Electromagnetic induction Note and Numerical : Physics Class 12

Magnetic flux (ϕ through any surface is defined as the number of magnetic lines of force passing through an area normally. This note provides us an information on magnetic flux.

Magnetic Flux

Magnetic flux (ϕ)(𝜙) through any surface is defined as the number of magnetic lines of forcecrossing through that surface .

Consider a small surface are A.a normal is drawn to the surface. If θ𝜃 be the angle between normal and uniform magnetic field as shown in figure, then the magnetic flux (ϕ)(𝜙) through the surface is defined as

$$ \begin{align*} \phi &= \vec B . \vec A= BA\cos \: \theta \\ \text {or,} \: \phi &=(B\: \cos \: \theta ) A \dots (i) \\ \end{align*} $$

Now \(B\cos \: \theta \) is the component of the magnetic field normal to the plane of the surface and can be represented by B_n­. so equation (i) becomes,

$$ \phi = B_n A \dots (ii) $$

Thus, magnetic flux over a given surface is defined as the product of the area of the surface and normal component of magnetic field.

Special cases

1. When \(\theta = 0^o \), i.e. uniform magnetic field is acting perpendicular to the plane of the suface, then,\(\phi = BA\cos \: 0^o\) = BA which is maximum.
2. When \(\theta = 90^o \) i.e.. uniform magnetic field is along the surface is then \(BA\cos \: 90^o = 0\) which is minimum.

Unit of Magnetic Flux

In SI, the magnetic flux measured in Weber (Wb).

$$ \begin{align*} \text {Since,} \: \phi = B_nA \\1 \: \text {Weber} = 1 \: \text {tesla} \times 1\: m^2 = 1 T m^2 \\ \text {In CGS-system, unit of magnetic flux is Maxwell.} \\ 1 \: \text {Maxwell} = 10^{-8} \: \text {Weber} \\ \end{align*} $$

Dimensional formula of Magnetic Flux

We know,

$$ \begin{align*} \phi = BA \: \cos\: \theta \\ \text {Since,} \: B = \frac {F}{qv} \: \text {and } \: \cos \: \theta \: \text { is dimensionless. So} \\ \phi &= \frac {FA}{QV} = \frac {FA}{ItV} \:\:\: [ \because I = \frac qt \rightarrow q = it ] \\\frac {[F][A]}{[I]\: [t]\:[v]} &= \frac {[MLT^{-2}][L^2]}{[A][T][LT^{-1}]} \\ \therefore \: [\phi ] &= [ML^2T^{-2} A^{-1}] \\\end{align*} $$

Magnetic flux \((\phi )\) is a scalar quantity.

Flux Linkages and Faraday’s Law of Electromagnetic Induction

Flux Linkages

The product of number of turns (N) of the coil and magnetic flux \((\phi \)linking the coil is called flux linkages.

$$ \text {i.e. Flux Linkage} = N \phi $$

Experiments show that the magnitude of e.m.f induced in a coil is directly proportional to the rate of change of flux linkages. If N be the number of turns of the coil and the magnetic flux linking the coil changes from \(\phi _1\) to \(\phi _2\) in time t, then

$$ \begin{align*}\text {induced e.m.f} \: (\epsilon) \: \alpha \: \text {Rate of change of flux linkages} \\ \therefore \: \epsilon \propto \frac {N\phi _2 – N\phi _1}{t} \\ \end{align*} $$

Magnetic Flux Density (β)

The number of magnetic lines of force passing through unit area normally or magnetic flux per unit area is called magnetic flux density.

$$ \begin{align*} B &= \frac {\phi }{A_1} \\ \phi &= BA_1 \\ &= BA\cos \theta \\ \therefore \phi &= \vec B .\vec A \\ \end{align*} $$

Special cases

1. If \(\theta = 0 \) i.e. flux is normal to the surface then \( \phi _{\text {max}} = BA \).
2. If \(\theta = 90 \) i.e. flux is parallel to the surface then \( \phi _{\text {min}} = 0 \).

