in it is given by

$$\mathrm{\varepsilon}\mathrm{}=\mathrm{}\mathrm{B}\mathrm{}\mathrm{l}\mathrm{}\mathrm{v}$$

**Proof**

Consider a straight conductor moving in a uniform and time independent magnetic field. There is a rectangular conductor PQRS in which the conductor PQ is free to move.

The rod PQ is moved towards the left with a constant velocity *v*** **as shown.

Assume that there is no loss of energy due to friction. PQRS forms a closed circuit enclosing an area that changes as PQ moves.

It is placed in a uniform magnetic field *B*** **which is perpendicular to the plane of this system. If the length RQ = *x *and RS = *l*, then magnetic flux *Φ*_{B} enclosed by the loop *PQRS* will be

$${\mathrm{\Phi}}_{\mathrm{B}}=\mathrm{}\mathrm{B}\mathrm{}\mathrm{l}\mathrm{}\mathrm{x}$$

Since *x *is changing with time, the rate of change of flux *Φ*_{B}** **will induce an emf given by,

$$\mathrm{\varepsilon}\mathrm{}=\mathrm{}\u2013\frac{\mathrm{d}\left(\mathrm{B}\mathrm{l}\mathrm{x}\right)}{\mathrm{d}\mathrm{t}}\mathrm{}=\mathrm{}\u2013\mathrm{B}\mathrm{l}\frac{\mathrm{d}\mathrm{x}}{\mathrm{d}\mathrm{t}}\mathrm{}$$

Now, taking

$$\frac{\mathrm{d}\mathrm{x}}{\mathrm{d}\mathrm{t}}=\mathrm{}\u2013\mathrm{v}$$

We get,

$$\mathrm{}\mathrm{\varepsilon}=\mathrm{}\mathrm{B}\mathrm{L}\mathrm{v}$$

**Alternatively, **

Lorentz magnetic force

$$\mathrm{F}\mathrm{}=\mathrm{}\mathrm{q}\mathrm{v}\mathrm{B}\mathrm{}$$

This force is in the direction from P to Q (use right hand rule for vector product). The work done in moving the charge from P to Q is,

$$\mathrm{W}\mathrm{}=\mathrm{}\mathrm{F}.\mathrm{l}\mathrm{}=\mathrm{}\mathrm{q}\mathrm{B}\mathrm{l}\mathrm{v}$$

$$\Rightarrow \mathrm{\varepsilon}=\frac{\mathrm{W}}{\mathrm{q}}=\mathrm{B}\mathrm{l}\mathrm{v}$$

Here we have considered area change due to movement of the conductor. However it is not obvious how an emf is induced when a conductor is stationary and the magnetic field is changing.

In the case of a stationary conductor, *v* = 0, therefore, the force on its charges is given by

$$\overrightarrow{\mathrm{F}}\mathrm{}=\mathrm{}\mathrm{q}\mathrm{}\left(\overrightarrow{\mathrm{E}}\mathrm{}+\mathrm{}\overrightarrow{\mathrm{v}}\mathrm{}\times \mathrm{}\overrightarrow{\mathrm{B}}\right)=\mathrm{}\mathrm{q}\overrightarrow{\mathrm{E}},$$

Thus, any force on the charge must arise from the electric field term $\overrightarrow{\mathrm{E}}$** **alone. This magnetic force has to be generated by the time-varying magnetic field.

We know that charges in motion (current) can exert force/torque on a stationary magnet. Conversely, a bar magnet in motion (or more generally, a changing magnetic field) can exert a force on the stationary charge. This is the fundamental significance of the Faraday’s discovery. Electricity and magnetism are thus related.

**Induced emf through a rotating rod**

If a metallic rod of length *l* rotates about one of its ends in a plane perpendicular to the magnetic field, then the induced emf produced across its ends is given by

$$\mathrm{\varepsilon}\mathrm{}=\frac{1}{2}\mathrm{B}\mathrm{\omega}{\mathrm{l}}^{2}=\mathrm{}\mathrm{B}\mathrm{A}\mathrm{\nu}$$

where,

*ω* = angular velocity of rotation,

*ν* = frequency of rotation and

*A* = *πl*^{2} = area of ring

**Proof**

As the rod is rotated, free electrons in the rod move towards the outer end due to Lorentz force. Thus, the resulting separation of charges produces an emf across the ends of the rod. At a certain value of emf, there is no more flow of electrons and a steady state is reached.

