Electromagnetic Induction Class 12 Physics | Notes

Electromagnetic induction:

The phenomenon in which an electromotive force will be set up on a coil due to relative motion between coil and magnet is known as electromagnetic induction. 

The electromotive force set up in the coil is known as induced emf and current due to the induced emf is known as induced current.

Magnetic flux:

The number of magnetic lines of force crossing through an area perpendicularly is known as magnetic flux.

It is denoted by $\phi $. Mathematically, magnetic flux is defined as the scalar product of magnetic field intensity and area vector. (effective area).

i.e. Magnetic flux ($\phi $)  =  $\overrightarrow{B}.\overrightarrow{A}$ 

or,    $\phi $   =  BA cos$\theta $

where $\theta $ is the angle between the direction of the magnetic field and normal to the plane of the conductor. 

If coil has N turns, then total magnetic flux,

$\phi $   =  NBA cos $\theta $

It’s unit is Tm2 or weber (Wb)

It is a scalar quantity.

Special cases

i. If $\theta $ = 0°

We have,

$\phi $ = BA cos$\theta $

$\phi $ = BA cos 0°

$\therefore $ $\phi $ = BA (max.)

ii. If $\theta $ = 90°

We have,

or, $\phi $ = BA cos$\theta $

or, $\phi $ = BA cos 90°

$\therefore $ $\phi $ = 0

Faraday’s laws of electromagnetic induction:

1. Faraday’s First law of electromagnetic induction:

Faraday’s First law of electromagnetic induction states that whenever there is change in magnetic flux linked with a coil, an amount of e.m.f. induced in it and it lasts so long as the change in magnetic flux is taking place. 

2. Faraday’s Second law of electromagnetic induction:

Faraday’s Second law of electromagnetic induction states that the induced emf in a coil is directly proportional to rate of change of magnetic flux linked with the coil.

i.e. Induced emf ($\varepsilon $ or E or e) $\propto $  $\frac{d\phi }{dt}$

Or,   $\varepsilon $  =  – k $\frac{d\phi }{dt}$

–ve sign indicates that induced emf is opposing nature and also known as back emf.

$\therefore $ $\varepsilon $  =  $\frac{-d\phi }{dt}$

Let ${{\phi}_{1}}$ be the initial magnetic flux linked with the coil. After time t, the magnetic flux linked with the coil becomes ${{\phi }_{2}}$. Now, above equation becomes, 

$\varepsilon $  =  $\frac{-({{\phi }_{2}}-{{\phi }_{1}})}{t}$

Lenz’s Law:

Lenz’s law states that, “the direction of induced current is such that it always opposes the cause which produced it.”

Electromagnetic Induction Class 12 Physics | Notes

Fig (a): When magnet is moving towards the coil

Electromagnetic Induction Class 12 Physics | Notes

Fig (b): When magnet is moving away from the coil

Explanation

Let us consider a coil AB is connected with a sensitivity galvanometer. Let a bar magnet is moving towards the coil with its north-pole pointing to end A of the coil. When a bar magnet is moved to the coil, the rate of change of magnetic flux is linked with the coil. Due to this, rate of change of magnetic flux, an amount of emf induced in the coil and induced current flows, in anticlockwise direction which is indicated by deflection of galvanometer. When induced current flows through the coil, magnetic field will be developed with N-pole at end A and south pole at end B. If we furthermore the bar magnet towards the coil we have to do extra work against the repulsive force between bar magnet and coil magnet. The expense of mechanical work is converted into electrical energy in the form of induced emf. 

Now the bar magnet is moved away from the coil. When it moves away from the coil, the rate of change of magnetic flux linked with the coil due to which an amount of emf induces in the coil and induced current flows in clockwise direction which is indicated by deflection of the galvanometer. When induced current flows through the coil. Magnetic field will be developed in it with north pole at end B and south pole at end A. If we further move the bar magnet away from the coil. We have to do extra work against the attractive force between bar magnet and coil magnet.

The expense of mechanical work is converted into electrical energy in the form of induced emf. In this way, the energy is conserved by Lenz’s law. 