Faraday’s Law of Electromagnetic Induction

There were various series of an experiment performed by Faraday to prove electromagnetic induction. Faraday’s laws are as follows:

1. Whenever there is a change in the magnetic flux associated with a close lobe, there is an induced e.m.f or current and the e.m.f remains as long as the change in flux takes place.
2. The rate of change of magnetic flux is directly proportional to the rate of change of flux.
   $$ \begin{align*} \text {i.e} \: \:\text {Induced e.m.f} \: \epsilon \propto \text {rate of change of flux linkage.} \\ \epsilon &\propto \frac {N\phi _2 – N\phi _1}{t} \\ \text {or,} \: \epsilon &= k\frac {(N\phi _2 – N\phi _1)}{t} \\ \text {where k is proportionality constant.} \\ \text {In SI-system,} \: K = 1 \\ \epsilon &= \frac {N\phi _2 – N\phi _1}{t} \\ \text {In differential form, we have,} \\ \epsilon &= N \frac {d\phi }{dt} \dots (i)\\ \end{align*} $$

Since direction of induced e.m.f. is acting in the opposite direction to the cause due to which it is produced. So equation (i) can be written as

$$ \epsilon =- N \frac {d\phi }{dt} $$

which is combined Faraday-Lenz law or simply Faraday’s law.

Lenz’s Law

Lenz’s law states that the induced e.m.f opposes the motion or change producing it. In other words, Lenz’s law states that the induced current opposes the flux change causing it.

Consider a case in which a current is induced in a coil due to change in magnetic flux. For this assume that N-pole of a magnet is approaching towards the end of the coil. If len’s law applies, the induced current should flow in the coil that makes near end of coil behave as N-pole and repels the N-pole of a magnet. So, opposes the flux change. In the same arrangement, if the magnet is moved away from the coil, the near end of the coil should behave like a S-pole and attract the N-pole of the magnet. So, opposes the flux change again. The induced current in the case is therefore in the opposite direction to that when the magnet approaches. Lenz’s law is added in the expression of Faraday’s law by inducing a negative sign which shows that the current due to induced e.m.f opposes flux change.

Lenz law is in accordance with the law of conservation of energy

If S-pole were produced at the near end of the coil when N-pole of the magnet is approaching then the magnet would accelerate towards the coil, gaining kinetic energy as well as generating electrical energy. Thus, energy will be created. However, work has to be done to overcome the forces. Thus mechanical energy is transferred into electrical energy.

Direction of Induced e.m.f and Current

The direction of induced e.m.f and hence current can be determined by one of the following two methods:
1. Lenz’s law which has already discussed

2. Fleming’s Right Hand Rule

Fleming’s Right Hand Rule

Stretch the forefinger, middle finger and the thumb of the right hand mutually perpendicular to each other. If the force finger points in the direction of the magnetic field, the thumb gives the direction of the motion of the conductor then the middle finger gives the direction of the induced current.

Methods of Producing Induced e.m.f

From Faraday’s law, induced e.m.f is given by

$$ E = -N \frac {d\phi }{dt} $$

It shows that e.m.f depends on the number of turns N and rate of change of magnetic flux\(\phi \). Since magnetic flux, \(\phi = BA\: \cos \: \theta \).

So magnetic flux can be changed by

1. Changing the intensity of magnetic field (B).
2. Changing the orientation \((\theta)\) of the coil with respect to the magnetic field.
3. Changing the area of the conducting circuit.

Induced e.m.f in a straight conductor moving in a uniform magnetic field: Motional e.m.f.

Consider a straight conductor PQ of length l moving at right angles to a uniform magnetic field B with a velocity v. suppose the conductor moves through a small distance x in time t. then area swept out by the conductor is given by

$$ \begin{align*} \text {Area swept} &=l \times x \\ \therefore \text {Magnetic flux cut,} \: \phi &= B \times \text {Area swept} \\ &= b\times l \times x \\ &= Blx \\\end{align*} $$

From Faraday’s law of electromagnetic inducton, the magnitude of induced e.m.f in the conductor is given by

$$ \begin{align*} \\ e &= N \frac {d\phi }{dt} = N \frac {d}{dt} (Bl\: x) \\ &= N\: Bl \frac {dx}{dt} \\ \therefore e &= Bl\:v \:\:\: \left [\therefore N = 1, \: \text {and} \: \frac {dx}{dt} = v \right ] \\ \end{align*} $$

Special case:

If the conductor moves at an angle \(\theta \) to which thw conductor moves across the field is \(v\: \sin\: \theta\).

$$\therefore \;\;\; \epsilon = Bv\:l\sin \theta $$

The direction of the induced e.m.f. can be determined by Fleming's right hand rule .

Induced E.m.f. in a coil rotating in a magnetic field

Consider a rectangular coil being rotated with constant angular velocity \(\omega \) in the uniform magnetic field about an axis perpendicular to the field.

