Electromagnetic Induction Class 12 MCQs Questions with Answers
Electromagnetic Induction Class 12 MCQ Question 1.
A square of side L meters lies in the x-y plane in a region where the magnetic field is given by B = B0 (2\(\hat{\mathbf{i}}\) + 3\(\hat{\mathbf{j}}\) + 4\(\hat{\mathbf{k}}\)) Tesla, where B0 is constant. The magnitude of flux passing through the square is
(A) 2B0L2 Wb.
(B) 3B0L2 Wb.
(C) 4B0L2 Wb.
(D) \(\sqrt{29} \mathrm{~B}_{0} \mathrm{~L}^{2}\) Wb.
Answer:
(C) 4B0L2 Wb.
Explanation:
Magnetic flux is defined as the total number of magnetic lines of force passing normally through an area placed in a magnetic field and is equal to the magnetic flux linked with that area. Square lies in X – Y plane in \(\vec{B}\) so \(\vec{A}\) = L2 \(\hat{\mathbf{k}}\)
Q = B.A
= B0(2\(\hat{\mathbf{i}}\) + 3\(\hat{\mathbf{j}}\)+ 4\(\hat{\mathbf{k}}\)).(L\(\hat{\mathbf{k}}\))
= B0[2 x \(\hat{\mathbf{i}}\)\(\hat{\mathbf{k}}\) + 3 x \(\hat{\mathbf{j}}\)\(\hat{\mathbf{i}}\) + 4 x \(\hat{\mathbf{k}}\)\(\hat{\mathbf{k}}\)]
= B0L2[0 + 0 + 4]
= 4B0L2Wb.
Class 12 Physics Chapter 6 MCQ Question 2.
A loop, made of straight edges has six corners at A(0, 0, 0), B(L, 0, 0) C(L, L, 0), D(0, L, 0), E(0, L, L) and F(0,0, L). A magnetic field B=B0 (\(\hat{\mathbf{i}}\) + \(\hat{\mathbf{k}}\)) Tesla is present in the region. The flux passing through the loop ABCDEFA (in that order) is –
(A) B0L2 Wb.
(B) 2B0L2 Wb.
(C) \(\sqrt{2} \mathrm{~B}_{0} \mathrm{~L}^{2}\)
(A) 4 B0L2Wb.
Answer:
(B) 2B0L2 Wb.
Explanation:
The loop can be considered in two planes:
(i) Plane of ABCDA is in X – Y plane. So its vector
\(\vec{A}\) is in Z – direction. Hence,
A1 = \(|A|\)\(\hat{\mathbf{k}}\) = L2\(\hat{\mathbf{k}}\)
(ii) Plane of DEFAD is in Y – Z p1ane
So A2 = \(|A|\)\( [latex]\hat{\mathbf{k}}\) =L2\(\hat{\mathbf{i}}\)
A = A1 – A2 = L2(\(\hat{\mathbf{i}}\) + \(\hat{\mathbf{k}}\))
B = B0(\(\hat{\mathbf{i}}\) + \(\hat{\mathbf{k}}\))
So, Q = B A = B0( \(\hat{\mathbf{i}}\) + \(\hat{\mathbf{k}}\) )L2(\(\hat{\mathbf{i}}\) + \(\hat{\mathbf{k}}\) = B0 L2[\(\hat{\mathbf{i}}\).\(\hat{\mathbf{i}}\) + \(\hat{\mathbf{i}}\).\(\hat{\mathbf{k}}\) + \(\hat{\mathbf{k}}\).\(\hat{\mathbf{i}}\) + \(\hat{\mathbf{k}}\)\(\hat{\mathbf{k}}\)
=B0L2[1 + 0 + 0 + 1 ] (∴cos 90°= 0)
= 2B0L2Wb
Electromagnetic Induction MCQ Chapter 6 Question 3.
A cylindrical bar magnet is rotated about its axis in the figure. A wire is connected from the axis and is made to touch the cylindrical surface through a contact. Then
(A) a direct current flows in the ammeter A.
(B) no current flows through the ammeter A.
(C) an alternating sinusoidal current flows through the ammeter A with a time period 2jt/o).