It is obvious that *R = l*

The magnitude of the emf generated, across a length *dr *of the rod as it moves at right angles to the magnetic field, is given by

$\mathrm{d}\mathrm{\epsilon}\mathrm{}=\mathrm{}\mathrm{B}\mathrm{v}\mathrm{}\mathrm{d}\mathrm{r}$* *.

Hence,

$$\mathrm{\epsilon}=\int \mathrm{d}\mathrm{\epsilon}\mathrm{}=\mathrm{}{\int}_{0}^{\mathrm{R}}\mathrm{B}\mathrm{v}\mathrm{}\mathrm{d}\mathrm{r}$$

$$={\int}_{0}^{\mathrm{R}}\mathrm{B}\mathrm{}\mathrm{\omega}\mathrm{r}\mathrm{}\mathrm{d}\mathrm{r}=\mathrm{}\frac{\mathrm{B}\mathrm{\omega}{\mathrm{R}}^{2}}{2}$$

$$=\frac{\mathrm{B}\times \mathrm{}2\mathrm{\pi}\mathrm{\nu}\mathrm{}\times {\mathrm{R}}^{2}}{2}=\mathrm{B}\mathrm{A}\mathrm{\nu}\mathrm{}$$

- The direction of induced current in any conductor can be obtained from Fleming’s right hand rule.
- When a rectangular coil moves linearly with constant velocity in a uniform magnetic field, flux and induced emf will be zero.
- If the rod moves at an angle θ with the direction of magnetic field with velocity
*v*,$$\mathrm{}\mathrm{\varepsilon}\mathrm{}=\mathrm{}\u2013\mathrm{}\mathrm{B}\mathrm{R}\mathrm{v}\mathrm{}\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{}\mathrm{\theta}$$

**Fleming’s right hand rule**

[Given direction of v and B find the direction of current]

If we stretch the thumb, the forefinger and the central finger of right hand in such a way that all three are perpendicular to each other, and if thumb represent the direction of motion, the forefinger represent the direction of magnetic field, then central finger will represent the direction of induced current.

**Energy consideration of induced emf**

Let *r *be the resistance of movable arm PQ of the rectangular conductor shown. Assume that the remaining arms QR, RS and SP have negligible resistances compared to *r*. Thus, the overall resistance of the rectangular loop is *r *and this does not change as PQ is moved. The current I* *in the loop is,

$$\mathrm{I}\mathrm{}=\frac{\mathrm{\varepsilon}}{\mathrm{r}}=\frac{\mathrm{B}\mathrm{l}\mathrm{v}}{\mathrm{r}}$$

The force on arm *PQ*,

$$\overrightarrow{\mathrm{F}}=\mathrm{I}\mathrm{}(\overrightarrow{\mathrm{l}}\mathrm{}\times \overrightarrow{\mathrm{B}})$$

This force is directed towards right in the direction opposite to the velocity of the rod. The magnitude of this force is,

$$\mathrm{F}=\mathrm{I}\mathrm{L}\mathrm{B}=\frac{{\mathrm{B}}^{2\mathrm{}}{\mathrm{l}}^{2\mathrm{}}\mathrm{v}}{\mathrm{r}}$$

The arm PQ is being pushed with a constant speed *v*, the power required to do this will be,

$$\mathrm{P}\mathrm{}=\mathrm{}\mathrm{F}\mathrm{}\mathrm{v}=\frac{{\mathrm{B}}^{2\mathrm{}}{\mathrm{l}}^{2\mathrm{}}{\mathrm{v}}^{2}}{\mathrm{r}}$$

This mechanical energy is dissipated as Joule heat, and the **power** is given by

$${\mathrm{P}}_{\mathrm{J}\mathrm{}}={\mathrm{I}}^{2\mathrm{}}\mathrm{r}$$

$$=\mathrm{}{\left(\frac{\mathrm{B}\mathrm{l}\mathrm{v}}{\mathrm{r}}\right)}^{2}\mathrm{r}=\frac{{\mathrm{B}}^{2\mathrm{}}{\mathrm{l}}^{2\mathrm{}}{\mathrm{v}}^{2}}{\mathrm{r}}$$

Now, the magnitude of the induced emf is

$$\left|\mathrm{\varepsilon}\right|=\frac{\mathrm{\Delta}{\mathrm{\Phi}}_{\mathrm{B}}}{\mathrm{\Delta}\mathrm{t}}\mathrm{}\mathrm{}$$

