Get summaries, questions, answers, solutions, notes, extras, workbook solutions, PDF and guide of chapter 10 Electro-magnetism: ICSE Class 10 Physics which is part of the syllabus of students studying under the Council for the Indian School Certificate Examinations board. These solutions, however, should only be treated as references and can be modified/changed.
Summary
Hans Oersted discovered in 1820 that an electric current flowing through a wire creates a magnetic field around it. This can be seen when a compass needle, placed near the wire, moves from its usual north-south direction. The direction the needle moves depends on the direction of the electric current and whether the compass is above or below the wire. If the current is stronger, the magnetic effect is also stronger.
The magnetic field around a straight wire forms invisible circles. We can find the direction of this field using the right-hand thumb rule: if your thumb points in the direction of the current, your curled fingers show the direction of the magnetic field. For a circular loop of wire carrying current, the magnetic field lines are circular near the wire and straighter towards the center of the loop. The loop itself acts like a flat magnet with a north pole and a south pole. The clock rule helps determine these poles: if you look at the face of the loop and the current flows anticlockwise, it’s a north pole; if it flows clockwise, it’s a south pole.
A solenoid is a coil of wire shaped like a spring. When current flows through it, it creates a magnetic field that is very much like the field of a bar magnet. Inside the solenoid, the magnetic field is strong and uniform. An electromagnet is made by winding a coil of wire around a piece of soft iron. It acts as a magnet only when current is flowing. Its strength can be increased by using more turns of wire, more current, or a soft iron core. Electromagnets can be I-shaped or U-shaped (horseshoe) and are used in devices like electric bells. An electric bell works by an electromagnet pulling a hammer to strike a gong; this action also breaks the circuit, releasing the hammer, and the process repeats, causing continuous ringing.
When a wire carrying current is placed in a magnetic field (not parallel to it), it experiences a force. The direction of this force can be found using Fleming’s left-hand rule: if your first finger points in the direction of the magnetic field and your second finger points in the direction of the current, your thumb will point in the direction of the force. This principle is used in DC electric motors, which change electrical energy into mechanical energy. A motor has a rotating coil (armature) in a magnetic field. A split-ring commutator reverses the current in the coil at the right moments to keep it spinning.
Michael Faraday discovered electromagnetic induction: if the magnetic field lines (magnetic flux) passing through a coil change, an electromotive force (e.m.f.), or voltage, is created across the coil. If the coil is part of a complete circuit, a current will flow. This induced current lasts as long as the magnetic flux is changing. The direction of the induced current is given by Lenz’s law, which states that the current flows in a direction that opposes the change causing it. Fleming’s right-hand rule also gives this direction: if your thumb points in the direction of motion of the wire, your first finger in the direction of the field, your second finger shows the induced current’s direction. This is how AC generators work, changing mechanical energy into electrical energy. An AC generator has a coil rotating in a magnetic field, producing an alternating current (AC) that changes direction periodically. Household electricity is AC, typically at 50 Hertz. Direct current (DC) flows in one direction only.
A transformer is a device that changes the voltage of an alternating current. It has two coils, a primary and a secondary, wound on a soft iron core. AC in the primary coil creates a changing magnetic field, which induces an AC voltage in the secondary coil. A step-up transformer has more turns in the secondary coil than in the primary, increasing the voltage. A step-down transformer has fewer turns in the secondary, decreasing the voltage. Transformers are used to adjust voltages for different appliances and for efficient long-distance power transmission.
Workbook Solutions (Concise/Selina)
Exercise (A)
MCQ
1. In the figure given below, when the key is pressed, a current passes in the wire in the direction from A to B (i.e. from south to north). The north pole of the compass needle will deflect towards :
(a) East
(b) North
(c) South
(d) West
Answer: (d) West
2. The strength of a magnetic field depends on :
(a) magnitude of current
(b) direction of current
(c) both (a) and (b)
(d) none of the above
Answer: (c) both (a) and (b)
3. In the figure given below, current is passed in a downward direction through a cardboard. The magnetic field lines would be :
(a) straight lines
(b) concentric circles
(c) elliptical lines
(d) squares
Answer: (b) concentric circles
4. According to the right hand thumb rule, the thumb points in the direction of ………. and fingers encircle the wire in direction of …………..
(a) magnetic field lines, current
(b) electric field lines, current
(c) current, magnetic field lines
(d) none of the above
Answer: (c) current, magnetic field lines
5. In the figure given below, the direction of current in the loop is ……… So, it behaves as a ………….
(a) anticlockwise, south pole
(b) anticlockwise, north pole
(c) clockwise, north pole
(d) clockwise, south pole
Answer: (b) anticlockwise, north pole
6. Which of the following statements is not true ?
(a) The strength of magnetic field due to a bar magnet can be changed.
(b) A solenoid behaves like a bar magnet i.e. it attracts iron filings.
(c) The magnetic field increases if a soft iron core is placed along the axis of solenoid.
(d) The magnetic field lines inside the solenoid are nearly straight and parallel to the axis of solenoid.
Answer: (a) The strength of magnetic field due to a bar magnet can be changed.
7. Which of the following statements is not true for an electromagnet ?
(a) It is made of soft iron.
(b) Its magnetic field strength can be changed.
(c) The polarity of an electromagnet cannot be changed.
(d) It can easily be demagnetised.
Answer: (c) The polarity of an electromagnet cannot be changed.
8. The magnetic field lines inside a solenoid are :
(a) parallel and circular
(b) straight and parallel to the axis of solenoid
(c) straight and perpendicular to the axis of solenoid
(d) circular
Answer: (b) straight and parallel to the axis of solenoid
9. On reversing the direction of current in a wire, the magnetic field produced by it :
(a) gets reversed in direction
(b) increases in strength
(c) decreases in strength
(d) remains unchanged in strength and direction
Answer: (a) gets reversed in direction
10. A strong permanent magnet is made of :
(a) soft iron
(b) steel
(c) nickel
(d) copper
Answer: (b) steel
11. Horse shoe magnets are used in:
(a) d.c. motor
(b) a.c. generator
(c) electric bell
(d) all of the above
Answer: (d) all of the above
Very Short Answer Type Questions
1. How is the magnetic field due to a straight current carrying wire affected if current in the wire is (a) decreased, (b) reversed?
Answer: If the current in the wire is (a) decreased, the magnetic field decreases. If the current in the wire is (b) reversed, the magnetic field reverses.
2. A student draws three magnetic field lines 1, 2 and 3 of a bar magnet with the help of a compass needle. Are all the lines of force correctly drawn? Give reason for your answer.
Answer: No, the field line labeled 3 is incorrect, while 1 and 2 are correct. This is because magnetic field lines must emerge from the North pole and enter the South pole outside the magnet. However, line 3 is shown going from South to North outside the magnet, which is incorrect.
3. The diagram given below shows the magnetic field pattern around a bar magnet. Identify the poles A and B.
Answer: Since magnetic field lines travel from the North pole to the South pole outside a magnet, and here the lines are emerging from pole B and entering pole A, it follows that:
A is the South Pole
B is the North Pole
4. The diagrams given below show some magnetic field patterns. Answer the following questions:
(a) Which pattern shows uniform magnetic field lines?
(b) Which pattern is for a straight current carrying wire?
(c) Which pattern is depicted by current in a loop?
Answer: (a) Pattern D shows uniform magnetic field lines, as the magnetic field lines inside a solenoid are nearly straight and parallel to the axis of the solenoid, meaning the magnetic field is uniform inside the solenoid.
(b) Pattern A is for a straight current carrying wire, where the magnetic field lines form concentric circles around the wire.
(c) Pattern B is depicted by current in a loop, showing the characteristic magnetic field lines for a circular coil.
5. A straight wire lying in a horizontal plane carries a current from north to south. (a) What will be the direction of magnetic field at a point just underneath it? (b) Name the law used to arrive at the answer in part (a).
Answer:
(a) The direction of the magnetic field at a point just underneath the wire will be towards east.