Let

$$ \begin{align*} N &= \text {number of turns of the coil} \\ A &= \text {area of each turn} \\ B &= \text {strength of magnetic field.} \\ \end{align*} $$

Suppose initially i.e. at t = 0, the plane of the coil is perpendicular to the direction of the magnetic field. Let the coil rotates in anticlockwise direction through an angle \(\theta = \omega t\) in time t. At this instant, the perpendicular to the plane of the coil makes an angle \(\theta \) with the direction of he field. Therefore, at this instant, the magnetic flux \(\phi \) through each turn of the coil is given by

$$ \begin{align*} \phi &= AB\: \cos\: \theta \\ &= AB\cos\: \omega t \\\end{align*} $$

Using Faraday's laws of electromagnetic induction, induced e.m.f in the coil is given by

$$ \begin{align*} E &= -N \frac {d\phi }{dt} \\ &= -N \frac {d}{dt} (AB\cos\: \omega t), \\ \text {where N is number of turns of the coil.} \\ &= -N\: AB \frac {d}{dt}(\cos\: \omega t) \\&= -N\: AB(-sin\: \omega t)\: \omega \\ \epsilon &= N\: AB\omega\: \sin\: \omega t \dots (i) \\ \end{align*}  $$ The magnitude of induced e.m.f. will be maximum (\(\epsilon _0\)) 

$$ \begin{align*}\\ \text {when} \: \sin\: \omega t = 1 \\ \text {i.e.} \: \epsilon_0 &= N\: AB\omega \\ \text {So equation} \: (i)\: \text {becomes} \\ \therefore \epsilon &= \epsilon _0\: \sin\: \omega t \dots (ii) \end{align*} $$

Thus a coil rotating with a constant angular velocity in a uniform magnetic field produces a sinusoidally alternating e.m.f. If a resistor of resistance R is connected across the coil, the resultant current will be sinusoidal and is given by

$$ \begin{align*} I &= \frac {\epsilon}{R} = \frac {\epsilon _0 \sin \: \omega t}{R} \\ \therefore I &= I_0\sin\: \omega t \\ \text {where}\: \frac {\epsilon _0}{R} = I_0 \: \text {is maximum value} \\ \end{align*} $$

Such a rotating coil in a uniform magnetic field is the basic operating principle of an a.c. generator.

AC Generator or AC Dynamo

AC Generator

An electrical machine used to convert mechanical energy into electrical energy is called A.C. generator.

Principle

It works on the principle of electromagnetic induction i.e. when a coil is rotated in a uniform magnetic field, an e.m.f is induced in it.

Construction

The main components of a.c. generator are:

1. Armature
   Armature coil (ABCD) consists of a large number of turns of insulated copper wire wound over a soft iron core.
2. Strong field magnet
   The armature is rotated in a strong uniform magnetic field provided by powerful permanent magnet NS. The axis of rotation is perpendicular to the field.
3. Slip rings
   The two ends of the armature are connected to rings R₁and R₂.
4. Brushes
   The two carbon brushes B₁and B₂are pressed against the slip rings. These brushes are connected to load and remain fixed while slip rings rotate along the armature.

Working

When the armature coil ABCD rotates in the magnetic field provided by the strong magnetic field. Magnetic flux is produced in the coil while cutting the magnetic lines of force. Hence e.m.f is induced in the coil. The direction of induced e.m.f or the current in the coil is determined by Fleming’s right-hand rule.

The current flows out through the brush B₁in one direction of half of the revolution and through the brush B₂in the next half revolution in the reverse direction. This process is repeated. Therefore, e.m.f produced is of alternating nature.

Theory

Consider the plane of the coil be perpendicular to the magnetic field \(\vec B\). Let the coil be rotated anticlockwise with a constant angular velocity \(\omega \). Then the angle between the normal to the coil and \(\vec B\) at any time t is given by

$$\theta = \omega t$$

The component of the magnetic field normal to the plane of the coil = B\cos\: \theta = B\: \cos\: \omega t.\)

Magnetic flux linked with a single coil \(= B\cos\:\omega t\) A where A is the area of the coil.

So, magnetic flux linked with N coils, \(\phi = NBA\: \cos\: \omega t\).

