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204
6.1 INTRODUCTION
Electricity and magnetism were considered separate and unrelated
phenomena for a long time. In the early decades of the nineteenth century,
experiments on electric current by Oersted, Ampere and a few others
established the fact that electricity and magnetism are inter-related. They
found that moving electric charges produce magnetic fields. For example,
an electric current deflects a magnetic compass needle placed in its vicinity.
This naturally raises the questions like: Is the converse effect possible?
Can moving magnets produce electric currents? Does the nature permit
such a relation between electricity and magnetism? The answer is
resounding yes! 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.
When Faraday first made public his discovery that relative motion
between a bar magnet and a wire loop produced a small current in the
latter, he was asked, “What is the use of it?” His reply was: “What is the
use of a new born baby?” The phenomenon of electromagnetic induction
Chapter Six
ELECTROMAGNETIC
INDUCTION
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Induction
205
is not merely of theoretical or academic interest but also
of practical utility. Imagine a world where there is no
electricity – no electric lights, no trains, no telephones and
no personal computers. 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.
6.2 THE EXPERIMENTS OF FARADAY AND
HENRY
The discovery and understanding of electromagnetic
induction are based on a long series of experiments carried
out by Faraday and Henry. We shall now describe some
of these experiments.
Experiment 6.1
Figure 6.1 shows a coil C
1
* connected to a galvanometer
G. When the North-pole of a bar magnet is pushed
towards the coil, the pointer in the galvanometer deflects,
indicating the presence of electric current in the coil. The
deflection lasts as long as the bar magnet is in motion.
The galvanometer does not show any deflection when the
magnet is held stationary. When the magnet is pulled
away from the coil, the galvanometer shows deflection in
the opposite direction, which indicates reversal of the
current’s direction. Moreover, when the South-pole of
the bar magnet is moved towards or away from the
coil, the deflections in the galvanometer are opposite
to that observed with the North-pole for similar
movements. Further, the deflection (and hence current)
is found to be larger when the magnet is pushed
towards or pulled away from the coil faster. Instead,
when the bar magnet is held fixed and the coil C
1
is
moved towards or away from the magnet, the same
effects are observed. It shows that it is the relative
motion between the magnet and the coil that is
responsible for generation (induction) of electric
current in the coil.
Experiment 6.2
In Fig. 6.2 the bar magnet is replaced by a second coil
C
2
connected to a battery. The steady current in the
coil C
2
produces a steady magnetic field. As coil C
2
is
* Wherever the term ‘coil’ or ‘loop’ is used, it is assumed that they are made up of
conducting material and are prepared using wires which are coated with insulating
material.
FIGURE 6.1 When the bar magnet is
pushed towards the coil, the pointer in
the galvanometer G deflects.
Josheph Henry [1797 –
1878] American experimental
physicist, professor at
Princeton University and first
director of the Smithsonian
Institution. He made important
improvements in electro-
magnets by winding coils of
insulated wire around iron
pole pieces and invented an
electromagnetic motor and a
new, efficient telegraph. He
discoverd self-induction and
investigated how currents in
one circuit induce currents in
another.
JOSEPH HENRY (1797 – 1878)
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moved towards the coil C
1
, the galvanometer shows a
deflection. This indicates that electric current is induced in
coil C
1
. When C
2
is moved away, the galvanometer shows a
deflection again, but this time in the opposite direction. The
deflection lasts as long as coil C
2
is in motion. When the coil
C
2
is held fixed and C
1
is moved, the same effects are observed.
Again, it is the relative motion between the coils that induces
the electric current.
Experiment 6.3
The above two experiments involved relative motion between
a magnet and a coil and between two coils, respectively.
Through another experiment, Faraday showed that this
relative motion is not an absolute requirement. Figure 6.3
shows two coils C
1
and C
2
held stationary. Coil C
1
is connected
to galvanometer G while the second coil C
2
is connected to a
battery through a tapping key K.
FIGURE 6.2 Current is
induced in coil C
1
due to motion
of the current carrying coil C
2
.
FIGURE 6.3 Experimental set-up for Experiment 6.3.
It is observed that the galvanometer shows a momentary deflection
when the tapping key K is pressed. The pointer in the galvanometer returns
to zero immediately. If the key is held pressed continuously, there is no
deflection in the galvanometer. When the key is released, a momentory
deflection is observed again, but in the opposite direction. It is also observed
that the deflection increases dramatically when an iron rod is inserted
into the coils along their axis.
6.3 MAGNETIC FLUX
Faraday’s great insight lay in discovering a simple mathematical relation
to explain the series of experiments he carried out on electromagnetic
induction. However, before we state and appreciate his laws, we must get
familiar with the notion of magnetic flux,
Φ
B
. Magnetic flux is defined in
the same way as electric flux is defined in Chapter 1. Magnetic flux through
Interactive animation on Faraday’s experiments and Lenz’s law:
http://micro.magnet.fsu.edu/electromagnet/java/faraday2/
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Induction
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a plane of area A placed in a uniform magnetic field B (Fig. 6.4) can
be written as
Φ
B
= B
.
A = BA cos
θ
(6.1)
where
θ
is angle between B and A. The notion of the area as a vector
has been discussed earlier in Chapter 1. Equation (6.1) can be
extended to curved surfaces and nonuniform 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
1 1 2 2
d d
Φ
= + +
B A B A
. .
