Selina Concise Class 10 Physics Solution Chapter No. 10 – ‘Electro-magnetism’ For ICSE Board Students.
Selina Concise Class 10 Physics Chapter 10 Electro-magnetism Exercise All Questions and Answers by Physics Teacher here in this post.
Exercise 10 (A)
Question: 1
By using a compass needle describe how can you demonstrate that there is a magnetic field around a current-carrying conductor.
Solution: Circuit below shows the compass needle, when key is open needle shows no deflection as there is no induced magnetic field through the circuit as there is no current is flowing through the conductor.
When the key is closed the magnetic needle shows the deflection. As the current flowing through the wire which induces the magnetic field in the wire the deflection is shown the below diagram.
Question: 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.
Solution:
Question: 3
How is the magnetic field due to a straight current-carrying wire affected if current in the wire is (a) decreased, (b) reversed?
Solution:
a.) When current is decreased magnetic field becomes rarer.
b.) When current is reversed magnetic field also reversed.
Question: 4
State a law, which determines the direction of the magnetic field around a current-carrying wire.
Solution: If we hold the current carrying conductor in our right hand and our thumb pointing towards the direction of the current then the fingers wrapped around the conductor shows the direction of the magnetic field. It is called as right hand thumb rule.
Question: 5
A straight wire lying in a horizontal plane carries a current from north to south.
(a.) What will be the direction of the magnetic field at a point just underneath it?
(b.) Name the law used to arrive at this answer in part (a).
Solution:
a.) The direction of the magnetic field at a point just underneath it is towards east.
b.) Right hand thumb rule.
Question: 6
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.
Solution: The needle gets deflected. As current flows through wire it produces magnetic field around it which cause the deflection in the magnetic needle.
Question: 7
Draw a labelled diagram showing the three magnetic field lines of a loop carrying current. Mark the direction of current and the direction of the magnetic field by arrows in your diagram.
Solution:
Question: 8
A wire, bent into a circle, carries a current in an anticlockwise direction. What polarity does this face of the coil exhibit?
Solution: North polarity.
Question: 9
What is the direction of the magnetic field at the centre of a coil carrying current in (i.) the clockwise, (ii.) the anticlockwise, direction?
Solution:
i.) The direction of the magnetic field is in the clockwise direction along the axis at the centre of the coil.
ii.) The direction of the magnetic field is in the anticlockwise direction along the axis at the centre of the coil.
Question: 10
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.
Solution:
Question: 11
Name and state the rule by which the polarity at the ends of a current carrying solenoid is determined.
Solution: If we hold the current carrying conductor in our right hand and our thumb pointing towards the direction of the current then the fingers wrapped around the conductor shows the direction of the magnetic field. It is called as right hand thumb rule.
Question: 12
The diagram in the figure 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.
Solution:
a.) Polarity of end A is north and polarity of end B is south.
b.) The magnetic will be repelled. As the polarity is facing in same direction, N pole faces N pole.
Question: 13
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.
Solution:
a.) Polarity of end A is north and polarity of end B is south.
b.) The needle will deflect to the west as the end A acts as N pole it deflects the needle to west.
Question: 14
State two ways by which the magnetic field due to a current-carrying solenoid can be made stronger.
Solution: to increase the magnetic field of the current carrying solenoid we have either increase the current flowing through it or we have to increase the winding of the solenoid.
Question: 15
Why does a current-carrying freely suspended solenoid rest along a particular direction? State the direction in which it rests.
Solution: A current carrying solenoid behave as bar magnetic because of magnetic field produced by it. If it is suspended freely it will come at rest in north-south direction.
Question: 16
What effect will there be on a magnetic compass when it is brought near a current-carrying solenoid?
Solution: Needle of the magnetic compass will rest In the direction of magnetic fled produced by the current-carrying solenoid.
Question: 17
How is the magnetic field due to a solenoid carrying current affected if a soft iron bar is introduced inside the solenoid?
Solution: Magnetic field of the solenoid will increase.
Question: 18
Complete the following sentences:
(a.) When current flows in a wire, it creates ____________
(b.) On reserving 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 rest in __________ direction.
