Self-Induction MCQ Quiz - Objective Question with Answer for Self-Induction - Download Free PDF

Last updated on May 7, 2025

Latest Self-Induction MCQ Objective Questions

Self-Induction Question 1:

Suppose a tightly wound toroid with 103 √e turns has a rectangular cross-section with a thickness of e cm. The outer radii (R) is related to inner radii (r) as R=e3 r. The self-inductance of this toroid is x/5 H. Find the value of x.

Answer (Detailed Solution Below) 3

Self-Induction Question 1 Detailed Solution

Calculation:
B × 2πr = μ0 NI

⇒ B = μ0 NI / 2πr

Flux through a small element:

Flux = μ0 NI / 2πr dr

⇒ Flux through a cross-section:

Flux = μ0 NI / 2π ∫ab (dr / r) = μ0 NIh / 2π ln(b / a)

⇒ Flux through ‘N’ turns:

Flux = μ0 N2 I h / 2π ln(b / a)

⇒ L = φ / I = μ0 N2 h / 2π ln(b / a)

Putting values, we get L = 0.6 =3/5

Hence, x = 3

Self-Induction Question 2:

When the current in a coil changes from 2 amp. to 4 amp. in 0.05 sec., an e.m.f. of 8 volt is induced in the coil. The coefficient of self inductance of the coil is

  1. 0.1 henry
  2. 0.2 henry
  3. 0.4 henry
  4. 0.8 henry

Answer (Detailed Solution Below)

Option 2 : 0.2 henry

Self-Induction Question 2 Detailed Solution

Calcultion:
The formula for the induced e.m.f. in a coil is given by:

e = -L × (ΔI / Δt)

Where:

  • e = induced e.m.f. = 8 V
  • L = self-inductance of the coil (the quantity we need to find)
  • ΔI = change in current = 4 A - 2 A = 2 A
  • Δt = time taken for the change in current = 0.05 s

Now, substituting the known values into the equation:

8 = L × (2 / 0.05)

8 = L × 40

L = 8 / 40 = 0.2 henry

The coefficient of self-inductance of the coil is 0.2 henry.

Self-Induction Question 3:

Consider I1 and I2 are the currents flowing simultaneously in two nearby coils 1 & 2, respectively. If L1 = self inductance of coil 1, M12 = mutual inductance of coil 1 with respect to coil 2, then the value of induced emf in coil 1 will be

  1. \(\varepsilon_{1}=-\mathrm{L}_{1} \frac{\mathrm{dI}_{1}}{\mathrm{dt}}+\mathrm{M}_{12} \frac{\mathrm{dI}_{2}}{\mathrm{dt}} \)
  2. \(\varepsilon_{1}=-\mathrm{L}_{1} \frac{\mathrm{dI}_{1}}{\mathrm{dt}}-\mathrm{M}_{12} \frac{\mathrm{dI}_{1}}{\mathrm{dt}} \)
  3. \(\varepsilon_{1}=-\mathrm{L}_{1} \frac{\mathrm{dI}_{1}}{\mathrm{dt}}-\mathrm{M}_{12} \frac{\mathrm{dI}_{2}}{\mathrm{dt}} \)
  4. \(\varepsilon_{1}=-\mathrm{L}_{1} \frac{\mathrm{dI}_{2}}{\mathrm{dt}}-\mathrm{M}_{12} \frac{\mathrm{dI}_{1}}{\mathrm{dt}} \)

Answer (Detailed Solution Below)

Option 3 : \(\varepsilon_{1}=-\mathrm{L}_{1} \frac{\mathrm{dI}_{1}}{\mathrm{dt}}-\mathrm{M}_{12} \frac{\mathrm{dI}_{2}}{\mathrm{dt}} \)

Self-Induction Question 3 Detailed Solution

Calculation:

ϕ1 = L1 I1 + M12 I2

\(\varepsilon_{1}=-\frac{\mathrm{d} \phi_{1}}{\mathrm{dt}}=-\mathrm{L}_{1} \frac{\mathrm{dI}_{1}}{\mathrm{dt}}-\mathrm{M}_{12} \frac{\mathrm{dI}_{2}}{\mathrm{dt}}\)

Self-Induction Question 4:

Regarding self-inductance :

A : The self-inductance of the coil depends on its geometry.

B : Self-inductance does not depend on the permeability of the medium.

C : Self-induced e.m.f. opposes any change in the current in a circuit.

D : Self-inductance is electromagnetic analogue of mass in mechanics.

E : Work needs to be done against self-induced e.m.f. in establishing the current.