B
... =
B A.
i i
d
all
∑
(6.2)
where ‘all’ stands for summation over all the area elements d
A
i
comprising the surface and B
i
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.
6.4 FARADAY’S LA
W OF INDUCTION
From the experimental observations, Faraday arrived at a
conclusion that an emf
is induced in a coil when magnetic flux
through the coil changes with time. Experimental observations
discussed in Section 6.2 can be explained using this concept.
The motion of a magnet towards or away from coil C
1
in
Experiment 6.1 and moving a current-carrying coil C
2
towards
or away from coil C
1
in Experiment 6.2, change the magnetic
flux associated with coil C
1
. The change in magnetic flux induces
emf in coil C
1
. It was this induced emf which caused electric
curr
ent to flow in coil C
1
and through the galvanometer. A
plausible explanation for the observations of Experiment 6.3 is
as follows: When the tapping key K is pressed, the current in
coil C
2
(and the resulting magnetic field) rises from zero to a
maximum value in a short time. Consequently, the magnetic
flux through the neighbouring coil C
1
also increases. It is the change in
magnetic flux through coil C
1
that produces an induced emf in coil C
1
.
When the key is held pressed, current in coil C
2
is constant. Therefore,
there is no change in the magnetic flux through coil C
1
and the current in
coil C
1
drops to zero. When the key is released, the current in C
2
and the
resulting magnetic field decreases from the maximum value to zero in a
short time. This results in a decrease in magnetic flux through coil C
1
and hence again induces an electric current in coil C
1
*. The common
point in all these observations is that the time rate of change of magnetic
flux through a circuit induces emf in it. Faraday stated experimental
observations in the form of a law called Faraday’s law of electromagnetic
induction. The law is stated below.
FIGURE 6.4 A plane of
surface area A placed in a
uniform magnetic field B.
FIGURE 6.5 Magnetic field B
i
at the i
th
area element. dA
i
represents area vector of the
i
th
area element.
* Note that sensitive electrical instruments in the vicinity of an electromagnet
can be damaged due to the induced emfs (and the resulting currents) when the
electromagnet is turned on or off.
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EXAMPLE 6.1
The magnitude of the induced emf in a circuit is equal
to the time rate of change of magnetic flux through the
circuit.
Mathematically, the induced emf is given by
d
–
d
B
t
Φ
ε
=
(6.3)
The negative sign indicates the direction of
ε
and hence
the direction of current in a closed loop. This will be
discussed in detail in the next section.
In the case of a closely wound coil of N turns, change
of flux associated with each turn, is the same. Therefore,
the expression for the total induced emf is given by
d
–
d
B
N
t
Φ
ε
=
(6.4)
The induced emf can be increased by increasing the
number of turns N of a closed coil.
From Eqs. (6.1) and (6.2), we see that the flux can be
varied by changing any one or more of the terms B, A and
θ
. In Experiments 6.1 and 6.2 in Section 6.2, the flux is
changed by varying B. The flux can also be altered by
changing the shape of a coil (that is, by shrinking it or
stretching it) in a magnetic field, or rotating a coil in a
magnetic field such that the angle
θ
between B and A
changes. In these cases too, an emf is induced in the
respective coils.
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.
Michael Faraday [1791–
1867] Faraday made
numerous contributions to
science, viz., the discovery
of electromagnetic
induction, the laws of
electrolysis, benzene, and
the fact that the plane of
polarisation is rotated in an
electric field. He is also
credited with the invention
of the electric motor, the
electric generator and the
transformer. He is widely
regarded as the greatest
experimental scientist of
the nineteenth century.
MICHAEL FARADAY (1791–1867)
EXAMPLE 6.2
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Induction
209
EXAMPLE 6.2
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
–2
0.1 10
Wb
2
×
=
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
(
)
– 0
B
t t
Φ
Φ
ε
∆
= =
∆ ∆
–3
10
= 1.0 mV
2 0.7
=
×
And the magnitude of the current is
–3
10 V
2 mA
0.5
I
R
ε
= = =
Ω
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 (initial)
= BA cos θ
= 3.0 × 10
–5
× (π ×10
–2
) × cos 0°
= 3π × 10
–7
Wb
Final flux after the rotation,
Φ
B (final)
= 3.0 × 10
–5
× (π ×10
–2
) × cos 180°
= –3π × 10
–7
Wb
Therefore, estimated value of the induced emf is,
N
t
Φ
ε
∆
=
∆
= 500 × (6π × 10
–7
)/0.25
= 3.8 × 10
–3
V
I = ε/R = 1.9 × 10
–3
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.
EXAMPLE 6.3
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6.5 LENZ
’S LA
W AND CONSERV
ATION OF ENERGY
In 1834, German physicist Heinrich Friedrich Lenz (1804-1865) deduced
a rule, known as
Lenz’s law which gives the polarity of the induced emf
in a clear and concise fashion. The statement of the law is:
The polarity of induced emf is such that it tends to produce a current
which opposes the change in magnetic flux that produced it.
The negative sign shown in Eq. (6.3) represents this effect. We can
understand Lenz’s law by examining Experiment 6.1 in Section 6.2.1. In
Fig. 6.1, we see that the North-pole of a bar magnet is being pushed
towards the closed coil. As the North-pole of the bar magnet moves towards
the coil, the magnetic flux through the coil increases. Hence current is
induced in the coil in such a direction that it opposes the increase in flux.