(D) a time varying non-sinusoidal current flows through the ammeter A.
Answer:
(B) no current flows through the ammeter A.
Explanation:
The phenomenon of electromagnetic induction is used in this problem. Whenever the number of magnetic lines of force (magnetic flux) passing through a circuit changes (or a moving conductor cuts the magnetic flux), an emf is produced in the circuit (or emf induces across the ends of the conductor) is called induced emf.
The induced emf persists only as long as there is a change or cutting of flux. When cylindrical bar magnet is rotated about its axis, no change in flux linked with the circuit takes place, consequently no emf induces and hence, no current flows through the ammeter A. Hence the ammeter shows no deflection.
Chapter 6 Physics Class 12 MCQ Question 4.
There are two coils A and B as shown in figure. A current starts flowing in B as shown, when A is moved towards B and stops when A stops moving. The current in A is counter clockwise. B is kept stationary when A moves. We can infer that
(A) there is a constant current in the clockwise direction in A.
(B) there is a varying current in A.
(C) there is no current in A.
(D) there is a constant current in the counter clockwise direction in A.
Answer:
(D) there is a constant current in the counter clockwise direction in A.
Explanation:
When coil A moves towards coil B with constant velocity, so rate of change of magnetic flux due to coil B in coil A will be constant that gives constant current in coil A in same direction as in coil B by Lenz’s law.
Physics Class 12 Chapter 6 MCQ Question 5.
Same as problem 4 except the coil A is made to rotate about a vertical axis figure. No current flows in B if A is at rest. The current in coil A, when the current in B (at t = 0) is counter-clockwise and the coil A is as shown at this instant, t – 0, is
(A) constant current clockwise.
(B) varying current clockwise.
(C) varying current counter-clockwise.
(D) constant current counter-clockwise.
Answer:
(A) constant current clockwise.
Explanation:
In this case, the direction of the induced electromotive force/induced current is determined by the Lenz’s law. According to the Lenz’s law, the direction of induced emf or current in a circuit is such that it oppose x the cause that produces it. This law is based upon law of conservation of energy.
When’the current in coil B (at t = 0) is counter-clockwise and the coil A is considered above it. he counter-clockwise flow of the current in coil B is equivalent to North Pole of magnet and magnetic field lines are eliminating upward to coil A. When coil A starts rotating at f = 0, the current in coil A is constant along clockwise direction by Lenz’s rule.
MCQ On Electromagnetic Induction Class 12 Question 6.
The polarity of induced emf is defined by
(A) Ampere’s circuital law.
(B) Biot-Savart law.
(C) Lenz’s law.
(D) Fleming’s right hand rule.
Answer:
(C) Lenz’s law.
Explanation:
According to Lenz’s law, the direction of an induced e.m.f. always opposes the change in magnetic flux that causes the e.m.f.
Ch 6 Physics Class 12 MCQ Question 7.
Lenz’s law is consequence of the law of conservation of –
(A) Charge
(B) mass
(C) energy
(D) momentum
Answer:
(C) energy
Explanation:
Lenz’s law is a consequence of the law of conservation of energy. Lenz law says that induced current always tends to oppose the cause which produces it. So work is done against opposing force. This j work is transformed into electrical energy. ; So it a consequence of law of conservation of energy.
Electromagnetic Induction MCQ Class 12 Chapter 6 Question 8.
The magnetic flux linked with a coil is given by an equation ø = 5t2 + 2t + 3. The induced e.m.f. in the coil at the third second will be –
(A) 32 units
(B) 54 units
(C) 40 units
(D) 65 units
Answer:
(A) 32 units
Explanation:
Induced e.m.f = – d$/dt
= (5t2 + 2t + 3) = – (10t + 2) = – 32
MCQ Of Electromagnetic Induction Chapter 6 Question 9.
The self-inductance of a solenoid of length l and area of cross-section A, with a fixed number of turns N increases as –
(A) l and A increase.
(B) l decreases and A increases.
(C) l increases and A decreases.
(D) both l and A decrease.
Answer:
(B) l decreases and A increases.