From Faraday’s laws of electromagnetic induction, the induced e.m.f. in the coil is given by

$$ \begin{align*} \epsilon &= - \frac {d\phi }{dt} = -\frac {d(NBA\: \cos \omega t)}{dt} \\ &= - NBA\frac {d}{dt}(\cos\: \omega t) = -NBA(\sin \: \omega t)\omega \\ \epsilon &= NBA\: \omega \: ]sin\: \omega t \dots (i) \\ \end{align*} $$

The magnitude of induced e.m.f. will be maximum i.e. \((\epsilon _0)\), when \(\sin\: \omega t = 1,\) so

$$ \begin{align*} \epsilon _0 &= NBA\omega \\ \text {Thus equation} \: (v)\: \text {becomes,} \\ \epsilon &= \epsilon _0 \sin \omega t \dots (i) \\ \text {The magnitude of induced e.m.f. is max when,}\: \sin\: \omega t = 1, \\ \epsilon _0 &= NBA\omega \\ \text {Equation} \: (v)\: \text {becomes,} \epsilon &= \epsilon _0 \sin\: \omega t \dots (ii) \\ \text {Instantaneous current is given by} \\ I &= \frac {\epsilon}{R} \\&= \frac {\epsilon _0 \sin\: \omega t}{R}\\ \text {where R is resistance of the coil} \\ \therefore I &= I_0 \sin \omega t \\ \text {where}\: \frac {\epsilon _0}{R} = I_0 \: \text {is maximum value} \\ \end{align*} $$

D.C. generator is a device which converts mechanical energy into electrical energy. This note provides us an information on DC generator or DC dynamo.

DC Generator or DC Dynamo

D.C. generator is a device which converts mechanical energy into electrical energy.

Principle

It is based on the principle of electromagnetic induction i.e. whenever the amount of magnetic flux linking coil changes, an e.m.f. is induced in the coil.

Construction

A D.C.generator has the same parts as that of A.C. generator except for the slip ring which is replaced by split rings or commutator.

The main components of d.c. generator are:

1. Armature
   Armature coil (ABCD) consists of a large number of turns of insulated copper wire wound over a soft iron core.
2. Strong field magnet
   The armature is rotated in a strong uniform magnetic field provided by powerful permanent magnet NS. The axis of rotation is perpendicular to the field.
3. split rings
   The two ends of the armature are connected to a split rings R₁and R₂. The split rings convert alternating voltage into direct voltage across the brushed
4. Brushes
   The two carbon brushes B₁and B₂are pressed against the split rings.

Working

Working of d.c. the generator is almost similar to a.c. generator. Let the plane of the armature coil is perpendicular to the magnetic field. At this position, the e.m.f. across the load is zero.

 

Now, the coil is rotated through 90^oanticlockwise, the brushes B₁and B₂are in contact with the split rings. The plane of the coil is along the magnetic field and hence maximum e.m.f. is produced in the coil. This e.m.f. is obtained across the load. The direction of induced current is anticlockwise according to Fleming’s right-hand rule.

Again the coil is rotated through 90^o. Now from the brushes B₁and B₂lose their contact from the split ring or commutator and the e.m.f. across the load is zero.

The coil is further rotated through 90^o. Now the brushes B₁is pressed against R₂and B₂against R₁. Thus, it is seen that the direction of induced e.m.f. or current at any position in the coil is same.

 

The e.m.f. or currently is not smooth as desired. However, the output can be made smooth if another coil of same dimensions as that of first is fixed on the same axis of rotation. These two coils are kept in such a way that their planes are at right angle to each other. Besides, each coil works independently and sends its own current into the load. Thus by increasing the number of coils and the corresponding split rings or commutator, nearly unidirectional smooth current can be obtained across the load.

 

Self Inductance

Self-inductance is the property of a coil by virtue of which it opposes the growth or decay of the current flowing through it.

Consider a coil connected to a battery through a key(K) as shown in the figure. As the key is on, current in the coil starts increasing. Due to it, the magnetic field and hence flux linkage around the coil also increases. The direction of induced e.m.f. is such that it opposes the growth of current in the coil. This delays the current to acquire the maximum value.

When the key (K) is switched off, the current in the coil starts decreasing. So the magnetic flux linked with the coil decreases as a result. Due to change in magnetic flux e.m.f is also produced. According to lenz’s law, the direction of induced e.m.f is such that it opposes the decay of current in the coil. This delays the current to acquire minimum or zero value.

Such property of the coil which opposes the growth or decay of the current in the circuit.