Choose the correct answer from the options given below: 

  1. A, B, C, D only
  2. A, C, D, E only
  3. A, B, C, E only
  4. B, C, D, E only 

Answer (Detailed Solution Below)

Option 2 : A, C, D, E only

Self-Induction Question 4 Detailed Solution

Concept:

 

Self-Induction

  • Every current-carrying coil produces a magnetic field when it is linked to the same coil.
  • When there is a change in the current flowing through the coil, the magnetic flux also changes and it is linked to the coil itself.
  • According to Faraday’s law of induction, the change in a magnetic field produces an induced emf which opposes the change.
  • Hence, the phenomenon in which an induced emf is produced in a coil due to a change in the current flowing in the coil is known as self-induction.
  • The induced emf is referred to as self-induced emf, which is equal to the rate of change of flux.
  • Mathematically, it is written as

\(E \propto \frac{-d\phi}{dt}\)

  • The negative sign denotes the induced emf that opposes the applied emf.

\(\frac{d\phi}{dt} \propto \frac{di}{dt}\)

\(\frac{d\phi}{dt} = L \frac{di}{dt}\)

Where 'L' is the constant of proportionality and it is called self-inductance of the coil.

\(E = -L \frac{di}{dt}\)

\(L = \frac{-E}{(\frac{di}{dt})}\)

  • Thus, the self inductance of a coil is numerically equal to the ratio of the induced emf to the rate of change of current in the coil.

 

Explanation:

Self inductance of coil

L = \(\frac{\mu_{0} \mathrm{~N}^{2} \mathrm{~A}}{2 \pi \mathrm{R}}\)

Self-Induction Question 5:

The current in a coil changes steadily from 3 A to 5 A in 0.2 s when an emf of 2μV is induced in it. The self-inductance of the coil is

  1. 0.2 mH
  2. 20 μH
  3. μH
  4. 0.2 μH

Answer (Detailed Solution Below)

Option 4 : 0.2 μH

Self-Induction Question 5 Detailed Solution

Concept:

Self-inductance is the property of a coil that opposes any change in the current flowing through it by inducing an emf (electromotive force) in response to the change in current.

The induced emf (E) in a coil due to a change in current is given by the formula:

E = - L (dI/dt)

E is the induced emf (in volts)

L is the self-inductance of the coil (in henries)

dI/dt is the rate of change of current (in amperes per second)

Calculation:

We are asked to find the self-inductance (L) of the coil. The given values are:

Induced emf, E = 2 μV = 2 × 10-6 V

Initial current, I₁ = 3 A

Final current, I₂ = 5 A

Time interval, Δt = 0.2 s

The rate of change of current is:

dI/dt = (I₂ - I₁) / Δt = (5 - 3) / 0.2 = 2 / 0.2 = 10 A/s

Now, using the formula for induced emf:

E = - L (dI/dt)

2 × 10-6 = L × 10

Solving for L:

L = (2 × 10-6) / 10 = 2 × 10-7 H

∴ The self-inductance of the coil is 0.2 μH.
Hence, the correct option is 4) 0.2 μH.

Top Self-Induction MCQ Objective Questions

If we apply law of conversion of energy to electromagnetic induction, electrical energy induced in a conductor comes from

  1. Potential energy
  2. Heat energy
  3. Kinetic energy
  4. Radiation energy

Answer (Detailed Solution Below)

Option 3 : Kinetic energy

Self-Induction Question 6 Detailed Solution

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CONCEPT:

  • Potential energy: Potential energy is the energy stored within an object, due to the object's position, arrangement or state.
  • Heat energy: Heat is the transfer of energy from one system to another, and it can affect the temperature of a singular system.
  • Kinetic energy: Kinetic energy is the energy of mass in motion. The kinetic energy of an object is the energy it has because of its motion.
  • Radiation energy: Radiation is the emission or transmission of energy in the form of waves or particles through space or through a material medium.

EXPLANATION:

  • The work done by moving coil (Change in kinetic energy) induces the electrical energy in the conductor.

This principle is used in electric generators.

So option 3 is correct.

When number of turns per unit length of a solenoid is doubled, its self inductance becomes:

  1. 4 times
  2. 8 times
  3. Same
  4. Doubled

Answer (Detailed Solution Below)

Option 1 : 4 times

Self-Induction Question 7 Detailed Solution

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CONCEPT:

  • Self-Induction: Whenever the electric current passing through a coil changes, the magnetic flux linked with it will also change.
    • As a result of this, in accordance with Faraday’s laws of electromagnetic induction, an emf is induced in the coil which opposes the change that causes it.
    • This phenomenon is called ‘self-induction’ and the emf induced is called back emf, current so produced in the coil is called induced current.