The above result shows that the cause of induced current is motion of the magnet but when induced current is produced it opposes the motion of the magnet. 

Emf induced in a rod (straight conductor) moving in a uniform magnetic field. 

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Fig: Emf induced in a rod (conductor) moving in uniform magnetic field.

Let us consider a conductor PQ of length l is moving in uniform magnetic field B in a direction perpendicular to that of the magnetic field. When the conductor moves in a magnetic field, the free electrons experience Lorentz force. Due to this Lorentz force, electrons are deflected downward i.e. toward Q leaving an equal amount of positive charge towards P i.e. upward. 

Suppose in a small time the conductor PQ moves through a small distance x.

Now,

Area swapped by the conductor (A) = lx

Magnetic flux linked with conductor ($\phi $)  =  BA cos$\theta $, there $\theta $ is the angle between magnetic field and area vector. 

Or, $\phi $  =  BA cos 0°  ($\because $$\theta $ =  0°)

Or, $\phi $  =  B l x  …………….. (i) ( $\because $  A = lx)

From Faraday’s second law of electromagnetic induction,

Induced emf ($\varepsilon $) = – $\frac{d\phi }{dt}$

Or, $\varepsilon $  =  $\frac{-d}{dt}$  (Blx)

Or, $\varepsilon $  =  Bl  $\frac{dx}{dt}$

$\therefore $ $\varepsilon $  =  B l V    [$\because $   v  =  $\frac{dx}{dt}$ ]

This is the required expression for emf induced in a rod (conductor) moving in a uniform magnetic field.

Emf induced in a coil rotating in a uniform magnetic field.

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Fig: Emf induced in a coil rotating in uniform magnetic field.

Let us consider a rectangular coil PQRS of area A and total no. of turns N is rotating in uniform magnetic field B with constant angular velocity $\omega $. Suppose at any instant of time the normal to the plane of the coil makes an angle θ with direction of magnetic field. When the coil rotates, the angle between normal to the plane of coil and magnetic field changes continuously due to which the magnetic flux linked with coil becomes variable and an amount of emf is induced in the coil. 

Magnetic flux linked with each turns ($\phi $)’  =  $\overrightarrow{B}.\overrightarrow{A}$

Or, $\phi $’  =  BA cos $\theta $

Total magnetic flux linked with coil ($\phi $)  =  N$\phi $’

Or, $\phi $  =  NBA cos$\theta $

Or, $\phi $  =  NBA cos $\omega $t.     [$\because $      $\theta $ = $\omega $t]

From Faraday’s 2nd law,

Induced emf ($\varepsilon $)  =  $\frac{-d\phi }{dt}$

or, $\varepsilon $  =  $\frac{-d}{dt}$  (NBA cos $\omega $t)

or, $\varepsilon $  =  – NBA $\frac{d}{dt}$  cos $\omega $t

or, $\varepsilon $  =  – NBA  (–$\omega $ sin $\omega $t)

$\therefore $ $\varepsilon $  =  NBA  $\omega $sin $\omega $t

$\varepsilon $  =  ${{\varepsilon }_{o}}$ sin $\omega $t

Where, ${{\varepsilon }_{o}}$ = NBA$\omega $ is maximum value of induced emf.

AC generator:

AC generator is an electrical instrument which is used to convert mechanical energy into electrical energy. 

Principle

It works upon the principle that “when a coil is rotated in a uniform magnetic field, an amount of emf will be induced in it. The induced emf is $\varepsilon $  =  ${{\varepsilon }_{o}}$ sin $\omega $t. 

Construction

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An AC generator consist of mainly following parts:

1. Armature

It consists of a rectangular coil (cdef) made up of insulated copper wire on a non-magnetic metallic frame with soft iron as a central core.

2. Field magnet

It consists of a concave pole piece of permanent magnet. The permanent magnet is used for small generators but electromagnet is used for large generators. 

3. Slip ring and brushes

In figure, R1 and R1 are slip rings and B1 and B2 are brushes. The slip rings are connected with two ends of armature. The slip ring rotates when the armature is rotated. The brushes B1 and B2 are in contact with slip rings which are used to transfer current from armature to external circuit. 