Demonstration of the Self Inductance Effect

1. Connect two lamps in parallel to each other, one through an ohmic resistor (R) and next one through a coil (N). A battery B with one-way key (K) is connected to them as shown in the figure. When the key is closed, it is found that the lamp B₁, glows immediately with brilliance but lamp B₂glows slowly.
   When current flows through the coil, an e.m.f. is induced in it, which opposes the growth of current in the circuit. Hence the glow of B₂ is slow. On the other hand, no induced e.m.f is produced in the resistor. Hence the lamp B₁receives maximum current as so as the key is closed and it glows at once.
   When the key is opened, lamp B₁ stops growing at once but the lamp B₂ takes some time to stop glowing. In this case, current begins to decrease through a coil, so again e.m.f. is induced in it. This induced e.m.f. opposes the decay of current in the circuit of lamp B₂.
2. Spark is produced in the electric switch when the light is switched off:
   When the current is switched off, the current in the circuit starts to decrease rapidly. So, large induced e.m.f is set up across the switch contacts, which tries to maintain the current. This e.m.f. is sufficient to break down the insulation of the air between the switch contacts and hence spark is produced.
3. Non-inductive Coil
   Each coil of resistance box is doubled back on itself while being coiled up. It is done to decrease the self-induction. The current in every part of the coil is equal and opposite. So the magnetic field around one part is canceled by the magnetic field around the opposite part. Thus self-induction becomes minimum.

Coefficient of Self Induction

Let I be the current flowing through a coil then the magnetic flux \(\phi \) linked with the coil is found to be proportional to the strength of current (I).

$$ \begin{align*} \: \text {i.e} \: \phi &\propto I \\ \text {or,}\: \phi &= LI \\ \end{align*} $$

where L is coefficient of self-inductance. From Faraday’s law of electromagnetic induction, the induced e.m.f. in the coil is given by

$$ \begin{align*} E &= -\frac {d\phi }{dt} \\ &= -\frac {d}{dt}(LI) = - L \frac {dI}{dt} \\ \therefore E &= -L \frac {dI}{dt} \dots (i) \\ \end{align*} $$

If \(\frac {dI}{dt} = I\), i.e. rate of decrease of current is unity then equation (i) becomes,

$$ E = L$$

Thus coefficient of self induction of a coil is the e.m.f. induced in the coil through which the rate of decrease of current is unity.

Dimensional Formula for Self Inductance (L)

$$ \begin{align*} E &= L \frac {dI}{dt} \\ \text {or,} \: L &= \frac {E}{dI/dt} = \frac {W/dq}{dI/dt}\:\:\:\:\: [\because \text {e.m.f.} = \frac {\text {work done}}{\text {charge}} ]\\ &= \frac {W}{\frac {dq}{dt} dI} = \frac {W}{IdI} \\ &= \frac {[ML^2T^{-2}]}{[A^2]} \\ &= \therefore [L] = [M^1L^2T{-2}A^{-2}] \end{align*} $$

An ideal inductance has high value of self inductance and zero ohmic resistance.

Self Inductance of a Solenoid

Let a long solenoid of length l, the area of cross-section A, the current flowing through solenoid I and a number of turns per unit length n.

The magnetic field inside the solenoid is uniform and given by

$$B = \mu _0nI$$

Total number of turns in the solenoid, N = nl. Now the magnetic flux linked with each turn of the solenoid\(= BA = \mu_0\: nIA\).

total magnetic flux with the whole solenoid, \(\phi \) = magnetic flux linked with each turn × number of turns in the solenoid.

$$ \begin{align*} \therefore \: \phi = \mu_0 n IA\times nl = \mu_o n^2 IAl \\ \text {or,} \: LI &= \mu_0 n^2 IAl \:\:\: [\because \phi = Li ] \\ \therefore L = \mu_0\: n^2 Al\dots (i) \\ \text {Since,}\: n = \frac Nl. \text {So equation} \:(i) \: \text {becomes,} \\ L &= \mu _0 \frac {N^2}{l^2}Al \\ L &= \mu_0\frac {N^2}{l}A \\ \end{align*} $$

Thus, self inductance of an air cored solenoid (L) depends on

1. Total number of turns (N) of the solenoid
2. Length of the solenoid
3. Area f cross-section of the solenoid (A).

Mutual Induction

Mutual induction is the phenomenon of inducing e.m.f in a coil due to rate of change of current or change in magnetic flux linked with nearby coil.

Consider a primary coil P connected to a battery through a key and another coil called secondary coil ‘s’ connected to a galvanometer is placed near the primary coil as shown in the figure. A key is pressed the current through primary coil begins to increase so that magnetic field around P increases as a result magnetic flux linking with secondary coil also changes. Due to it, e.m.f. is induced in the secondary coil. Hence current flows through the secondary coil which is indicated by the deflection in the galvanometer. This phenomenon of inducing e.m.f is called mutual inductance.

Coefficient of Mutual Inductance

It is found that the magnetic flux linked with the secondary coil is directly proportional to the current flowing though the primary coil.

$$ \begin{align*} \text {i.e} \: \phi _s &\propto I_p \\ \phi _s &= M\: I_p \dots (i) \\ \end{align*} $$

where M is proportionality constant called coefficient of mutual induction.