(b) The law used to arrive at the answer is the right hand thumb rule.
6. A wire, bent into a circle, carries a current in an anticlockwise direction. What polarity does this face of the coil exhibit?
Answer: This face of the coil exhibits north polarity.
7. What is the direction of magnetic field at the centre of a coil carrying current in (i) the clockwise, (ii) the anticlockwise, direction?
Answer:
(i) In the clockwise direction, the magnetic field at the centre of the coil is along the axis of the coil inwards.
(ii) In the anticlockwise direction, the magnetic field at the centre of the coil is along the axis of the coil outwards.
8. How is the magnetic field due to a solenoid carrying current affected if a soft iron bar is introduced inside the solenoid?
Answer: If a soft iron bar is introduced inside the solenoid, the strength of the magnetic field increases.
9. Name one device that uses an electromagnet.
Answer: An electric bell is one device that uses an electromagnet.
10. Complete the following sentences:
(a) When current flows in a wire, it creates ……….
(b) On reversing the direction of current in a wire, the magnetic field produced by it gets ……….
(c) A current carrying solenoid behaves like a………
(d) A current carrying solenoid when freely suspended, it always rests in ………. direction.
Answer:
(a) When current flows in a wire, it creates a magnetic field around it.
(b) On reversing the direction of current in a wire, the magnetic field produced by it gets reversed.
(c) A current carrying solenoid behaves like a bar magnet.
(d) A current carrying solenoid when freely suspended, it always rests in the north-south direction.
Short Answer Type Questions
1. State a law which determines the direction of magnetic field around a current carrying wire.
Answer: The right hand thumb rule determines the direction of the magnetic field around a current carrying wire. The rule states: If we hold the current carrying conductor in our right hand such that the thumb points in the direction of flow of current, then the fingers encircle the wire in the direction of the magnetic field lines.
2. What will happen to a compass needle when the compass is placed below a wire with needle parallel to it and a current is made to flow through the wire? Give a reason to justify your answer.
Answer: When a current is made to flow through the wire, the north pole of the compass needle will deflect. For instance, if the wire is lying in the north-south direction and current passes from south to north, and the compass is placed below it, the north pole of the needle deflects towards the west.
The reason for this deflection is that when an electric current is passed through the conducting wire, a magnetic field is produced around it. The magnetic needle of the compass experiences a torque in this magnetic field, so it deflects to align itself in the direction of the resultant magnetic field at that point.
3. Name and state the rule by which the polarity at the ends of a current carrying solenoid is determined.
Answer: The polarity at the ends of a current carrying solenoid is determined by the Clock rule.
The Clock rule states: Looking at the face of the loop (or an end of the solenoid), if the current in the wire around that face is in an anticlockwise direction, the face has the north polarity. If the current at that face is in a clockwise direction, the face has the south polarity. For example, in a solenoid, the end at which the direction of current is anticlockwise behaves as a north pole (N), while the end at which the direction of current is clockwise behaves as a south pole (S).
4. The diagram shows a small magnet placed near a solenoid AB with its north pole N near the end A. Current is switched on in the solenoid by pressing the key K.
(a) State the polarity at the ends A and B.
(b) Will the magnet be attracted or repelled? Give a reason for your answer.
Answer:
(a) At end A of the solenoid, the polarity will be a north pole, and at end B, the polarity will be a south pole.
(b) The magnet will be repelled. This is because the end A of the solenoid becomes the north pole as current at this face is anti-clockwise, and it repels the north pole of the magnet.
5. State two ways by which the magnetic field due to a current carrying solenoid can be made stronger.
Answer: Two ways by which the magnetic field due to a current carrying solenoid can be made stronger are:
(i) By increasing the current in the solenoid, the magnetic field lines become denser, meaning a stronger magnetic field is obtained.
(ii) By increasing the number of turns in the solenoid of a given length, the magnetic field is increased.
(iii) By placing a soft iron rod (core) along the axis of the solenoid, the magnetic field is also increased.
6. A magnet kept at the centre of two coils A and B is moved to and fro as shown in the diagram given below. The two galvanometers show deflection. State with a reason whether: x > y or x < y.
Answer: In this case, x < y.
The reason is that the magnitude of the induced e.m.f., and therefore the current and the galvanometer deflection, is increased by increasing the number of turns in the coil. Since coil B has more turns than coil A (as shown in the diagram), the deflection ‘y’ in the galvanometer connected to coil B will be greater than the deflection ‘x’ in the galvanometer connected to coil A, assuming the rate of change of magnetic flux and other factors are the same for both coils.
7. Why does a current carrying freely suspended solenoid rest along a particular direction? State the direction in which it rests.
Answer: A current carrying freely suspended solenoid rests along a particular direction because a current carrying solenoid behaves like a bar magnet. It rests in the geographic north-south direction.
8. What effect will there be on a magnetic compass when it is brought near a current carrying solenoid?
Answer: When a magnetic compass is brought near a current carrying solenoid, the needle of the compass will rest in the direction of the magnetic field due to the solenoid at that point.
9. The diagram shows a coil wound around a soft iron bar XY. (a) State the polarity at the ends X and Y as the switch is pressed. (b) Suggest one way of increasing the strength of electromagnet so formed.
Answer: (a) As the switch is pressed, assuming the winding direction shown in typical diagrams like Fig 10.7 (I-shaped electromagnet) where current flows to create specific poles, if the current at end X makes it behave like an anticlockwise current face, X will be a north pole, and Y will be a south pole. (The exact polarity depends on the direction of winding and current flow which should be inferred from the diagram conventions). Based on the answer key, X is a north pole and Y is a south pole.
(b) One way of increasing the strength of the electromagnet so formed is by reducing the resistance of the circuit by means of a rheostat to increase the current in the coil.
10. What is an electromagnet? Name two factors on which the strength of magnetic field of an electromagnet depends and state how does it depend on the factors stated by you.
Answer: An electromagnet is a temporary strong magnet made by passing current in a coil wound around a piece of soft iron. It is an artificial magnet.
Two factors on which the strength of the magnetic field of an electromagnet depends are:
(i) The number of turns of winding in the solenoid: The magnetic field strength increases by increasing the number of turns of winding.
(ii) The current through the solenoid: The magnetic field strength increases by increasing the current through the solenoid.
11. State two ways by which the strength of an electromagnet can be increased.
Answer: The strength of an electromagnet can be increased by:
(i) Increasing the number of turns of winding in the solenoid.
(ii) Increasing the current through the solenoid.
(iii) Inserting a soft iron core within the coil, as soft iron has high magnetic permeability.
12. State two advantages of an electromagnet over a permanent magnet.
Answer: Two advantages of an electromagnet over a permanent magnet are:
(i) An electromagnet can produce a strong magnetic field, often much stronger than a permanent magnet of similar size.
(ii) The strength of the magnetic field of an electromagnet can easily be changed by changing the current or the number of turns in its solenoid.
(iii) The polarity of the electromagnet or the direction of the field produced by it can be reversed by reversing the direction of current in its solenoid.
13. State two differences between an electromagnet and a permanent magnet.
Answer: Two differences between an electromagnet and a permanent magnet are:
(i) An electromagnet is made of soft iron, while a permanent magnet is typically made of steel.
(ii) An electromagnet produces a temporary magnetic field (only so long as current flows in its coil), whereas a permanent magnet produces a permanent magnetic field.
(iii) The magnetic field strength of an electromagnet can be changed, while the strength of a permanent magnet cannot be easily changed.
(iv) The polarity of an electromagnet can be reversed, but the polarity of a permanent magnet cannot be reversed.
(v) An electromagnet can be easily demagnetised by switching off the current, while a permanent magnet cannot be easily demagnetised.
14. Why is soft iron used as the core of the electromagnet in an electric bell?
Answer: Soft iron is used as the core of the electromagnet in an electric bell for two main reasons:
(i) Soft iron has high magnetic permeability, which means it increases the strength of the magnetic field of the solenoid significantly when current passes through the coil.