This is possible only if the current in the coil is in a counter-clockwise
direction with respect to an observer situated on the side of the magnet.
Note that magnetic moment associated with this current has North polarity
towards the North-pole of the approaching magnet. Similarly, if the North-
pole of the magnet is being withdrawn from the coil, the magnetic flux
through the coil will decrease. To counter this decrease in magnetic flux,
the induced current in the coil flows in clockwise direction and its South-
pole faces the receding North-pole of the bar magnet. This would result in
an attractive force which opposes the motion of the magnet and the
corresponding decrease in flux.
What will happen if an open circuit is used in place of the closed loop
in the above example? In this case too, an emf is induced across the open
ends of the circuit. The direction of the induced emf can be found
using Lenz’s law. Consider Figs. 6.6 (a) and (b). They provide an easier
way to understand the direction of induced currents. Note that the
direction shown by
and indicate the directions of the induced
currents.
A little reflection on this matter should convince us on the
correctness of Lenz’s law. Suppose that the induced current was in
the direction opposite to the one depicted in Fig. 6.6(a). In that case,
the South-pole due to the induced current will face the approaching
North-pole of the magnet. The bar magnet will then be attracted
towards the coil at an ever increasing acceleration. A gentle push on
the magnet will initiate the process and its velocity and kinetic energy
will continuously increase without expending any energy. If this can
happen, one could construct a perpetual-motion machine by a
suitable arrangement. This violates the law of conservation of energy
and hence can not happen.
Now consider the correct case shown in Fig. 6.6(a). In this situation,
the bar magnet experiences a repulsive force due to the induced
current. Therefore, a person has to do work in moving the magnet.
Where does the energy spent by the person go? This energy is
dissipated by Joule heating produced by the induced current.
FIGURE 6.6
Illustration of
Lenz’s law.
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EXAMPLE 6.4
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.
FIGURE 6.7
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.
EXAMPLE 6.5
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EXAMPLE
6.5
FIGURE 6.8
(d) Predict the polarity of the capacitor in the situation described by
Fig. 6.9.
FIGURE 6.9
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.
6.6 MOTIONAL ELECTROMOTIVE FORCE
Let us consider a straight conductor moving in a uniform and time-
independent magnetic field. Figure 6.10 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:
( )
– d d
–
d d
B
Blx
t t
Φ
ε
= =
=
d
–
d
x
Bl Blv
t
=
(6.5)
FIGURE 6.10 The arm PQ is moved to the left
side, thus decreasing the area of the
rectangular loop. This movement
induces a current I as shown.
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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 Eq. (6.5)
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,
W
q
ε
=
= Blv
This equation gives emf induced across the rod PQ and is identical
to Eq. (6.5). 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 (6.6)
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. In Chapter 4, we learnt that
charges in motion (current) can exert force/torque on a stationary magnet.
Conversely, a bar magnet in motion (or more generally, a changing
magnetic field) can exert a force on the stationary charge. This is the
fundamental significance of the Faraday’s discovery. Electricity and
magnetism are 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. 6.11). 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?
EXAMPLE 6.6
Interactive animation on motional emf:
http://ngsir.netfirms.com/englishhtm/Induction.htm
http://webphysics.davidson.edu/physlet_resources/bu_semester2/index.html
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EXAMPLE
6.6
FIGURE 6.11
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
d d
Bv r
ε
=
. Hence,
ε ε
= =
∫∫
d dBv r
R
0
= =
∫
B r r
B R
R
ω
ω
d
2
0
2
Note that we have used v =
ω
r. This gives
ε
2
1
1.0 2 50 (1 )
2
= × × π × ×
= 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
2 2
1
2 2
R R
θ
θ
π × =
π
where R is the radius of the circle. Hence, the induced emf is
ε
=
B
t
R×
d
d
1
2
2
θ
=
2
2
1 d
2 d 2
θ ω
=
B R
BR
t
[Note:
d
2
dt
θ
ω ν
= = π
]
This expression is identical to the expression obtained by Method I
and we get the same value of
ε
.
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EXAMPLE 6.7
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.
6.7 ENERGY CONSIDERATION: A QUANTITATIVE STUDY
In Section 6.5, we discussed qualitatively that Lenz’s law is consistent with
the law of conservation of energy. Now we shall explore this aspect further
with a concrete example.
Let r be the resistance of movable arm PQ of the rectangular conductor
shown in Fig. 6.10. We assume that the remaining arms QR, RS and SP
have negligible resistances compared to r. Thus, the overall resistance of
the rectangular loop is r and this does not change as PQ is moved. The
current I in the loop is,
I
r
ε
=
=
B l v
r
(6.7)
On account of the presence of the magnetic field, there will be a force
on the arm PQ. This force I (l
××
××
×
B), is directed outwards in the direction
opposite to the velocity of the rod. The magnitude of this force is,
F = I l B =
2 2
B l v
r
where we have used Eq. (6.7). Note that this force arises due to drift velocity
of charges (responsible for current) along the rod and the consequent
Lorentz force acting on them.
