6.0 Preview:

We define Electromagnetic induction as the phenomenon by which, electric current is generated by varying magnetic fields.

We shall begin our discussion with Experiments of Faraday and Henry in which they induced current by a magnet, by a coil and by changing current itself to analyse the emf induced. Emf refers to the electromotive force that is responsible for flow of electrons and hence, the current flow.

Next, we shall look at magnetic flux as the quantity of magnetism taking into account, the strength of magnetic field and the extent to which it acts. We shall obtain an expression for the same.

Further, we will study Farady’s laws of induction in the form of two laws— Faraday’s first law which states that, whenever there is change in flux associated with a coil, an emf and hence a current is induced in the coil. This induced emf and current lasts only as long as there is change in the flux associated with the coil; Faraday’s second law which states that, the magnitude of induced emf is directly proportional to the rate of change of flux.

We shall state another important law called, The Len’s law (to explain the polarity of the induced emf with respect to the change of flux) as: “the direction of induced current is always such that it opposes the cause which is producing it”.

Then, we will obtain an important relation called, Neumannn’s relation for current induced by a magnet. We will also obtain an expression for motional-emf, that is, for a straight conductor moving in a magnetic field.

Next, we will look at a special type of induced current called, Eddy Current; Eddy currents are the induced, circular currents developed on the surface of a thick metal block, which is placed in a changing magnetic field.

Then comes, Inductance; Inductance is the property of an electric conductor that causes an electromotive force to be generated by a change in the current flowing through it. Consequently, we shall study about mutual induction as the inductance arising from the interaction of two solenoids; Self-induction as the emf induced in a single isolated coil due to change of flux through the coil by means of varying the current through the same coil.

We shall come to end of this chapter with the study of working of an AC generator, otherwise called, Dynamo—an electrical device used to convert mechanical energy into electrical energy.

6.1 Introduction

The experiments of Michael Faraday in England and Joseph Henry in USA, conducted around 1830, demonstrated conclusively that electric currents were induced in closed coils when subjected to changing magnetic fields. In this chapter, we will study the phenomena associated with changing magnetic fields and understand the underlying principles. The phenomenon in which electric current is generated by varying magnetic fields is appropriately called electromagnetic induction.

The pioneering experiments of Faraday and Henry have led directly to the development of modern day generators and transformers. Today’s civilisation owes its progress to a great extent to the discovery of electromagnetic induction.

Questions for section 6.1

1. Define electromagnetic induction.
2. Give a brief history of the understanding of electromagnetic induction in the scientific community.

6.2 Experiments to demonstrate EMI by Faraday and Henry

6.2.1 Experiment 1: Current induced by magnet

Apparatus:

1. A coil connected to a sensitive galvanometer

2. Bar magnet placed at its side.

Experiment and observation:

1. Both coil and magnet are kept stationary. No deflection is found in the galvanometer connected to the coil.

2. One end of the magnet is moved towards one face of the coil. A small deflection is found in the galvanometer in one direction.

3. The magnet is taken out from the coil and the deflection is found to be in the opposite direction.

4. Keeping the magnet stationary, if the arrangement of coil and galvanometer is taken towards or away from the magnet, there will be deflections in the galvanometer in opposite directions.

Interpretation of the results

2. The deflection in the galvanometer shows that there is current flowing in the coil.
3. The fact is that for current to flow in a circuit, there must be emf (electromotive force) behind it.
4. Because in our experiment there is no external source connected, the current produced is called induced current and the emf responsible for it is called induced emf.
5. The relative motion between the magnet and the coil is the reason behind induced emf and induced current.
6. The factors which affect the magnitude of induced emf are:

• Using powerful magnet
• Increasing the relative rate of speed between magnet and the coil
• Increasing the number of turns of the coil
• Introducing a soft iron cylinder into the coil

6.2.2 Experiment 2: Current induced by coil

Apparatus:

1. Primary circuit (coil connected to a battery)
2. Secondary circuit (coil connected to a sensitive galvanometer)

Experiment and observation:

1. When both primary and secondary circuits are kept stationary, there is no deflection in the galvanometer connected to the secondary circuit.
2. Let us now bring the primary circuit near the secondary circuit. A deflection is found in the galvanometer of secondary circuit.
3. When the primary circuit is taken away from the secondary circuit, the deflection is found to be in the opposite direction.
4. Keeping the primary circuit stationary, if the secondary circuit is moved towards or away from the primary circuit, the deflections is found to be in the opposite directions.

Interpretation of the results

1. The deflection in the galvanometer shows that there is current flowing in the coil.
2. The fact is that for current to flow in a circuit, there must be emf (electromotive force) behind it.
3. Because in our experiment there is no external source connected, the current produced is called induced current and the emf responsible for it is called induced emf.
4. Battery in the primary circuit allows current to flow through it. Due to this, magnetic field is developed along its axis which increases the flux associated with it.
5. When the primary circuit is brought near the secondary circuit, the flux associated with the secondary coil goes on increasing giving rise to induced emf and induced current.
6. When the primary circuit is taken away, there will be decrease in the flux which causes deflection indicating that there is induced emf and current in the secondary coil.
7. The factors that affect magnitude of induced emf are

• Use of powerful battery
• Increasing the relative speed between the circuits
• Increasing the number of turns in the primary coil
• Introducing a soft iron cylinder in the secondary coil.