(ii) Soft iron has low retentivity, meaning it gets demagnetised as soon as the current is switched off. This temporary magnetism is essential for the make-and-break mechanism of the electric bell, allowing the hammer to repeatedly strike the gong.
15. How is the working of an electric bell affected, if alternating current be used instead of direct current?
Answer: If an a.c. source is used in place of the battery for an electric bell, the core of the electromagnet will still get magnetised, but the polarity at its ends will change with the frequency of the a.c. However, since the attraction of the armature does not depend on the polarity of the electromagnet, the bell will still ring on pressing the switch K.
16. Name the material used for making the armature of an electric bell. Give a reason for your answer.
Answer: The material used for making the armature of an electric bell is soft iron.
The reason is that soft iron is a ferromagnetic material that is easily and strongly attracted by the electromagnet when it is magnetised. This attraction pulls the armature towards the electromagnet, causing the hammer to strike the gong.
Long Answer Type Questions
1. By using a compass needle describe how can you demonstrate that there is a magnetic field around a current carrying conductor.
Answer:
- When the key is open (i.e., no current flows through the wire), the compass needle aligns itself in the north–south direction along the Earth’s magnetic field, with the north pole of the needle pointing towards the geographical north. In this state, the needle is parallel to the wire, as shown in Fig. (a).
- When the key is pressed, allowing current to flow through the wire from A to B (i.e., from south to north), the north pole (N) of the compass needle deflects towards the west, as shown in Fig. (b).
- If the direction of current is reversed (by switching the battery terminals), making the current flow from B to A, the north pole (N) of the needle deflects towards the east, as shown in Fig. (c).
- When the compass needle is placed above the wire, the north pole (N) deflects towards the east if the current flows from A to B [Fig. (d)], and towards the west if the current flows from B to A [Fig. (e)].
Conclusion: From this experiment, we can conclude that an electric current passing through a conducting wire produces a magnetic field around it.
2. Draw a diagram showing the direction of three magnetic field lines due to a straight wire carrying a current. Also show the direction of current in the wire.
Answer: The current is flowing upwards, therefore the magnetic field lines will be anticlockwise when viewed from above.
3. Draw a labelled diagram showing the three magnetic field lines of a loop carrying current. Mark the direction of current and the direction of magnetic field by arrows in your diagram.
Answer:
4. Draw a diagram to represent the magnetic field lines along the axis of a current carrying solenoid. Mark arrows to show the direction of current in the solenoid and the direction of magnetic field lines.
Answer:
5. The diagram shows a spiral coil wound on a hollow cardboard tube AB. A magnetic compass is placed close to it. Current is switched on by closing the key.
(a) What will be the polarity at the ends A and B?
(b) How will the compass needle be affected? Give reason.
Answer: (a) Assuming the current direction from the battery connection in the figure, if the current at end A of the coil is such that it appears anticlockwise when viewed from that end, then end A will be a north pole, and end B will be a south pole. Based on the answer key, end A is a north pole and end B is a south pole.
(b) The north pole of the compass needle will deflect towards the west. The reason is that end A of the coil behaves like a north pole, which repels the north pole of the compass needle, causing it to deflect towards the west.
6. You are required to make an electromagnet from a soft iron bar by using a cell, an insulated coil of copper wire and a switch. (a) Draw a circuit diagram to represent the process. (b) Label the poles of the electromagnet.
Answer:
7. (a) What name is given to a cylindrical coil of diameter less than its length? (b) If a piece of soft iron is placed inside the coil mentioned in part (a) and current is passed in the coil from a battery, what name is then given to the device so obtained? (c) Give one use of the device mentioned in part (b).
Answer: (a) A cylindrical coil whose diameter is less in comparison to its length is called a solenoid.
(b) If a piece of soft iron is placed inside such a coil (solenoid) and current is passed through it from a battery, the device so obtained is called an electromagnet.
(c) One use of an electromagnet is in an electric relay, or an electric bell, or for lifting heavy iron scrap.
8. Show with the aid of a diagram how a wire is wound on a U-shaped piece of soft iron in order to make it an electromagnet. Complete the circuit diagram and label the poles of the electromagnet.
Answer:
9. The figure shows the current flowing in the coil of wire wound around the soft iron horse shoe core.
(a) State the polarities developed at the ends A and B.
(b) How will the polarity at the ends A and B change on reversing the direction of current?
(c) Suggest one way to increase the strength of the magnetic field produced.
Answer: (a) Based on the direction of current shown flowing into the coil in the figure, and applying the clock rule to the winding, if the current at end A results in a clockwise view, end A will be a south pole. If the current at end B results in an anticlockwise view, end B will be a north pole. (The answer key states at A—south and at B—north).
(b) On reversing the direction of current, the polarities at the ends A and B will also get reversed. So, A will become a north pole and B will become a south pole.
(c) One way to increase the strength of the magnetic field produced is by increasing the current flowing through the coil.
10. The incomplete diagram of an electric bell is given. Draw winding of coil on the core and complete the electric circuit in the diagram.
Answer:
Exercise (B)
MCQ
1. A charge moving in a magnetic field in a direction other than the direction of the magnetic field experiences a force called :
(a) Lorentz force
(b) Nuclear force
(c) Electric force
(d) Gravitational force
Answer: (a) Lorentz force
2. In fleming’s left hand rule, the forefinger indicates the direction of:
(a) force
(b) current
(c) motion
(d) magnetic field
Answer: (d) magnetic field
3. In an electric motor, the energy transformation is from:
(a) electrical to mechanical
(b) mechanical to electrical
(c) chemical to light
(d) electrical to chemical
Answer: (a) electrical to mechanical
4. The magnitude of force acting on a current carrying wire placed in a magnetic field in a direction perpendicular to its length depends on :
(a) The current flowing in the wire
(b) The strength of the magnetic field
(c) The length of the wire
(d) all of the above
Answer: (d) all of the above
5. The speed of rotation of coil can be increased by :
(a) decreasing the strength of the magnetic field
(b) decreasing the area of the coil
(c) increasing the area of the coil
(d) decreasing the number of turns in the coil
Answer: (c) increasing the area of the coil
6. The function of a split ring commutator is to :
(a) increase the magnetic field
(b) increase the current
(c) reverse the direction of the current
(d) none of the above
Answer: (c) reverse the direction of the current
Very Short Answer Type Questions
1. How will the direction of force be changed, if the current is reversed in the conductor placed in a magnetic field?
Answer: If the current is reversed in the conductor placed in a magnetic field, the direction of force is reversed.
2. State the unit of magnetic field in terms of the force experienced by a current carrying conductor placed in a magnetic field.
Answer: The S.I. unit of magnetic field in terms of the force experienced by a current carrying conductor is N A⁻¹ m⁻¹, which is newton per ampere per metre.
3. What energy conversion does take place during the working of a d.c. motor?
Answer: During the working of a d.c. motor, the electrical energy is converted into mechanical energy.
4. A current flows through a wire in a south to north direction and the magnetic field is directed downwards. In what direction will the force act on the wire?
Answer: The force on the wire will act towards the west.
5. A proton moving towards east enters in a uniform magnetic field directed downwards. Find the direction of the force acting on it.
Answer: The direction of the force acting on the proton will be towards the north.
Short Answer Type Questions
1. Name three factors on which the magnitude of force on a current carrying conductor placed in a magnetic field depends and state how does the force depend on the factors stated by you.
Answer: The magnitude of force on a current carrying conductor placed in a magnetic field depends on the following three factors:
(i) The force F is directly proportional to the strength of magnetic field B (F ∝ B).
(ii) The force F is directly proportional to the current I in the conductor (F ∝ I).
(iii) The force F is directly proportional to the length l of conductor (F ∝ l).
2. State condition in each case for the magnitude of force on a current carrying conductor placed in a magnetic field to be (a) zero, and (b) maximum.
Answer: (a) The magnitude of force on a current carrying conductor placed in a magnetic field is zero when the current in the conductor is in the direction of the magnetic field.