Alternatively, the arm PQ is being pushed with a constant speed v,
the power required to do this is,
P F v
=
=
2 2 2
B l v
r
(6.8)
The agent that does this work is mechanical. Where does this
mechanical energy go? The answer is: it is dissipated as Joule heat, and
is given by
2
J
P I r
=
=
Blv
r
r
2
2 2 2
B l v
r
=
which is identical to Eq. (6.8).
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EXAMPLE
6.8
Thus, mechanical energy which was needed to move the arm PQ is
converted into electrical energy (the induced emf) and then to thermal energy.
There is an interesting relationship between the charge flow through
the circuit and the change in the magnetic flux. From Faraday’s law, we
have learnt that the magnitude of the induced emf is,
B
t
Φ
ε
∆
=
∆
However,
Q
Ir r
t
ε
∆
= =
∆
Thus,
B
Q
r
Φ
∆
∆ =
Example 6.8 Refer to Fig. 6.12(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.
(a)
FIGURE 6.12
Solution Let us first consider the forward motion from x = 0 to x = 2b
The flux
Φ
B
linked with the circuit SPQR is
B
0
B l x x b
Φ
= ≤ <
2
B l b b x b
= ≤ <
The induced emf is,
B
d
d
t
Φ
ε
= −
0
Blv x b
= − ≤ <
0 2
b x b
= ≤ <
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EXAMPLE 6.8
When the induced emf is non-zero, the current I is (in magnitude)
Bl v
I
r
=
(b)
FIGURE 6.12
The force required to keep the arm PQ in constant motion is I lB. Its
direction is to the left. In magnitude
2 2
0
0 2
B l v
F x b
r
b x b
= ≤ <
= ≤ <
The Joule heating loss is
2
J
P I r
=
2 2 2
0
0 2
B l v
x b
r
b x b
= ≤ <
= ≤ <
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. 6.12(b).
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6.8 EDDY CURRENTS
So far we have studied the electric currents induced in well defined paths
in conductors like circular loops. Even when bulk pieces of conductors
are subjected to changing magnetic flux, induced currents
are produced in them. However, their flow patterns resemble
swirling eddies in water. This effect was discovered by physicist
Foucault (1819-1868) and these currents are called eddy
currents.
Consider the apparatus shown in Fig. 6.13. A copper plate
is allowed to swing like a simple pendulum between the pole
pieces of a strong magnet. It is found that the motion is damped
and in a little while the plate comes to a halt in the magnetic
field. We can explain this phenomenon on the basis of
electromagnetic induction. Magnetic flux associated with the
plate keeps on changing as the plate moves in and out of the
region between magnetic poles. The flux change induces eddy
currents in the plate. Directions of eddy currents are opposite
when the plate swings into the region between the poles and
when it swings out of the region.
If rectangular slots are made in the copper plate as shown
in Fig. 6.14, area available to the flow of eddy currents is less.
Thus, the pendulum plate with holes or slots reduces
electromagnetic damping and the plate swings more freely.
Note that magnetic moments of the induced currents (which
oppose the motion) depend upon the area enclosed by the
currents (recall equation m = IA in Chapter 4).
This fact is helpful in reducing eddy currents in the metallic
cores of transformers, electric motors and other such devices in
which a coil is to be wound over metallic core. Eddy currents are
undesirable since they heat up the core and dissipate electrical
energy in the form of heat. Eddy currents are minimised by using
laminations of metal to make a metal core. The laminations are
separated by an insulating material like lacquer. The plane of the
laminations must be arranged parallel to the magnetic field, so
that they cut across the eddy current paths. This arrangement
reduces the strength of the eddy currents. Since the dissipation
of electrical energy into heat depends on the square of the strength
of electric current, heat loss is substantially reduced.
Eddy currents are used to advantage in certain applications like:
(i) 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.
(ii) 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.
FIGURE 6.13 Eddy currents are
generated in the copper plate,
while entering
and leaving the region of
magnetic field.
FIGURE 6.14 Cutting slots
in the copper plate reduces
the effect of eddy currents.
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219
(iii) 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.
(iv) 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.
You can observe the rotating shiny disc in the power meter of your
house.
ELECTROMAGNETIC DAMPING
Take two hollow thin cylindrical pipes of equal internal diameters made of aluminium and
PVC, respectively. Fix them vertically with clamps on retort stands. Take a small cylinderical
magnet having diameter slightly smaller than the inner diameter of the pipes and drop it
through each pipe in such a way that the magnet does not touch the sides of the pipes
during its fall. You will observe that the magnet dropped through the PVC pipe takes the
same time to come out of the pipe as it would take when dropped through the same height
without the pipe. Note the time it takes to come out of the pipe in each case. You will see that
the magnet takes much longer time in the case of aluminium pipe. Why is it so? It is due to
the eddy currents that are generated in the aluminium pipe which oppose the change in
magnetic flux, i.e., the motion of the magnet. The retarding force due to the eddy currents
inhibits the motion of the magnet. Such phenomena are referred to as electromagnetic damping.
Note that eddy currents are not generated in PVC pipe as its material is an insulator whereas
aluminium is a conductor.
6.9 INDUCTANCE
An electric current can be induced in a coil by flux change produced by
another coil in its vicinity or flux change produced by the same coil. These
two situations are described separately in the next two sub-sections.