Solution:
(a.) a magnetic field around it.
(b.) reversed.
(c.) bar magnet.
(d.) north-south direction.
Question: 19
You are required to make an electromagnet from a soft iron bar by using a cell, an insulated coil of copper and a switch. (a) Draw a circuit diagram to represent the process. (b) Label the poles of the electromagnet.
Solution:
Question: 20
The diagram in the figure shows a coil wound around a soft iron bar XY. (a) State the polarity at the end X and Y as the switch is pressed. (b)Suggest one way of increasing the strength of electromagnet so formed.
Solution:
a.) Polarity of end X is north and polarity of end Y is south.
b.) To increase the magnetic field we have increase the current flowing through the coil. Which can be done using rheostat to decrease the resistance of the circuit.
Question: 21
(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).
Solution:
a.) It is called solenoid.
b.) This device is called as electromagnet.
c.) It can be used in electric bell.
Question: 22
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.
Solution:
Question: 23
What is an electromagnet? Name two factors on which the strength of the magnetic field of an electromagnet depends and state how does it depend on the factors stated by you.
Solution: A temporary magnet produced by passing current in a coil wound around a piece of soft iron is called as electromagnet. to increase the magnetic field of a electromagnet we have either increase the current flowing through it or we have to increase the winding of the coil.
Question: 24
Figure shows the current flowing in the coil of wire wound around the soft iron horseshoe 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 increase the strength of magnetic field produce.
Solution:
End A is south pole and end B north pole.
- If the polarity is reversed. End A becomes north pole and end B becomes south pole.
- We can increase magnetic field by increasing the winding of the coil.
Question: 25
State two ways through which the strength of an electromagnet can be increased.
Solution: To increase the magnetic field of the electromagnet we have either increase the current flowing through it or we have to increase the winding of the electromagnet.
Question: 26
Name one device that uses an electromagnet.
Solution: Electromagnet is used in electric relay.
Question: 27
State two advantages of an electromagnet over a permanent magnet.
Solution: An electromagnet is strong magnet and strength of an electromagnet can be altered according to the use. These are the advantage of electromagnet over a permanent magnet.
Question: 28
State two differences between an electromagnet and a permanent magnet.
Solution:
| Electromagnet | Permanent magnet |
| It is made up of soft iron. | It is made up of steel. |
| Magnetic field can be altered. | Magnetic field is fixed. |
Question: 29
Why is soft iron used as the core of the electromagnet in an electric bell?
Solution: Working of electromagnet in an electric bell is based on the electromagnetism induced in coil, when current is passed through it. As soft iron bar acquires magnetic properties only when current is passed through it. Hence, soft iron is used as electromagnet in an electric bell.
Question: 30
How is the working of an electric bell affected, if alternating current be used instead of direct current?
Solution: The use of A.C. current will only reserve the polarity. Which will not affect the attraction force. Hence, even if A.C. current used there will be no change in working of the electric bell.
MULTIPLE CHOICE TYPE
Question: 1
The presence of the magnetic field at a point can be detected by:
(a) A strong magnet
(b) A solenoid
(c) A compass needle
(d) A current-carrying wire
Solution: (c) A compass needle
Question: 2
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
Solution:(a) Gets reversed in direction
Exercise 10 (b)
Question: 1
Name three factors on which the magnitude of the 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.
Solution: The magnitude of the force on a current-carrying conductor placed in a magnetic field depends on strength of the magnetic field, current in the conductor and length of the conductor. Force is directly proportional to all this factors that is current, magnetic field and length.
Question: 2
State condition in each case for the magnitude of the force on a current-carrying conductor placed in a magnetic field is (a) zero, (b) maximum
Solution:
- When current flows in the direction of the magnetic field force becomes zero.
- When the current flows through the conductor is perpendicular to the magnetic field force becomes maximum.
Question: 3
How will the direction of force be changed, if the current is reversed in the conductor placed in a magnetic field?
Solution: If the direction of the current is reserved the direction of the force is also reversed.
Question: 4
Name and state the law which is used to determine the direction of the force on a current-carrying conductor placed in a magnetic field.
