Advance Physics MCQs – Electromagnetic Induction ( Advance Physics ) MCQs
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Advance Physics MCQs – Electromagnetic Induction ( Advance Physics ) MCQs
The most occurred mcqs of Electromagnetic Induction ( ) in past papers. Past papers of Electromagnetic Induction ( Advance Physics ) Mcqs. Past papers of Electromagnetic Induction ( Advance Physics ) Mcqs . Mcqs are the necessary part of any competitive / job related exams. The Mcqs having specific numbers in any written test. It is therefore everyone have to learn / remember the related Electromagnetic Induction ( Advance Physics ) Mcqs. The Important series of Electromagnetic Induction ( Advance Physics ) Mcqs are given below:
Faraday and Henry Experiments
1. What is the instrument used in Faraday’s experiment?
a) Galvanometer
b) Ammeter
c) Voltmeter
d) Meter Bridge
Answer: a
Explanation: Faraday used a galvanometer and connected it to a coil. A bar magnet was pushed towards the coil, such that the north-pole is pointing towards the coil. As the bar magnet is shifted, the pointer in the galvanometer gets deflected, thus indicating the presence of current in the coil.
2. Which among the following affects the deflection in the galvanometer?
a) Area of the coil
b) Current passing through the coil
c) Speed with which the bar magnet is pulled towards or away from the coil
d) Resistance offered for current flow
Answer: c
Explanation: The deflection of the pointer is larger or smaller depending upon the speed with which the bar magnet is pulled towards or away from the coil. Moreover, the direction of deflection of the pointer depends upon the direction of motion of the bar magnet.
3. In the second experiment, Faraday replaced ‘X’ with ‘Y’ for checking the deflection in the galvanometer. Identify X and Y.
a) X ➔ Current carrying coil; Y ➔ Bar magnet
b) X ➔ Bar magnet; Y ➔ Current carrying coil
c) X ➔ Bar magnet; Y ➔ Battery
d) X ➔ Current carrying coil; Y ➔ Battery
Answer: b
Explanation: In the second experiment, Faraday replaced the bar magnet by a second current carrying coil that was connected to a battery. As we move the second coil towards the primary coil, the pointer in the galvanometer undergoes deflection, which indicates the presence of the electric current in the first coil.
4. The galvanometer shows deflection even when the bar magnet is stationary.
a) True
b) False
Answer: b
Explanation: No, this statement is false. When the bar magnet is stationary, the pointer shows no deflection and the motion lasts only till the magnet is in motion. This shows that only the relative motion between the magnet and the coil is responsible for the generation of current in the coil.
5. Find the true statement.
a) The relative motion between the coils was not compulsory for the current in the primary to be generated
b) The direction of deflection of the pointer does not depend upon the direction of motion of the secondary coil towards or away from the primary coil in the second experiment
c) When the south-pole of the bar magnet is moved towards or away from the coil, the deflections in the galvanometer are same to that observed with the north-pole for similar movements
d) There is a difference in effect when the bar magnet is kept stationary and the coil is in motion
Answer: a
Explanation: The relative motion between the coils was not really necessary for the current in the primary to be generated. In the third experiment, Faraday placed two stationary coils and connected one of them to the galvanometer and the other to a battery, through a push button. As the button was pressed, the galvanometer in the other coil showed a deflection, indicating the presence of current in that coil. All the other statements are not valid.
Magnetic Flux
1. Pick out the SI unit of magnetic flux.
a) Ampere
b) Tesla meter
c) Weber
d) Maxwell
Answer: c
Explanation: The magnetic flux through a surface is the surface integral of the normal component of the magnetic field flux density passing through that surface. The SI unit of magnetic flux is weber (Wb). One weber is the flux produced when a uniform magnetic field of one tesla acts normally over an area of 1 cm2.
2. Identify the correct dimensions of magnetic flux.
a) [M3 L2 A-1 T-2]
b) [M L2 A-1 T-2]
c) [M2 L2 A-1 T-2]
d) [M L2 A-1 T2]
Answer: b
Explanation: Magnetic flux = Magnetic field × Area.