However, in both the cases, 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,
d
d
d d
B
I
t t
Φ
∝
For a closely wound coil of N turns, the same magnetic flux is linked
with all the turns. When the flux
Φ
B
through the coil changes, each turn
contributes to the induced emf. Therefore, a term called flux linkage is
used which is equal to N
Φ
B
for a closely wound coil and in such a case
N
Φ
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
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and intrinsic material properties. This aspect is akin to capacitance which
for a parallel plate capacitor depends on the plate area and plate separation
(geometry) and the dielectric constant K of the intervening medium
(intrinsic material property).
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.
6.9.1 Mutual inductance
Consider Fig. 6.15 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
Φ
1
. The corresponding flux linkage
with solenoid S
1
is
N
1
1 12 2
M I
Φ
=
(6.9)
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
µ
0
n
2
I
2
. The resulting flux linkage
with coil S
1
is,
(
)
(
)
(
)
2
1 1 1 1 0 2 2
N n l r n I
Φ µ
= π
2
0 1 2 1 2
n n r l I
µ
= π
(6.10)
where n
1
l is the total number of turns in solenoid S
1
. Thus, from Eq. (6.9)
and Eq. (6.10),
M
12
=
µ
0
n
1
n
2
πr
2
1
l (6.11)
Note that we neglected the edge effects and considered
the magnetic field
µ
0
n
2
I
2
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
2
Φ
2
= M
21
I
1
(6.12)
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
(
)
(
)
(
)
2
2 2 2 1 0 1 1
N n l r n I
Φ µ
= π
FIGURE 6.15 Two long co-axial
solenoids of same
length l.
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221
EXAMPLE 6.9
where n
2
l is the total number of turns of S
2
. From Eq. (6.12),
M
21
=
µ
0
n
1
n
2
πr
2
1
l (6.13)
Using Eq. (6.11) and Eq. (6.12), we get
M
12
= M
21
= M (say) (6.14)
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
1
Φ
1
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
µ
r
had been present,
the mutual inductance would be
M =
µ
r
µ
0
n
1
n
2
π
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
0
I
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
1
B
2
2
0 1
2
2
2
r
I
r
µ
π
=
= M
12
I
2
Thus,
2
0 1
12
2
2
r
M
r
µ
π
=
From Eq. (6.14)
2
0 1
12 21
2
2
r
M M
r
µ π
= =
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
1
2
. However, we
can accept this value because r
1
<< r
2
.
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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
1
Φ
1
= MI
2
For currents varrying with time,
(
)
(
)
1 1 2
d d
d d
N MI
t t
Φ
=
Since induced emf in coil C
1
is given by
(
)
1 1
d
–
d
N
t
Φ
ε
1
=
We get,
2
d
–
d
I
M
t
ε
1
=
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-inductance
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
B
N I
Φ
∝
B
L
N I
Φ
=
(6.15)
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 emf is induced
in the coil. Using Eq. (6.15), the induced emf is given by
(
)
B
d
–
d
N
t
Φ
ε
=
d
–
d
I
L
t
ε
=
(6.16)
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
n I (neglecting edge
effects, as before). The total flux linked with the solenoid is
(
)
(
)
(
)
0B
N nl n I A
Φ µ
=
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223
IAln
2
0
µ
=
where nl is the total number of turns. Thus, the self-inductance is,
L
I
Β
ΝΦ
=
2
0
n Al
µ
=
(6.17)
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 permeability),
then,
2
0r
L n Al
µ µ
=
(6.18)
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
d
d
W
I
t
ε
=
If we ignore the resistive losses and consider only inductive effect,
then using Eq. (6.16),
d d
d d
W I
L I
t t
=
Total amount of work done in establishing the current I is
W W L I I
I
= =
∫ ∫
d d
0
Thus, the energy required to build up the current I is,
2
1
2
W LI
=
(6.19)
This expression reminds us of mv
2
/2 for the (mechanical) kinetic energy
of a particle of mass m, and shows that L is analogous to m (i.e., L is
electrical inertia and opposes growth and decay of current in the circuit).
Consider the general case of currents flowing simultaneously in two
nearby coils. The flux linked with one coil will be the sum of two fluxes
which exist independently. Equation (6.9) would be modified into
N
1
1 11 1 12 2
M I M I
Φ
= +
where M
11
represents inductance due to the same coil.
Therefore, using Faraday’s law,
1 2
1 11 12
d d
d d
I I
M M
t t
ε
= − −
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EXAMPLE
6.10
M
11
is the self-inductance and is written as L
1
. Therefore,
1 2
1 1 12
d d
d d
I I
L M
t t
ε
= − −
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
2
1
2
B
U LI
=
=
=
( )
1
2
2
L
B
n
nI
µ
µ
0
0
Bsince for a solenoid,
=
1
2
0
2
0
2
( )
µ
µ
n Al
B
n
[from Eq. (6.17)]
2
0
1
2
B Al
µ
=
(b) The magnetic energy per unit volume is,
B
B
U
u
V
=
(where V is volume that contains flux)
B
U
Al
=
2
0
2
B
µ
=
(6.20)
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),
2
0
1
2
u E
Ε
ε
=
(2.77)
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.