From Faraday’s law of electromagnetic induction,

$$ \begin{align*} E &= -\frac {d\phi }{dt}, \\ \text {so} \: E_s &=\frac {d\phi _s}{dt} \\ \text {or,} \: E_s &= - \frac {d}{dt}(M\:I_p) \\ \therefore E_s &= -M \frac {dI_p}{dt} \dots (ii) \\ \text {If} \: -\frac {dI_p}{dt} = 1, \: \text {then equation}\: (ii)\: \text {becomes} \\ E_s = M \\ \end{align*} $$

Thus, the coefficient of mutual induction is defined as the e.m.f.induced in the secondary when the rate of change of current in the primary is unity. SI unit of M is Henry. The negative sign shows that M and e.m.f. are opposite in sign.

 

Mutual Inductance of Two long Co-axial Solenoids

Consider two solenoids S₁ and S₂ such that solenoid S₂ completely surrounds the solenoid S₁. The two solenoids are so closely wound that they have the same area of cross-section A. Let N₁ and N₂ be the total number of turns of solenoids S₁ and S₂ respectively.

Let current I₁ flows through solenoids S₁. Then magnetic field inside the solenoid S₁ is given by

$$ \begin{align*} B_1 &= \mu_0n_1I_1 \\ &= \mu_0\frac {N_1}{l}I_1 \\ \text {where}\: n_1 &= \frac {N_1}{l} \\\therefore \text {Magnetic flux linked with each turn of solenoid}\: S_2 \: \text {is} \\ \phi _2 &= N_2(B_1A) = N_2\mu_0 \frac {N_1}{l}I_1A \\ \text {or,} \: \phi_2 &= \mu_0 \frac {N_1N_2I_1A}{l} \dots (i) \\ \text {But}\: \phi_2 &= M_{12}I_1 \dots (ii) \\ \end{align*} $$

where \(M_{12}\) is the mutual inductance when current changes in solenoid S₁ to change the magnetic flux linking with solenoid S₂ or M₁₂ is the mutual inductance of S₁ with respect to S₂.

$$ \begin{align*} \text {From equation}\: (i)\: \text {and} \: (ii,)\: \text {we get} \\ M_{12} I_1 &= \frac {\mu_0N_1N_2I_1A}{l} \\ \therefore M_{12} &= \frac {\mu_0N_1N_2A}{l}\\ \dots (iii) \\ \text {Similarly,} \\ M_{21} &= \frac {\mu_0N_1N_2A}{l}\\ \dots (iv) \\ \end{align*} $$

where M₂₁ is the mutual inductance when current changes is solenoid S₂ to change the magnetic flux linking with Solenoid S₁ or mutual inductance of S₂ with respect to S₁.

$$ \begin{align*} \text {From equation}\:(iii)\: \text {and} (iv), \: \text {we get} \\ M_{12} = M_{21} = M \end{align*} $$

Thus the mutual inductance between the two coils is same, no matter which of the two coils carries the current.

$$\therefore M = \frac {\mu_0N_1N_2A}{l} $$

Induced Electric Fields and Energy Stored in an Inductor

Induced Electric Fields

When a conductor moves in a magnetic field, emf is induced. Let us consider a long, thin solenoid of cross-sectional area (A), n number of turns per unit length encircled at its centre by a circular conducting loop. The galvanometer G is used to measure current in the loop. When current I passes through the solenoid, magnetic field is produced that is \(B = \mu_0 nI\), when \(\mu_0\) is the permeability of free space. If we neglect the small field outside the solenoid and take area vector \(\vec A\) to point in the same direction as be \(\vec B\)m then magnetic flux through the loop is given by,

$$\phi = BA = \mu_0 nIA$$

From Faraday's law, the induced emf is given by

$$ \begin{align*} \epsilon &= -\frac {d\phi }{dt} = -\frac {d}{dt}(\mu_0\: nIA)\dots (i) \\ \text {or,} \: \epsilon &= -\mu _0nA\frac {dI}{dt} \dots (ii) \\ \end{align*} $$

If \(\epsilon \) be the induced emf, R be the total resistance of the loop, the induced current is

$$\text {i.e.} \: \:\: \text {Induced current} = \frac {\epsilon }{R} $$

The change in magnetic flux is a source of electric field. When a charge goes once around the loop, the total work done on it by the electric field must be equal to q times the emf \(\epsilon \). We can conclude that the electric field in the loop is not conservative because line integral of \(\vec E\) around a closed path is not zero. This line integral representing the work done by the induced \(\vec E\) field per unit charge is equal to the induced emf \(\epsilon \).

$$ \begin{align*} \text {i.e.} \:\oint \vec E.\vec {dl} &= \epsilon \dots (iii) \\ \text {From equations} \:(i) \: \text {and} \: (iii),\: \text {we get} \\ \oint \vec E.\vec {dl} &= -\frac {d\phi }{dt} \dots (iv) \\ \end{align*} $$

Equation (iv) is true only when the path around which we integrate is stationary.