(b) The magnitude of force on a current carrying conductor placed in a magnetic field is maximum when the current in the conductor is normal to the magnetic field.
3. Name and state the law which is used to determine the direction of force on a current carrying conductor placed in a magnetic field.
Answer: The law used to determine the direction of force on a current carrying conductor placed in a magnetic field is Fleming’s left hand rule.
Fleming’s left hand rule states: Stretch the forefinger, central finger and the thumb of your left hand mutually perpendicular to each other. If the forefinger indicates the direction of magnetic field and the central finger indicates the direction of current, then the thumb will indicate the direction of motion of conductor (i.e., force on conductor).
4. State Fleming’s left hand rule.
Answer: Fleming’s left hand rule states: Stretch the forefinger, central finger and the thumb of your left hand mutually perpendicular to each other. If the forefinger indicates the direction of magnetic field and the central finger indicates the direction of current, then the thumb will indicate the direction of motion of conductor (i.e., force on conductor).
5. What is an electric motor ? State its principle.
Answer: An electric motor is a device which converts the electrical energy into the mechanical energy.
Its principle is that on passing electric current through a conductor placed normally in a magnetic field, a force acts on the conductor as a result of which the conductor begins to move and thus mechanical energy (or work) is obtained. The direction of force on the conductor is obtained with the help of Fleming’s left hand rule.
6. State two ways by which the speed of rotation of an electric motor can be increased.
Answer: Two ways by which the speed of rotation of an electric motor can be increased are:
(1) by increasing the strength of current in the coil, and
(2) by increasing the number of turns in the coil.
Other ways include increasing the area of the coil, and increasing the strength of the magnetic field.
7. Name two appliances in which an electric motor is used.
Answer: Two appliances in which an electric motor is used are a fan and a washing machine. Other examples include juicers, mixers, and grinders.
8. A copper wire is held between the poles of a magnet. There is a switch provided to reverse the current in the wire. Also, the polarity can be changed. In how many directions can the force act on the wire ?
Answer: .
According to Fleming’s Left-Hand Rule, the direction of the force on a current-carrying conductor placed in a magnetic field is determined as follows:
- The forefinger points in the direction of the magnetic field (from North to South).
- The middle finger indicates the direction of the current.
- The thumb shows the direction of the force or the motion of the conductor.
In this setup, the current in the wire can flow in two possible directions — either forward or in the reverse direction (controlled by the switch). Similarly, the magnetic field can also be oriented in two directions, either from North to South or from South to North, depending on the polarity of the magnets.
Therefore, since both the current and the magnetic field can each have two possible directions, the resulting force on the wire can act in:
2 (current directions) × 2 (field directions) = 4 different directions.
Thus, the force on the wire can act in four possible directions.
Long Answer Type Questions
1. A flat coil ABCD is freely suspended between the poles of a U-shaped permanent magnet with the plane of coil parallel to the magnetic field.
(a) What happens when a current is passed in the coil?
(b) When will the coil come to rest?
(c) When will the couple acting on the coil be (i) maximum, and (ii) minimum ?
(d) Name an instrument which makes use of the principle stated above.
Answer: (a) When a current is passed in the coil, the coil experiences a torque due to which it rotates.
(b) The coil will come to rest when its plane becomes normal to the magnetic field.
(c) (i) The couple acting on the coil will be maximum when the plane of coil is parallel to the magnetic field.
(ii) The couple acting on the coil will be minimum (zero) when the plane of coil is normal to the magnetic field.
(d) An instrument which makes use of this principle is a d.c. motor.
2. A coil ABCD mounted on an axle is placed between the poles N and S of a permanent magnet as shown in the figure.
(a) In which direction will the coil begin to rotate when current is passed through the coil in direction ABCD by connecting a battery between the ends A and D of the coil ?
(b) Why is a commutator necessary for the continuous rotation of coil?
(c) Complete the diagram with commutator, etc. for the flow of current in the coil.
Answer: (a) When current is passed through the coil in the direction ABCD, the coil will begin to rotate in an anticlockwise direction.
(b) A commutator is necessary for the continuous rotation of the coil because after half rotation, the arms AB and CD get interchanged, so the direction of torque on the coil reverses. To keep the coil rotating in the same direction, a commutator is needed to reverse the direction of current in the coil after each half rotation of coil.
(c) To complete the diagram for the flow of current in the coil, the ends A and D of the coil should be connected to split ring commutators, which press against carbon brushes connected to a d.c. source.
3. Draw a labelled diagram of a DC motor showing its main parts.
Answer: The main parts of a DC motor are the armature coil ABCD mounted on an axle, the split ring commutator (split parts S₁ and S₂ of a ring), a pair of carbon (or copper) brushes B₁ and B₂, a horse-shoe electromagnet NS, and a d.c. source (i.e., battery).
Exercise (C)
MCQ
1. Electromagnetic induction is the phenomenon in which emf is induced in the coil if there is a change in the …………………. linked with the coil.
(a) force
(b) electric flux
(c) magnetic flux
(d) all of the above
Answer: (c) magnetic flux
2. Electromagnetic induction is based on :
(a) Faraday’s law
(b) Lenz’s law
(c) Right hand thumb rule
(d) Fleming’s left hand rule
Answer: (a) Faraday’s law
3. In electromagnetic induction, energy transformation is from:
(a) Electrical to mechanical
(b) Mechanical to electrical
(c) Electrical to chemical
(d) None of the above
Answer: (b) Mechanical to electrical
4. The magnitude of induced emf depends on the:
(a) number of turns in the coil
(b) rate of change of magnetic flux linked with each turn
(c) current
(d) Both (a) and (b)
Answer: (d) Both (a) and (b)
5. Lenz’s law is based on:
(a) conservation of momentum
(b) conservation of energy
(c) conservation of force
(d) none of the above
Answer: (b) conservation of energy
6. In Fleming’s right hand rule, the central finger indicates the direction of:
(a) the magnetic field
(b) the motion of conductor
(c) the induced current
(d) all of the above
Answer: (c) the induced current
7. In an a.c. generator the coil makes n rotations per second. The magnitude of induced emf at any instant t is given by:
(a) e = i₀ sin 2πnt
(b) e = e₀ sin 2πnt
(c) e = e₀ sin 2πn/t
(d) e = e₀ cos 2πnt
Answer: (b) e = e₀ sin 2πnt
8. In our houses, the frequency of electric supply is :
(a) 100 Hz
(b) 70 Hz
(c) 50 Hz
(d) 40 Hz
Answer: (c) 50 Hz
9. A current of constant magnitude and unique direction is called …………………. while a current changing its magnitude and direction periodically is called ………………….
(a) a.c., d.c.
(b) d.c., a.c.
(c) a.c., a.c.
(d) d.c., d.c.
Answer: (b) d.c., a.c.
10. The direction of induced current is obtained by :
(a) Fleming’s left hand rule
(b) Clock rule
(c) Right hand thumb rule
(d) Fleming’s right hand rule
Answer: (d) Fleming’s right hand rule
11. In a step up transformer:
(a) Ns = Np
(b) Ns < Np
(c) Ns > Np
(d) nothing can be said
Answer: (c) Ns > Np
12. A step up transformer ………….. the current whereas a step down transformer …………………. the current.
(a) increases, increases
(b) decreases, decreases
(c) decreases, increases
(d) increases, decreases
Answer: (c) decreases, increases
13. Out of the following, which statement is incorrect?
(a) The lamination of core prevents loss of energy due to eddy currents.
(b) The core is made of soft iron.
(c) A transformer can be used with d.c.
(d) A closed core is used so as to provide a closed path for magnetic field lines.
Answer: (c) A transformer can be used with d.c.
14. Assertion (A): A current carrying solenoid when suspended freely sets itself in the north-south direction just like a bar magnet.
Reason (R): The end of the solenoid where the direction of current is anti-clockwise behaves as a north pole and the end where the direction of the current is clockwise behaves as a south pole.