6.10 AC GENERATOR
The phenomenon of electromagnetic induction has been technologically
exploited in many ways. An exceptionally important application is the
generation of alternating currents (ac). The modern ac generator with a
typical output capacity of 100 MW is a highly evolved machine. In this
section, we shall describe the basic principles behind this machine. The
Yugoslav inventor Nicola Tesla is credited with the development of the
machine. As was pointed out in Section 6.3, one method to induce an emf
Interactive animation on ac generator:
http://micro.magnet.fsu.edu/electromag/java/generator/ac.html
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Electromagnetic
Induction
225
or current in a loop is through a change in the
loop’s orientation or a change in its effective area.
As the coil rotates in a magnetic field B, the
effective area of the loop (the face perpendicular
to the field) is A cos
θ
, where
θ
is the angle
between A and B. This method of producing a
flux change is the principle of operation of a
simple ac generator. An ac generator converts
mechanical energy into electrical energy.
The basic elements of an ac generator are
shown in Fig. 6.16. It consists of a coil mounted
on a rotor shaft. The axis of rotation of the coil
is perpendicular to the direction of the magnetic
field. The coil (called armature) is mechanically
rotated in the uniform magnetic field by some
external means. The rotation of the coil causes
the magnetic flux through it to change, so an
emf is induced in the coil. The ends of the coil
are connected to an external circuit by means
of slip rings and brushes.
When the coil is rotated with a constant
angular speed
ω
, the angle
θ
between the magnetic field vector B and the
area vector A of the coil at any instant t is
θ
=
ω
t (assuming
θ
= 0° at t = 0).
As a result, the effective area of the coil exposed to the magnetic field lines
changes with time, and from Eq. (6.1), the flux at any time t is
Φ
B
= BA cos
θ
= BA cos
ω
t
From Faraday’s law, the induced emf for the rotating coil of N turns is
then,
d
d
– – (cos )
dt d
B
N NBA t
t
Φ
ε ω
= =
Thus, the instantaneous value of the emf is
ε ω ω
=
NBA sin t
(6.21)
where NBA
ω
is the maximum value of the emf, which occurs when
sin
ω
t = ±1. If we denote NBA
ω
as
ε
0
, then
ε
=
ε
0
sin
ω
t (6.22)
Since the value of the sine fuction varies between +1 and –1, the sign, or
polarity of the emf changes with time. Note from Fig. 6.17 that the emf
has its extremum value when
θ
= 90° or
θ
= 270°, as the change of flux is
greatest at these points.
The direction of the current changes periodically and therefore the current
is called alternating current (ac). Since
ω
= 2π
ν
, Eq (6.22) can be written as
ε
=
ε
0
sin 2π
ν
t (6.23)
where
ν
is the frequency of revolution of the generator’s coil.
Note that Eq. (6.22) and (6.23) give the instantaneous value of the emf
and
ε
varies between +
ε
0
and –
ε
0
periodically. We shall learn how to
determine the time-averaged value for the alternating voltage and current
in the next chapter.
FIGURE 6.16 AC Generator
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EXAMPLE
6.11
In commercial generators, the mechanical energy required for rotation
of the armature is provided by water falling from a height, for example,
from dams. These are called hydro-electric generators. Alternatively, water
is heated to produce steam using coal or other sources. The steam at
high pressure produces the rotation of the armature. These are called
thermal generators. Instead of coal, if a nuclear fuel is used, we get nuclear
power generators. Modern day generators produce electric power as high
as 500 MW, i.e., one can light up 5 million 100 W bulbs! In most
generators, the coils are held stationary and it is the electromagnets which
are rotated. The frequency of rotation is 50 Hz in India. In certain countries
such as USA, it is 60 Hz.
Example 6.11 Kamla peddles a stationary bicycle. The pedals of the
bicycle are attached to a 100 turn coil of area 0.10 m
2
. 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
ν
= 0.5 Hz; N =100, A = 0.1 m
2
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.
FIGURE 6.17 An alternating emf is generated by a loop of wire rotating in a magnetic field.
2020-21
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227
SUMMARY
1. The magnetic flux through a surface of area A placed in a uniform magnetic
field B is defined as,
Φ
B
= B.A = BA cos
θ
where
θ
is the angle between B and A.
2. Faraday’s laws of induction imply that the emf induced in a coil of N
turns is directly related to the rate of change of flux through it,
B
d
d
N
t
Φ
ε
= −
Here
Φ
Β
is the flux linked with one turn of the coil. If the circuit is
closed, a current I = ε/R is set up in it, where R is the resistance of the
circuit.
3. Lenz’s law states that the polarity of the induced emf is such that it
tends to produce a current which opposes the change in magnetic flux
that produces it. The negative sign in the expression for Faraday’s law
indicates this fact.
4. When a metal rod of length l is placed normal to a uniform magnetic
field B and moved with a velocity v perpendicular to the field, the
induced emf (called motional emf) across its ends is
ε
= Bl v
5. Changing magnetic fields can set up current loops in nearby metal
(any conductor) bodies. They dissipate electrical energy as heat. Such
currents are eddy currents.
6. Inductance is the ratio of the flux-linkage to current. It is equal to N
Φ
/I.