Solution: Stretch the forefinger, central finger and the thumb of your left hand mutually perpendicular to each other. If the forefinger indicates the direction of the magnetic field and the central finger indicates the direction of current, then the thumb will indicate the direction of motion of the conductor. This law is called as Flemings left hand rule.
Question: 5
State Fleming’s left handle rule.
Solution: Stretch the forefinger, central finger and the thumb of your left hand mutually perpendicular to each other. If the forefinger indicates the direction of the magnetic field and the central finger indicates the direction of current, then the thumb will indicate the direction of motion of the conductor.
Question: 6
State the unit of magnetic field in terms of the force experienced by a current-carrying conductor placed in a magnetic field.
Solution: The force experienced by a current-carrying conductor placed in a magnetic field is N/Am.
Question: 7
A flat coil ABCD is freely suspended between the pole pieces 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 coil come to rest?
(c.) When will the couple acting on the coil be (i) maximum, (ii) minimum?
(d.) Name an instrument which makes use of the principle stated above.
Solution:
a.) The coil will rotate due magnetic field produces torque in the coil.
b.) When the coil comes in perpendicular to the magnetic field, the coil comes to rest.
c.) (i) The couple will be maximum when its parallel to the magnetic field.
(ii) The couple will be minimum when its perpendicular to the magnetic field.
d.) D.C. motor
Question: 8
A coil ABCD mounted on an axle is placed between the poles N and S of a permanent magnet as shown in 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 at the ends A and D of the coil?
(b.)Why is a commutator necessary for the continuous rotation of the coil?
(c.) Complete the diagram with the commutator, etc. for the flow of current in the coil.
Solution:
a.) The coil will rotate in anticlockwise direction due to force acting on it.
b.) The commutator is needed as after half the rotation the direction of the torque on coil is reversed. Commutator is needed to reverse the direction of the current, to keep the continuous rotation of the coil.
c.)
Question: 9
What is an electric motor? State its principle.
Solution: A machine which converts the electrical energy to mechanical energy is called as electric motor.Principle of D.C. motor is that when an electric current is passed 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 is obtained.
Question: 10
Draw a labelled diagram of a d.c motor showing its main parts.
Solution:
Question: 11
What energy conversion does take place during the working of a d.c motor?
Solution: Electrical energy is converted into mechanical energy.
Question: 12
State two ways by which the speed of rotation of an electric motor can be increased.
Solution: Speed of rotation of motor can be increased by increasing the strength of the current in coil and by increasing number of turns in the coil.
Question: 13
Name two appliances in which an electric motor is used.
Solution: Electric motor is used in many appliances like electric pumps, fans, electric vehicles etc.
MULTIPLE CHOICE TYPE
Question: 1
In an electric motor, the energy transformation is:
(a) From electrical to chemical
(b) From chemical to light
(c) From mechanical to electrical
(d) From electrical to mechanical
Solution:(d) From electrical to mechanical
Exercise 10 (C)
Question: 1
(a) What is electromagnetic induction?
(b) Describe one experiment to demonstrate the phenomenon of electromagnetic induction.
Solution:
a.) Electromotive force developed between the ends of the conductor which lasts as long as there is a change in the number of the magnetic field lines through the conductor. It is called as electromagnetic induction.
b.)
Above diagram shows the movement of a bar magnet around a solenoid connected to the galvanometer. The experiment goes like when the bar magnet is stationary the galvanometer shows no deflection and when the bar magnet is moved back and forth the galvanometer shows deflection. Which proves the change in the deflection is due to change in the magnetic field.
Question: 2
State Faraday’s laws of electromagnetic induction.
Solution: Electromotive force is produced due to change magnetic flux linked with the coil and it lasts till there is change in magnetic flux linked with the coil.
The magnitude of electromotive force induced is directly proportional to rate of change of magnetic flux linked with the coil.
Question: 3
State two factor on which the magnitude of induced e.m.f in a coil depend.
Solution: The number of change of magnetic flux with each turn and number of turns in the coil are the two factors electromotive force’s magnitude depends upon.
Question: 4
(a) What kind of the energy change takes place when a magnet is moved towards a coil having a galvanometer between its ends?