Let us consider the stationary circular loop of radius r. The electric field \(\vec E\) has the same magnitude at every point on the circle and is tangent to it at each point due to cylindrical symmetry. The line integral of equation (iv) becomes simply the magnitude E times the circumference \(2\pi r\) of the loop.

$$ \begin{align*} \text {i.e.}\: \oint \vec E.\vec {dl} &= 2\pi rE \\ \text {or,} \: -\frac {d\phi }{dt} &= -2\pi r\:E \\ \therefore E &= \frac {1}{2\pi r} \frac {d\phi }{dt} \\ \end{align*} $$

Here E is the magnitude of induced electric field.

Energy Stored in an Inductor

Consider an inductor of inductance L having initially zero current. It is assumed that an inductor has zero resistance so that there is no dissipation of energy with in inductor. Let I be the current at any instant of time so that di/dt is the rate of change of current. Here current is increasing. The voltage between the terminals a and b of the inductor at this instant is \(V_{ab} = L\frac {di}{dt} \) and the rate P at which energy is being delivered to the inductor is given by

$$ P = v_{ab}I = Li\frac {di}{dt} $$

The energy supplied to the inductor in small amount of time is small which is written as

$$ \begin{align*} dU &= P\:dt \\ &= Li \frac {di}{dt}dt \\ dU &= Lidt \dot (i) \end{align*} $$

To obtain total amount of energy stored in the inductor, we have to integrate it from 0 to 1.

$$ \begin{align*} \text {i.e}\: U &= \int_0^I L.i.di \\ \text {or,}\: U &= L\left [\frac {i^2}{2}\right ]_0^I = \frac 12 LI^2 \\\therefore \text {Total energy stored in an inductor,} \:U = \frac 12 LI^2 \end{align*} $$

When current has reached final steady value I, \(\frac {di}{dt} =0 \) so no more energy is input to the inductor. Where there is no charge then energy is not stored, the energy is \(\frac LI^2\).

Energy stored in Toroid

The self inductance of the toroidal solenoid with vacuum within its coils is

$$ \begin{align*} L &= \frac {\mu_0 N^2A}{2\pi r} \\ \text {Thus energy stored in the toroidal solenoid} \\ U &= \frac 12LI^2 = \frac 12\frac {\mu_0N^2A}{2\pi r}I^2 \\ \therefore U &= \frac 12 \frac {\mu_0N^2A}{2\pi r}I^2 \\\end{align*} $$

Theory

When an a.c. source of e.m.f E_p is connected to the primary coil, an alternating current flows through it so that alternating magnetic field is produced and hence magnetic flux linked with the coil and N_s be the number of turns in the primary and secondary coils respectively. The iron coil is capable of coupling all of the magnetic flux \(\phi \) produced by the turns of the primary coil with the secondary coil.

$$ \begin{align*} \epsilon_p &= -N_p\frac {d\phi }{dt} \dots (i) \\ \text {The induced e.m.f. in the secondary coil,} \\ \epsilon_s &= N_s \frac {d\phi }{dt} \dots (ii)\\ \text {Dividing equation}\: (ii)\:\text {by}\:(i), \: \text {we get}\\ \frac {\epsilon _s}{\epsilon _p} &= \frac {N_s}{N_p} \\ \end{align*} $$

$$ \begin{align*} \text {where}\: \frac {N_s}{N_p} &= \sqrt {\frac {L_P}{L_S}} = k \: \text {transformation ratio.} \\ \text {Here,}\\ L_P &= \text {coefficient of self induction of primary coil} \\ L_S &= \text {coefficient of self induction of secondary coil} \\ \therefore {\epsilon _s}{\epsilon _p} &= \frac {N_s}{N_p} = K \dots (iii) \\\end{align*} $$

$$ \begin{align*} \text {For step-down transformer,} \: K< 1\\ \text {For step-up transformer} \: K >1 \\ \text {For an ideal transformer} \\ \text {Output power} &= \text {Input power} \\ \text {or,} \: {\epsilon _s}{\epsilon _p} &= \frac {I_p}{I_s} \dots (iv) \\ \text {In general,}\: \epsilon \propto \frac 1I \\ \end{align*} $$

A step-up transformer increases the alternating voltage by decreasing the alternating current and a step-down transformer decreases the alternating voltage by increasing the alternating current.