Working

When the coil rotates, the angle between normal to the plane of coil and magnetic field changes continuously due to which the magnetic flux linked with the coil becomes variable and an amount of emf is induced in the coil. The emf induced in the coil in different time is calculated below: 

i. At t = 0

We have,

$\varepsilon $  =  ${{\varepsilon }_{o}}$ sin $\omega $t

Or, $\varepsilon $  =  ${{\varepsilon }_{o}}$ sin ( $\frac{2\pi }{T}$ . 0)

Or, $\varepsilon $  =  ${{\varepsilon }_{o}}$ sin 0

$\therefore $ $\varepsilon $  =  0 …..…. (i)

ii. At t = $\frac{T}{4}$

We have,

$\varepsilon $  =   ${{\varepsilon }_{o}}$sin $\omega $t

Or, $\varepsilon $  =  ${{\varepsilon }_{o}}$ sin $\frac{2\pi }{T}$ . $\frac{T}{4}$

Or, $\varepsilon $  =  ${{\varepsilon }_{o}}$ sin $\frac{\pi }{2}$

$\therefore $ $\varepsilon $  =  ${{\varepsilon }_{o}}$ ….. (ii)

iii. At t = $\frac{T}{2}$

We have,

$\varepsilon $  =  ${{\varepsilon }_{o}}$ sin $\omega $t

Or, $\varepsilon $  = ${{\varepsilon }_{o}}$ sin $\frac{2\pi }{T}$ . $\frac{T}{2}$

Or, $\varepsilon $  =  ${{\varepsilon }_{o}}$ sin $\pi $

$\therefore $ $\varepsilon $  =  0  ………………. (iii)

iv. At t = $\frac{3T}{4}$

We have,

$\varepsilon $  =  ${{\varepsilon }_{o}}$ sin $\omega $t

Or, $\varepsilon $  =  ${{\varepsilon }_{o}}$ sin $\frac{2\pi }{T}$ . $\frac{3T}{4}$

Or, $\varepsilon $  =  ${{\varepsilon }_{o}}$ sin $\frac{3\pi }{2}$

$\therefore $ $\varepsilon $  =  –${{\varepsilon }_{o}}$  ……………….. (iv)

v. At t = T

We have,

$\varepsilon $  =  ${{\varepsilon }_{o}}$ sin $\omega $t

Or, $\varepsilon $  = ${{\varepsilon }_{o}}$ sin $\frac{2\pi }{T}$ . T

Or, $\varepsilon $  =  ${{\varepsilon }_{o}}$ sin 2$\pi $

$\therefore $ $\varepsilon $  =  0  ………………. (v)

The graph between induced emf and time is
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Self-induction:

The phenomenon of inducing emf in a coil by passing changing current through it is known as self-induction. 

Coefficient of self-induction:

The property of a coil by virtue of which it opposes the changing current flowing through it is known as coefficient of self-induction or self-inductance. 

It is denoted by ‘L’
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Let us consider a coil of total number of turns N, no. of turns per unit length ‘n’ is connected with an AC source. When the coil is connected with an AC source, changing current flows through it. Due to which, magnetic flux linked with the coil changes and an amount of emf induced in the coil. Here the current forms a closed loop, so current flows through it.

According to Faraday’s second law, the emf induced in the coil is

Induced emf ($\varepsilon $) $\propto $ $\left( \frac{d\phi }{dt} \right)$ ………….. (i)

Since we know that the magnetic flux linked with the coil is directly proportional to current flowing through it.

i.e.  $\phi $  $\propto $  I  ………… (ii) 

From equation (i) and (ii);

$\varepsilon $  $\propto $  $\frac{dI}{dt}$

$\varepsilon $  =  – L $\frac{dI}{dt}$

Where L is proportionality constant and known as coefficient of self induction or self inductance. The negative sign shows that induced emf is opposing in nature. 

Or, L  =  – $\frac{\varepsilon }{\frac{dI}{dt}}$

If  $\frac{dI}{dt}$  =  1 A s–1 then

L  =  – $\varepsilon $

Now, coefficient of self induction or self inductance can also be defined as the emf induced in a coil due to flow of changing current 1 A s–1 through it.

