Slides for Electromagnetic Induction

1. 3.3 Analysing Electromagnetic Induction
Force will act on a current-carrying conductor in magnetic field.
In the reverse process current can be produced by moving a
conductor in a magnetic field (Discovered by Michael Faraday -
invention of dynamo/generator)
Electromagnetic Induction-Production of electric current by a
changing magnetic field

2. Induced current is produced only when there is a relative motion
between the conductor/coil and the magnetic field lines.
The direction of induced current is influence by the direction of
relative motion

3. Experiment to observed electromagnetic induction in a straight wire
and a solenoid
Action Observation
Wire-downwards Current flows
Wire-upwards Current flows-opposite direction
Wire-horizontally No current generated
Magnet-upwards Current flows
Magnet & wire move together No current generated
Wire-stationary No current generated

4. Action Observation
N-pole into solenoid Current flows
Magnet moved out of solenoid Current flows-opposite direction
Solenoid into stationary magnet No current generated
Magnet held stationary in magnet No current generated
Magnet & wire move together
In the same direction
No current generated

5. Induced Current and Induced Electromotive Force
When a wire moves and cuts the magnetic field lines, there is a
change of magnetic flux, an e.m.f. is induced across the wire
Magnetic flux is the number of magnetic lines passing through the
surface of a conductor

7. The magnitude of the induced e.m.f. can be determined by Faraday’s
Law
The direction of the induced e.m.f. can be determined by Lenz’s
Law and also Fleming’s right-hand rule or right hand slap rule

8. Laws of Electromagnetic Induction
Faraday’s Law:
The magnitude of the induced e.m.f. is directly proportional to the
rate of change of magnetic flux or the rate of cutting of the magnetic
field lines.
For a solenoid, the magnitude of induced e.m.f. can be increased by
(a) Increasing the relative speed of motion of wire/solenoid
(b) Increasing the number of turns
(c) Using a stronger magnet

10. For a straight wire, the magnitude of induced e.m.f.
can be increased by
(a) Increasing the speed of motion of wire
(b) Increasing the strength of the magnetic field
The magnitude of the induced current depends on
(a) The magnitude of the induced e.m.f.
(b) The resistance of the circuit

11. Lenz’s Law:
The induced current always flows in such a direction so as to
oppose the change (or motion) producing it.

13. Lenz’s Law is a form of the law of conservation of energy
Electrical energy cannot be created without any form of work being
done.
When a magnet moves towards or away from a coil , work must be
done to overcome the opposing force
The work done (mechanical energy) is converted into electrical
energy

14. Application of Electromagnetic Induction
Dynamic Microphone (Moving coil microphone)
Sounds waves vibrates diaphragm which is attracted to a coil.
The motion of the coil relative to the magnet induces a current in
the coil.
The induced current has the same characteristic as the sound
wave received.

15. Bicycle dynamo
It consists of a cylindrical permanent magnet with poles on opposite
sides placed within concave poles of a soft iron core where a solenoid is
wound
It needs no slip rings and commutators and also it doesn’t need carbon
brushes

16. Direct Current Generator (D.C. Dynamo)
Ends of coil are connected to split-ring commutator
Two halves of the split-ring exchange contact with the carbon
brushes every half rotation
Output current flows in one direction through the load resistance, R

19. Alternating Current Generator (A.C. Dynamo)
Ends of coil are connected to two slip rings
Each slip ring is always in contact with the same carbon brush
Output current flows to and fro in the opposite direction
Increasing the speed of rotation of the coil also increases the
frequency of the output voltage.

22. Direct Current (D.C.)
(1) Current that flows in one direction only in a circuit
(2) Magnitude may be (a) constant (b) changes with time
(3) Current-time graph

24. (4) Current cannot flow through a capacitor
(5) No effect on moving coil loudspeaker

25. Alternating Current (A.C.)
(1) Current which flows to and fro in two opposite directions in a
Circuit. It changes its direction periodically
(2) Magnitude may be (a) constant (b) changes with time
(3) Current-time graph
A.C. with constant magnitude A.C. with varying magnitude

27. (4) Current can flow through a capacitor
(5) Vibrating effect on moving coil loudspeaker