Flux = ([MLT−2].[L2])(AL)
Flux = [ML2T−2][A]
Flux = [M L2 A-1 T-2].
3. Which of the following is a unit of magnetic flux?
a) Maxwell
b) Gauss
c) Tesla
d) Ampere
Answer: a
Explanation: The CGS unit of magnetic flux is Maxwell (Mx). One Maxwell is the flux produced when a uniform magnetic field of one gauss acts normally over an area of 1 cm2.
4. Find out the correct relation from the following.
a) 1 Wb = 1 G cm
b) 1 Wb = 1 Tm
c) 1 Wb = 1 G cm2
d) 1 Wb = 108 Maxwell
Answer: d
Explanation: 1 Wb = 1 T × 1 m2
Wb = 104 G × 104 cm2.
1 Wb = 108 Maxwell.
5. When is the magnetic flux said to be positive?
a) θ = 180o
b) θ = 360o
c) θ = 0o
d) θ = 90o
Answer: c
Explanation: A normal to a plane can be drawn from either side. If the normal drawn to a plane points out in the direction of the field, then θ = 0o and the flux are taken as positive. If the flux is positive and increasing the voltage will be negative.
6. The total number of magnetic lines of force crossing the surface placed in a magnetic field normally is called the magnetic induction.
a) True
b) False
Answer: b
Explanation: The magnetic flux through any surface place in a magnetic field is the total number of magnetic lines of force crossing this surface normally. It is measured as the product of the component of the magnetic field normal to the surface and the surface area.
7. Calculate the magnetic flux when the magnetic field is perpendicular to the surface area.
a) Minimum
b) Maximum
c) Zero
d) Depends on the surface area
Answer: b
Explanation: Magnetic flux = B ΔS cos 00
Magnetic flux = B ΔS.
It follows that magnetic flux linked with a surface is maximum when the direction of the magnetic field is perpendicular to the surface area.
8. Which type of physical quantity is magnetic flux?
a) Scalar
b) Vector
c) Isotropic
d) Isentropic
Answer: a
Explanation: The magnetic flux is measured as the product of the component of the magnetic field normal to the surface and the surface area.
Magnetic flux is a scalar quantity.
9. Calculate the magnetic flux produced when the magnetic field is parallel to the surface area.
a) Maximum
b) Minimum
c) Zero
d) Depends on the magnetic field
Answer: c
Explanation: Magnetic flux = B ΔS cos 90o
Magnetic flux = 0.
It follows that magnetic flux linked with a surface is zero when the direction of the magnetic field is parallel to the surface area.
10. When is the magnetic flux said to be negative?
a) θ = 180o
b) θ = 360o
c) θ = 90o
d) θ = 0o
Answer: a
Explanation: If the normal points in the opposite direction of the field, then θ = 180o and the flux is taken as negative. If the magnetic field is pointing opposite the direction of the surface area then the value of this flux is negative.
Faraday’s Law of Induction
1. Which among the following is true about Faraday’s law of Induction?
a) An emf is induced in a conductor when it cuts the magnetic flux
b) An emf is induced in a conductor when it moves parallel to the magnetic field
c) An emf is induced in a conductor when it moves perpendicular to the magnetic field
d) An emf is induced in a conductor when it is just entering a magnetic field
Answer: a
Explanation: According to Faraday’s law of electromagnetic induction, an emf is induced in a conductor when it cuts across the flux of a magnetic field. If the two ends of the conductor are connected to an outside circuit, the induced emf causes current to flow in the circuit.
2. What is proportional to the magnitude of the induced emf in the circuit?
a) Rate of change of current in the circuit
b) Rate of change of resistance offered
c) Rate of change of magnetic flux
d) Rate of change of voltage
Answer: c
Explanation: The magnitude of induced emf is equal is equal to the time rate of change of magnetic flux. It is mathematically expressed as:
ε = −dϕdt
The negative sign indicates the direction of the emf induced. This is Faraday’s second law of electromagnetic induction.