Explanation:
As we know that,
L = μrμ0\(\frac{\mathrm{N}^{2} \mathrm{~A}}{l}\)
As L is constant for a coil,
L ∝ A and L ∝ –
As μr and N are constant here so, to increase L for a coil, area A must be increased and l must be decreased. So answer (B) is correct.
Important point:
The self and mutual inductance of capacitance and resistance depend on the geometry of the devices as well as permittivity/permeability of the medium.
MCQ Questions For Class 12 Physics Chapter 6 Question 10.
An iron-cored solenoid has self inductance 2.8H. When the core is removed, the self inductance becomes 2 mH. The relative permeability of the material of the core is –
(A) 1400
(B) 1200
(C) 2800
(D) 2000
Answer:
(A) 1400
Explanation:
μr = L/L0 = 2.8/(2 x 10-3) = 1400
Class 12 Physics Ch 6 MCQ Question 11.
In which of the following application, eddy current has no role to play?
(A) Electric power meters
(B) Induction furnace
(C) LED lights
(D) Magnetic brakes in trains
Answer:
(C) LED lights
Explanation:
LED is a p-n junction diode and emits light when forward biased.
Physics Chapter 6 Class 12 MCQ Question 12.
Which one of the following statements is wrong?
(A) Eddy currents are produced in a steady magnetic field.
(B) Eddy current is used to produce braking force in moving trains.
(C) Eddy currents is minimized by using a laminated core.
(D) Induction furnace uses eddy current to produce heat
Answer:
(A) Eddy currents are produced in a steady magnetic field.
Explanation:
Eddy current is produced when a metal is kept in a time-varying magnetic field.
Electromagnetic Induction MCQs Class 12 Question 13.
If the back e.m.f. induced in a coil, when current changes from 1A to zero in one millisecond, is 5 volts, the self-inductance of the coil is –
(A) 5 H
(B) 1 H
(C) 5 x 10-3 H
(D) 5 x 103H
Answer:
(C) 5 x 10-3 H
Explanation:
e = – L \(\frac {di}{dt}\)
5 = -L x \(\frac{0-1}{10^{-3}}\)
∴ L = 5 x 10-3H
Physics Class 12 MCQ Chapter 6 Question 14.
Magnetic field energy stored in a coil is
(A) Li2
(B) 1/2 Li
(C) Li
(D) 1/2 Li2
Answer:
(D) 1/2 Li2
Explanation:
If current I flows through a coil of self-inductance L, then magnetic field energy stored in it is 1/2 Li2
Class 12 Electromagnetic Induction MCQ Question 15.
If two coils of self inductance L1 and L2 are coupled together, their mutual inductance becomes
(A) M = k \(\sqrt{\mathrm{L}_{1} \mathrm{~L}_{2}}\)
(B) M = k \(\sqrt{\frac{\mathrm{L}_{1}}{\mathrm{~L}_{2}}}\)
(C) M = k \(\sqrt{\mathrm{L}_{1}+\mathrm{L}_{2}}\)
(D) None of the above
Answer:
(A) M = k \(\sqrt{\mathrm{L}_{1} \mathrm{~L}_{2}}\)
Explanation:
If two coils of self inductance Lj and L2 are coupled together, their mutualb inductance becomes M = \(\sqrt{\mathrm{L}_{1} \mathrm{~L}_{2}}\) where k = coupling constant whose value lies between 0 and 1.
MCQ Of Chapter 6 Physics Class 12 Question 16.
An inductor and a bulb are connected in series with a dc source. A soft iron core is then inserted in the inductor. What will happen to intensity of the bulb?
(A) Intensity of the bulb remains the same.
(B) Intensity of the bulb decreases.
(C) Intensity of the bulb increases.
(D) The bulb ceases to glow.
Answer:
(B) Intensity of the bulb decreases.
Explanation:
When a soft iron core is inserted in the inductor, the magnetic flux increases. According to Lenz’s law, it will be resisted by reducing the current Since the current reduces, the intensity of the bulb decreases.
Assertion And Reason Based MCQs ‘(1 Mark each)
Directions: In the following questions, A statement of Assertion (A) is followed by a statement of Reason (R). Mark the correct choice as.