Also,

$$\left|\mathrm{\varepsilon}\right|=\mathrm{I}\mathrm{r}=\mathrm{}\frac{\mathrm{\Delta}\mathrm{Q}}{\mathrm{\Delta}\mathrm{t}}\mathrm{}\mathrm{r}$$

Therefore,

$$\frac{\mathrm{\Delta}{\mathrm{\Phi}}_{\mathrm{B}}}{\mathrm{\Delta}\mathrm{t}}=\mathrm{}\frac{\mathrm{\Delta}\mathrm{Q}}{\mathrm{\Delta}\mathrm{t}}\mathrm{}\mathrm{r}$$

$$\Rightarrow \mathrm{}\mathrm{\Delta}\mathrm{Q}=\frac{\mathrm{\Delta}{\mathrm{\Phi}}_{\mathrm{B}}}{\mathrm{r}}\mathrm{}$$

**EDDY CURRENTS**

If a piece of metal is rotated with high speed in a uniform magnetic field, then induced currents are set up in the metal piece, which are like whirlpools in water. Thses currents are called **eddy currents**.

The magnitude of eddy currents is given by

$$\mathrm{i}\mathrm{}=\mathrm{}\u2013\frac{\mathrm{\epsilon}}{\mathrm{R}}=\frac{\mathrm{d}{\mathrm{\Phi}}_{\mathrm{B}}}{\mathrm{d}\mathrm{t}}\times \frac{1}{\mathrm{R}}$$

Where R is the resistance of the metal.

If a copper plate is allowed to swing like a simple pendulum between the pole pieces of a strong magnet, then it is found that the motion of plate is damped and in a little while the plate comes to a halt in the magnetic field.

**Magnetic flux** associated with the plate **keeps on changing** as the plate moves in and out of the region between magnetic poles. The **flux change induces eddy currents** in the plate.

**Directions **of eddy currents are **opposite** when the plate swings **into the region** between the poles and when it swings **out of the region.**

- Eddy currents are
**undesirable**since they heat up the core and dissipate electrical energy in the form of heat.

**Minimizing eddy currents**

**Eddy currents are minimised by **using laminations (or small rectangular pieces of plate) of metal to make a metal core.

The laminations are separated by an insulating material like lacquer.

The plane of the laminations must be arranged parallel to the magnetic field, so that they cut across the eddy current paths.

Since the dissipation of electrical energy into heat depends on the square of the strength of electric current, **heat loss is substantially reduced**.

**Uses of eddy currents**

**Magnetic braking in trains**Strong electromagnets are situated above the rails in some electrically powered trains. When the electromagnets are activated, the eddy currents induced in the rails oppose the motion of the train. As there are no mechanical linkages, the braking effect is smooth.

**Electromagnetic damping**Certain galvanometers have a fixed core made of nonmagnetic metallic material. When the coil oscillates, the eddy currents generated in the core oppose the motion and bring the coil to rest quickly.

**Induction furnace**Induction furnace can be used to produce high temperatures and can be utilised to prepare alloys, by melting the constituent metals. A high frequency alternating current is passed through a coil which surrounds the metals to be melted. The eddy currents generated in the metals produce high temperatures sufficient to melt it.

**Electric power meters**The shiny metal disc in the electric power meter (analogue type) rotates due to the eddy currents. Electric currents are induced in the disc by magnetic fields produced by sinusoidally varying currents in a coil.

**INDUCTANCE**

The phenomenon of production of induced emf in a coil due to change in current flowing in itself or the current flowing through other coil, is called induction.

The flux through a coil is proportional to the current. That is,

*Φ*_{B}* ∝ I*

**Flux linkage** is the magnetic flux running through, or "linked" to a coil, which is equal to *NΦ*_{B} for a closely wound coil with *N* turns.

When the flux *Φ*_{B} through the coil, changes each turn contributes to the induced emf.

*NΦ*_{B}* ∝ I*

The magnetic flux linked (**flux linkage**) with a coil can, therefore, be written as,

*NΦ*_{B}* = L I*

Where,* L* is the constant of proportionality, called **inductance**. Inductance is a scalar quantity. It has the dimensions of [M L^{2} T^{–2} A^{–2}] and SI unit is Henry (H)

Further, if the geometry of the coil does not vary with time then,

$$\mathrm{N}\frac{\mathrm{d}{\mathrm{\Phi}}_{\mathrm{B}}}{\mathrm{d}\mathrm{t}}=\mathrm{}\mathrm{L}\frac{\mathrm{d}\mathrm{I}}{\mathrm{d}\mathrm{t}}$$

**Mutual Inductance**

The phenomenon of production of induced emf in a circuit due to the change in the current through its neighbouring circuit is called **mutual induction**.