3. Faraday’s laws are result of the conservation of which quantity?
a) Momentum
b) Energy
c) Charge
d) Magnetic field
Answer: b
Explanation: Faraday’s laws are result of the conservation of energy. These laws are based on the conversion of electrical energy into mechanical energy. Mechanical energy can be converted into electrical energy such as in the example of a dynamo. In the same way, electrical energy can be converted into mechanical energy such as in the example of electric motor. Both of the above examples work on the principle of Faraday’s law.
4. The induced emf persists only as long as the change in magnetic flux continues.
a) True
b) False
Answer: a
Explanation: According to Faraday’s first law, whenever the amount of magnetic flux linked with a circuit changes, an emf is induced in the circuit. This induced emf persists as long as he change in magnetic flux continues. Therefore, this is a true statement.
5. The magnetic flux in a closed circuit of resistance 20 Ω varies with time t as Φ = 4t3 + 2t2 – 15t + 3. Calculate the magnitude of induced emf at t = 1s.
a) 3 V
b) 4 V
c) 5 V
d) 6 V
Answer: b
Explanation: Given: Resistance = 20 Ω; Φ = 4t3 + 2t2 – 15t + 3
Required equation ➔ ε = −dΦdt
ε=−d(4t3+2t2–15t+3)dt = -(12t2 + 4t – 15t + 3)
= -12 – 4 + 15 – 3 (Since t = 1s)
= 4 V
Therefore, the magnitude of induced emf is 4V.
Lenz’s Law and Conservation of Energy
1. A closed-loop move normal to the constant electric field between the plates of a large capacitor. What is the amount of current produced when it is wholly inside the region between the capacitor plates?
a) Maximum
b) Minimum
c) Zero
d) Independent of current
Answer: c
Explanation: No current is produced when it is wholly inside the region between the capacitor plates. Current cannot be induced by changing electric flux. So, the amount of current produced when it is wholly inside the region between the capacitor plates is zero.
2. Which law is used in finding the direction of current in an a.c. generator?
a) Maxwell’s law
b) Lenz’s law
c) Corkscrew law
d) Ampere circuital law
Answer: b
Explanation: In an a.c. generator, induced current due to a change of magnetic flux linked with a closed circuit can be found out using Lenz’s law. Lenz’s law, in electromagnetism, statement that an induced electric current flows in a direction such that the current opposes the change that induced it.
3. Which of the following statement is valid?
a) Lenz’s law is a consequence of the law of conservation of energy
b) Lenz’s law is a consequence of the law of conservation of momentum
c) Lenz’s law is a consequence of the law of conservation of force
d) Lenz’s law is a consequence of the law of conservation of mass
Answer: a
Explanation: Whether a magnet is moved towards or away from a closed coil, the induced current always opposes the motion of the magnet, as predicted by Lenz’s law. Work has to be done in moving the magnet closer to the coil against this force of repulsion. Thus Lenz’s law is valid and is a consequence of the law of conservation of energy.
4. The current in a wire passing normally through the center of a conducting loop is increasing at a constant rate. What is the net current induced in the loop?
a) Indefinite
b) Maximum
c) Minimum
d) Zero
Answer: d
Explanation: The magnetic lines of force due to current are parallel to the plane of the loop. The flux linked with the loop is zero. Hence no current is induced in the loop.
5. Which of the following is found using Lenz’s law?
a) Induced emf
b) Induced current
c) The direction of induced emf
d) The direction of alternating current
Answer: c
Explanation: Lenz’s law is a general law for determining the direction of induced emf and hence that of induced current in a circuit. Lenz’s law states that the direction of an induced emf will be such that if it were to cause a current to flow in a conductor in an external circuit, then that current would generate a field that would oppose the change that created it.
6. Lenz’s law is invalid. State true or false.
a) True
b) False
Answer: b
Explanation: When a magnet is moved towards or away from a closed coil, the induced current always opposes the motion of the magnet, as predicted by Lenz’s law. Thus Lenz’s law is valid and is a consequence of the law of conservation of energy.