(a) both A and R are true and R is the correct explanation of A
(b) both A and R are True and R is not the correct explanation of A
(c) assertion is false but reason is true
(d) assertion is true but reason is false
Answer: (b) both A and R are True and R is not the correct explanation of A
15. Assertion (A): The strength of the magnetic field produced due to a current carrying circular coil increases with an increase in the strength of the current.
Reason (R): Magnetic field strength is inversely proportional to the current flowing in a coil.
(a) both A and R are true and R is the correct explanation of A
(b) both A and R are True and R is not the correct explanation of A
(c) assertion is false but reason is true
(d) assertion is true but reason is false
Answer: (d) assertion is true but reason is false
16. Assertion (A): A transformer can increase the voltage of an alternating e.m.f.
Reason (R): This is achieved by keeping the number of turns in the secondary coil less than the number of turns in the primary coil.
(a) both A and R are true and R is the correct explanation of A
(b) both A and R are True and R is not the correct explanation of A
(c) assertion is false but reason is true
(d) assertion is true but reason is false
Answer: (d) assertion is true but reason is false
17. Assertion (A) : A transformer cannot be used with a direct current.
Reason (R) : The lamination of core prevents the loss of energy due to induced currents in the core.
(a) both A and R are true and R is the correct explanation of A
(b) both A and R are True and R is not the correct explanation of A
(c) assertion is false but reason is true
(d) assertion is true but reason is false
Answer: (b) both A and R are True and R is not the correct explanation of A
18. Assertion (A): The magnitude of induced e.m.f. becomes maximum when the magnetic flux linked with the coil reduces to zero from its maximum value.
Reason (R): This happens when the plane of the coil lies in the direction of the magnetic field.
(a) both A and R are true and R is the correct explanation of A
(b) both A and R are True and R is not the correct explanation of A
(c) assertion is false but reason is true
(d) assertion is true but reason is false
Answer: (a) both A and R are true and R is the correct explanation of A
Very Short Answer Type Questions
1. (a) What kind of energy change takes place when a magnet is moved towards a coil having a galvanometer between its ends? (b) Name the phenomenon.
Answer:
(a) When a magnet is moved towards a coil having a galvanometer between its ends, the mechanical energy spent in moving the magnet is transformed into electrical energy in the form of current in the coil.
(b) The phenomenon is called electromagnetic induction.
2. Complete the following sentence: The current is induced in a closed circuit only if there is ……………..
Answer: The current is induced in a closed circuit only if there is a change in the number of magnetic field lines linked with the circuit.
3. In which of the following cases e.m.f. is induced?
(i) A current is started in a wire held near a loop of wire.
(ii) The current is switched off in a wire held near a loop of wire.
(iii) A magnet is moved through a loop of wire.
(iv) A loop of wire is held near a magnet.
Answer: E.m.f. is induced in cases (i), (ii), and (iii).
4. A conductor is moved in a varying magnetic field. Name the law which determines the direction of current induced in the conductor.
Answer: Fleming’s right hand rule determines the direction of current induced in the conductor.
5. What determines the frequency of a.c. produced in a generator?
Answer: The number of rotations of the coil in one second (or the speed of rotation of the coil) determines the frequency of a.c. produced in a generator.
6. Complete the sentence: An a.c. generator changes the …………….. energy to …………….. energy.
Answer: An a.c. generator changes the mechanical energy to electrical energy.
7. What energy conversion does take place in a generator when it is in use?
Answer: In a generator, when it is in use, the mechanical energy changes into electrical energy.
8. State one advantage of using an a.c. over the d.c.
Answer: One advantage of using an a.c. over d.c. is that it is cheaper and easy to generate a.c. than d.c. (Other advantages include: The efficiency of an a.c. generator is higher than that of a d.c. generator; It is easy to change a.c. into d.c.; The magnitude of a.c. voltage can easily be increased or decreased by the use of step up or step down transformer; a.c. can be transmitted over a long distance without much loss in energy in the line wires.)
9. How are the e.m.f. in the primary and secondary coils of a transformer related with the number of turns in these coils?
Answer: The e.m.f. in the primary (Eₚ) and secondary (Eₛ) coils of a transformer are related to the number of turns in these coils (Nₚ and Nₛ respectively) by the relation: Eₛ / Eₚ = Nₛ / Nₚ.
10. Complete the following sentences:
(i) In a step up transformer, the number of turns in the primary are …………….. than the number of turns in the secondary.
(ii) The transformer is used in …………….. current circuits.
(iii) In a transformer, the frequency of a.c. voltage …………….. (increase/decreases/remains same).
Answer:
(i) In a step up transformer, the number of turns in the primary are less than the number of turns in the secondary.
(ii) The transformer is used in alternating current circuits.
(iii) In a transformer, the frequency of a.c. voltage remains same.
11. Name the material of the core in (a) an electric bell, (b) electromagnet, (c) a d.c. motor, (d) an a.c. generator, and (e) a transformer.
Answer: The material of the core in each case is soft iron.
(a) an electric bell: soft iron core
(b) electromagnet: soft iron core
(c) a d.c. motor: soft iron core
(d) an a.c. generator: soft iron core
(e) a transformer: soft iron (laminated sheets)
Short Answer Type Questions
1. State Faraday’s laws of electromagnetic induction.
Answer: Faraday’s laws of electromagnetic induction are:
(1) Whenever there is a change in magnetic flux linked with a coil, an e.m.f. is induced. The e.m.f. induced lasts so long the magnetic flux linked with the coil changes.
(2) The magnitude of e.m.f. induced is directly proportional to the rate of change of magnetic flux linked with the coil. If magnetic flux changes at a constant rate, a steady e.m.f. is produced.
2. State two factors on which the magnitude of induced e.m.f. in a coil depend.
Answer: The magnitude of induced e.m.f. in a coil depends on:
(i) the rate of change in magnetic flux linked with each turn of the coil, and
(ii) the number of turns in the coil.
3. (a) How would you demonstrate that a momentary current can be obtained by the suitable use of a magnet, a coil of wire and a galvanometer? (b) What is the source of energy associated with the current obtained in part (a)?
Answer: (a) To demonstrate that a momentary current can be obtained, wind an insulated copper wire to form a coil and connect its ends to a centre-zero galvanometer. When a magnet is moved towards the coil, the galvanometer shows a momentary deflection, indicating a current. When the magnet is stopped, the deflection becomes zero. When the magnet is moved away from the coil, the galvanometer again shows a momentary deflection but in the opposite direction.
(b) The source of energy associated with the current obtained in part (a) is the mechanical energy spent in moving the magnet towards the coil.
4. State Fleming’s right hand rule.
Answer: Fleming’s right hand rule states: Stretch the thumb, central finger and forefinger of your right hand mutually perpendicular to each other. If the forefinger indicates the direction of magnetic field and the thumb indicates the direction of motion of the conductor, then the central finger will indicate the direction of induced current.
5. What is Lenz’s law?
Answer: Lenz’s law states that the direction of induced e.m.f. (or induced current) is such that it opposes the cause which produces it.
6. Why does it become more difficult to move a magnet towards a coil when the number of turns in the coil has been increased?
Answer: It becomes more difficult to move a magnet towards a coil when the number of turns in the coil has been increased because the e.m.f. induced in the coil, and hence the magnetic field induced by it, becomes more when the number of turns in the coil are increased. The stronger field induced opposes the motion of the magnet towards the coil.
7. Explain why an induced current must flow in such a direction so as to oppose the change producing it.
Answer: An induced current must flow in such a direction so as to oppose the change producing it because this ensures that some mechanical energy is spent in producing the change, which then changes into electrical energy in the form of the induced current, thereby conserving energy.
8. Name and state the principle of a simple a.c. generator. What is its use?
Answer:
Name: A.C. generator.