It’s unit is Vs A–1 or Henry. (H). 

Relation between $\phi$ and I.

According to Faraday’s second law, the induced in a coil is

Induced emf ($\varepsilon $)  =  $\frac{-d\phi }{dt}$  …………….. (i) 

Also, from the expression of coefficient of self-induction, the emf induced in a coil is 

Induced emf ($\varepsilon $)  =  – L $\frac{dI}{dt}$

Or,     $\varepsilon $  =  – $\frac{d(LI)}{dt}$ ………………… (ii) 

From equation (i) and (ii)

$\phi $  =  LI

Or, L = $\frac{\phi }{I}$

Factors affecting self-inductance of a coil:

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Let us consider a coil (solenoid) having length l, total number of turns (N), no. of turns per unit length n, area of cross section A is connected with an AC source. Let I be the instantaneous value of current flowing the coil. The magnetic field developed in the coil is solenoid. 

B  =  ${{\mu }_{o}}$ n I    

Or, B  =  $\frac{{{\mu }_{o}}NI}{l}$ $\left( \because n=\frac{N}{l} \right)$

Magnetic flux linked with each turn of coil ($\phi ‘$)  =  BA  =$\frac{{{\mu }_{o}}NIA}{l}$

Total magnetic flux linked with coil ($\phi $)  =  N$\phi ‘$

$\therefore $ $\phi $  = $\frac{{{\mu }_{o}}{{N}^{2}}IA}{l}$

By the definition of self-inductance 

L  =  $\frac{\phi }{I}$

L  =  $\frac{\frac{{{\mu }_{o}}{{N}^{2}}IA}{l}}{I}$

$\therefore $ L  = $\frac{{{\mu }_{o}}{{N}^{2}}A}{l}$

The above result shows that self-inductance of a coil is

i. Directly proportional to magnetic permeability of medium. 

i.e.  L $\propto $  $\mu $  (if N, A and l are constant)

ii. Directly proportional to square of number of turns. 

i.e. L $\propto $  N2  (if $\mu $, A and l are constant)  

iii. Directly proportional to the area of cross-section of the coil.

i.e. L $\propto $  A (if $\mu $, l and N are constant)

iv. Inversely proportional to length of coil. 

i.e.  L $\propto $  $\frac{1}{l}$  (if N, A and $\mu $ are constant)

Mutual induction:

The phenomena of inducing emf in a coil by passing changing current through a neighboring coil is known as mutual induction. 

Coefficient of mutual induction / Mutual inductance. 

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Let us consider two coils C1 and C2 are placed very close to each other without metallic contact. The coil C1 is connected with AC source (changing current) and output is taken from coil C2. Two coils are placed in such a way that when changing current from AC source is passed through the first coil C1, the magnetic flux linked with the coil C1 changes which enhances the change in magnetic flux through C2. Due to the change in magnetic flux linked with the coil C2 and emf is induced in it.

According to Faraday’s second law of electromagnetic induction,

Induced emf ($\varepsilon $) $\propto $  $\frac{d{{\phi }_{2}}}{dt}$  ………….. (i)

Let I1 be the instantaneous value of current flowing through coil C1. Since magnetic flux linked with the second coil C2 is directly proportional to the current through C1

i.e.  ${{\phi }_{2}}$  $\propto $ I1  ………………. (ii)

From equation (i) and (ii);

$\varepsilon $  $\propto $  $\frac{d{{I}_{1}}}{dt}$

Or, $\varepsilon $  =  – M $\frac{d{{I}_{1}}}{dt}$

Where, M is proportionality constant and known as coefficient of mutual induction or mutual inductance and negative sign shows that the induced emf opposes the cause inducing it.

M  =  – $\frac{\varepsilon }{\frac{d{{I}_{1}}}{dt}}$

If  $\frac{d{{I}_{1}}}{dt}$ =  1 As–1  then,

M  =  – $\varepsilon $

The coefficient of mutual induction is defined as the emf induced in a coil due to flow rate of change of current 1 A s–1 through neighboring coil.

