In the previous section we saw the basics about mutual induction and transformers. In this section, we will see self induction.
Let us do an activity:
1. Take two insulated copper wires
• Both should have the same length (about 3 m)
• Both should have the same diameter
2. Keep one wire as such
• Wind the other wire in the form of a coil
3. Connect the first wire to a 3 W bulb and a 6 V battery. This is 'circuit 1' shown in fig.11.29(a) below
• Connect the coil to a 3 W bulb and a 6 V battery. This is 'circuit 2' shown in fig.11.29(b) below:
Note: The lengths of copper wires in both the circuits are the same. But in circuit 2, it is in a coiled shape. So it occupies a lesser space.
4. Now we can begin the trials
Trial 1:
(i) Turn on the switch in circuit 1.
• Observe the intensity of the light emitted by the bulb
(ii) Turn on the switch in circuit 2.
• Observe the intensity of the light emitted by the bulb
■ We can observe the following points:
• At the instant when switch is turned on, the light is of a low intensity
• But after that, the light is of the same intensity as in circuit 1
Trial 2:
• Change the type of power source. That is., remove the 6 V batteries and connect to 6 V AC
This is shown in fig.11.30 below:
(i) Turn on the switch in circuit 1
• Observe the intensity of the light emitted by the bulb
(ii) Turn on the switch in circuit 2.
• Observe the intensity of the light emitted by the bulb
■ We can observe the following points:
• In circuit 1, the emitted light has the same intensity as trial 1
• In circuit 2, the emitted light has a lower intensity than in trial 1
Let us write a summary:
(i) In trial 1, the intensity in circuit 2 is decreased only at the instant when the switch was turned on
(ii) In trial 2, the intensity in circuit 2 is decreased during the entire time when switch is kept in on position
Let us analyse and find the reason for the 'decrease in intensity of light':
1. In trial 2, we used AC current. So the direction of current through the coil changes at regular intervals.
2. We know that:
(i) Whenever current flows through a coil, a magnetic field is produced around it
(ii) When the current flowing is AC, the magnetic field changes direction
(iii) Such a changing magnetic field can produce a voltage in another coil which is kept near by. This is the basis of the working of a transformer
3. But such a changing magnetic field can produce a voltage not only in a nearby coil, but also in the same coil itself.
4. This 'newly produced emf in the same coil itself' is in a direction opposite to the applied emf. So it is called back emf
5. So the net emf will decrease. That is the reason for the decreased intensity of the emitted light
• In trial 1, at the instant when switch is turned on, there is a change in flux. So a back emf is created at that instant
♦ When the switch is kept on, there is no change in flux, and hence, no back emf
■ Now, in trial 2, if we place a soft iron core inside the coil, the intensity of light will become very low. The reason can be written as follows:
1. When the soft iron core is present, the 'magnetic flux intensity' is increased
2. So the 'change in magnetic flux intensity' is also increased
3. Thus the 'back emf' will be higher
4. As the 'back emf' is in an opposite direction to the applied emf, the intensity of light will be very low.
• When current flows through a solenoid, a magnetic flux is created in the solenoid
• If the current flowing is AC, the magnetic flux created will be a 'changing magnetic flux'
• This changing magnetic flux will produce an emf in the same solenoid in which the AC is flowing
• This phenomenon is called self induction
• The emf thus created will be in a direction, opposite to the 'direction of applied emf'
Now we will see a practical application of self induction:
1. Inductors are special devices used in some electric circuits. Let us see it's details:
• It is an insulated copper wire wound in a helical shape Some images can be seen here.
2. When AC current passes through it, a back emf will be produced in it
• This will oppose the applied emf
• So the net emf in the circuit will decrease
3. As a result, the current will decrease
• So, whenever we want to reduce the current in an AC circuit, we can use an inductor
4. For example, choke coil is used in fluorescent lamps.
• This is because, the fluorescent lamps do not require high currents for it's working.
• In fact, high currents will cause damage to fluorescent lamps.
5. We can use a resistor to reduce the current. But in resistors, there will be loss of electrical energy in the form of heat.
