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.”

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

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. 

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.

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 2^nd 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

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

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’

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 V_s 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:

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. 

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.

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):

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 

P_ins  =  I$\varepsilon $

Or, P_ins  =  I . L $\frac{dI}{dt}$

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

dw  =  P_ins 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

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 N_p and N_s be the total number of turns of primary and secondary coil respectively.

Now, the total magnetic linked with the primary coil is 

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

Similarly, the total magnetic linked with the secondary coil is 

${{\phi }_{S}}$ = N_s $\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, ${{\varepsilon }_{S}}$ = – N_s  $\frac{d\phi }{dt}$ ………  (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 L_p and L_s 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. L_p $\propto $  N_p²  ………. (iv) $\because $ L = $\frac{{{\mu }_{o}}{{N}^{2}}A}{l}$

Similarly for secondary coil,

L_s $\propto $  N_s² ….……. (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 I_p and I_s 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 (P_in)  =  I_P ${{\varepsilon }_{P}}$

Output power (P_out)  = I_S ${{\varepsilon }_{S}}$

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

I_p ${{\varepsilon }_{P}}$  =  I_s ${{\varepsilon }_{S}}$

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

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. 

Numerical problems

Electromagnetic Induction

Q.1       The magnetic flux passing perpendicular to the plane of coil is given by $\phi $= 4t² + 5t + 2 , where f is in weber and t is in second. Calculate the magnitude of instantaneous emf induced in the coil when t = 2 sec.

Ans: 21V

Q.2       The magnetic flux passing perpendicular to the plane of a coil is given by $\phi $= 4t² + 5t +2 , where f is in Weber and t is in seconds. Calculate the magnitude of instantaneous emf induced in the coil when t = 3 sec.

Ans: 29 V

Q. extra-1 A magnetic field of flux density 10 T acts normal to a 50 turn coil of 100 cm² area. Find the emf induced in it if the coil is removed from the field in 0.1 sec.

Ans: 50 volt

Q.extra-2 A coil of wire containing 500 circular loops with radius 4 cm is placed between the poles of a large electromagnet, where the magnetic field is uniform and at an angle of 60^o with the plane of coil. The field decreases at the rate of 0.2 Ts^–1 . What is the magnitude of the induced emf ?

0.435 V

Q.3       A straight conductor of length 25 cm is moving perpendicular to its length with a uniform speed of 10 m/s making an angle of 45° with a uniform magnetic field of 10 T. Calculate the emf induced across its length.

Ans: 17.67 V

Q.4       A straight conductor of length 15 cm is moving with uniform speed of 10 ms^-1 making an angle of 30° with uniform magnetic field of 10^-4 Tesla. Calculate the emf induced across the length.

Ans: 7.5×10^-5 V

Q.5       A coil of 100 turns, each of area 2 x 10^-3 m² has a resistance of 12 W. It lies in a horizontal plane in a vertical magnetic flux density of 3×10^-3 Wbm^-2. What charge circulates through the coil if its ends are short- circuited and the coil is rotated through 180° about a diametrical axis?

Ans: 1×10^-4 C

Concept of Angle of dip

Q.6       A jet plane is flying due west at the speed of 1800 km/hr. What is the voltage difference developed between the ends of the wings 25m long of the earths’ magnetic field at that location is 5×10⁴ T and the angle of dip is 45°?

Ans: 4.419 V

Q.7       A metal aircraft with a wing span of 40 m flies with a speed 1000 km hr1 in a direction due east at constant altitude in a region of the northern hemisphere where the horizontal component of the earth’s magnetic field is 1.6×10^-5T and the angle of dip is 41°. Find the potential difference developed between the tips of the wing.

Ans: 0.154V

Q.8       An aircraft with a wingspan of 40 m files with a speed of 1080 kmhr^–1 in the eastward direction at a constant altitude in the northern hemisphere, where the vertical component of earth’s magnetic field is 1.75 ×10^-5 T. Find the emf that develops between the tips of the wings. Ans: 0.21 V

Q.9       A jet plane is traveling due west at the speed of 1800 km/h. What is the voltage difference developed between the ends of the wings 25m long, if the earth’s magnetic field at the location is 5.0×10^-4 T and the dip angle is 30°?

Ans: 3.125 V

Rotation

Q.10     When a wheel with metal spokes 1.2m long is rotated in a magnetic field of flux density 5×10^-5 T normal to the plane of wheel, an emf of 10^-2V is induced between the rim and axle. Find the rate of rotation of the wheel.

Ans: 44.21 rev/S

Q.11     A circular metal disc of area 3.0×10^-3 m² is rotated at 50 rev/s about an axis through its centre perpendicular to its plane. The disc is in a uniform magnetic field of flux density 5.0×10^-3T in the direction of the axle. What is the value of induced emf?

Ans: 7.5×10^-4 V

Q.12     Find the emf induced in a straight conductor of length 25 cm, on the armature of a dynamo and 12 cm from the axis, when the conductor is moving in a uniform radial magnetic field of 0.5 T. The armature is rotating at 1000 revolutions per minute.

Ans: 1.57 V

Q.13     A rectangular coil of 100 turns has dimensions 15×10 cm. It is rotated at the rate of 300 revolutions per minute in a uniform magnetic field of flux density 0.6 T. Calculate the maximum emf induced in it.

Ans: 28.27V

Self-Inductance & Mutual-Inductance

Q.14     A long solenoid of 1000 turns and cross sectional area 2×10^-3 m² carries a current of 2A and produces a flux density 52×10^-3 T inside it. Calculate the self-inductance of the coil.

Ans: 0.052H

Q.15     A long solenoid of 1000 turns and cross- sectional area 2×10^-3 m² carries a current of 2A and produces a flux density 52×10^-3 T in the middle of the coil. Assuming this value of flux density at all sections of the solenoid, calculate its self-inductance.

Ans: 0.052H

Q.16     A plane circular coil has 200 turns and its radius is 0.10m. It is connected to a battery. After switching on the circuit a current of 2A is set up in the coil. Calculate the energy stored in the coil. (μo = 4π×10^-7 Hm^-1)

Ans: 0.016 J

Q.17     A long solenoid with 15 turns per cm has a small loop of area 2cm² placed inside, normal to the axis of the solenoid. If the current carried by solenoid changes steadily from 2A to 4A in 0.1 second, what is the induced voltage in the loop, while the current is changing?

Ans: 7.5×10^–6 V

Q.18     An air-cored solenoid having a diameter of 4 cm and a length of 60 cm is wound with 4000 turns. Find the inductance of the solenoid. What will be the inductance of the solenoid if it has an iron core of relative permeability 400? (μ_о = 4p×10^-7 Н/m).

Ans: 0.042H, 16.84H

Q.19     Two plane coils having number of turns 1000 and 2000, and radii 5 cm and 10 cm respectively are placed co-axially in the same plane. Calculate their mutual inductance. (μ_o = 4π×10^-7 H/m)

Ans: M₂₁ = 0.79 H, M₁₂ = 0.099 H

Transformer

Q.20     A transformer has 500 turns in the primary coil and 100 turns in the secondary coil. What is the output voltage if the input voltage is 4000 volts? If the transformer is assumed to have an efficiency of 100%, what primary current is required to draw 2000 watts from the secondary?

Ans: 800V and 0.5 A

Q.21     A step down transformer transforms a supply line voltage 220 volts into 100 volts. Primary coil has 500 turns. The efficiency and power transmitted by the transformer are 80% and 80kw. Find (a) the number of turns in the secondary coil (b) power supplied.

Ans: 227, 10⁵ W

Also Read: Alternating Current Notes Class 12