(b) Name the phenomenon.
Solution:
a.) Mechanical energy changes to electrical energy when a magnet is moved toward a coil having a galvanometer between its ends.
b.) Electromagnetic induction.
Question: 5
(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)?
Solution:
a.) When the magnet is moved towards coil magnetic flux linked with it changes. As the change in magnetic flux lines changes electromotive force is induced in the coil. And galvanometer shows deflection.
b.) Mechanical energy is source of energy.
Question: 6
(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?
Solution:
Below diagram shows the movement of a bar magnet around a solenoid connected to the galvanometer. The experiment goes like when the bar magnet is stationary the galvanometer shows no deflection and when the bar magnet is moved back and forth the galvanometer shows deflection. Which proves the change in the deflection is due to change in the magnetic field.
a.) Magnitude of induced electromotive force is depends upon change in magnetic flux.
b.) Increase or decrease in the magnetic flux determines the direction of induced electromotive force.
Question: 7
Complete the following sentences:
The current is induced in a closed circuit only if there is _________
Solution: change in a number of magnetic flux lines linked with the circuit.
Question: 8
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.
Solution:
i.) Yes
ii.) Yes
iii.) Yes
iv.) No
Question: 9
A conductor is moved in a varying magnetic field. Name the law which determines the direction of current induced in the conductor.
Solution: Fleming’s right hand rule.
Question: 10
State Fleming’s right-hand rule.
Solution: Stretch the thumb, central finger and forefinger of your right hand mutually perpendicular to each other. If the forefinger indicates the direction of the magnetic field and the thumb indicates the direction of motion of the conductor, then the central finger will indicate the direction of induced current. Is called as Fleming’s right hand rule.
Question: 11
What is Lenz’s law?
Solution:The direction of induced electromotive force (or induced current) is such that it opposes the cause which produces it, is stated by Lenz’s law.
Question: 12
Why does it become more difficult to move a magnet towards a coil when the number of turns in the coil has been increased?
Solution: when the number of turns increases the induced electromotive force increases. By Lenz’s law the electromotive force opposes the movement of magnet as it produces it.
Question: 13
Explain why an induced current must flow in such a direction so as to oppose the change producing it.
Solution: As induced current must flow in such a direction so as to oppose the change producing it. Mechanical energy exerted by change in magnetic flux is converted into electrical energy. Hence, more the mechanical force more the electrical energy.
Question: 14
Explain how does the Lenz’s law show the conservation of energy in electromagnetic induction.
Solution: Lenz’s law states that mechanical energy used against opposing force experienced by the moving magnet is converted into electrical energy. Which is conservation of energy. Hence, Lenz’s law is based on conservation of energy.
Question: 15
The diagram shows a coil of several turns of copper wire near a magnet NS. The coil is moved in the direction of the 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?
Solution:
i.) Current flows from A to B
ii.) Lenz’s law.
Iii.) (a) current is directly proportional to number of turns, if the turns are doubled current will also be doubled.
(b) current induced is directly proportional to the change in magnetic flux lines. Hence, if the coil moves three times fast current is tripled.
Question: 17
Name and state the principle of a simple a.c. generator. What is its use?
Solution: In a generator, if a coil is rotated in a magnetic field, then due to rotation, the magnetic flux linked with the coil changes and therefore an electromotive force. is induced between the ends of the coil. Thus, a generator acts as a source of current in an external circuit containing load when connected between the ends of its coil. Is the principle of A.C. generator. The A.C. generator is used supply power in alternating current.
Question: 18
What determines the frequency of a.c. produced in a generator?
Solution: The frequency of A.C. generator depends upon number of rotations of coil per second.
Question: 19
Complete the sentence:
An a.c. generator changes the ___________ energy to ___________ energy.
Solution: mechanical energy, electrical energy.
Question: 20
Draw a labelled diagram of a simple a.c generator.
Solution:
Question: 21
In an a.c generator the speed at which the coil rotates is doubled. How would this affect
(a) the frequency of output voltage
(b) the maximum output voltage.