Relation between ${{\phi}_{2}}$ and I1.

According to Faraday’s second law of electromagnetic induction the emf induced in the second coil is 

Induced emf ($\varepsilon $)  =  –$\left( \frac{d{{\phi }_{2}}}{dt} \right)$   …………………… (i) 

From the expression of coefficient of mutual induction, the emf induced in a coil is 

Induced emf ($\varepsilon $)  =  – M$\frac{d{{I}_{1}}}{dt}$

Or, $\varepsilon $  =  $\frac{-d}{dt}$(MI1)  ………………….. (ii) 

From equation (i) and (ii);

${{\phi }_{2}}$  =  MI1

Or, M  =  $\frac{\phi {}_{2}}{{{I}_{1}}}$

Factors affecting mutual inductance 

‘OR’

Expression for coefficient of mutual induction. 

‘OR’

Mutual inductance of a coil with respect to another coil.

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Let us consider two coils C1 and C2 of same length and same cross-section area are placed very close to each other without metallic contact. Let N1 and N2 be the total number of turns and n1 and n2 be the no. of turns per unit length of coil C1 and C2 respectively. 

Initially the coil C1 is connected with AC source, let I1 be the instantaneous current flowing through the coil C1 due to which magnetic field developed in the coil C1 is 

B1 = ${{\mu }_{o}}$ n1 I1 

Or, B1  = $\frac{{{\mu }_{o}}{{N}_{1}}{{I}_{1}}}{l}$ $\because $ n1 = $\frac{{{N}_{1}}}{l}$

Magnetic flux linked with each turn of coil is (${{\phi }_{2}}$’)  =  B1A

${{\phi }_{2}}$’= $\frac{{{\mu }_{o}}{{N}_{1}}{{I}_{1}}A}{l}$

Total magnetic flux linked with coil C2 is 

${{\phi }_{2}}$  = N2 ${{\phi }_{2}}$’

${{\phi }_{2}}$  =   $\frac{{{\mu }_{o}}{{N}_{1}}{{N}_{2}}{{I}_{1}}A}{l}$  ……….. (i) 

Now, mutual inductance of coil C2 w.r.t. coil C1 is

M21 =  $\frac{{{\phi }_{2}}}{{{I}_{1}}}$  ……… (ii)

From equation (i) and (ii) ;

M21 =  $\frac{\frac{{{\mu }_{o}}{{N}_{1}}{{N}_{2}}{{I}_{1}}A}{l}}{{{I}_{1}}}$

$\therefore $ M21 =$\frac{{{\mu }_{o}}{{N}_{1}}{{N}_{2}}A}{l}$

Now, the coil C2 is connected to the AC source. Let I2 be the instantaneous value of current flowing through the coil C2 due to which magnetic field develop in the coil C2 is

B2 =  ${{\mu }_{O}}{{n}_{2}}{{I}_{2}}$  

Or, B2 =  $\frac{{{\mu }_{o}}{{N}_{2}}{{I}_{2}}}{l}$ $\because $ ${{n}_{2}}=\frac{{{\mu }_{o}}{{N}_{2}}{{I}_{2}}}{l}$

Magnetic flux linked with each turn of coil C1 is (${{\phi }_{1}}$’)  =  B2 A

= $\frac{{{\mu }_{o}}{{N}_{2}}{{I}_{2}}A}{l}$

Total magnetic flux linked with coil C1 is

${{\phi }_{1}}$  =  N1 ${{\phi }_{1}}$’

${{\phi }_{1}}$ = $\frac{{{\mu }_{o}}{{N}_{1}}{{N}_{2}}{{I}_{2}}A}{l}$ ………(iii)

Now, 

Mutual inductance of coil C1 w.r.t coil C2 is 

M12  =  $\frac{{{\phi }_{1}}}{{{I}_{2}}}$  …………. (iv) 

From equation (iii) and (iv);

M12  =$\frac{\frac{{{\mu }_{o}}{{N}_{1}}{{N}_{2}}{{I}_{2}}A}{l}}{{{I}_{2}}}$

$\therefore $ M12  =$\frac{{{\mu }_{o}}{{N}_{1}}{{N}_{2}}A}{l}$

Energy stored in an inductor (coil):