MIGRATION
OF
BIRDS
The migratory pattern of birds is one of the mysteries in the field of biology, and indeed all
of science. For example, every winter birds from Siberia fly unerringly to water spots in the
Indian subcontinent. There has been a suggestion that electromagnetic induction may
provide a clue to these migratory patterns. The earth’s magnetic field has existed throughout
evolutionary history. It would be of great benefit to migratory birds to use this field to
determine the direction. As far as we know birds contain no ferromagnetic material. So
electromagnetic induction seems to be the only reasonable mechanism to determine
direction. Consider the optimal case where the magnetic field B, the velocity of the bird v,
and two relevant points of its anatomy separated by a distance l, all three are mutually
perpendicular. From the formula for motional emf, Eq. (6.5),
ε
= Blv
Taking B = 4 × 10
–5
T, l = 2 cm wide, and v = 10 m/s, we obtain
ε
= 4 × 10
–5
× 2 × 10
–2
× 10 V = 8 × 10
–6
V
= 8
µ
V
This extremely small potential difference suggests that our hypothesis is of doubtful
validity. Certain kinds of fish are able to detect small potential differences. However, in
these fish, special cells have been identified which detect small voltage differences. In birds
no such cells have been identified. Thus, the migration patterns of birds continues to remain
a mystery.
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POINTS TO PONDER
1. Electricity and magnetism are intimately related. In the early part of the
nineteenth century, the experiments of Oersted, Ampere and others
established that moving charges (currents) produce a magnetic field.
Somewhat later, around 1830, the experiments of Faraday and Henry
demonstrated that a moving magnet can induce electric current.
2. In a closed circuit, electric currents are induced so as to oppose the
changing magnetic flux. It is as per the law of conservation of energy.
However, in case of an open circuit, an emf is induced across its ends.
How is it related to the flux change?
3. The motional emf discussed in Section 6.5 can be argued independently
from Faraday’s law using the Lorentz force on moving charges. However,
Quantity Symbol Units Dimensions Equations
Magnetic Flux
Φ
B
Wb (weber) [M
L
2
T
–2
A
–1
]
Φ
B
=
B A
EMF
ε
V (volt) [M
L
2
T
–3
A
–1
]
ε
=
B
d( )/ d
N t
Φ
−
Mutual Inductance M H (henry) [M L
2
T
–2
A
–2
]
ε
1
(
)
12 2
d / d
M I t
= −
Self Inductance L H (henry) [M
L
2
T
–2
A
–2
]
(
)
d /d
L I t
ε
= −
7. A changing current in a coil (coil 2) can induce an emf in a nearby coil
(coil 1). This relation is given by,
2
1 12
d
d
I
M
t
ε
= −
The quantity M
12
is called mutual inductance of coil 1 with respect to
coil 2. One can similarly define M
21
. There exists a general equality,
M
12
= M
21
8. When a current in a coil changes, it induces a back emf in the same
coil. The self-induced emf is given by,
d
d
I
L
t
ε
= −
L is the self-inductance of the coil. It is a measure of the inertia of the
coil against the change of current through it.
9. The self-inductance of a long solenoid, the core of which consists of a
magnetic material of relative permeability
µ
r
, is given by
L =
µ
r
µ
0
n
2
Al
where A is the area of cross-section of the solenoid, l its length and n
the number of turns per unit length.
10. In an ac generator, mechanical energy is converted to electrical energy
by virtue of electromagnetic induction. If coil of N turn and area A is
rotated at
ν
revolutions per second in a uniform magnetic field B, then
the motional emf produced is
ε
= NBA (2π
ν
) sin (2π
ν
t)
where we have assumed that at time t = 0 s, the coil is perpendicular to
the field.
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EXERCISES
6.1 Predict the direction of induced current in the situations described
by the following Figs. 6.18(a) to (f).
even if the charges are stationary [and the q (v × B) term of the Lorentz
force is not operative], an emf is nevertheless induced in the presence of a
time-varying magnetic field. Thus, moving charges in static field and static
charges in a time-varying field seem to be symmetric situation for
Faraday’s law. This gives a tantalising hint on the relevance of the principle
of relativity for Faraday’s law.
4. The motion of a copper plate is damped when it is allowed to oscillate
between the magnetic pole-pieces. How is the damping force, produced by
the eddy currents?
FIGURE 6.18
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230
6.2 Use Lenz’s law to determine the direction of induced current in the
situations described by Fig. 6.19:
(a) A wire of irregular shape turning into a circular shape;
(b) A circular loop being deformed into a narrow straight wire.
FIGURE 6.19
6.3 A long solenoid with 15 turns per cm has a small loop of area 2.0 cm
2
placed inside the solenoid normal to its axis. If the current carried
by the solenoid changes steadily from 2.0 A to 4.0 A in 0.1 s, what is
the induced emf in the loop while the current is changing?
6.4 A rectangular wire loop of sides 8 cm and 2 cm with a small cut is
moving out of a region of uniform magnetic field of magnitude 0.3 T
directed normal to the loop. What is the emf developed across the
cut if the velocity of the loop is 1 cm s
–1
in a direction normal to the
(a) longer side, (b) shorter side of the loop? For how long does the
induced voltage last in each case?
6.5 A 1.0 m long metallic rod is rotated with an angular frequency of
400 rad s
–1
about an axis normal to the rod passing through its one
end. The other end of the rod is in contact with a circular metallic
ring. A constant and uniform magnetic field of 0.5 T parallel to the
axis exists everywhere. Calculate the emf developed between the
centre and the ring.