6. So inductors can be effectively used to reduce current with out loss of energy.
7. It may be noted that, we can use inductors only in AC circuits. This is because, only AC will produce back emf. In DC, there is no change in magnetic flux and hence no back emf
Let us do an activity:
1. Take two insulated copper wires
• Both should have the same length (about 3 m)
• Both should have the same diameter
2. Keep one wire as such
• Wind the other wire in the form of a coil
3. Connect the first wire to a 3 W bulb and a 6 V battery. This is 'circuit 1' shown in fig.11.29(a) below
• Connect the coil to a 3 W bulb and a 6 V battery. This is 'circuit 2' shown in fig.11.29(b) below:
 |
| Fig.11.29 |
4. Now we can begin the trials
Trial 1:
(i) Turn on the switch in circuit 1.
• Observe the intensity of the light emitted by the bulb
(ii) Turn on the switch in circuit 2.
• Observe the intensity of the light emitted by the bulb
■ We can observe the following points:
• At the instant when switch is turned on, the light is of a low intensity
• But after that, the light is of the same intensity as in circuit 1
Trial 2:
• Change the type of power source. That is., remove the 6 V batteries and connect to 6 V AC
This is shown in fig.11.30 below:
 |
| Fig.11.30 |
• Observe the intensity of the light emitted by the bulb
(ii) Turn on the switch in circuit 2.
• Observe the intensity of the light emitted by the bulb
■ We can observe the following points:
• In circuit 1, the emitted light has the same intensity as trial 1
• In circuit 2, the emitted light has a lower intensity than in trial 1
Let us write a summary:
(i) In trial 1, the intensity in circuit 2 is decreased only at the instant when the switch was turned on
(ii) In trial 2, the intensity in circuit 2 is decreased during the entire time when switch is kept in on position
Let us analyse and find the reason for the 'decrease in intensity of light':
1. In trial 2, we used AC current. So the direction of current through the coil changes at regular intervals.
2. We know that:
(i) Whenever current flows through a coil, a magnetic field is produced around it
(ii) When the current flowing is AC, the magnetic field changes direction
(iii) Such a changing magnetic field can produce a voltage in another coil which is kept near by. This is the basis of the working of a transformer
3. But such a changing magnetic field can produce a voltage not only in a nearby coil, but also in the same coil itself.
4. This 'newly produced emf in the same coil itself' is in a direction opposite to the applied emf. So it is called back emf
5. So the net emf will decrease. That is the reason for the decreased intensity of the emitted light
• In trial 1, at the instant when switch is turned on, there is a change in flux. So a back emf is created at that instant
♦ When the switch is kept on, there is no change in flux, and hence, no back emf
■ Now, in trial 2, if we place a soft iron core inside the coil, the intensity of light will become very low. The reason can be written as follows:
1. When the soft iron core is present, the 'magnetic flux intensity' is increased
2. So the 'change in magnetic flux intensity' is also increased
3. Thus the 'back emf' will be higher
4. As the 'back emf' is in an opposite direction to the applied emf, the intensity of light will be very low.
• When current flows through a solenoid, a magnetic flux is created in the solenoid
• If the current flowing is AC, the magnetic flux created will be a 'changing magnetic flux'
• This changing magnetic flux will produce an emf in the same solenoid in which the AC is flowing
• This phenomenon is called self induction
• The emf thus created will be in a direction, opposite to the 'direction of applied emf'
Now we will see a practical application of self induction:
1. Inductors are special devices used in some electric circuits. Let us see it's details:
• It is an insulated copper wire wound in a helical shape Some images can be seen here.
2. When AC current passes through it, a back emf will be produced in it
• This will oppose the applied emf
• So the net emf in the circuit will decrease
3. As a result, the current will decrease
• So, whenever we want to reduce the current in an AC circuit, we can use an inductor
4. For example, choke coil is used in fluorescent lamps.
• This is because, the fluorescent lamps do not require high currents for it's working.
• In fact, high currents will cause damage to fluorescent lamps.
5. We can use a resistor to reduce the current. But in resistors, there will be loss of electrical energy in the form of heat.
6. So inductors can be effectively used to reduce current with out loss of energy.
7. It may be noted that, we can use inductors only in AC circuits. This is because, only AC will produce back emf. In DC, there is no change in magnetic flux and hence no back emf
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