Let us consider two long co-axial solenoids each of length *l*. We denote the radius of the inner solenoid *S*_{1} by *r*_{1} and the number of turns per unit length by *n*_{1}. The corresponding quantities for the outer solenoid *S*_{2} are *r*_{2} and *n*_{2}, respectively. Let *N*_{1} and *N*_{2} be the total number of turns of coils *S*_{1} and *S*_{2}, respectively.

When a current I_{2} is set up through S_{2}, it in turn sets up a magnetic flux through S_{1}. Let us denote it by Φ_{1}. The corresponding flux linkage with solenoid S_{1} is

*N*_{1}*Φ*_{1} = *M*_{12}*I*_{2}

Where M_{12} is called the **mutual inductance*** *of solenoid S_{1} with respect to solenoid S_{2}. It is also referred to as the **coefficient of mutual induction.**

The magnetic field due to the current *I*_{2 }*in S*_{2}* is *

$${\mathrm{B}}_{2}=\mathrm{}{\mathrm{\mu}}_{\mathrm{o}}{\mathrm{n}}_{2}{\mathrm{I}}_{2}.\mathrm{}$$

The resulting flux linkage, due to this with coil *S*_{1} is,

$${\mathrm{\Phi}}_{1}={\mathrm{A}}_{1}{\mathrm{B}}_{2}$$

Therefore,

$${{{\mathrm{N}}_{1}{\mathrm{\Phi}}_{1}=(\mathrm{n}}_{1}\mathrm{l}\left)\mathrm{}\left(\mathrm{\pi}{{\mathrm{r}}_{1}}^{2}\right)\mathrm{}\right(\mathrm{\mu}}_{\mathrm{o}}{\mathrm{n}}_{2}{\mathrm{I}}_{2})$$

$$\Rightarrow {\mathrm{N}}_{1}{\mathrm{\Phi}}_{1}={{{\mathrm{\mu}}_{\mathrm{o}}\mathrm{n}}_{1}{\mathrm{n}}_{2}\mathrm{\pi}{{\mathrm{r}}_{1}}^{2}\mathrm{}\mathrm{l}\mathrm{}\mathrm{I}}_{2}$$

Comparing this equation with,

*N*_{1}*Φ*_{1} = *M*_{12}*I*_{2}

We get,

$${\mathrm{M}}_{12}={{\mathrm{\mu}}_{\mathrm{o}}\mathrm{n}}_{1}{\mathrm{n}}_{2}\mathrm{\pi}{{\mathrm{r}}_{1}}^{2}\mathrm{l}$$

- We have neglected the edge effects and considered the magnetic field $\mathrm{\mu}}_{\mathrm{o}}{\mathrm{n}}_{2}{\mathrm{I}}_{2$ to be uniform throughout the length and width of the solenoid
*S*_{2}. This is ok if the solenoid is long, i.e., implying*l*>> r_{2}.

Further the flux linkage with coil S_{2 }due to_{ }a current I_{1} passed through the solenoid S_{1} is,

*N*_{2}*Φ*_{2} = *M*_{21}*I*_{1}

M_{21} is called the mutual inductance of solenoid S_{2} with respect to solenoid S_{1}.

Now, the flux due to the current *I*_{1} in *S*_{1} can be assumed to be confined solely inside *S*_{1} since the solenoids are very long. Thus, flux linkage with solenoid *S*_{2} is

$${{{\mathrm{N}}_{2}{\mathrm{\Phi}}_{2}=(\mathrm{n}}_{2}\mathrm{l}\left)\mathrm{}\left(\mathrm{\pi}{{\mathrm{r}}_{1}}^{2}\right)\mathrm{}\right(\mathrm{\mu}}_{\mathrm{o}}{\mathrm{n}}_{1}{\mathrm{I}}_{1})={{{\mathrm{\mu}}_{\mathrm{o}}\mathrm{n}}_{1}{\mathrm{n}}_{2}\mathrm{\pi}{{\mathrm{r}}_{1}}^{2}\mathrm{}\mathrm{l}\mathrm{}\mathrm{I}}_{1}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}$$

$$\Rightarrow {\mathrm{M}}_{21}={{\mathrm{\mu}}_{\mathrm{o}}\mathrm{n}}_{1}{\mathrm{n}}_{2}\mathrm{\pi}{{\mathrm{r}}_{1}}^{2}\mathrm{l}$$

We can see that

*M*_{12} =* M*_{21} =*M*

**Flux linkage-Faraday’s experiment**

Consider two coils C_{1} and C_{2} held stationary. Coil C_{1} is connected to galvanometer G while the second coil C_{2} is connected to a battery through a tapping key K.