6.2.3 Current induced by changing current

Apparatus:

1. Primary circuit (A coil connected to a battery and a tap key)
2. Secondary circuit (coil connected to sensitive galvanometer)

Experiment and observation:

1. When the primary circuit is open, no current flows through it and hence there is no deflection in the galvanometer of the secondary coil.

2. The tap key of the primary coil is pressed. Momentarily, the deflection is found to be in the secondary.

3. When the key is released, momentarily there is deflection but in opposite direction.

4. Whereas, when the key is continuously pressed, no deflections id found in the galvanometer of the secondary coil.

Interpretation of the results

1. The deflection that is momentarily found in the galvanometer indicates that there is some current flowing in the secondary coil.
2. Since the secondary circuit is not connected to any external source, the emf is induced emf and the current is induced current.
3. When the key of the primary is pressed, the current starts flowing through it producing a magnetic field along the axis, so, the flux associated with it goes on increasing which cause corresponding increase in the flux associated with the secondary.
4. The factors affecting the magnitude of induced emf are

• Using a powerful battery
• Operating tap key at a faster rate
• Increasing number of turns in the primary circuit
• Introducing a soft iron cylinder into the secondary circuit

Questions for section 6.2

1. Describe how current can be induced by a magnet.

2. Explain how current can be induced by coil.

3. Explain how current can be induced by changing current.

6.3 Magnetic flux

Magnetic flux is a measure of quantity of magnetism taking into account the strength of magnetic field and the extent to which it acts.

It is measured by total number of magnetic lines of force passing through any surface which is held in a magnetic field.

Magnetic flux through a plane of area A placed in a uniform magnetic field B can be written as
$$ Φ B = B . A = BA \space cos θ $$

Where θ is angle between B and A. The notion of the area as a vector has been discussed earlier. Equation can be extended to curved surfaces and non-uniform fields. If the magnetic field has different magnitudes and directions at various parts of a surface as shown in Fig. 6.5, then the magnetic flux through the surface is given by

$$ Φ _B =B_1.dA_1+B_2.dA_2+…..= \Sigma{all} B_i.dA_i $$

where ‘all’ stands for summation over all the area elements $dA_i$ comprising the surface and Bi is the magnetic field at the area element $dA_i$. The SI unit of magnetic flux is weber (Wb) or tesla meter squared ($T m^2$). Magnetic flux is a scalar quantity.

Questions for section 6.3

1. Define magnetic flux.
2. Derive an equation for magnetic flux.

6.4 Faraday’s law of induction

From the analysis made in the experiments to demonstrate the phenomenon of electromagnetic induction, the conclusion derived is that emf and current can be found in the coil only when there is change in magnetic flux associated with the coil. Based on this, the following two laws were formed.

Faraday’s first law

Whenever there is change in flux associated with a coil, an emf and hence a current is induced in the coil. This induced emf and current lasts only as long as there is change in the flux associated with the coil.

Faraday’s second law

The magnitude of induced emf is directly proportional to the rate of change of flux.

If Φ be the flux and ε be the induced emf then $ ε = \frac {d Φ}{dt} $

If there are N turns in a coil then $ ε = N \frac {d Φ}{dt} $

Example 6.1 Consider Experiment 6.2. (a) What would you do to obtain a large deflection of the galvanometer? (b) How would you demonstrate the presence of an induced current in the absence of a galvanometer?

Solution: (a) To obtain a large deflection, one or more of the following steps can be taken: (i) Use a rod made of soft iron inside the coil$ C_2$, (ii) Connect the coil to a powerful battery, and (iii) Move the arrangement rapidly towards the test coil $C_1$.
(b) Replace the galvanometer by a small bulb, the kind one finds in a small torch light. The relative motion between the two coils will cause the bulb to glow and thus demonstrate the presence of an induced current.
In experimental physics one must learn to innovate. Michael Faraday who is ranked as one of the best experimentalists ever, was legendary for his innovative skills.

Example 6.2 A square loop of side 10 cm and resistance 0.5 Ω is placed vertically in the east-west plane. A uniform magnetic field of 0.10 T is set up across the plane in the north-east direction. The
magnetic field is decreased to zero in 0.70 s at a steady rate. Determine the magnitudes of induced emf and current during this time-interval.

Solution: The angle θ made by the area vector of the coil with the magnetic field is 45°. From Eq. (6.1), the initial magnetic flux is
Φ = BA cos θ

$$ = \frac {0.1×10^{-2}}{ \sqrt 2} Wb $$

Final flux, $Φ{min}$ = 0 The change in flux is brought about in 0.70 s.