7. ‘X’ states that the direction of induced current in a circuit is such that it opposes the cause or the change which produces it. Identify X.
a) Faraday’s law
b) Lenz’s law
c) Maxwell’s law
d) Ampere’s law
Answer: b
Explanation: Lenz’s law states that the direction of induced current in a circuit is such that it opposes the cause or the change which produces it. It is a general law for determining the direction of induced emf and hence that of induced current in a circuit.
8. Identify the law used to find the direction of eddy currents.
a) Lenz’s law
b) Maxwell’s law
c) Ampere’s law
d) Faraday’s law
Answer: a
Explanation: Eddy currents are the currents induced in solid metallic masses when the magnetic flux threading through them changes. Eddy currents also oppose the change in magnetic flux, so their direction is given Lenz’s law.
Motional Electromotive Force
1. Identify the expression for the motional electromotive force from the following?
a) E = -vLB
b) E = vLB
c) E = vLB
d) E = LBv
Answer: a
Explanation: Motional electromotive is the emf induced by the motion of the conductor across the magnetic field. The expression for motional electromotive force is given by:
E = -vLB
This equation is true as long as the velocity, magnetic field, and length are mutually perpendicular to each other. The negative sign is associated with Lenz’s law.
2. A bar of length 0.7 m slides along metal rails at a speed of 1 m/s. The bar and rails are in a magnetic field of 20 T, pointing out into the page. Calculate the motional emf.
a) 0.7 V
b) 7 V
c) 14 V
d) 1.4 V
Answer: c
Explanation: Length (L) = 0.7 m; Speed (v) = 1 m/s; Magnetic field (B) = 20 T
The required equation E = -vLB (The negative sign only applies to the direction)
E = 1 × 0.7 × 20
E = 0.7 × 20
E = 14 V
Therefore, the motional emf in the bar and rails is 14 V.
3. A bar of length 0.15 m slides along metal rails at a speed of 5 m/s. The bar and rails are in a magnetic field of 40 T, pointing out into the page. The resistance of two resistors in parallel is both 20 Ω, and the resistance of the bar is 5 Ω. What is the current in the bar?
a) 1 A
b) 2 A
c) 3 A
d) 5 A
Answer: b
Explanation: L = 0.15 m; B = 40 T; v = 5 m/s; R1 = 10 Ω; R2 = 10 Ω; R3 = 5 Ω
Emf (E) = vLB = 5 × 0.15 × 40
E = 5 × 6
E = 30 V
R1 and R2 are in parallel ➔ 1R=1R1+1R2=120+120=220=110 ➔ R = 10 Ω
RTOT = R + R3 = 10 + 5 = 15 Ω
Therefore, current (I) = ER
I = 3015
I = 2 A
4. Induced emf and motional emf are exactly the same.
a) True
b) False
Answer: b
Explanation: No. they are not exactly the same. According to Faraday’s Law, an induced emf is created whenever there’s a changing magnetic flux through a loop. If the changing emf is due to some kind motion of a conductor in a magnetic field, then it would be called as motional emf.
5. A bar of length 2m is said to fall freely in a magnetic field of magnitude 50 T. What is the motional emf in the bar when it has fallen 40 meters?
a) 700 V
b) 2100 V
c) 2800 V
d) 1400 V
Answer: c
Explanation: Given: L = 2 m; B = 50 T;
The bar is said to be falling freely, so ➔ u = 0
According to Newton’s third equation ➔ v2 – u2 = 2gh
v2 = 2gh; h = 40 m
v = (2×9.8×40)−−−−−−−−−−−√
v = 28 m/s
Motional emf (E) = vLB
E = 28 × 2 × 50
E = 2800 V
6. A metal rod is forced to move with constant velocity along two parallel metal rails, connected with a strip of metal at one end across a magnetic field (B) of 0.5 T, pointing out of the page. The rod is of length 45 cm and the speed of the rod is 70 cm/s. The rod has a resistance of 10 Ω and the resistance of the rails and connector is negligible. What is the rate at which energy is being transferred to thermal energy?