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Let us consider an inductor of inductance (L) is connected with an AC source as shown in figure. When changing current passes through the coil, the magnetic flux linked with coil changes which ultimately results in the induction of emf in the coil. The induced emf can be written as  

$\varepsilon $  =  L $\frac{dI}{dt}$…..(i) [Taking magnitude only]

Let I be the instantaneous value of current flowing through the inductor. The instantaneous power developed in the inductor is 

Pins  =  I$\varepsilon $

Or, Pins  =  I . L $\frac{dI}{dt}$

Let dw be the small amount of work done in small time dt i.e. 

dw  =  Pins dt

Or, dw  =  L I  $\frac{dI}{dt}$  dt

Or, dw  =  L I d I  …… (ii) 

Now, the total amount of work done on increasing the current from O to I can be obtained by integrating equation (ii) from limit O to I. 

W  =  $\int\limits_{0}^{I}{dw}$  =  L $\int\limits_{0}^{I}{IdI}$

Or, W  =  L $\left| \frac{{{I}^{2}}}{2} \right|_{0}^{I}$

Or, W  =  $\frac{1}{2}$  L I2

This work done is stored as energy in the inductor. 

$\therefore $ i.e. Energy stored (E)  =  $\frac{1}{2}$  L I2

Transformer:

It is a device which is used to convert voltage at high AC to high voltage at low AC and vice-versa.

A transformer which is used to convert to voltage at high AC to high voltage at low AC is known as step-up transformer which is used to convert high voltage at low AC to low voltage at high AC is known as step-down transformer.

Principle

A transformer works on the principle that if an amount of changing current is passed through a coil, emf is induced in a neighboring coil. 

Construction

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A transformer consists of two coils made up of insulated copper wire. A coil which is connected with AC source is known as primary coil and another coil from which output is taken is known as secondary coil. The coil wound on a non-magnetic metallic frame with soft iron as a central core. The soft iron consists of a large number of thin rectangular iron strips coated with varnish and joined together so as to form a single block. The iron strips are insulated from each other. The current through primary coil is primary current and through secondary coil is secondary current. 

Working

The construction of transformer is such that there is no loss of magnetic flux and hence the magnetic flux linked with each coil of the primary is equal to the magnetic flux linked with each coil of secondary i.e.

${{\phi }_{P}}$’ = ${{\phi }_{S}}$’ =$\phi $ (say)

Let Np and Ns be the total number of turns of primary and secondary coil respectively.

Now, the total magnetic linked with the primary coil is 

${{\phi }_{P}}$ = Np $\phi $

From faraday’s second law of EMI, induced emf in the primary coil is given by

${{\varepsilon }_{P}}$ = – $\frac{d{{\phi }_{P}}}{dt}$

Or, ${{\varepsilon }_{P}}$ = – $\frac{d({{N}_{P}}\phi )}{dt}$ 

Or, ${{\varepsilon }_{P}}$ = – NP $\frac{d\phi }{dt}$ ………  (i) 

Similarly, the total magnetic linked with the secondary coil is 

${{\phi }_{S}}$ = Ns $\phi $ 

From faraday’s second law of EMI, induced emf in the secondary coil is given by

${{\varepsilon }_{S}}$ = – $\frac{d{{\phi }_{s}}}{dt}$

Or, ${{\varepsilon }_{S}}$ = – $\frac{d({{N}_{s}}\phi )}{dt}$ 

Or, ${{\varepsilon }_{S}}$ = – Ns  $\frac{d\phi }{dt}$ ………  (ii)  

Dividing equation (ii) and (i), we get

$\frac{{{\varepsilon }_{s}}}{{{\varepsilon }_{P}}}$  =  $\frac{{{N}_{s}}}{{{N}_{P}}}$…..(iii)

The ratio $\frac{{{N}_{s}}}{{{N}_{P}}}$ is constant and known as transformation ratio, which determines whether the transformer is step up or step down. 

i. If k > 1 then ${{\varepsilon }_{S}}>{{\varepsilon }_{P}}$ $\Rightarrow $  Transformer is step up. 

ii. If k < 1 then ${{\varepsilon }_{S}}<{{\varepsilon }_{P}}$  $\Rightarrow $ Transformer is step down.