From Eq. (6.3), the magnitude of the induced emf is given by $$ε =\frac {|ΔΦ_B|}{Δt}=\frac { |\big( Φ-0 \big) |}{Δt} = \frac {10^{-3} \space V }{\sqrt 2× 0.7 } = 1.0 \space m \space V $$

And the magnitude of the current is $ I =\frac{ε}{R} = \frac {10^{-3} \space V }{O.5Ω } = 2m A $

Note that the earth’s magnetic field also produces a flux through the loop. But it is a steady field (which does not change within the time span of the experiment) and hence does not induce any emf.

Example 6.3 A circular coil of radius 10 cm, 500 turns and resistance 2 Ω is placed with its plane perpendicular to the horizontal component of the earth’s magnetic field. It is rotated about its vertical diameter through 180° in 0.25 s. Estimate the magnitudes of the emf and current induced in the coil. Horizontal component of the earth’s magnetic field at the place is $3.0 × 10^{–5} $ T.

Solution

Initial flux through the coil, $$ Φ{B \big( initial \big) } = BA \space cosθ $$
$$ = 3.0 ×10^{-5} × (π ×10^{-2} ) × cos \space 0^o $$
$$ = 3π × 10^{–7} Wb $$

Final flux after the rotation,
$$ Φ_{B \big( final \big) } = 3.0 ×10^{-5} × (π ×10^{-2} ) × cos \space 180^o $$
$$ =- 3π × 10^{–7} Wb $$

Therefore, estimated value of the induced emf is,
$$ ε = N \frac {ΔΦ}{Δt} $$
$$ =500 × (6π × 10^{–7})/0.25 $$
$$ =3.8 ×10^{-3} \space V $$
$$ I = \frac {ε}{R} = 1.9 × 10^{-3} \space A $$

Note that the magnitudes of ε and I are the estimated values. Their instantaneous values are different and depend upon the speed of rotation at the particular instant.

Questions for section 6.4

1. State Faraday’s first law.
2. State and prove Faraday’s second law.

6.5 Lenz’s law and conservation of energy

This law explains the polarity of the induced emf with respect to the change of flux.

The Lenz’s law states that the direction of induced current is always such that it opposes the cause which is producing it.

Let us consider the following experiment to understand the concept.

6.5.1 Current induced by magnet

When the north pole of the magnet is brought towards one end of the coil, the face of the coil that is facing the north pole of the magnet develops polarity. Then, due to the mutual repulsion, the mechanical energy associated with magnet goes on decreasing. Simultaneously, due to increase of flux associated with the coil, electrical energy increases in the form of induced current. Thus, decrease in one form of energy gives rise to another which explains the conservation of energy principle.

Neumannn’s relation

According to Faraday’s second law, we know that $ e \propto \frac { dΦ }{dt} $

Where e is induced emf and Φ being flux.

From Lenz’s law, we have:

The direction of induced emf and current is opposite to the direction of change of flux.

This is represented by assigning negative sign to the expression for induced emf. That is $ e \propto – \frac { dΦ }{ dt} $ . This is known as Neumann’s relation.

Methods of producing induced emf

1. Keeping the field constant and moving a conductor across it, emf can be induced in the conductor due to the change of flux associated with it.

The emf induced in a conductor by virtue of its motion is called motional emf.

2. Keeping the conductor stationary and changing the magnetic field surrounding it.

Example 6.4

Figure 6.7 shows planar loops of different shapes moving out of or into a region of a magnetic field which is directed normal to the plane of the loop away from the reader. Determine the direction of induced current in each loop using Lenz’s law.

Solution

(i) The magnetic flux through the rectangular loop abcd increases, due to the motion of the loop into the region of magnetic field, The induced current must flow along the path bcdab so that it opposes
the increasing flux.
(ii) Due to the outward motion, magnetic flux through the triangular loop abc decreases due to which the induced current flows along bacb, so as to oppose the change in flux.
(iii) As the magnetic flux decreases due to motion of the irregular shaped loop abcd out of the region of magnetic field, the induced current flows along cdabc, so as to oppose change in flux.
Note that there are no induced current as long as the loops are completely inside or outside the region of the magnetic field.

Example 6.5
(a) A closed loop is held stationary in the magnetic field between the north and south poles of two permanent magnets held fixed. Can we hope to generate current in the loop by using very strong magnets?

(b) A closed loop moves normal to the constant electric field between the plates of a large capacitor. Is a current induced in the loop.

(i) when it is wholly inside the region between the capacitor plates

(ii) when it is partially outside the plates of the capacitor? The electric field is normal to the plane of the loop.

(c) A rectangular loop and a circular loop are moving out of a uniform magnetic field region (Fig. 6.8) to a field-free region with a constant velocity v. In which loop do you expect the induced emf to be
constant during the passage out of the field region? The field is normal to the loops.