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false and R is true
Chapter 6 Class 12 Physics MCQ Question 1.
Assertion(A): Faraday’s laws of electromagnetic induction are consequences of law of conservation of energy.
Reason (R): The parameter LR in a L-R circuit has the dimension of time.
Answer:
(C) A is true but R is false
Explanation:
In electromagnetic induction, magnetic energy, mechanical energy are converted into electrical energy. So, Faraday’s laws of electromagnetic induction are direct consequences of law of conservation of energy. Hence assertion is true.
In a L-R circuit, L/R parameter has the dimension of time. Hence the reason is false.
Class 12 Chapter 6 Physics MCQ Question 2.
Assertion (A): When two identical loops of copper and Aluminium are rotated with same speed in the same magnetic field, the induced e.m.f. will be same.
Reason (R): Resistance of the two loops are equal.
Answer:
(C) A is true but R is false
Explanation:
Induced e.m.f. in a rotating loop in a magnetic field depends on the area of the loop, number of turns, speed of rotation and magnetic field strength. It does not depend on the material of the coil. So, when two identical loops of copper and aluminium are rotated with same speed in the same magnetic field, the induced e.m.f. will be same. So, the assertion is true. Resistance of the two loops cannot be equal. Resistance of copper loop is less than that of the aluminium loop. So, the reason is false.
Electromagnetic Induction Class 12 MCQs Question 3.
Assertion (A): Lenz’s law does not violets the principle of conservation of energy.
Reason (R): Induce e.m.f. never opposes the change in magnetic flux that causes the e.m.f.
Answer:
(C) A is true but R is false
Explanation:
Lenz’s law is based on principle of conservation of energy. So/ the assertion is true. Induced e.m.f. always opposes the change in magnetic flux that causes the e.m.f. So, the – reason is also false.
Class 12 Physics Chapter 6 MCQ Questions Question 4.
Assertion (A): If the number of turns of a coil is increased, it becomes more difficult to push a bar magnet towards the coil.
Reason (R): The difficulty faced is according to Lenz’s law.
Answer:
(A) Both A and R are true and R is the correct explanation of A
Explanation:
As it is tried to push a bar magnet towards a coil, magnetic flux increases. According to Faraday’s law induced e.m.f. is generated. As the number of turns increases, induced e.m.f. increases. According to Lenz’s law. Induced e.m.f. always opposes the change in magnetic flux that causes the induction of e.m.f. So, the induced e.m.f. will oppose the motion of the bar magnet towards the coil. As the number of turns increases, opposition increases. Hence both assertion and reason are true and the reason explains the assertion properly.
Question 5.
Assertion (A): When the magnetic flux changes around a metallic conductor, the eddy current is produced.
Reason (R): Electric potential determines the flow of charge.
Answer:
(B) Both A and R are true but R is NOT the correct explanation of A
Explanation:
Change in flux induces emf in conductor which generates eddy current. So assertion is true.
Electric potential determines the flow of charge. So reason is also true. But reason is not the proper explanation of generation of eddy current.
Question 6.
Assertion (A): The cores of electromagnets are made of soft iron.
Reason (R): Coercivity of soft iron is small.
Answer:
(A) Both A and R are true and R is the correct explanation of A
Explanation:
The core of an electromagnet should be such that it gets magnetized easily. Also, it loses magnetism easily as soon as the magnetizing field is removed. Soft iron has this property. So, soft iron is used as the core electromagnet. So the assertion is true. Coercivity is a measure of the ability of a ferromagnetic substance to withstand external magnetic field without becoming demagnetized. For soft iron, it should be very low. Coercivity is low for soft iron. So, reason is also true. Also, reason properly explains the assertion.
Question 7.
Assertion (A): Mutual inductance becomes maximum when coils are wound on each other.
Reason (R): Mutual inductance is independent of orientation of coils.
Answer:
(C) A is true but R is false
Explanation:
Mutual inductance depends on size, number of turns, relative position and relative orientation of the 2 coils. So, when coils are wound on each other, the mutual inductance will be maximum. So, assertion is true, But the reason is false.
Question 8.