Solution:
a.) The frequency of A.C. generator is directly proportional to number of rotations of coil per second. Hence, when coil rotates at double speed frequency is doubled.
b.) The maximum output is also doubled.
Question: 22
State two ways to produce a higher e.m.f. in an a.c generator.
Solution: To produce a higher electromotive force in an A.C. generator we have either increase speed of rotation of the coil or we have increase the number of turns in the coil.
Question: 23
What energy conversion does take place in a generator when it is in use?
Solution: Mechanical energy is converted into electrical energy.
Question: 24
State (i) two dis-similarities, and (ii) two similarities between a D.C. motor and an A.C. generator.
Solution:
i.) Difference between D.C. motor and A.C. generator.
| D.C. motor | A.C. generator |
| It converts electrical energy to mechanical energy. | It converts mechanical energy into electrical energy. |
| It depends on principle of force acting on a current-carrying conductor placed in a magnetic field | It depends on principle of magnetic induction. |
ii.) Similarities between D.C. motor and A.C. generator.
In both the machine have coil rotating in magnetic field.
In both the machine converts one form of energy to other form of energy.
Question: 25
State one advantage of using a.c. over d.c.
Solution: The A.C. current is more efficient as A.C. current is reduces loss of electrical energy as heat. Hence A.C. current is used for transmission of electricity over longer distance.
Question: 26
For what purpose are the transformers used? Can they be used with a direct current source?
Solution: Transformers are used to step-up or step-down the voltage. It can not used with D.C. source.
Question: 27
State two factors on which the magnitude of an induced e.m.f. in the secondary coil of a transformer depends.
Solution: induced electromotive force in secondary coil depends upon ratio of number of turn in secondary coil to the number of turn in primary coil. And magnitude of electromotive force applied in the primary coil.
Question: 28
How are the e.m.f. in the primary and secondary coils of a transformer related with the number of turns in these coils?
Solution:
Electromotive force in secondary coil x number of turn in primary coil = electromotive force in primary coil x number of turn in secondary coil.
Question: 29
Draw a labelled diagram to show the various components of a step-up transformer.
Solution:
Question: 30
Name the device used to transform 12 V a.c. to 200 V a.c. Name the principle on which it works.
Solution: Step-up transformer. It works on principle of electromagnetic induction.
MULTIPLE CHOICE TYPE
Question: 1
The direction of the 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
Solution: (d) Fleming’s right-hand rule
Question: 2
In a step-up transformer:
(a) Ns = Np
(b) Ns < Np
(c) Ns > Np
(d) nothing can be said
Solution:(c) Ns > Np
NUMERICAL
Question: 1
The magnetic flux through a coil having 100 turns decreases from 5 milli weber to zero in 5 seconds.
Calculate the e.m.f. induced in the coil
Solution: Given
Number of turns = 100
Change in magnetic flux = 0.005 – 0 = 0.005 weber
Time = 5 s
E.M.F. induced = number of turn x rate of change in magnetic flux
= 100 x 0.005/5
= 0,1 V
Question: 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?
Solution: Given
Number of turns in primary coil, N1 = 800
Number of turns in secondary coil, N2 = 8
Input supply voltage,E1 = 220 V
As we know,
E1 / E2 = N1 / N2
E2 = E1 x N2 / N1
E2 = 220 x 8 / 800
E2 = 1760 / 800
E2 = 2.2 V
Question: 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?
Solution: Given
Voltage output. V1 = 8V
Voltage input, V2 = 240 V
Number of turn in primary coil, N1 = 4800
As we know,
V2 / V1 = N1 / N2
N2 = N1 x V1 / V2
N2 = 4800 x 8 / 240
N2 =38400 / 240
N2 = 160 turns
Question: 4
The input and output voltages of a transformer are 220 V and 44V respectively. Find
(a) the turns ratio
(b) the current in input circuit if the output current is 2 A.
Solution: Given
Voltage output. V1 = 44V
Voltage input, V2 = 220 V
Output current, I = 2A
ratio of number of turns = V1 / V2
= 44 / 240
= 1/5
And as we know,
Input current = V1 x I / V2
Input current = 44 x 2 / 240
Input current = 0.4 A