Solution
(a) No. However strong the magnet may be, current can be induced only by changing the magnetic flux through the loop.
(b) No current is induced in either case. Current can not be induced by changing the electric flux.
(c) The induced emf is expected to be constant only in the case of the rectangular loop. In the case of circular loop, the rate of change of area of the loop during its passage out of the field region is not constant, hence induced emf will vary accordingly.

(d) The polarity of plate ‘A’ will be positive with respect to plate ‘B’ in the capacitor.

Questions for section 6.5

1. What is Neumann’s relation?
2. What are various methods of producing induced emf?

6.6 Expression for motional EMF

Let us consider a straight conductor moving in a uniform and time independent magnetic field. Figure shows a rectangular conductor PQRS in which the conductor PQ is free to move. The rod PQ is moved towards the left with a constant velocity v as shown in the figure.

Assume that there is no loss of energy due to friction. PQRS forms a closed circuit enclosing an area that changes as PQ moves. It is placed in a uniform magnetic field B which is perpendicular to the plane of this system. If the length RQ = x and RS = l, the magnetic flux $Φ_B$ enclosed by the loop PQRS will be $ Φ_B = Blx $

Since x is changing with time, the rate of change of flux $Φ_B$ will induce an emf given by:
$$ ε = \frac {- d Φ_B}{dt} = – \frac {d}{dt} (Blx) $$
$$ = – Bl \frac {dx}{dt} = Blv $$

where we have used dx/dt = –v which is the speed of the conductor PQ. The induced emf Blv is called motional emf. Thus, we are able to produce induced emf by moving a conductor instead of varying the magnetic field, that is, by changing the magnetic flux enclosed by the circuit.

It is also possible to explain the motional emf expression in equation by invoking the Lorentz force acting on the free charge carriers of conductor PQ. Consider any arbitrary charge q in the conductor PQ. When the rod moves with speed v, the charge will also be moving with speed v in the magnetic field B. The Lorentz force on this charge is qvB in magnitude, and its direction is towards Q. All charges experience the same force, in magnitude and direction, irrespective of their position in the rod PQ.

The work done in moving the charge from P to Q is,

W = qvBl

Since emf is the work done per unit charge,

$$ ε = \frac {W}{q} = Blv $$

This equation gives emf induced across the rod PQ and is identical to equation. We stress that our presentation is not wholly rigorous. But it does help us to understand the basis of Faraday’s law when the conductor is moving in a uniform and time-independent magnetic field.
On the other hand, it is not obvious how an emf is induced when a conductor is stationary and the magnetic field is changing – a fact which Faraday verified by numerous experiments. In the case of a stationary conductor, the force on its charges is given by F = q (E + v × B) = qE

since v = 0. Thus, any force on the charge must arise from the electric field term E alone. Therefore, to explain the existence of induced emf or induced current, we must assume that a time-varying magnetic field generates an electric field. However, we hasten to add that electric fields produced by static electric charges have properties different from those produced by time-varying magnetic fields.

Conversely, a bar magnet in motion can exert a force on the stationary charge. This is the fundamental significance of the Faraday’s discovery. Electricity and magnetism are related.

Example 6.6 A metallic rod of 1 m length is rotated with a frequency of 50 rev/s, with one end hinged at the centre and the other end at the circumference of a circular metallic ring of radius 1 m, about an axis passing through the centre and perpendicular to the plane of the ring
(Fig. 6.10). A constant and uniform magnetic field of 1 T parallel to the axis is present everywhere. What is the emf between the centre and the metallic ring?

Solution

Method I

As the rod is rotated, free electrons in the rod move towards the outer end due to Lorentz force and get distributed over the ring. Thus, the resulting separation of charges produces an emf across the ends of the rod. At a certain value of emf, there is no more flow of electrons and a steady state is reached. Using Eq. (6.5), the magnitude of the emf generated across a length dr of the rod as it moves at right angles to the magnetic field is given by

dε = Bvdr Hence,

$$ ε = \int dε = \int_{0}^{R} Bv \space dr = \int_{0}^{R} B \space ωr \space dr = \frac {B \space ω R^2}{2} $$

Note that we have used v = ω r. This gives $ ε = \frac {1}{2} ×1.0 ×2π ×50 × (1 ^2) $
= 157 V

Method II

To calculate the emf, we can imagine a closed loop OPQ in which point O and P are connected with a resistor R and OQ is the rotating rod. The potential difference across the resistor is then equal to the induced emf and equals B × (rate of change of area of loop). If θ is the angle between the rod and the radius of the circle at P at time t, the area of the sector OPQ is given by
$$ π R^2× \frac {θ}{2 π} = \frac {1}{2} R^2 θ $$

where R is the radius of the circle. Hence, the induced emf is

$$ ε = B× \frac {d}{dt} \bigg[ \frac{1}{2} R^2 θ \bigg] = \frac {1}{2} BR^2 \frac {d θ }{dt} = \frac {B ω R^2}{2} $$

[Note: $ \frac {d θ}{dt}= ω=2 π \space ν ] $

This expression is identical to the expression obtained by Method I and we get the same value of ε.