Let Lp and Ls be the self-inductance of primary and secondary coil, respectively. Since, we know that the self-inductance is directly proportional to the square of number of turns.

So for primary coil,

i.e. Lp $\propto $  Np2  ………. (iv) $\because $ L = $\frac{{{\mu }_{o}}{{N}^{2}}A}{l}$

Similarly for secondary coil,

Ls $\propto $  Ns2 ….……. (v)

From equation (iv) and (v);

$\frac{{{L}_{s}}}{{{L}_{P}}}$  =  $\frac{{{N}_{S}}^{2}}{{{N}_{P}}^{2}}$

Or, $\frac{{{N}_{S}}}{{{N}_{P}}}$ = $\sqrt{\frac{{{L}_{S}}}{L{}_{P}}}$  ………… (vi)

From equation (iii) and (vi);

$\frac{{{\varepsilon }_{s}}}{{{\varepsilon }_{P}}}$ = $\frac{{{N}_{S}}}{{{N}_{P}}}$ =  $\sqrt{\frac{{{L}_{S}}}{L{}_{P}}}$

Let Ip and Is be the current through primary and secondary coil. Since, input is given to the primary coil and out is taken through secondary coil, so we can write

Input power (Pin)  =  IP ${{\varepsilon }_{P}}$

Output power (Pout)  = IS ${{\varepsilon }_{S}}$

Then,

Efficiency ($\eta $)  =  $\frac{{{P}_{out}}}{{{P}_{in}}}$  × 100%

$\therefore $   $\eta $  = $\frac{{{I}_{s}}{{\varepsilon }_{s}}}{{{I}_{P}}{{\varepsilon }_{P}}}$ × 100%

For ideal transformer,

Input power = Output power

Ip ${{\varepsilon }_{P}}$  =  Is ${{\varepsilon }_{S}}$

$\therefore \text{ }\frac{{{I}_{P}}}{{{I}_{S}}}=\frac{{{\varepsilon }_{S}}}{{{\varepsilon }_{P}}}$

It gives, I $\propto $ $\frac{1}{\varepsilon }$

Why is AC economical for long range transmission?

For AC, I $\propto $ $\frac{1}{\varepsilon }$ , low current high voltage so less heat is produced (H = I2Rt). If it is supplied in the form of high current, large heat is developed, energy lost and wire can melt. 

Different types of loss in transformers. 

1. Copper loss

Primary and secondary coil has copper insulated). Due to resistance of Cu, heat is developed. So, low resistance Cu need to be used. 

2. Hysteresis loss

Alternate cycle of magnetization and demagnetization causes energy loss which is hysteresis loss. It can be minimized with the use of soft iron because it has low hysteresis loop area.

3. Flux loss

Total rate of change of magnetic flux in primary coil cannot be transferred to secondary coil. Some loss occur which is flux loss. It can be minimized by perfect coupling with the help of rectangular iron strip. 

4. Humming loss

Energy loss due to vibration of wires and materials used in transformer in the form of sound. It can be minimized by use of varnish. 

5. Loss due to eddy current

Current circulated in the whole metal block is eddy current. To minimize Cu loss, low resistance material is used. When emf is induced a large amount of I is developed which in turn produces more heat. So, it can be minimized by using an insulated iron strip. 

Eddy currents (Focault’s currents):

The circulating currents induced in the metal itself due to the change in magnetic flux linked with the metal are known as eddy currents. This is also known as Focult’s current.

The undesirable effects of eddy currents are;

(i) loss of electrical energy in the form of heat energy.

(ii) breaking of insulation in the electrical machines/ appliances due to production of heat.

(iii) cause of damping  effect

The applications of eddy current are induction furnace, diathermy, speedometer, energy meter, electromagnetic break, etc. 

Also Read: Alternating Current Notes Class 12

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