a) 0.225 W
b) 22.55 W
c) 2.25 × 10-4 W
d) 2.25 × 10-3 W
Answer: d
Explanation: Given: B = 0.5 T; v = 70 cm/s = 70 × 10-2 m/s; L = 45 cm = 45 × 10-2 m; R = 10 Ω
Motional emf (E) = vLB = 70 × 10-2 × 45 × 10-2 × 0.5
E = 0.15 V
Current (I) = ER
I = 0.1510
I = 0.015 A
Rate of energy or power (P) = I2R
P = 0.0152 × 10
P = 2.25 × 10-3 W
Therefore, the rate of energy transfer is 2.25 × 10-3 W.
7. Find the true statement.
a) Motional emf is inversely proportional to speed of electric conductor
b) Motional emf is inversely proportion to the length of the conductor
c) Motional emf is directly proportional to the magnetic field
d) Motional emf is inversely proportional to the magnetic field
Answer: c
Explanation: Motional emf is directly proportional to the magnetic field across which the electric conductor moves. It is also directly proportional to the velocityspeed of the electric conductor as well as to the length of the conductor. All the other statements are not valid.
Quantitative Study
1. Identify the significance of energy consideration from the following.
a) Energy consideration provides a link between the force and energy
b) Energy consideration provides a link between the current and energy
c) Energy consideration provides a link between the mechanical and thermal energy
d) Energy consideration provides a link between the voltage and energy
Answer: a
Explanation: Energy consideration provides a link the force and energy. Newton’s problems can also be easily solved with the help of energy consideration. This is the significance of energy consideration.
2. Pick out the expression for power in a rectangular conductor from the following.
a) P=B2l2vR
b) P=B2lv2R
c) P=B2l2v2R
d) P=B2l2v2R2
Answer: c
Explanation: In a rectangular conductor, the emf (E) is given ➔ Blv
So, current (I) = ER=BlvR
Power (P) = I2R
P=B2l2v2R
The work done is mechanical and this mechanical energy is dissipated as Joule heat.
3. A circular coil having an area of 3.14 × 10-2 m2 and 30 turns is rotated about its vertical diameter with an angular speed of 70 rad-1 in a uniform horizontal magnetic field of magnitude 5 × 10-2 T. If the coil forms a closed loop of resistance 15 Ω, what is the average power loss due to Joule heating?
a) 0.360
b) 0.362
c) 0.724
d) 0.726
Answer: b
Explanation: Given: N = 30; A = 3.14 × 10-2 m2 ; Angular velocity (ω) = 70 rad-1; B = 5 × 10-2 T; R = 15 Ω
Emf (E) = NABω
E = 30 × 3.14 × 10-2 × 5 × 10-2 × 70
E = 3.297 V
Current (I) = ER
I = 3.29715
I = 0.2198 A
Average power loss (P) = I2R2
P = 0.21982 × 152
P = 0.362 W
Therefore, the average power loss due to Joule heating is 0.362 W.
4. Mechanical energy is initially converted into thermal energy in the case of motional emf.
a) True
b) False
Answer: b
Explanation: No, this statement is false. The work done is mechanical and the energy is dissipated as mechanical energy. Then it is initially converted into electric energy and then only finally converted into thermal energy.
Moving Charges And Magnetism MCQs
5. What is the magnitude of the force on a current carrying conductor of length L in a perpendicular magnetic field B?
a) F=IlB
b) F=IlB2
c) F=I2 lB
d) F=IlB
Answer: d
Explanation: The current passing through conductor in a perpendicular magnetic field is given as:
I=εr
I=Blvr
Since it is a perpendicular magnetic field, the force ➔ I(l×B) is directed outwards in the direction opposite to the velocity of the rod. The magnitude of the force is given as:
F=IlB
We can also write it as ➔ F=IlB=B2l2vr
6. Find the true statement.
a) The force on a current carrying conductor in a perpendicular magnetic field is due to the thermal energy dissipated
b) The force on a current carrying conductor in a perpendicular magnetic field is due to the mechanical energy dissipated
c) The force on a current carrying conductor in a perpendicular magnetic field is due to the current passing through it
d) The force on a current carrying conductor in a perpendicular magnetic field is due to the electrical energy initially converted from mechanical energy
Answer: d
Explanation: The force on a current carrying conductor in a perpendicular magnetic field is due to drift velocity of charges along the current carrying conductor and the consequent Lorentz force acting on it. So, it is due to the electric energy which was the energy that mechanical energy dissipated in the system was initially converted to.