Assertion (A): Self inductance may be called the inertia of electricity.
Reason (R): Due to self inductance, opposing induced e.m.f. is generated in a coil as a result of change in current or magnetic flux linked with the coil.
Answer:
(B) Both A and R are true but R is NOT the correct explanation of A
Explanation:
Inertia is defined as the tendency of an object to resists its change of state of motion. Induced e.m.f. in a coil is changed by the change in current or magnetic flux. The property by which a coil opposes these parameters to incur any change in induced e.m.f. is known as self-inductance. Hence, self inductance may be called the inertia of electricity. So, the assertion and reason both are true but reason cannot explain why so happens.
Case-Based MCQs
Attempt any 4 sub-parts out of 5. Each sub-part carries 1 mark.
I. Read the following text and answer the following questions on the basis of the same:
Bottle Dynamo:
A bottle dynamo is a small generator to generate electricity to power the bicycle light.
Is is not a dynamo. Dynamo generates DC but a bottle dynamo generates AC. Newer models are now available with a rectifier. The available DC can power the light and small electronic gadgets. This is also known as sidewall generator since it operates using a roller placed on the sidewall of bicycle tyre. When the bicycle is in motion, the dynamo roller is engaged and electricity is generated as the type spins the roller.
When engaged, a dynamo requires the bicycle rider to exert more effort to maintain a given speed than would otherwise be necessary when the dynamo is not present or disengaged. Bottle dynamos can be completely disengaged during day time when cycle light is not in use. In wet conditions, the roller on a bottle dynamo can slip against the surface of the type, which interrupts the electricity generated. This cause the lights to go out intermittently.
Question 1.
Why bottle dynamo is not a dynamo ?
(A) It generates AC only
(B) It generates DC only
(C) It looks like a bottle
(D) It requires no fuel to operate
Answer:
(A) It generates AC only
Explanation:
Dynamo generates DC. But bottle dynamo generates AC. So, it is not a dynamo in that sense. But, it generates electricity for bicycle light.
Question 2.
Can you recharge the battery of your mobile phone with the help of bottle dynamo ?
(A) Yes
(B) No
(C) Yes, when a rectifier is used
(D) Yes, when a transformer is used
Answer:
(C) Yes, when a rectifier is used
Explanation:
Newer models of bottle generators are now available with a rectifier. DC available from such bottle generator can be used directly for charging mobile phone. Otherwise with the old models, a rectifier is to be attached to convert AC to DC.
Question 3.
Bottle generator generates electricity:
(A) when fuel is poured in the bottle.
(B) when cycle is in motion.
(C) when it is mounted properly.
(D) when wind blows.
Answer:
(B) when cycle is in motion.
Explanation:
Bottle generator is also known as sidewall generator since it operates using a roller placed on the sidewall of bicycle tyre. When the bicycle is in motion, the dynamo roller is engaged and electricity is generated as the tyre spins the roller.
Question 4.
Bulb of bicycle light glows:
(A) with AC supply only.
(B) with DC supply only.
(C) with both AC and DC supply.
(D) only when AC supply is rectified.
Answer:
(C) with both AC and DC supply.
Explanation:
Normal lamps work with both AC and DC. So, bottle generators of older model or newer model can be directly used for bicycle lamp.
Question 5.
Which one of the following is not an advantages of newer model of bottle dynamo ?
(A) Works intermittently when it roller slips on type
(B) Small electronic gadgets can be charged
(C) Can be easily disengaged during day time
(D) Requires no fuel
Answer:
(A) Works intermittently when it roller slips on type
Explanation:
In wet conditions, the roller on a bottle dynamo (old model or new model) can slip against the surface of the tyre, which interrupts the electricity generated. This causes the lights to go out intermittently. This is not an advantage.
II. Read the following text and answer the following questions on the basis of the same:
Electromagnetic damping:
Take two hollow thin cylindrical pipes of equal internal diameters made of aluminium and PVC, respectively. Fix them vertically with clamps on retort stands. Take a small cylindrical magnet having diameter slightly smaller than the inner diameter of the pipes and drop it through each pipe in such a way that the magnet does not touch the sides of the pipes during its fall.