6.6 A circular coil of radius 8.0 cm and 20 turns is rotated about its
vertical diameter with an angular speed of 50 rad s
–1
in a uniform
horizontal magnetic field of magnitude 3.0 × 10
–2
T. Obtain the
maximum and average emf induced in the coil. If the coil forms a
closed loop of resistance 10 Ω, calculate the maximum value of current
in the coil. Calculate the average power loss due to Joule heating.
Where does this power come from?
6.7 A horizontal straight wire 10 m long extending from east to west is
falling with a speed of 5.0 m s
–1
, at right angles to the horizontal
component of the earth’s magnetic field, 0.30 × 10
–4
Wb m
–2
.
(a) What is the instantaneous value of the emf induced in the wire?
(b) What is the direction of the emf?
(c) Which end of the wire is at the higher electrical potential?
6.8 Current in a circuit falls from 5.0 A to 0.0 A in 0.1 s. If an average emf
of 200 V induced, give an estimate of the self-inductance of the circuit.
6.9 A pair of adjacent coils has a mutual inductance of 1.5 H. If the
current in one coil changes from 0 to 20 A in 0.5 s, what is the
change of flux linkage with the other coil?
6.10 A jet plane is travelling towards west at a speed of 1800 km/h. What
is the voltage difference developed between the ends of the wing
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231
having a span of 25 m, if the Earth’s magnetic field at the location
has a magnitude of 5 × 10
–4
T and the dip angle is 30°.
ADDITIONAL EXERCISES
6.11 Suppose the loop in Exercise 6.4 is stationary but the current
feeding the electromagnet that produces the magnetic field is
gradually reduced so that the field decreases from its initial value
of 0.3 T at the rate of 0.02 T s
–1
. If the cut is joined and the loop
has a resistance of 1.6 Ω, how much power is dissipated by the
loop as heat? What is the source of this power?
6.12 A square loop of side 12 cm with its sides parallel to X and Y axes is
moved with a velocity of 8 cm
s
–1
in the positive x-direction in an
environment containing a magnetic field in the positive z-direction.
The field is neither uniform in space nor constant in time. It has a
gradient of 10
–3
T cm
–1
along the negative x-direction (that is it increases
by 10
–3
T cm
–1
as one moves in the negative x-direction), and it is
decreasing in time at the rate of 10
–3
T s
–1
. Determine the direction and
magnitude of the induced current in the loop if its resistance is 4.50 mΩ.
6.13 It is desired to measure the magnitude of field between the poles of a
powerful loud speaker magnet. A small flat search coil of area 2 cm
2
with 25 closely wound turns, is positioned normal to the field
direction, and then quickly snatched out of the field region.
Equivalently, one can give it a quick 90° turn to bring its plane
parallel to the field direction). The total charge flown in the coil
(measured by a ballistic galvanometer connected to coil) is
7.5 mC. The combined resistance of the coil and the galvanometer is
0.50 Ω. Estimate the field strength of magnet.
6.14 Figure 6.20 shows a metal rod PQ resting on the smooth rails AB
and positioned between the poles of a permanent magnet. The rails,
the rod, and the magnetic field are in three mutual perpendicular
directions. A galvanometer G connects the rails through a switch K.
Length of the rod = 15 cm, B = 0.50 T, resistance of the closed loop
containing the rod = 9.0 mΩ. Assume the field to be uniform.
(a) Suppose K is open and the rod is moved with a speed of 12 cm s
–1
in the direction shown. Give the polarity and magnitude of the
induced emf.
FIGURE 6.20
(b) Is there an excess charge built up at the ends of the rods when
K is open? What if K is closed?
(c) With K open and the rod moving uniformly, there is no net
force on the electrons in the rod PQ even though they do
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experience magnetic force due to the motion of the rod. Explain.
(d) What is the retarding force on the rod when K is closed?
(e) How much power is required (by an external agent) to keep
the rod moving at the same speed (=12cm s
–1
) when K is closed?
How much power is required when K is open?
(f) How much power is dissipated as heat in the closed circuit?
What is the source of this power?
(g) What is the induced emf in the moving rod if the magnetic field
is parallel to the rails instead of being perpendicular?
6.15 An air-cored solenoid with length 30 cm, area of cross-section 25 cm
2
and number of turns 500, carries a current of 2.5 A. The current is
suddenly switched off in a brief time of 10
–3
s. How much is the average
back emf induced across the ends of the open switch in the circuit?
Ignore the variation in magnetic field near the ends of the solenoid.
6.16 (a) Obtain an expression for the mutual inductance between a long
straight wire and a square loop of side a as shown in Fig. 6.21.
(b) Now assume that the straight wire carries a current of 50 A and
the loop is moved to the right with a constant velocity, v = 10 m/s.
Calculate the induced emf in the loop at the instant when x = 0.2 m.
Take a = 0.1 m and assume that the loop has a large resistance.
FIGURE 6.21
6.17 A line charge
λ
per unit length is lodged uniformly onto the rim of a
wheel of mass M and radius R. The wheel has light non-conducting
spokes and is free to rotate without friction about its axis (Fig. 6.22).
A uniform magnetic field extends over a circular region within the
rim. It is given by,
B = – B
0
k (r ≤ a; a < R)
= 0 (otherwise)
What is the angular velocity of the wheel after the field is suddenly
switched off?
FIGURE 6.22
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