7. Identify the relation between charge flow and change in magnetic flux.
a) ∆Q = ∆ΦB×r
b) ∆Q = 2∆Φ;B r
c) ∆Q = △ΦBr
d) ∆Q = △ΦB2r
Answer: c
Explanation: There is a relationship between the charge flow through the circuit and the change in the magnetic flux. According to Faraday’s Law, the induced emf is given by:
|ε|=△ΦB△t
But, |ε|=Ir=(△Q△t)r
Therefore, the relation is given by:
∆Q = △ΦBr
Eddy Currents
1. Name the current induced in solid metallic masses when the magnetic flux threading through them changes.
a) Ampere currents
b) Faraday currents
c) Eddy currents
d) Solenoidal currents
Answer: c
Explanation: Eddy currents are the currents induced in solid metallic masses when the magnetic flux threading through them changes. These currents look like eddies or whirlpools in water and so they are known as eddy currents.
2. Identify the type of commercial motor which works as a consequence of eddy currents.
a) Compressors
b) Induction motors
c) Turbines
d) Hydropowered motors
Answer: b
Explanation: A rotating magnetic field is produced employing two single-phase currents. A metallic rotor placed inside the rotating magnetic field starts rotating due to large eddy currents produced in it. These motors are commonly used in fans.
3. Which of the following is not an application of eddy currents?
a) Ammeter
b) Dead beat galvanometer
c) Speedometer
d) Energy meters
Answer: a
Explanation: Although eddy currents are undesirable, still they find applications in the following devices: Induction furnace, Electromagnetic damping, Electric brakes, Speedometers, Induction motor, Electromagnetic shielding, and energy meters.
4. What is measured by the eddy currents induced in energy meters?
a) Electric potential
b) Electric induction
c) Electric power
d) Electric energy
Answer: d
Explanation: In energy meters used for measuring electric energy, the eddy currents induced in an aluminum disc are made use of. Therefore, electric energy is measured by the eddy currents induced in energy meters.
5. Eddy currents can be used to heat localized tissues of the human body. What is this branch of science referred to as?
a) Inductology
b) Eddy dynamics
c) Inductothermy
d) Eddy mechanics
Answer: c
Explanation: Eddy currents can be used to heat localized tissues of the human body. This branch is called inductothermy. Inductothermy is a method of electrotherapeutics in which certain parts of the body of the patient are heated by an alternating, predominantly high-frequency electromagnetic field, which induces eddy currents in body tissues.
6. Eddy currents may not be suitable for electrostatic shielding.
a) True
b) False
Answer: b
Explanation: No, Eddy currents can be used for electromagnetic shielding. Electromagnetic shielding is the practice of reducing the electromagnetic field in a space by blocking the field with barriers made of conductive or magnetic materials. The higher the conductivity of the sheet used, the better the shielding of the transient magnetic field.
7. ‘X’ set up in the drum exert a torque on the drum to stop the train. Identify X.
a) Induction
b) Eddy currents
c) Electromagnetic oscillations
d) Air pressure
Answer: b
Explanation: A strong magnetic field is applied to the rotating drum attached to the wheel. The eddy currents set up in the drum exert a torque on the drum to stop the train. That is the reason eddy currents are set up in the drum of the train wheel.
8. Identify the law which is used to find out the direction of eddy currents.
a) Lenz’s law
b) Faraday’s law
c) Ampere circuital law
d) Maxwell’s law
Answer: a
Explanation: Eddy currents are the currents induced in solid masses when the magnetic flux threading through them changes. Eddy currents also oppose the change in magnetic flux, so their direction is given by Lenz’s law.