Principle: A simple a.c. generator works on the principle of electromagnetic induction. In a generator, a coil is rotated in a magnetic field. Due to rotation, the magnetic flux linked with the coil changes and therefore an e.m.f. is induced between the ends of the coil. Thus, a generator acts as a source of current in an external circuit containing load when it is joined between the ends of its coil.
Use: It is used as an alternative source of electricity when electric supply from mains is not available, for example, in houses or factories.
9. In an a.c. generator, the speed at which the coil rotates is doubled. How would this affect (a) the frequency of the output voltage, (b) the maximum output voltage.
Answer:
(a) If the speed at which the coil rotates is doubled, the frequency of the output voltage is doubled.
(b) If the speed at which the coil rotates is doubled, the maximum output voltage is doubled.
10. State two ways to produce a higher e.m.f. in an a.c. generator.
Answer: Two ways to produce a higher e.m.f. in an a.c. generator are:
(i) by increasing the speed of rotation of the coil, and
(ii) by increasing the number of turns in the coil.
(Other ways include increasing the area of cross section of coil, and increasing the strength of the magnetic field).
11. State (i) two dis-similarities, and (ii) two similarities between a d.c. motor and an a.c. generator.
Answer:
(i) Two dis-similarities between a d.c. motor and an a.c. generator are:
1. A generator converts mechanical energy into electrical energy, whereas a d.c. motor converts electrical energy into mechanical energy.
2. A generator works on the principle of electromagnetic induction, whereas a d.c. motor works on the principle of force acting on a current carrying conductor placed in a magnetic field. (Also, an a.c. generator makes use of two separate coaxial slip rings, while a d.c. motor makes use of two parts of a slip ring, i.e., split rings, which act as a commutator.)
(ii) Two similarities between a d.c. motor and an a.c. generator are:
1. Both in an a.c. generator and d.c. motor, a coil rotates in a magnetic field between the pole pieces of a powerful electromagnet.
2. Both in an a.c. generator and d.c. motor, there is a transformation of energy from one form to the other form.
12. For what purpose are the transformers used? On which type of current do transformers work?
Answer: Transformers are used to step up or step down the a.c. voltage, i.e., to change the magnitude of an alternating e.m.f. Transformers work on alternating current (a.c.).
13. State two factors on which the magnitude of an induced e.m.f. in the secondary coil of a transformer depends.
Answer: The magnitude of an induced e.m.f. in the secondary coil of a transformer depends on:
(i) the turns ratio (the ratio of the number of turns in the secondary coil to the number of turns in the primary coil), and
(ii) the magnitude of the e.m.f. applied in the primary coil (the input voltage).
14. Name the device used to transform 12 V a.c. to 200 V a.c. Name the principle on which it works.
Answer: The device used to transform 12 V a.c. to 200 V a.c. is a step-up transformer. It works on the principle of electromagnetic induction.
15. Name the coil of which the wire is thicker in a (i) step up, (ii) step down transformer. Give reason to your answer.
Answer:
(i) In a step up transformer, the current in the primary coil is more than in the secondary coil (or Ip > Is), so the wire in the primary coil is kept thicker than in the secondary coil.
(ii) In a step down transformer, the current in the secondary coil is more than in the primary coil (i.e., Is > Ip), so the wire in the secondary coil is kept thicker than in the primary coil.
The reason for using a thicker wire is that its use reduces the wire’s resistance and therefore reduces the loss of energy as heat in that coil. This energy loss due to heat is known as copper loss.
16. The output current of a transformer in which the voltage is stepped down is usually higher than the input current. Explain why.
Answer:
For an ideal transformer, when there is no energy loss, the output power will be equal to the input power, i.e., power in secondary coil = power in primary coil, or E_s I_s = E_p I_p.
In a transformer where the voltage is stepped down, the output voltage (E_s) is less than the input voltage (E_p). For the output power (E_s I_s) to be equal to the input power (E_p I_p) under this condition, the output current (I_s) must be higher than the input current (I_p). As indicated by the relationship I_p/I_s = E_s/E_p, if E_s < E_p, then I_p < I_s, meaning the current in the secondary coil (output current) is more than in the primary coil (input current).
17. Why is the iron core of a transformer made laminated (thin sheets) instead of being in one solid piece?
Answer:
The iron core of a transformer is made laminated because the lamination of the core prevents the loss of energy due to induced (or eddy) currents in the core.
18. Name two kinds of energy losses in a transformer. How are they minimised?
Answer: Two kinds of energy losses in a transformer are:
- Loss of energy due to induced (or eddy) currents in the core: This is minimised by the lamination of the core.
- Loss of energy due to hysteresis in the core: This gets minimised because the core is made of soft iron.
Another loss is copper loss, which is the loss of energy as heat in the coils. This is minimised by using a thicker wire for the coil carrying higher current, as thicker wire reduces resistance.
19. Give two points of difference between a step up and a step down transformer.
Answer: Two points of difference between a step up and a step down transformer are:
- A step up transformer increases the a.c. voltage and decreases the current (i.e., E_s > E_p and I_s < I_p), whereas a step down transformer decreases the a.c. voltage and increases the current (i.e., E_s < E_p and I_s > I_p).
- In a step up transformer, the turns ratio N_s/N_p > 1, meaning it has more number of turns in the secondary coil than in the primary coil. In a step down transformer, its turns ratio N_s/N_p < 1, meaning it has less number of turns in the secondary coil than in the primary coil.
20. Name the transformer used in the (i) power generating station, (ii) power sub-station. State the function of each transformer.
Answer:
(i) At the power generating station, a step up transformer is used. Its function is to step up the voltage in the transmission of electric power.
(ii) At the power sub-station, a step down transformer is used. Its function is to step down the voltage before its distribution to the consumers
Long Answer Type Questions
1. (a) What is electromagnetic induction? (b) Describe one experiment to demonstrate the phenomenon of electromagnetic induction.
Answer: (a) Electromagnetic induction is the phenomenon in which an e.m.f. is induced in the coil if there is a change in the magnetic flux linked with the coil.
(b) To demonstrate the phenomenon of electromagnetic induction, wind an insulated copper wire in the form of a spiral on a paper (or wooden) cylinder to form a coil (solenoid). Connect a centre-zero galvanometer G between the two ends of the solenoid. Place a bar magnet NS at some distance along the axis of the solenoid.
Observations:
(i) When the magnet is stationary, there is no deflection in the galvanometer, and its pointer stays at zero, as shown in figure (a).
(ii) When the magnet with its north pole facing the solenoid is moved towards it, the galvanometer deflects to the right, indicating that a current flows in the solenoid from B to A, as shown in figure (b).
(iii) When the motion of the magnet stops, the galvanometer pointer returns to zero, as shown in figure (c). This shows that current in the solenoid is produced only while the magnet is in motion.
(iv) If the magnet is moved away from the solenoid, the galvanometer deflects to the left, as shown in figure (d), indicating that current again flows but now in the opposite direction—from A to B. The current becomes zero as soon as the magnet stops moving.
(v) When the magnet is moved more rapidly, the galvanometer shows a greater deflection, though in the same direction, indicating that a stronger current is produced.
(vi) If the magnet is brought towards the solenoid with its south pole facing it, the galvanometer deflects to the left, as shown in figure (e), indicating that the current flows from A to B, opposite to the direction shown in figure (b).
2. (a) Describe briefly one way of producing an induced e.m.f. (b) State one factor that determines the magnitude of induced e.m.f. in part (a) above. (c) What factor determines the direction of induced e.m.f. in part (a) above?
Answer: (a) One way of producing an induced e.m.f. is by causing a relative motion between a coil and a magnet. For instance, if a bar magnet is moved towards or away from a coil whose ends are connected to a galvanometer, an e.m.f. is induced in the coil, indicated by a deflection in the galvanometer.
(b) One factor that determines the magnitude of the induced e.m.f. is the speed of the relative motion between the magnet and the coil; a more rapid motion results in a larger induced e.m.f. Another factor is the number of turns in the coil; more turns lead to a larger e.m.f.