Let Φ_{1} be the flux through coil C_{1} (say of N_{1} turns) when current in coil C_{2} is I_{2}.

Then, *N*_{1}*Φ*_{1}* = MI*_{2}

For currents varrying with time,

$$\frac{\mathrm{d}\left({\mathrm{N}}_{1}{\mathrm{\Phi}}_{1}\right)}{\mathrm{d}\mathrm{t}}=\mathrm{}\frac{\mathrm{d}\left(\mathrm{M}{\mathrm{I}}_{2}\right)}{\mathrm{d}\mathrm{t}}$$

But

$$\frac{\mathrm{d}\left({\mathrm{N}}_{1}{\mathrm{\Phi}}_{1}\right)}{\mathrm{d}\mathrm{t}}=\mathrm{}{\mathrm{\varepsilon}}_{1}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}$$

$$\Rightarrow \mathrm{}\mathrm{}\mathrm{}{\mathrm{\varepsilon}}_{1}\mathrm{}=\mathrm{}\frac{\mathrm{d}\left(\mathrm{M}{\mathrm{I}}_{2}\right)}{\mathrm{d}\mathrm{t}}$$

Therefore, varying current in a coil can **induce emf in a neighbouring coil**. The magnitude of the **induced emf** depends upon the **rate of change of current** and **mutual inductance of the two coils.**

**Self Inductance**

The phenomenon of production of **induced emf **in a coil due to change in current flowing through itself is called self induction.

Induced emf in the coil,

$$\mathrm{\varepsilon}=-\frac{\mathrm{d}{\mathrm{\Phi}}_{\mathrm{B}}}{\mathrm{d}\mathrm{t}}\mathrm{}=\mathrm{}\u2013\mathrm{}\mathrm{L}\frac{\mathrm{d}\mathrm{I}}{\mathrm{d}\mathrm{t}}$$

where, *L* = coefficient of self induction or self inductance.

But** **

*B = μ*_{o}* n I *

And

$$\mathrm{}\mathrm{N}\mathrm{}=\mathrm{}\mathrm{n}\mathrm{}\mathrm{l}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}$$

Therefore,

$$\mathrm{N}{\mathrm{\Phi}}_{\mathrm{B}}\mathrm{}=\mathrm{}\mathrm{n}\mathrm{}\mathrm{l}\mathrm{}{\mathrm{\mu}}_{\mathrm{o}}\mathrm{}\mathrm{n}\mathrm{}\mathrm{I}\mathrm{}\mathrm{A}\mathrm{}$$

$$={\mathrm{\mu}}_{\mathrm{o}}\mathrm{}{\mathrm{n}}^{2}\mathrm{}\mathrm{A}\mathrm{}\mathrm{l}\mathrm{}\mathrm{I}$$

Comparing this equation with $\mathrm{N}{\mathrm{\Phi}}_{\mathrm{B}}\mathrm{}=\mathrm{L}\mathrm{}\mathrm{I}$, we get,

*L = μ*_{o}* n*^{2}* A l*

For medium **other than vaccum or air, **

*L = μ*_{o}* μ*_{r}* n*^{2}* A l*

**Important points to be noted**

- The self-inductance of the coil depends on its
**geometry**of the coil and on the**permeability of the medium.** - The
**self-induced emf**is also called the**back emf,**as it opposes any change in the current in a circuit. Physically, the**self-inductance***plays the role of*. It is the electromagnetic analogue of mass in mechanics*inertia*

**general case of currents**flowing simultaneously in two nearby coils. The flux linked with one coil will be the sum of two fluxes which exist independently.

