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Electromagnetic Induction
electromagnetic
induction
with
the
help
of
Faraday’s expt.
Ans:
Change in the magnetic flux linked with the coil
produces emf in the coil, which causes electric
current to flow through it. This emf is called
induced emf and the current is called induced
current.
Production of induced emf in the coil due to
change in the magnetic field associated with it is
called electromagnetic induction.
Coil and Magnet experiment
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Faraday’s laws of electromagnetic induction.
Ans:
1. Whenever there is change in the magnetic
field associated with a coil, an emf in induced
in it.
2. The
magnitude
of
this
emf
is
directly
proportional to the rate of change of flux
associated with it.
d
dt
e
e
or e
d
k
dt
d
dt
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Lenz’s law.
Ans
“The direction of induced emf is such as to
oppose the change in the magnetic flux which
produces it.”
e
d
dt
Thus, faraday’s law gives the magnitude and
Lenz’s law gives the direction of induced emf.
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theoretical proof of laws of electromagnetic
induction.
Ans:
The mathematical statement of the laws of
induction is
e
d
dt
F = q v × B,
to it. Hence, it is acted up on by a force F’ given
by F’ = Biℓ sin θ . But as θ = π/2,
sin θ = 1 and the force is F’ = Biℓ
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the work dW done in this process is
dW = -F’dx
( -ve sign as the force is applied
opposite to F’)
∴ dW = - Biℓ × dx
But, ℓ × dx is the area A of the frame swept by the
force.
∴ dW = - BiA
If dφ is the magnetic flux through area A,
B = dφ ℓ A
∴ dφ = BA
∴ dW = - i dφ------(1)
If e is the induced emf and i is the induced current,
the electrical power generated in the frame is
p = ei
If this power is generated in time dt,
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electrical work done = eidt-----(2)
This work must be equal to the work done by force
F’
∴ eidt = - idφ----- from 1 &2
e
d
dt
Hence, the laws of electromagnetic induction are
proved.
As e
d
dt
BA
dt
dx
B
dt
8
Bv
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2. Induction Furnace
i.
fast operation
ii. better control over the temperature
iii. surface heating
iv. less pollution etc.
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M.C.Q.
A coil having an area of 2m2 is placed in a
magnetic field which changes from 2Wb/m2 to
5Wb/m2 in 3s. The e.m.f. induced in the coil will
Be (a.6) 4V
(b.6) 3V
(c.6) 2V
(d.6) 1V
Q.7 The magnetic flux in a coil is Ф= 4t2 + 4t + 4.
What is the magnitude of induced e.m.f. at
t=3sec?
(a.7) 14V (b.7) 28V (c.7) 7V
(d.7) 35V
Q.6
Q.8
The two rails of a railway track separated by
1metre and insulated from each other, are
connected to a millivoltmeter. What is the
reading of the millivoltmeter when a train
passes at a speed of 180km/hour along the
track? [The vertical component of earth’s
magnetic field is 0.2×10-4Wb/m2]
(a.8) 1volt (b.8) 100mV
(c.8) 1mV (d.8) 10mV
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Self inductance.
Ans:
Production of induced emf in a coil due to
changing magnetic flux produced due the varying
current in the coil itself is called self induction.
Flux φ associated with the coil is proportional to
current i in the coil.
i.e. φ α i
or φ = Li where L is a constant which depends
on size and shape of the coil. The induced emf
generated in the coil is given as
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d Li
d
dt
e
L
dt
di
dt
The constant L is called self inductance of the
coil. If di/dt = 1, them e = - L
“Thus, self inductance of a coil is equal to the emf
induced in the coil, due to unit rate of change
of current through it”.
L
e
di / dt
The unit of self inductance is henry (H).
1 henry
1 volt
1 ampere / 1 second
1 henry :The self inductance of a coil is 1 henry, if the emf
of 1V is induced in the coil, when the
current through the coil changes at 1 ampere per
second.
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Mutual inductance.
Ans:
Production of induced emf in a coil due to
changes in current in the second coil near it is
called mutual induction.
i.e. e2
dIA
dt
dI A
M
dt
e2
When dIA /dt = 1,
e2 = - M
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Thus, mutual inductance between two coils is
equal to the emf induced in the second coil due to
unit rate of change of current in the first coil.
M
e2
dIA / dt
The unit of mutual inductance is henry (H).
1 henry
1 volt
1 ampere / 1 second
1 henry :
The mutual inductance between two coils is 1
henry, if the emf of 1V is induced in the second
coil, when the current through the first coil
changes at 1 ampere per second.
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E2
E1
N2
N1
E.M.F. induced across secondary
E.M.F. induced across primary
Number of turns of the secondary coil
Number of turns of the primary coil
Power output = Power input
E2I2 = E1I1
E2
E1
I1
I2
N2
N1
% efficiency
Output power
Input power
19
100%