9. How can electromagnetic damping be increased?
a) Increasing the magnetic density
b) Reduce viscosity
c) Winding the coil on the aluminum frame
d) Increase the temperature of the coil
Answer: c
Explanation: The electromagnetic damping can be increased by winding the coil on a lighter copper or aluminum frame. As the frame moves in the magnetic field, eddy currents are set up in the frame which resists the motion of the coil.
Electromagnetic Inductance
1. Give the SI unit of self-inductance.
a) Farad
b) Ampere
c) Henry
d) Maxwell
Answer: c
Explanation: The self-inductance of a coil is said to be one henry if an induced emf of one volt is set up in it when the current in it changes at the rate of one ampere per second. Self inductance is defined as the induction of a voltage in a current-carrying wire when the current in the wire itself is changing.
2. Write the dimensions of self-inductance.
a) [M3 L2 T-2 A-2]
b) [M L2 T-2 A-2]
c) [M2 L2 T-2 A-2]
d) [M L2 T2 A-2]
Answer: b
Explanation: Self-inductance = MagneticfluxCurrent
[ML2T−2][A2]
Self-inductance = [M L2 T-2 A-2]
3. Identify the factor on which self-inductance does not depend.
a) Number of turns
b) Area of cross-section
c) Permeability of the core material
d) The permittivity of the core material
Answer: d
Explanation: Self-inductance is directly proportional to the number of turns and the area of cross-section. The self-inductance of a solenoid increases μr times if it is wound over an iron core of relative permeability μr.
4. Which of the following is the unit of mutual inductance?
a) VsA-2
b) V3sA2
c) V2s
d) VsA-1
Answer: d
Explanation: Mutual Inductance is the interaction of one coils magnetic field on another coil as it induces a voltage in the adjacent coil. Unit of Mutual inductance = VAs−1.
SI unit = VsA-1.
5. Identify the factor on which mutual inductance does not depend.
a) Relative separation
b) The relative orientation of the two coils
c) Reciprocity
d) Permeability of the core material
Answer: c
Explanation: The mutual inductance of two coils is the property of their combination. It does not matter which one of them functions as the primary or the secondary coil. Hence, mutual inductance does not depend on reciprocity.
6. Mutual inductance is called the inertia of electricity.
a) True
b) False
Answer: b
Explanation: Self-induction of a coil is that the property by which it tends to take care of the magnetic flux linked with it and opposes any change within the flux by inducing a current in it. This is the reason why self-induction is named inertia of electricity.
7. Calculate the mutual inductance between two coils if a current 10 A in the primary coil changes the flux by 500 Wb per turn in the secondary coil of 200 turns.
a) 10 H
b) 104 H
c) 1000 H
d) 100 H
Answer: b
Explanation: NΦ = MI
200 × 500 = M × 10
M = 104 H
8. What is the relative permeability of an air-core inductor, if its self-inductance increases from 0.5 mH to 50 mH due to the introduction of an iron core into it?
a) 100
b) 0.1
c) 10000
d) 0.0001
Answer: a
Explanation: μr = μμ0
μr = LL0
μr = 500.5
μr = 100
9. What is the self-inductance of the coil, if the magnetic flux of 10 microwebers is linked with a coil when a current of 5 mA flows through it?
a) 20 mH
b) 5 mH
c) 2 mH
d) 250 mH
Answer: c
Explanation: Self-inductance = MagneticfluxCurrent
Self inductance = 10×10−65×10−3
Self inductance = 2 × 10-3 H
Self inductance = 2mH
10. Determine the self-inductance of a coil, which has a magnetic flux of 50 milliwebers that is produced when a current of 5 A flows through it?
a) 1 × 10-2 Wb
b) 1 × 10-3 Wb
c) 100 Wb
d) 1 × 103 Wb
Answer: a
Explanation: Self-inductance = MagneticfluxCurrent
Self-inductance = 50×10−35
Self-inductance = 1 × 10-2 Wb