Important notes JEE - Physics - Electromagnetic Induction Part 2

1. 9011041155 / 9011031155
• Live Webinars (online lectures) with
recordings.
• Online Query Solving
• Online MCQ tests with detailed
solutions
• Online Notes and Solved Exercises
• Career Counseling
1

2. 9011041155 / 9011031155
Electromagnetic Induction
Transformer action.
E1
and E2
E2
E1
N2
N1
d
N1
dt
d
N1
dt
N1
N2
d
dt
d
dt
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
2

3. 9011041155 / 9011031155
E2
E1
I1
I2
N2
N1
% efficiency
Output power
Input power
3
100%

4. 9011041155 / 9011031155
Emf induced in a coil rotating in a uniform
magnetic field.
The magnetic flux passing through the coil is
given by,
φ = nAB cos θ = nAB cos ωt
The induced emf is given by,
e
d
dt
d
n AB cos t
dt
d
cos t
dt
n AB sin t
nAB
If f is the frequency of rotation of the coil, ω = 2πf
∴ e = 2πfnABsin ωt As 2πfnAB are constant
4

5. 9011041155 / 9011031155
∴ e = E0 sin ωt
where E0 = 2πfnAB
θ = ωt 0
π/2
π
3π/2 2π
5π/2 3π
E
E0
0
-E0
E0
0
5
0
0

6. 9011041155 / 9011031155
Applied ac voltage and current to resistor
The figure shows a resistor of value R connected
across an ac source of emf e.
∴ e = E0 sin ωt and current i is given as
i
E0
sin
R
t
I0 sin
t
where I0 = E0 / R is called peak value of current.
Hence, current is in phase of emf. The variation of
current and emf with respect to time is shown
below
6

7. 9011041155 / 9011031155
i.
Peak value :
The
magnitude
of
maximum
value
of
alternating emf or current is called its peak
value. The peak values of emf and currents
are E0 and I0 respectively. The time interval
between two successive positive or negative
peak values is called periodic time T of the
emf or current.
ii. Average (mean) value:
The constant value of current or voltage read
by a dc ammeter or voltmeter over one
complete cycle of the current or emf is called
its average value.
Mathematically,
the
average
alternating emf is given by,
E av
E0 sin
t dt
T
7
0
T
0
value
of

8. 9011041155 / 9011031155
Similarly, Iav = 0. Hence, the dc meters read
zero emf or current.
iii. Root Mean Square (rms) value :
The rms value of alternating current is that
steady current, when flows through a given
resistor, for a given time, produces that same
heat that is produced by the alternating
current, in the same resistor in same time.
Mathematically it is given as,
2
E RMS
e dt
T
1/ 2
Solving it, E RMS
ERMS
E0 / 2
0.707 E0
Similarly, I RMS
I0 /
8
2 or I RMS
0.707 I0

9. 9011041155 / 9011031155
Q.43 In an ideal transformer, the primary and the secondary voltages always have
(a.43) Equal magnitude (b.43) The same phase (c.43) A phase difference of
900
(d.43) A phase difference of 1800
Q.44 An a.c. ammeter measures the
(a.44) Peak value of current
(b.44) Average value of current
(c.44) r.m.s. value of current
(d.44) Mean square current
Q.37 A step down transformer works on 220volts a.c. mains. It is used to light a 100W,
20V bulb. The main current is 0.5A. What is the efficiency of the transformer?
(a.37) 91% (b.37) 80% (c.37) 71% (d.37) 51%
Inductive reactance of a pure inductor
e'
L
di
dt
L
d
I0 sin
dt
t
∴ e’ = - ωL I0 cos ωt
∴ e’ = - E0 cos ωt where E0 = ωL I0 is the peak
value of emf.
For ideal inductor e’ = e
9

10. 9011041155 / 9011031155
e
E0 sin
found to be
2.
t
2
Thus, the value of φ is
emf leads current by
2
.
As E0 = ωL I0,
L
E0
I0
ERMS
I RMS
XL
XL is called inductive reactance. As ω = 2πf,
XL = 2πfL
If L is expressed in henry and frequency is
expressed in hertz, XL is expressed in ohm.
10

11. 9011041155 / 9011031155
Capacitive reactance of a pure capacitor
Application of A. C. to a capacitor:
11

12. 9011041155 / 9011031155
q
C
V
This is equal to the instantaneous value of the
applied e. m. f. (e)
e
q
C
q
Ce
q
C e0 sin t
dq
dt
i
....... 1
d
C e0 sin t
dt
C e0 cos t
C e0 cos t
This is equation of the instantaneous current.
If cos ωt = 1 then i = imax = i0 = peak value of
current = e0 ωC
i
i0cos t
i0sin
Where i0 = e0 ωC
t
2
…….(2)
12

13. 9011041155 / 9011031155
Comparing e = e0 sinωt with equation (2) we
conclude that
i.
e. m. f. or voltage across the capacitor
lags behind the current by phase angle
2
rad or current leads the e. m. f. or voltage
by
2
rad.
ii. Current and e. m. f. both are sinusoidal in
nature of same frequency.
13

14. 9011041155 / 9011031155
We know that i0
1
C
1
C
C
e0
i0
e0
2
i0
2
e0
e rms
i rms
“capacitive reactance” (XC)
XC
1
C
1
2 fC
1
Using equation,
C
14
…….(3)
e rms
i rms

15. 9011041155 / 9011031155
Capacitive reactance is defined as the ratio of
r. m. s. voltage across the capacitor to the r.
m. s. current. S. I. unit of XC is ohm.
Q.23 The coefficient of mutual induction of two coils is 10mH. If the current flowing in
one coil is 4A then the induced e.m.f. in the second coil will be
(a.23) 40mV
(b.23) 20mV
(c.23) Zero (d.23) 10mV
Q.24 The current in a coil decreases from 5A to 0 in 0.1sec. If the average e.m.f.
induced in the coil is 50V, then the self inductance of the coil is
(a.24) 0.25H
(b.24) 0.5H (c.24) 1H
15
(d.24) 2H

16. 9011041155 / 9011031155
16

17. 9011041155 / 9011031155
•
Ask Your Doubts
•
For inquiry and registration, call 9011041155 /
9011031155.
17