(c) The factor that determines the direction of the induced e.m.f. is the direction of the relative motion between the magnet and the coil (or the direction of change of magnetic flux). For example, if the magnet is moved towards the coil, the current is in one direction, and if it is moved away, the current is in the opposite direction. Lenz’s law or Fleming’s right hand rule can be used to determine this direction.
3. Explain how does the Lenz’s law show the conservation of energy in the phenomenon of electromagnetic induction.
Answer: Lenz’s law states that the direction of the induced current is such that it opposes the cause producing it. This opposition means that mechanical work must be done to produce the change in magnetic flux (e.g., to move the magnet against the opposing magnetic force created by the induced current). This mechanical energy spent in doing work against the opposing force is transformed into electrical energy in the form of the induced current flowing in the solenoid. Thus, Lenz’s law is a consequence of the law of conservation of energy, as electrical energy is produced at the expense of mechanical energy.
4. The diagram shows a coil of several turns of copper wire near a magnet NS. The coil is moved in the direction of arrow shown in the diagram.
(i) In what direction does the induced current flow in the coil?
(ii) Name the law used to arrive at the conclusion in part (i).
(iii) How would the current in coil be altered if (a) the coil has twice the number of turns, (b) the coil was made to move three times fast?
Answer: (i) As the coil is moved towards the N-pole of the magnet (in the direction of the arrow), the face of the coil approaching the N-pole will acquire north polarity to oppose this motion (according to Lenz’s law). For this face to become a north pole, the current, when viewed from the magnet side, must be anticlockwise. If A and B are terminals, the current will flow from A to B.
(ii) The law used to arrive at the conclusion in part (i) is Lenz’s law. (Fleming’s Right Hand Rule can also be used).
(iii) (a) If the coil has twice the number of turns, the induced e.m.f., and hence the current in the coil, becomes twice, assuming the resistance of the coil does not change significantly or the external circuit resistance is dominant.
(b) If the coil was made to move three times fast, the rate of change of magnetic flux would be three times greater, so the induced e.m.f., and hence the current in the coil, becomes three times.
5. The diagram shows a fixed coil of several turns connected to a centre zero galvanometer G and a magnet NS which can move in the direction shown in the diagram.
(a) Describe the observation in the galvanometer if (i) the magnet is moved rapidly, (ii) the magnet is kept stationary after it has moved into the coil, (iii) the magnet is then rapidly pulled out of the coil.
(b) How would the observation in (i) of part (a) change if a more powerful magnet is used?
Answer: (a) (i) If the magnet (N-pole first) is moved rapidly into the coil, the galvanometer will show a momentary deflection (e.g., towards the right), indicating an induced current. The deflection will be larger due to the rapid motion.
(ii) If the magnet is kept stationary after it has moved into the coil, there is no change in magnetic flux linked with the coil, so the galvanometer will show no deflection; its pointer will be at zero.
(iii) If the magnet is then rapidly pulled out of the coil, the galvanometer will show a momentary deflection in the opposite direction (e.g., towards the left), indicating an induced current in the reverse direction. The deflection will be larger due to the rapid motion.
(b) If a more powerful magnet is used in (i) of part (a), the magnetic flux and its rate of change would be greater. Therefore, the deflection observed in the galvanometer would be increased.
6. Draw a labelled diagram of a simple a.c. generator.
Answer: A simple a.c. generator consists of an armature coil (ABCD) wound on a soft iron core, placed between the poles (N and S) of a strong horse-shoe electromagnet. The ends of the coil are connected to two separate coaxial metallic slip rings (S₁ and S₂) which rotate with the coil. Two carbon brushes (B₁ and B₂) press gently against the slip rings and act as terminals to connect to an external load. An axle is used to rotate the coil.
7. Draw a labelled diagram to show the various components of a step up transformer.
Answer: A step-up transformer consists of a primary coil (P) and a secondary coil (S) wound on a laminated soft iron core. For a step-up transformer, the number of turns in the secondary coil (Nₛ) is greater than the number of turns in the primary coil (Nₚ). The input a.c. voltage is applied across the primary coil, and the output (stepped-up) a.c. voltage is obtained across the secondary coil. The wire of the primary coil is thicker than that of the secondary coil.
8. Draw a labelled diagram of a step up transformer and explain how does it work. State two characteristics of the primary coil as compared to its secondary coil.
Answer: Diagram: A step-up transformer has a laminated soft iron core. The primary coil (P) with fewer turns (Nₚ) is wound on one arm, and the secondary coil (S) with more turns (Nₛ) is wound on another arm (or over the primary). Input a.c. is connected to the primary, and output is taken from the secondary.
Working: When an alternating e.m.f. (input voltage) is applied to the primary coil, a varying current flows through it. This varying current produces a varying magnetic field in the core of the transformer. The magnetic flux linked with the secondary coil (which is also wound over the core) therefore varies. Due to this change in magnetic flux, an e.m.f. is induced in the secondary coil. Since the number of turns in the secondary coil (Nₛ) is greater than in the primary coil (Nₚ) for a step-up transformer, the induced e.m.f. in the secondary coil (Eₛ) is greater than the applied e.m.f. in the primary coil (Eₚ), according to the relation Eₛ/Eₚ = Nₛ/Nₚ. The frequency of the induced e.m.f. is the same as that of the applied e.m.f.
Two characteristics of the primary coil as compared to its secondary coil in a step-up transformer are:
- The number of turns in the primary coil is less than the number of turns in the secondary coil (Nₚ < Nₛ).
- The wire of the primary coil is thicker than the wire of the secondary coil (because the current in the primary coil, Iₚ, is greater than the current in the secondary coil, Iₛ, for an ideal step-up transformer, assuming power output equals power input).
9. Draw a labelled diagram of a device you would use to transform 200 V a.c. to 15 V a.c. Name the device and explain how does it work. Give its two uses.
Answer:
The name of the device is Step-down transformer. A step-down transformer has a laminated soft iron core. The primary coil (P) with more turns (Nₚ) is wound on one arm, and the secondary coil (S) with fewer turns (Nₛ) is wound on another arm. Input 200 V a.c. is connected to the primary, and output 15 V a.c. is taken from the secondary. The wire of the secondary coil is thicker than that of the primary coil.
When an alternating e.m.f. of 200 V is applied to the primary coil, a varying current flows, producing a varying magnetic field in the core. This changing magnetic flux links with the secondary coil, inducing an e.m.f. in it. Since it is a step-down transformer, the number of turns in the secondary coil (Nₛ) is less than the number of turns in the primary coil (Nₚ).
Therefore, the induced e.m.f. in the secondary coil (Eₛ = 15 V) is less than the applied e.m.f. in the primary coil (Eₚ = 200 V), as per Eₛ/Eₚ = Nₛ/Nₚ.
Two uses of a step-down transformer:
- Used with electric bells, which typically require a low voltage.
- Used in mobile phone chargers to step down the mains voltage to a suitable low voltage for charging the battery.
10. (a) Complete the following diagram of a transformer and name the parts labelled A and B.
(b) Name the part you have drawn to complete the diagram in part (a).
(c) What is the material of the part named above?
(d) Is this transformer a step up or step down? Give reason.
Answer: (a)
(b) The part drawn to complete the diagram is the laminated soft iron core.
(c) The material of the part named above (the core) is soft iron.
(d) Looking at the incomplete coil A (input) appears to have fewer turns than coil B (output). If this visual representation is accurate and A is the primary coil and B is the secondary coil, then this transformer, once completed, would be a step-up transformer. Reason: The number of turns in the secondary coil (B) is greater than the number of turns in the primary coil (A), which results in an output voltage higher than the input voltage. (If the number of turns were reversed, it would be step-down).
11. The diagram shows the core of a transformer and its input and output connections.
(a) State the material used for the core and describe its structure.
(b) Complete the diagram of the transformer and connections by labelling all parts joined by you.
(c) Name the transformer: step up or step down?