Example 6.7

A wheel with 10 metallic spokes each 0.5 m long is rotated with a speed of 120 rev/min in a plane normal to the horizontal component of earth’s magnetic field $H_E$ at a place. If $H_E$ = 0.4 G at the place, what is the induced emf between the axle and the rim of the wheel? Note that 1 G = $10^{–4} T$.

Solution
$$ Induced emf = (1/2) ω B R^2 $$
$$ = (1/2) × 4π × 0.4 × 10^{–4} × (0.5)^2 $$
$$ = 6.28 × 10^{–5} V $$
The number of spokes is immaterial because the emf’s across the spokes are in parallel.

Questions for section 6.6

1. Derive an expression for motional emf.
2. How are electricity and magnetism related?

6.7 Energy consideration: A quantitative study

Let us consider a conductor of length l moving across a field of strength B with a speed v. The motional emf induced in the conductor is given by

ϵ = Blv

let us assume the resistance of the conductor to be r. Then, the current through the conductor is given by

$$ I = \frac { ϵ}{r} = \frac {Blv}{r}………(1) $$

Due to the presence of the magnetic field , the electron carrying conductor experiences a force given by F = BI l

Substituting for I , $ F = \frac {Bl.Blv}{r} = \frac { B^2l^2 v}{r}……(2) $

Since the conductor is being pushed with a constant speed v across the field , the power required to push it is given by,

$$ P = Fv= \frac { B^2l^2 v^2}{r}……from \space equation (2) $$

So, the mechanical energy that is used to push the conductor is converted into electrical energy. Further, this electrical energy gets converted into heat energy due to the resistance of the conductor called Joule’s heat. It is given by

$$ P_J = I^2r = \bigg( \frac {Blv}{r} \bigg)^2 r = \frac {B^2l^2v^2}{r} $$

Relation between charge induced and change in magnetic flux

Acoording to Faraday’s law , $ e = \frac {d Φ}{dt} $

Also , e = Ir = $ \frac {d Φ}{dt} $ where r is the resistance.

$$ \therefore \bigg( \frac {dq}{dt} r \bigg) = \frac {d Φ}{dt} \space or \space dq = \frac { d Φ}{r} $$

Example 6.8 Refer to Fig. 6.11(a). The arm PQ of the rectangular conductor is moved from x = 0, outwards. The uniform magnetic field is perpendicular to the plane and extends from x = 0 to x = b and is zero for x > b. Only the arm PQ possesses substantial resistance r. Consider the situation when the arm PQ is pulled outwards from x = 0 to x = 2b, and is then moved back to x = 0 with constant speed v. Obtain expressions for the flux, the induced emf, the force necessary to pull the arm and the power dissipated as Joule heat. Sketch the variation of these quantities with distance.

Solution:

Let us first consider the forward motion from x = 0 to x = 2b

The flux $Φ_B$ linked with the circuit SPQR is

$$ Φ_B =Blx \hspace{10mm} 0≤x < b $$

$$ =Blb \hspace{10mm} b≤x < 2b $$

The induced emf is,

$$ ε = – \frac {d Φ_B}{dt} $$

$$ = – Blv \hspace{10mm} 0≤x < b $$

$$ = 0 \hspace{10mm} b≤x < 2b $$

When the induced emf is non-zero, the current I is (in magnitude)

$$ I = \frac {Bl \space v}{r} $$

The force required to keep the arm PQ in constant motion is I lB. Its direction is to the left. In magnitude

$$ F = \frac {B^2l^2v}{r} \hspace{10mm} 0≤x < b $$

$$ = 0 \hspace{10mm} b≤x < 2b $$

The Joule heating loss is $ P_J =I^2r $
$$ = \frac {B^2l^2v}{r} \hspace{10mm} 0≤x < b $$
$$ = 0 \hspace{10mm} b≤x < 2b $$

One obtains similar expressions for the inward motion from x = 2b to x = 0. One can appreciate the whole process by examining the sketch of various quantities displayed in Fig. 6.11(b).

Questions for section 6.7

1. Derive an expression for the Joule’s heat.
2. Derive a relationship between charge induced and change in magnetic flux.

6.8 Eddy currents

Definition box Eddy currents are the induced, circular currents developed on the surface of a thick metal block, which is placed in a changing magnetic field.

Let us consider an experiment to demonstrate the presence of eddy currents.

• The set up for the experiment is as shown in the figure. A copper plate is connected to a copper rod and is made to oscillate in a magnetic field.
• It is observed that the oscillation of the plate goes on decreasing and comes to a rest state.
• The decrease in the oscillations of the plate is called dampening effect which proves the presence of eddy currents.

Fact box When the plate enters into the field from one direction, the flux associated with it will increase giving rise to induced currents on its surface. As the plate leaves the field from the other direction, flux will decrease giving rise to induced currents in opposite direction. These oppositely induced currents restrict the motion of the plate. So, the oscillations decrease and finally the plate comes to a rest.