$${\mathrm{N}}_{1}\mathrm{}{\mathrm{\Phi}}_{1}\mathrm{}=\mathrm{}{\mathrm{M}}_{11}\mathrm{}{\mathrm{I}}_{1}\mathrm{}+\mathrm{}{\mathrm{M}}_{12}\mathrm{}{\mathrm{I}}_{2}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}$$

$$\Rightarrow \mathrm{}{\mathrm{\varepsilon}}_{1}=\mathrm{}-\mathrm{}{\mathrm{M}}_{11}\mathrm{}\frac{\mathrm{d}{\mathrm{I}}_{1}}{\mathrm{d}\mathrm{t}}\mathrm{}-{\mathrm{M}}_{12}\mathrm{}\frac{\mathrm{d}{\mathrm{I}}_{2}}{\mathrm{d}\mathrm{t}}$$

where M_{11} represents inductance due to the same coil and is the self-inductance.

Hence,

$${\mathrm{\varepsilon}}_{1}=\mathrm{}-\mathrm{}{\mathrm{L}}_{1}\mathrm{}\frac{\mathrm{d}{\mathrm{I}}_{1}}{\mathrm{d}\mathrm{t}}\mathrm{}-{\mathrm{M}}_{12}\mathrm{}\frac{\mathrm{d}{\mathrm{I}}_{2}}{\mathrm{d}\mathrm{t}}$$

**Energy stored in an inductor**

Work needs to be done against the **back emf **

*(*$\mathrm{\varepsilon}$

*)*in establishing the current. This work done is stored as magnetic potential energy. For the current

**at an instant in a circuit, the rate of work done is**

*I*$$\frac{\mathrm{d}\mathrm{W}}{\mathrm{d}\mathrm{t}}=\left|\mathrm{\varepsilon}\right|\mathrm{}\mathrm{I}$$

If **resistive losses** are ignored then

$$\frac{\mathrm{d}\mathrm{W}}{\mathrm{d}\mathrm{t}}=\mathrm{L}\mathrm{}\mathrm{I}\mathrm{}\frac{\mathrm{d}\mathrm{I}}{\mathrm{d}\mathrm{t}}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}$$

$$\Rightarrow \mathrm{}\mathrm{W}=\mathrm{}\underset{0}{\overset{\mathrm{I}}{\int}}\mathrm{L}\mathrm{I}\mathrm{}\mathrm{d}\mathrm{I}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}$$

$$\Rightarrow \mathrm{}\mathrm{W}=\mathrm{}\frac{1}{2}\mathrm{}\mathrm{L}\mathrm{}{\mathrm{I}}^{2}$$

This equation is analogous to the equation for kinetic energy, i.e.

$$\mathrm{E}\mathrm{}=\frac{1}{2}\mathrm{}\mathrm{m}{\mathrm{v}}^{2}$$

hence *L* behaves as electrical inertia like *m* and opposes the current in the circuit.

**AC GENERATOR**

**Structure of AC generator**

It consists of a coil mounted on a rotor shaft. The axis of rotation of the coil is perpendicular to the direction of the magnetic field.

There are four main parts of an AC generator

**Armature -**It is rectangular coil of insulated copper wire having a large number of turns.**Field Magnets**- These are two pole pieces of a strong electromagnet.**Slip Rings -**These are two hollow metallic rings.**Brushes -**These are two flexible metals or carbon rods, which remains slightly in contact with slip rings.

**Note - **A **DC** generator or dynamo contains split rings or **commutator **instead o**f slip rings**.

**Working of AC generator**

The coil (called armature) is mechanically rotated in the uniform magnetic field by some external means.

The rotation of the coil causes the magnetic flux through it to change, so an emf is induced in the coil.

The ends of the coil are connected to an external circuit by means of slip rings and brushes.

When the coil is rotated with a constant angular speed ω, the angle θ between the magnetic field vector B** **and the area vector A** **of the coil at any instant *t *is θ = ω*t *(assuming θ = 0º at *t *= 0).