You will observe that the magnet dropped through the PVC pipe takes the same time to come out of the pipe as it would take when dropped through the same height without the pipe. Now instead of PVC pipe use an aluminium pipe. Note the time it takes to come out of the pipe in each case. You will see that the magnet takes much longer time in the case of aluminium pipe.
Why is it so ? It is due to the eddy currents that are generated in the aluminium pipe which oppose the change in magnetic flux, i.e., the motion of the magnet. The retarding force due to the eddy currents inhibits the motion of the magnet. Such phenomena are referred to as electromagnetic damping. Note that eddy currents are not generated in PVC pipe as its material is an insulator whereas aluminium is a conductor. This effect was discovered by physicist Foucault (1819-1868).
Question 1.
Eddy current is generated in a:
(A) metallic pipe.
(B) PVC pipe.
(C) glass pipe.
(D) wooden pipe.
Answer:
(A) metallic pipe.
Explanation:
Eddy currents are not generated in non-conductor/insulator. Eddy currents are generated in conductor/metal.
Question 2.
Eddy current was first observed by:
(A) Helmhotlz
(B) Foucault
(C) D’Arsonval
(D) Shock ley
Answer:
(B) Foucault
Explanation:
The generation of eddy current was discovered by physicist Foucault (1819-1869).
Question 3.
What is electromagnetic damping ?
(A) Generation of electromagnetic wave during the passage of a magnet through a metal pipe
(B) Change of the direction of propagation of electromagnetic wave due to a variable magnetic flux
(C) Change of the frequency of electromagnetic wave due to a variable magnetic flux
(D) To slow down the motion of a magnet moving through a metal pipe due to electromagnetically induced current.
Answer:
(D) To slow down the motion of a magnet moving through a metal pipe due to electromagnetically induced current.
Explanation:
The retarding force due to the eddy currents inhibits the motion of the magnet in a metal pipe. This phenomena is known as electromagnetic damping.
Question 4.
To observe electromagnetic damping a magnet should be dropped through a metal pipe and:
(A) the magnet should not touch inner wall of the pipe.
(B) the magnet should touch the inner wall of the pipe.
(C) it does not matter whether the magnet touches the inner wall of the pipe or not.
(D) the magnet should be larger in size than the diameter of the pipe.
Answer:
(A) the magnet should not touch inner wall of the pipe.
Explanation:
To observe electromagnetic damping, a magnet should be dropped through a metal pipe and the magnet should not touch the inner wall of the pipe.
Question 5.
A piece of wood and a bar magnet of same dimension is dropped through an aluminium pipe. Which of the following statements is true ?
(A) The piece of wood will take more time to come out from the pipe.
(B) The bar magnet will take more time to come out from the pipe.
(C) Both will take same time to come out from the pipe.
(D) The time required will depend on the mass of the wooden piece and the mass of the bar magnet.
Answer:
(B) The bar magnet will take more time to come out from the pipe.
Explanation:
When a piece of wood and a bar magnet of same dimension is dropped through an aluminium pipe, the bar magnet will take more time to come out from the pipe due to electromagnetic damping.
II. Read the following text and answer the following questions on the basis of the same:
Spark coil:
The principle of electromagnetic induction was discovered by Michael Faraday in 1831. Induction coils were used widely in electrical experiments and for medical therapy during the last half of the 19th century, eventually leading to the development of radio in the 1890’s. The spark coil designed on the principle of electromagnetic induction was the heart of the earliest radio transmitters.
Marconi used a spark coil designed by Heinrich Rhumkorff in his early experiments. An induction coil or “spark coil” is a type of electrical transformer used to produce high-voltage pulses from a low-voltage (DC) supply. To create the flux changes necessary to induce voltage in the secondary coil, the direct current in the primary coil is repeatedly interrupted by a vibrating mechanical contact called interrupter.
The spark scoil consists of two coils of insulated wire wound around a common iron core. One coil, called the primary coil, is made from relatively few (tens or hundreds) turns of coarse wire. The other coil, the secondary coil typically consists of up to a million turns of fine wire (up to 40 gauge). An electric current is passed through the primary, creating a magnetic field. Because of the common core, most of the primary’s flux couples with the secondary. When the primary current is suddenly interrupted, the magnetic field rapidly collapses.