Disadvantages of eddy currents

1. A part of electrical energy is lost in the form of heat energy.

2. Due to the developed heat, the insulation on the device melts causing damage to the device.

3. Due to dampening effect, a device may stop working.

Methods to minimize eddy currents

1. Laminated plates

The effect of eddy currents can be minimized by introducing the solid plates in electrical devices. If the plate is installed as a whole, eddy currents will be developed on the surface heating it up. So, the plates are cut into a number of thin plates and laminated on both sides. Then they are bundled together so that the eddy currents so developed are confined to each plate and will not spread across. So, the effects of eddy currents can be minimized.

2. Slotted plate

To minimize the dampening effect in an oscillating plate, the eddy currents so developed will be oppositely directed in each limb of the plate so that the total effect will be minimum.

Applications of eddy currents

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

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

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

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

Questions for section 6.8

1. What are eddy currents? Explain an experiment to demonstrate its presence.
2. What are the disadvantages of eddy currents? How can it be minimized?
3. Explain in brief some of the applications of eddy currents.

6.9 Inductance

The flux through a coil is proportional to the current. That is, $Φ_B α I$. Further, if the geometry of the coil does not vary with time then,
$$ \frac {d Φ_B}{dt} α \frac {dI}{dt} $$

For a closely wound coil of N turns, the same magnetic flux is linked with all the turns. When the flux ΦB through the coil changes, each turn contributes to the induced emf. Therefore, a term called flux linkage is used which is equal to$ NΦ_B$ for a closely wound coil and in such a case
$$ NΦ_B∝ I $$

The constant of proportionality, in this relation, is called inductance. We shall see that inductance depends only on the geometry of the coil and intrinsic material properties.

Inductance is a scalar quantity. It has the dimensions of $[M L^2 T^{–2} A^{–2}]$ given by the dimensions of flux divided by the dimensions of current. The SI unit of inductance is Henry and is denoted by H. It is named in honour of Joseph Henry who discovered electromagnetic induction in USA, independently of Faraday in England.

Definition box What is Inductance? Inductance is the property of an electric conductor that causes an electromotive force to be generated by a change in the current flowing through it.

6.9.1 Mutual inductance

Consider figure which shows two long co-axial solenoids each of length l. We denote the radius of the inner solenoid $S_1$ by $r_1$ and the number of turns per unit length by $n_1$. The corresponding quantities for the outer solenoid $S_2$ are $r_2$ and $n_2$, respectively. Let $N_1$ and $N_2$ be the total number of turns of coils $S_1$ and $S_2$, respectively.

When a current $I_2$ is set up through $S_2$, it in turn sets up a magnetic flux through $S_1$. Let us denote it by $Φ1$. The corresponding flux linkage with solenoid $S_1$ is $$ N_1Φ_1 =M{12}I_2 $$

$M_{12}$ is called the mutual inductance of solenoid $S_1$ with respect to solenoid $S_2$. It is also referred to as the coefficient of mutual induction. For these simple co-axial solenoids it is possible to calculate $M_{12}$. The magnetic field due to the current $I_2$ in $S_2$ is $μ0n_2I_2$. The resulting flux linkage with coil $S_1$ is, $$ N_1Φ_1 = (n_1l)( πr_1^2)( μ_0n_2I_2)= μ_0n_1n_2 πr_1^2lI_2 $$ where $n_1l$ is the total number of turns in solenoid S1. Thus, from equations $$ M{12} = μ0n_1n_2 πr_1^2l $$ Note that we neglected the edge effects and considered the magnetic field μ0n2I2 to be uniform throughout the length and width of the solenoid $S_2$. This is a good approximation keeping in mind that the solenoid is long, implying l >> $r_2$. We now consider the reverse case. A current $I_1$ is passed through the solenoid $S_1$ and the flux linkage with coil $S_2$ is, $$ N_2Φ_2 = M{21} I_1 $$

$M_{21}$ is called the mutual inductance of solenoid $S_2$ with respect to solenoid $S_1$.
The flux due to the current $I_1$ in $S_1$ can be assumed to be confined solely inside $S_1$ since the solenoids are very long. Thus, flux linkage with solenoid $S_2$ is $ N_2Φ2 = (n_2l)( πr_1^2)( μ_0n_1I_1) $$ $$ M{21} = μ0n_1n_2πr^2_1l $$ We get , $$ M{12}= M_{21}= M (say) $
We have demonstrated this equality for long co-axial solenoids.

However, the relation is far more general. Note that if the inner solenoid was much shorter than (and placed well inside) the outer solenoid, then we could still have calculated the flux linkage $N_1Φ1$ because the inner solenoid is effectively immersed in a uniform magnetic field due to the outer solenoid. In this case, the calculation of $M{12}$ would be easy. However, it would be extremely difficult to calculate the flux linkage with the outer solenoid as the magnetic field due to the inner solenoid would vary across the length as well as cross section of the outer solenoid. Therefore, the calculation of $M_{21}$ would also be extremely difficult in this case. The equality $M_{12}=M_{21}$ is very useful in such situations.