As a result, the effective area of the coil exposed to the magnetic field lines changes with time, and the flux at any time *t *is

*Φ*_{B}* = BA cos θ = BA cos ωt*

Then

$$\mathrm{\epsilon}\mathrm{}=\mathrm{}-\mathrm{N}\frac{\mathrm{d}{\mathrm{\Phi}}_{\mathrm{B}}}{\mathrm{d}\mathrm{t}}=-\mathrm{N}\mathrm{B}\mathrm{A}\mathrm{}\frac{\mathrm{d}}{\mathrm{d}\mathrm{t}}\mathrm{}\left(\mathit{cos}\mathrm{\omega}\mathrm{t}\right)$$

$$\Rightarrow \mathrm{\epsilon}\mathrm{}=\mathrm{}\mathrm{N}\mathrm{B}\mathrm{A}\mathrm{\omega}\mathrm{}\mathrm{sin}\mathrm{\omega}\mathrm{t}$$

$$\Rightarrow \mathrm{}\mathrm{\epsilon}\mathrm{}=\mathrm{}{\mathrm{\epsilon}}_{\mathrm{o}}\mathrm{sin}\mathrm{\omega}\mathrm{t}\mathrm{}$$

where *NBAω = ε*_{o} is the maximum value of the emf, which occurs when sin ωt *= *±1.

The emf has its extremum value when θ = 90º** **or θ = 270º, as the change of flux is greatest at these points.

The direction of the current changes periodically and therefore the current is called **alternating current (ac)**.

Since

$$\mathrm{\omega}\mathrm{}=\mathrm{}2\mathrm{\pi}\mathrm{\nu}\mathrm{}$$

$$\Rightarrow \mathrm{}\mathrm{\epsilon}\mathrm{}=\mathrm{}{\mathrm{\epsilon}}_{\mathrm{o}}\mathrm{}\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{}2\mathrm{\pi}\mathrm{\nu}\mathrm{t}$$

where ν is the frequency of revolution of the generator’s coil.

If R is the electrical resistance of the circuit, then induced current in the circuit is given by

$$\mathrm{I}=\frac{\mathrm{\varepsilon}}{\mathrm{R}}$$

The induced current produced in a coil rotated in uniform magnetic field,

$$\mathrm{I}=\mathrm{}\frac{\mathrm{N}\mathrm{B}\mathrm{A}\mathrm{\omega}\mathrm{sin}\mathrm{\omega}\mathrm{t}}{\mathrm{R}}=\mathrm{}{\mathrm{I}}_{\mathrm{o}}\mathrm{sin}\mathrm{\omega}\mathrm{t}$$

where,

*I*_{o}* = NBA ω* = peak value of induced current,

N = number of turns in the coil,

B= magnetic induction,

ω = angular velocity of rotation and

A = area of cross-section of the coil.

**Different Orientations of Armature and Magnetic Field**

**The coefficient of coupling of inductances**

Two coils are said to be coupled if full or part of the flux produced by one links with the other.

$$\mathrm{K}\mathrm{}=\frac{\mathrm{M}}{\sqrt{{\mathrm{L}}_{1}{\mathrm{L}}_{2}}}$$

Where, *L*_{1} and *L*_{2} are coefficients of self-inductance of the two coils, and *M* is coefficient of mutual induction of the two coils.

Coefficient of coupling is maximum (*K* = 1), when coils are coaxial

Coefficient of coupling is minimum (*K* = 0), when coils are placed at right angles to each other.

**Series combination of inductances**When coefficient of coupling

*K*= 0,*L = L*_{1}*+ L*_{2}When coefficient of coupling

*K*= 1,(a) If currents in two coils are in the same direction, then

*L = L*_{1}*+ L*_{2}*+*2*M*(b) If currents in two coils are in opposite directions, then

*L = L*_{1}*+ L*_{2}*–*2*M***Parallel combination of inductance**When the coefficient of coupling

*K*= 0, then$$\mathrm{L}\mathrm{}=\frac{1}{{\mathrm{L}}_{1}}+\frac{1}{{\mathrm{L}}_{2}}$$

When the coefficient of coupling

*K*= 1, then(a) If currents in two coils are in same direction, then

$$\mathrm{L}\mathrm{}=\frac{{\mathrm{L}}_{1}{\mathrm{L}}_{2}\mathrm{}\u2013\mathrm{}{\mathrm{M}}^{2}}{{\mathrm{L}}_{1}\mathrm{}+\mathrm{}{\mathrm{L}}_{2}\mathrm{}+\mathrm{}2\mathrm{M}}$$

(b) If currents in two coils are in opposite directions, then

$$\mathrm{L}\mathrm{}=\frac{{\mathrm{L}}_{1}{\mathrm{L}}_{2}\mathrm{}\u2013\mathrm{}{\mathrm{M}}^{2}}{{\mathrm{L}}_{1}\mathrm{}+\mathrm{}{\mathrm{L}}_{2}-\mathrm{}2\mathrm{M}}$$