Answer: (a) The material used for the core is soft iron. Its structure consists of thin rectangular laminated sheets of soft iron, often of T and U shape, placed alternately one above the other and insulated from each other by a paint or varnish coating. This lamination helps to reduce energy loss due to eddy currents.
(b) To complete the diagram, the primary coil should be wound on one arm of the core and connected to the “INPUT 220 V A.C. MAINS”. The secondary coil should be wound on the other arm of the core (or over the primary) and connected to the “OUTPUT 44 V A.C.” The coils should be labelled as primary coil and secondary coil.
(c) The transformer is a step-down transformer. This is because the input voltage is 220 V A.C. and the output voltage is 44 V A.C.; the output voltage is less than the input voltage.
12. The diagram below shows a magnetic needle kept just below the conductor AB which is kept in North South direction.
(a) In which direction will the needle deflect when the key is closed ?
(b) Why is the deflection produced ?
(c) What will be the change in the deflection if the magnetic needle is taken just above the conductor AB ?
(d) Name one device which works on this principle.
Answer: (a) Assuming current flows from A to B, and the wire AB is oriented from South (A) to North (B) as in Oersted’s typical experiment setup, when the key is closed, the north pole (N) of the needle will deflect towards the west.
(b) The deflection is produced because when an electric current is passed through a conducting wire, a magnetic field is produced around it. The magnetic needle, being in this magnetic field, experiences a torque due to which it deflects to align itself in the direction of the resultant magnetic field. The deflection of the magnetic needle on passing current in the wire clearly indicates the creation of a magnetic field around the wire.
(c) If the magnetic needle is taken just above the conductor AB, with the current still flowing from A to B (South to North), the north pole (N) of the needle will deflect towards the east. The direction of deflection reverses.
(d) One device which works on this principle (the magnetic effect of current) is an electric bell. The working of an electric bell is based on the magnetic effect of current, as it uses an electromagnet.
Numericals
1. The magnetic flux through a coil having 100 turns decreases from 5 milli weber to zero in 5 second. Calculate the e.m.f. induced in the coil.
Answer:
Given:
Number of turns (N) = 100
Initial magnetic flux (Φ_initial) = 5 milli weber = 5 × 10⁻³ Wb
Final magnetic flux (Φ_final) = 0 Wb
Time taken (Δt) = 5 s
To find:
Induced e.m.f. (ε) = ?
Solution:
According to Faraday’s Law of electromagnetic induction, the induced e.m.f. (ε) is given by the rate of change of magnetic flux through the coil. The formula is:
ε = -N * (ΔΦ / Δt)
Where,
N = Number of turns
ΔΦ = Change in magnetic flux
Δt = Change in time
First, we calculate the change in magnetic flux (ΔΦ):
ΔΦ = Final magnetic flux – Initial magnetic flux
=> ΔΦ = Φ_final – Φ_initial
=> ΔΦ = 0 – (5 × 10⁻³ Wb)
=> ΔΦ = -5 × 10⁻³ Wb
Now, we substitute the given values into the formula for induced e.m.f.:
ε = -N * (ΔΦ / Δt)
=> ε = -100 * ((-5 × 10⁻³ Wb) / 5 s)
=> ε = -100 * (-1 × 10⁻³ V)
=> ε = 100 × 10⁻³ V
=> ε = 0.1 V
Therefore, the e.m.f. induced in the coil is 0.1 Volts.
2. The primary coil of a transformer has 800 turns and the secondary coil has 8 turns. It is connected to a 220 V a.c. supply. What will be the output voltage?
Answer:
Given:
Number of turns in the primary coil (N_p) = 800
Number of turns in the secondary coil (N_s) = 8
Input voltage (V_p) = 220 V
To find:
Output voltage (V_s) = ?
Solution:
The relationship between the voltage and the number of turns in a transformer is given by the formula:
V_s / V_p = N_s / N_p
To find the output voltage (V_s), we can rearrange the formula:
=> V_s = (N_s / N_p) × V_p
Now, we substitute the given values into the formula:
=> V_s = (8 / 800) × 220
=> V_s = (1 / 100) × 220
=> V_s = 220 / 100
=> V_s = 2.2 V
Therefore, the output voltage will be 2.2 V.
3. A transformer is designed to give a supply of 8 V to ring a house-bell from the 240 V a.c. mains. The primary coil has 4800 turns. How many turns will be in the secondary coil?
Answer:
Given:
Here is the table with the given information:
Primary Voltage (Vp) = 240 V
Secondary Voltage (Vs) = 8 V
Number of turns in the primary coil (Np) = 4800
To find:
Number of turns in the secondary coil (Ns) = ?
Solution:
The relationship between the voltages and the number of turns in a transformer is given by the transformer equation:
Vs / Vp = Ns / Np
To find the number of turns in the secondary coil (Ns), we can rearrange the formula:
Ns = Np * (Vs / Vp)
Now, we substitute the given values into the equation:
=> Ns = 4800 * (8 / 240)
=> Ns = 4800 * (1 / 30)
=> Ns = 4800 / 30
=> Ns = 160
Therefore, there will be 160 turns in the secondary coil.
Here is the updated table with the final answer:
4. The input and output voltages of a transformer are 220 V and 44 V respectively. Find: (a) the turns ratio, (b) the current in input circuit if the output current is 2 A.
(a) Find the turns ratio.
Answer:
Given:
Input voltage (V_p) = 220 V
Output voltage (V_s) = 44 V
To find:
Turns ratio (n_s / n_p) = ?
Solution:
The relationship between the voltages and the number of turns in the primary (p) and secondary (s) coils of a transformer is given by the formula:
V_s / V_p = n_s / n_p
The turns ratio is the ratio of the number of turns in the secondary coil (n_s) to the number of turns in the primary coil (n_p).
=> Turns ratio = V_s / V_p
=> Turns ratio = 44 / 220
=> Turns ratio = 1 / 5
Thus, the turns ratio is 1/5 or 1:5.
(b) Find the current in input circuit if the output current is 2 A.
Answer:
Given:
Input voltage (V_p) = 220 V
Output voltage (V_s) = 44 V
Output current (I_s) = 2 A
To find:
Input current (I_p) = ?
Solution:
Assuming the transformer is 100% efficient (an ideal transformer), the input power is equal to the output power.
Input Power = Output Power
=> V_p × I_p = V_s × I_s
We can rearrange the formula to solve for the input current (I_p).
=> I_p = (V_s × I_s) / V_p
=> I_p = (44 × 2) / 220
=> I_p = 88 / 220
=> I_p = 0.4 A
Therefore, the current in the input circuit is 0.4 A.
5. A student in a laboratory wants to light a bulb of 6 W, 11 V rating with a supply of 220 V. Mention the kind of transformer he should use. Also calculate the turn ratio for the transformer recommended by you.
Answer:
Given:
Power of the bulb, P_s = 6 W
Voltage rating of the bulb (Secondary Voltage), V_s = 11 V
Supply voltage (Primary Voltage), V_p = 220 V
To find:
(i) The kind of transformer to be used.
(ii) The turn ratio for the transformer.
Solution:
(i) Kind of Transformer
The input voltage from the supply is the primary voltage, V_p = 220 V.
The output voltage required to light the bulb is the secondary voltage, V_s = 11 V.
Since the input voltage (220 V) is greater than the required output voltage (11 V), the voltage needs to be decreased or “stepped down”.
Therefore, the student should use a step-down transformer.
(ii) Turn Ratio
The relationship between the voltages and the number of turns in the primary coil (N_p) and the secondary coil (N_s) of a transformer is given by the formula:
V_p / V_s = N_p / N_s
Here, the turn ratio is the ratio of the number of turns in the primary coil to the number of turns in the secondary coil (N_p : N_s).
Substituting the given values into the formula:
=> 220 / 11 = N_p / N_s
=> 20 = N_p / N_s
=> N_p / N_s = 20 / 1
Thus, the turn ratio (N_p : N_s) for the transformer should be 20:1. This means for every 20 turns in the primary coil, there should be 1 turn in the secondary coil.
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