We explained the above example with air as the medium within the solenoids. Instead, if a medium of relative permeability $μr$ had been present, the mutual inductance would be $$ M =μ_r μ_0 n_1n_2π r^2_1 l $$ It is also important to know that the mutual inductance of a pair of coils, solenoids, etc., depends on their separation as well as their relative orientation. Example 6.9 Two concentric circular coils, one of small radius $ r_1$ and the other of large radius $r_2$, such that $r_1 << r_2$, are placed co-axially with centres coinciding. Obtain the mutual inductance of the arrangement. Solution : Let a current $I_2$ flow through the outer circular coil. The field at the centre of the coil is $B_2 = μ-0I_2 / 2r_2$. Since the other co-axially placed coil has a very small radius, $B_2$ may be considered constant over its cross-sectional area. Hence, $$ Φ_1 = πr^ 2_1B_2 $$ $$ = \frac { μ_0 π r^2_1}{2r_2} I_2 $$ $$ = M{12} I_2 $$
Thus,
$$ M_{12}= \frac { μ0 π r^2_1}{2r_2} $$ From Eq. (6.14) $$ M{21} = M_{12}= \frac { μ0 π r^2_1}{2r_2} $$ Note that we calculated $ M{12} $ from an approximate value of $Φ_1$, assuming the magnetic field $B_2$ to be uniform over the area $π r^2_1$. However, we can accept this value because $r_1 <<r_2$.

Now, let us recollect Experiment 6.3 in Section 6.2. In that experiment, emf is induced in coil $C_1$ wherever there was any change in current through coil $C_2$. Let $Φ_1$ be the flux through coil $C_1$ (say of $N_1$ turns) when current in coil $C_2$ is $I_2$.
Then, from Eq. (6.9), we have $ N_1Φ_1 = MI_2 $

For currents varrying with time,
$$ \frac {d \big(N_1Φ_1 \big) }{dt} = \frac {d \big( MI_2 \big) }{dt} $$

Since induced emf in coil C1 is given by
$$ ε_1 = -\frac {d \big(N_1Φ_1 \big) }{dt} $$
We get,
$$ ε_1 = -M \frac{dI_2}[dt} $$

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

6.9.2 Self – Induction

In the previous sub-section, we considered the flux in one solenoid due to the current in the other. It is also possible that emf is induced in a single isolated coil due to change of flux through the coil by means of varying the current through the same coil. This phenomenon is called self-induction.

In this case, flux linkage through a coil of N turns is proportional to the current through the coil and is expressed as $ NΦB ∝I $$ $$ NΦ_B = LI $ where constant of proportionality L is called self-inductance of the coil. It is also called the coefficient of self-induction of the coil. When the current is varied, the flux linked with the coil also changes and an e mf is induced in the coil. Using equation the induced emf is given by $ ε = – \frac {d(NΦ_B)}{dt} $

$ ε = -L \frac {dI}{dt} $ Thus, the self-induced emf always opposes any change (increase or decrease) of current in the coil. It is possible to calculate the self-inductance for circuits with simple geometries. Let us calculate the self-inductance of a long solenoid of cross-sectional area A and length l, having n turns per unit length. The magnetic field due to a current I flowing in the solenoid is $B = μ_0 \space n I$ (neglecting edge effects). The total flux linked with the solenoid is

$$ NΦ_B = (nl)( μ_onI)(A) $$

$$ = μ_on^2AlI $$

where nl is the total number of turns.

Thus, the self-inductance is, $

$L = \frac { NΦ_B}{I} $$

$$ = μ_on^2Al $$

If we fill the inside of the solenoid with a material of relative permeability $μ_r$ (for example soft iron, which has a high value of relative permiability),then, $ L=μ_r μ_o n^2Al $ The self-inductance of the coil depends on its geometry and on the permeability of the medium. The self-induced emf is also called the back emf as it opposes any change in the current in a circuit. Physically, the self-inductance plays the role of inertia. It is the electromagnetic analogue of mass in mechanics. So, work needs to be done against the back emf (ε ) in establishing the current. This work done is stored as magnetic potential energy. For the current I at an instant in a circuit, the rate of work done is $ \frac {dW}{dt}= | ε| I $ If we ignore the resistive losses and consider only inductive effect, then $ \frac {dW}{dt} = L\space I \frac {dI}{dt} $ Total amount of work done in establishing the current I is $ W = \int d W = \int{0}^{I} L \space I \space dI $

Thus, the energy required to build up the current I is,
$$ W = \frac {1}{2} LI^2 $$

This expression reminds us of $mv^2/2$ for the (mechanical) kinetic energy of a particle of mass m, and shows that L is analogus to m.