For a transformer,

$$ \begin{align*}\text {efficiency,}\: \eta &= \frac {\text {output power}}{\text {input power}} = \frac {E_SI_S}{E_PE_P} \\ \end{align*} $$

Energy Losses in a transformer

1. Copper Losses
   Energy lost in windings of the transformer is known as copper loss. When current flows through copper wires, there is loss in power. This can be reduced by using thick wires for windings.
2. Flux losses
   In the actual transformer, the coupling between primary and secondary coil is not perfect. So certain amount of electrical energy supplied to the primary coil is wasted.
3. Iron losses
   a)Eddy current losses:
   When a changing magnetic flux links with the iron core of the transformer, eddy currents are set-up. These eddy currents in the iron core produce heat which leads to wastage of energy. This can be reduced by using laminated core.
   b. Hysteresis losses
   When alternating current passes through the primary coil of the transformer, the iron core of the transformer is magnetized and demagnetized over a complete cycle. Some energy is lost during magnetizing and demagnetizing the iron core. The energy loss in a complete cycle is equal to the area of the hysteresis loop. This can be minimized by using the suitable material having narrow hysteresis loop for the core of a transformer.
4. Losses due to vibration of core
   A transformer produces humming noise due to magnetostriction effect. Some electrical energy is lost in the form of mechanical energy to produce vibration in the core.

Use of Transformer for Long Distance A.C. Supply Power System

Electricity produced at a power station is transmitted to the consumer through wires. These wires have resistance R so when the current I flows through these wires for certain time t, the electrical energy is wasted in the form of heat (I²Rt). Power loss during the long distance supply of alternating current can be reduced by either reducing resistance of transmission or by decreasing the magnitude of current flowing in the resistance can be reduced by using very thick wires but it is not economical as the cost of manufacturing and installing such long and thick wires will be very high and to decrease the current flowing in the resistance can be reduced by using long and thick wires. The use of long and thick wires is not economical so transformers are used to decrease the current flowing through the wires.

The electric power generated at the power station is fed to the primary coil of a step-up transformer. Step-up transformer makes the voltage high and low alternating current. This low current at high voltage is carried by the transmission wire to the sub-station which will reduce the power loss. After transmission to the substation, a step-down transformer is used to decrease voltage and increase the alternating current.

Eddy Currents

The induced circulating currents produced in a metal itself due to change in magnetic flux linked with the metal are called eddy currents.

When a metallic piece is placed in a changing magnetic field, the induced currents are set-up in the metal piece which is eddy currents. The direction of eddy currents is given by Lenz's law.

Drawbacks of Eddy Currents

1. The production of eddy currents in a metallic block leads to the loss of electrical energy in the form of heat.
2. The heat produced due to eddy currents break the insulation used in the electrical machine.
3. Eddy currents may cause unwanted dampening effect.

Application of Eddy Current

1. Induction Furnace
   It is based on the heating effect of eddy currents. The metallic block to be melt is placed in a high frequency changing magnetic field. Strong eddy current is produced in a block. A large amount of heat is produced in the block due to high resistance of the metal. The metallic block melts due to heat. So the induction furnace is used to separate metals from their ores and to make some alloys.
2. Diathermy
   Eddy currents are used for the localized heating of tissues in the human body which is called diathermy.
3. Speedometer
   Speedometer is a device used to measure the instantaneous speed of vehicle. A small magnet is attached to the axle of wheel of the speedometer. Due to the rotation of the magnet magnetic flux linked with the aluminium drum changes and hence eddy currents are produced in it. A pointer attached to the drum is deflected in the direction of the rotation of the drum. This speed is measured which corresponds to the deflection of the pointer.
4. Eddy current damping or dead beat galvanometer
   According to Lenz's law, eddy currents always flow in such a direction as to oppose the motion which has produced them. So they can reduce the oscillations of a vibrating system. The coil of galvanometer is wound on a metal frame. As the coil swings in the magnetic field of the instrument, eddy currents are induced in the frame. These eddy currents oppose the motion of the coil and hence the pointer attached to it. The pointer quickly attains the final position without overshooting or oscillating violently. Thus eddy currents reduce the oscillations of the pointer.
5. Electromagnetic brakes
   Eddy current breaking can be used to control the speed of electric trains. In order to reduce the speed of train, an electromagnet is turned on that applies its field to the wheels. Large eddy currents are set up which produce the retarding effect.
6. Energy meters
   Energy meters uses the concept of eddy current to record the consumption of electricity.

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