This causes a high voltage pulse to be developed across the secondary terminals due to electromagnetic induction. Because of the large number of turns in the secondary coil, the secondary voltage pulse is typically many thousands of volts. This voltage is sufficient to create an electric spark, to jump across an air gap separating the secondary’s output terminals. For this reason, this induction coils are also called spark coils.
To operate the coil continually, the DC supply current must be repeatedly connected and disconnected. To do that, a magnetically activated vibrating arm called an interrupter is used which rapidly connects and breaks the current flowing into the primary coil. The interrupter is mounted on the end of the coil next to the iron core. When the power is turned on, the produced magnetic field attracts the armature.
When the armature has moved far enough, contacts in the primary circuit breaks and disconnects the primary current. Disconnecting the current causes the magnetic field to collapse and create the spark. A short time later the contacts reconnect, and the process repeats. An arc which may form at the interrupter contacts is undesirable. To prevent this, a capacitor of 0.5 to 15 IUF is connected across the primary coil.
Question 1.
The heart of the radio transmitters of Marconi was a
(A) spark coil.
(B) toroid.
(C) RF tuning coil.
(D) Van de Graff generator.
Answer:
(A) spark coil.
Explanation:
The spark coil designed on the principle of electromagnetic induction was the heart of the earliest radio transmitters. Marconi usecl a spark coil designed by Heinrich Rhumkorff in his early experiments.
Question 2.
Spark coil is a type of –
(A) electrical generator.
(B) electrical transformer.
(C) static electricity generator.
(D) large capacitor.
Answer:
(B) electrical transformer.
Explanation:
A spark coil is a type of 1 electrical transformer used to produce high- 1 voltage pulses from a low-voltage (DC) g supply. To create the flux changes necessary to induce voltage in the secondary coil, the direct current in the primary coil is repeatedly interrupted by a vibrating mechanical contact called interrupter.
Question 3.
Which of the following statements is correct?
(A) Spark coil consists of two coils of insulated wire. Primary coil, is made from relatively few turns of fine wire. The secondary coil consists of up to a million turns of coarse wire.
(B) Spark coil consists of two coils of insulated wire. Primary coil, is made from a (tens or million turns of coarse wire. The secondary coil consists of up to a few turns of fine wire.
(C) Spark coil consists of two coils of insulated wire. Primary coil, is made from relatively few turns of coarse wire. The secondary coil consists of up to a million turns of fine wire.
(D) Spark coil consists of two coils of insulated wire. Both primary and secondary coil, is made from a million turns of fine wire.
Answer:
(C) Spark coil consists of two coils of insulated wire. Primary coil, is made from relatively few turns of coarse wire. The secondary coil consists of up to a million turns of fine wire.
Explanation:
The spark coil consists of two coils of insulated wire wound around a common iron core. One coil, called the primary coil, is made from relatively few (tens or hundreds) turns of coarse wire. The other coil, the secondary coil typically consists of up to a million turns of fine wire (up to 40 gauge).
Question 4.
Why most of the primary’s flux couples with the secondary in spark coil?
(A) Since the primary coil is wound on the secondary coil
(B) Since the primary coil is of thick wire
(C) Since the core is common
(D) None of the above
Answer:
(C) Since the core is common
Explanation:
The spark coil designed on the principle of electromagnetic induction was die heart of the earliest radio transmitters. Marconi used a spark coil designed by Heinrich Rhumkorff in his early experiments.
Question 5.
What is the function of interrupter in a spark coil?
(A) To rapidly connect and break the current flowing into the primary coil
(B) To rapidly connect and break the current flowing into the secondary coil
(C) to control the formation of spark
(D) None of the above
Answer:
(A) To rapidly connect and break the current flowing into the primary coil
Explanation:
To operate the coU continually, the DC supply current must be repeatedly connected and disconnected. To do that, a magnetically activated vibrating arm called an interrupter is used which rapidly connects and breaks the current flowing into the primary coil.