28. Comparison between d.c. and a.c.
d.c. a.c.
Currents flows in 1 direction Current flows to and fro in opposite
direction
Produces fixed magnetic poles
when it passes though a solenoid
Produces changing magnetic poles
when it passes though a solenoid
Cannot pass through a capacitor Can pass through a capacitor
Cannot produce vibrating effect
on moving coil loudspeaker
Can produce vibrating effect on
moving coil loudspeaker
Magnitude cannot be step up of
step down by using transformer
Magnitude can be step up of step
down by using transformer
Can be measured by moving-
coil meters, hot-wire meters or
moving iron meters
Can be measure by hot wire and
moving –iron meters only
Difficult to be converted to a.c. Can be easily converted to d.c.
using rectifiers

29. Peak Value and Root-Mean-Square Value
The root-mean-square-current Ir.m.s. of an a.c. is the effective value of
that a.c. current
Definition: Ir.m.s. Is the same as the steady direct current which
passing through the same resistor, produces heat at the same rate as
that of the a.c. current

30. (1) Initially switch is closed on the a.c. circuit and the brightness
of the bulb is measured using light intensity meter . The display on
CRO is observed and the peak voltage Vo is noted.
(2) After that switch is push to the d.c. circuit. Rheostat is adjusted
until the bulb gives the same brightness as before. The display on
the CRO and the value of d.c. is noted.
(3) The above activities is repeated using 2.5 V bulb
Bulb V0 Vd.c. Vd.c..
/ V0
Bulb 1 ( 3.8 V ) 4.0 2.8 0.7
Bulb 2 ( 2.5 V ) 3.0 2.1 0.7

31. Vd.c.. = 0.7 Vo
Therefore Vr.m.s. = 0.7 Vo = Vo / √2
For a.c. source, Voltage Vr.m.s. = Vo / √2
For a.c. source, Current Ir.m.s. = Io /√2
Example:-
The Voltage supply to our house is 240V. What is the
maximum voltage of the power supply.
Answer:-
Using V r.m.s = Vo / √2
VO = √2 x 240 =339V

32. Example:
The graph shows the output current of an alternating current
Generator. Which statement is true about the graph?
A.The peak current is 8A
B. The root mean square current is 2 A
C. The time for one cycle is 0.4 S
D. The frequency of the current generated is 25 Hz
Solution:-
Io =4A, Ir.m.s. Io/ √2 = 2.8 A,
T =0.4 s f=1/T =2.5 Hz
Answer: C

33. Example:
State two (a) similarities (b) differences
Between a d.c. generator and an a.c. generator
Answer:-
(a) (i) Produces an e.m.f. using electromagnetic induction
(ii) The output current flows out through carbon brushes
(b).
D.C. generator A.C. generator
Ends of coil connected to split-
ring commutator
Ends of coil connected to two
slip rings
Output is a direct current Output is a alternating current

34. Example:
The induced currents flow in the directions as shown in the
diagrams below. Determine
(a) The poles at P and Q
(b) The pole at X.
Answer:
(a) P=south Q=north
(b) X=south

35. Example:
A bar magnet is placed between two coils as shown in the following
diagram
When the bar magnet is withdrawn, what are the poles at P and Q ?
In what direction will the induced current flow along wire AB? Give
an explanation.
Answer:
P is south pole and Q is north pole. Because the number of turns in
the solenoid QW is more than the number of turns in the solenoid
VP, the induced e.m.f. in QW is greater than the induced e.m.f. in
VP. The resultant current flows from A to B

36. Meters to measure d.c and a.c.

37. Example of converting a milliammeter into ammeter
A multiplier is connected in parallel to a milliammeter to adapt it as an
ammeter. Consider the milliammeter which has a resistance of 5Ω and a
full-scale deflection current of 10 mA. Calculate the shunt resistance
required for the meter to be adapted to measure up to 5.0A
Solution:
p.d. across milliammeter= p.d. across the shunt
10 x 10-3
(5) = (5-0.01)Rs Rs =0.01 x 5/4.99 =0.01Ω

38. Example of converting a milliammeter into a voltmeter
A multiplier is connected in series to a with a milliammeter to adapt
it as a voltmeter measure up to 5V. If the coil resistance of the
milliammeter is 5 Ω and the full scale deflection current is 10 mA,
determine the resistance of the multiplier
Solution:
When a p.d. of 5 V is applied across the multiplier and the meter,
current of 10 mA(0.01 A) flows
5= 0.01(R+5) R+5 =5/0.01 =500 R =495 Ω