Example 6.10 (a) Obtain the expression for the magnetic energy stored in a solenoid in terms of magnetic field B, area A and length l of the solenoid. (b) How does this magnetic energy compare with the electrostatic energy stored in a capacitor?
Solution:
(a) From Eq. (6.19), the magnetic energy is
$$ U_B = \frac {1}{2} LI^2 $$
$$ = \frac {1}{2} L \bigg( \frac {B}{μ_On} \bigg) ^2 $$ ( since B = $μ_O$ nI , for a solenoid )
$$ = \frac {1}{2} \big(μ_On^2Al \big) \bigg( \frac { B}{ μ_On} \bigg)^2 $$ [from Eq. (6.17)]
$$ = \frac {1}{2μ_O} B^2 Al $$
(b) The magnetic energy per unit volume is,
$ u_B = \frac {U_B}{V} \space (where \space V \space is \space volume \space that \space contains \space flux) $
$$ = \frac {U_B}{Al} $$
$$ = \frac {B^2}{2μ_O} $$

We have already obtained the relation for the electrostatic energy stored per unit volume in a parallel plate capacitor (refer to Chapter 2,Eq. 2.77),

$$ u_E = \frac {1}{2} ε_o E^2 $$

In both the cases energy is proportional to the square of the field strength. Equations (6.20) and (2.77) have been derived for special cases: a solenoid and a parallel plate capacitor, respectively. But they are general and valid for any region of space in which a magnetic field or/and an electric field exist.

Questions for section 6.9

1. Explain the concept of mutual inductance.
2. Explain the concept of self-inductance.

6.10 AC generator (dynamo/alternator)

An AC generator is an electrical device used to convert mechanical energy into electrical energy.

• The setup of an AC generator is as shown in the figure.
• ARMATURE is the coil ABCD which consists of large number of turns of insulated copper wire wound on a soft iron core.
• SLIP RINGS are $R_1$ and $R_2$ to which the ends of armature coil is connected. It is made of brass which rotates along with the coil.
• The two carbon BRUSHES $B_1$ and $B_2$ are connected to the rings to load the resistor through which the input is obtained.
• The armature coil is placed between two cylindrical magnetic pole pieces to establish a magnetic field perpendicular to the axis of rotation of coil.

The armature coil is rotated mechanically about its vertical axis. The flux linked with the coil changes as it cuts the magnetic lines of force during rotation. Hence, emf an current are induced.

The current flows through out brush $B_1$ in one half rotation and through $B_2$ in other half rotation. So, it reverses its direction in one complete rotation. Hence it is sinusoidal in nature.

Theory : Let the rectangular coil ABCD consists of n turns and A be the area of each turn. Let the coil be rotated about its vertical axis with an angular speed $ \ omega $.

At time t = 0 , the plane of the coil is placed perpendicular to the direction of the field and made to rotate. After an interval of t seconds , let the coil rotate through an angle $ \theta $. Then, the flux associated with the coil in the rotated position is given by $ Φ = nBA \space cos \theta $

Since the coil is rotating with an angular speed $ \omega $ , the displacement after time t is $ \theta = \omega t $
$$ \therefore Φ = nBA \space cos \omega t…….(1)$$

Due to the change of flux linked with the coil an emf is induced in it. According to Faraday Lenz law ,
$$ e = – \frac {d Φ}{dt} = – \frac {d}{dt} ( nBA \space cos \omega t ) $$
$$ = -nBA(-sin \omega t. \omega) $$
$$ e = nBA \omega \space sin \omega t $$
$$ Let \space nBA \omega = e_o $$

Then , $ e =e_o \space sin \omega t……(2) $

Equation (2) represents that induced emf is sinusoidal in nature. It becomes maximum twice at $90^o$ and $270^o$ and minimum twice at $180^o$ and $360^o$ in once complete cycle. When the voltage across the coil is maximum , the plane of the coil will be parallel to the direction of the field.

If R be the resistance of the coil , then instantaneous current through the coil is given by

$$ I = \frac {e}{R} = \frac { e_o \space sin \omega t}{R} $$

$$ Let \space \frac {e_o}{R} = I_o $$

$$ \therefore I = I_o \space sin \omega t …..(3) $$

Hence, current through the coil is also sinusoidal in nature as shown in figure.

Example 6.11 Kamla peddles a stationary bicycle the pedals of the bicycle are attached to a 100 turn coil of area 0.10 m2. The coil rotates at half a revolution per second and it is placed in a uniform magnetic field of 0.01 T perpendicular to the axis of rotation of the coil. What is the maximum voltage generated in the coil?

Solution: Here f = 0.5 Hz; N =100, A = 0.1 m2 and B = 0.01 T. Employing

Eq. (6.21)

$$ ε_0 = NBA (2 π ν) $$
= 100 × 0.01 × 0.1 × 2 × 3.14 × 0.5

= 0.314 V

The maximum voltage is 0.314 V.

We urge you to explore such alternative possibilities for power generation.

Questions for section 6.10

1. Explain the construction and working of an AC generator.
2. Give a brief theory for the sinusoidal waves.