Self-inductance of a solenoid is given by:

\(L=\frac{{{\mu }_{o}}{{N}^{2}}A}{l}\) -- (1)

Where μo = Absolute permeability, N = Number of turns, l = length of the solenoid, and A = Area of the solenoid.

EXPLANATION:

 μo ,I, A are constants. So, we can say that 

L = k N2 -- (2)

k is constant, N is the number of turns

L is directly proportional to the square of a number of turns.

If number of turns becomes N' = N2, then inductance 

L' = k N' 2 

⇒ L' = k (2N)2 = k 4 N2 = 4 K N2

⇒ L' = 4 L

So, the inductance is increased by 4 times.

Hence the correct option is 4 times.

Important Points

Induced e.m.f can be given as

\(⇒ e =- L\frac{{di}}{{dt}}\)

If we increase the current in an inductor, self inductance of the inductor will __________.

  1. decrease
  2. increase
  3. remains same
  4. first decrease then increase

Answer (Detailed Solution Below)

Option 3 : remains same

Self-Induction Question 8 Detailed Solution

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CONCEPT:

Self-Induction

  • Whenever the electric current passing through a coil changes, the magnetic flux linked with it will also change.
  • As a result of this, in accordance with Faraday’s laws of electromagnetic inductionan emf is induced in the coil which opposes the change that causes it.
  • This phenomenon is called ‘self-induction’ and the emf induced is called back emfcurrent so produced in the coil is called induced current.
  • Self-inductance of a solenoid is given by 

\(L = \frac{{{μ _o}{N^2}A}}{l}\)

EXPLANATION:

  • Self-inductance of a solenoid is given by 
\(L = \frac{{{μ _o}{N^2}A}}{l}\)
  • From the above equation, it is clear that the self-inductance of a solenoid depends on the geometry and magnetic permeability of the core material.  
  • It is independent of current, so if we increase the current in an inductor, there will be no change in the self-inductance of the inductor. Therefore the correct answer is option 3.

A coil of self-inductance L is connected in series with a bulb B and an AC source. Brightness of the bulb decreases when:

  1. frequency of the AC source is decreased.
  2. number of turns in the coil is reduced.
  3. a capacitance of reactance XC = XL is included in the same circuit.
  4. an iron rod is inserted in the coil.

Answer (Detailed Solution Below)

Option 4 : an iron rod is inserted in the coil.

Self-Induction Question 9 Detailed Solution

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CONCEPT:

  • Self Inductance: Self Inductance is the capacity of an inductor to store energy in form of a magnetic field.
  • The expression for self-inductance is given as 

L = μN2 A / l

μ is permeability

N is the number of turns

I is the length of the inductor

ϕ is flux in the inductor

  • The impedance for the inductor circuit is given as

XL = 2πfL.

  • If L  increases the impedance will increase and the current will reduce.
  • The inductance can be increased by inserting an iron rod in the coil as the permeability of the iron rod is more than free space.

EXPLANATION:

Since the inductance is increased by inserting a rod, the current will decrease and the blub will glow less. Thus, inserting a rod will reduce the glow.

Important Points

  •  If the number of turns is reduced the inductance will decrease, thus impedance will decrease, and hence the current will increase. The bulb will glow more.
  • If a capacitance of reactance XC = XL is included in the same circuit, the overall impedance will reduce, and hence current will increase. 
  • The impedance of the circuit with inductance is given as XL = 2πfL. If frequency f will decrease, the impedance will get reduced and the bulb will glow more.

A current in a coil of inductance 5 H decreases at a rate of 2 Amp/sec. The induced emf is

  1. 2 V
  2. 5 V
  3. 10 V
  4. - 10 V

Answer (Detailed Solution Below)

Option 3 : 10 V

Self-Induction Question 10 Detailed Solution

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CONCEPT:

Self-Induction:

  • Whenever the electric current passing through a coil changes, the magnetic flux linked with it will also change.
  • As a result of this, in accordance with Faraday’s laws of electromagnetic induction, an emf is induced in the coil which opposes the change that causes it.
  • This phenomenon is called ‘self-induction’ and the emf induced is called back emf, current so produced in the coil is called induced current.
  • Self-inductance of a solenoid is given by –

\(L=\frac{{{\mu }_{o}}{{N}^{2}}A}{l}\)

Where μo = Absolute permeability, N = Number of turns, l = length of the solenoid, the resistance of the coil (R) = 4Ω, and A = Area of the solenoid.

Induced e.m.f can be given as,

\(⇒ e =- L\frac{{di}}{{dt}}\)

CALCULATION:

Given -

di/dt = -2 Amp/sec and L = 5 H (Since the current decreases at a rate of 2 Amp/sec).

The emf induced across the inductor is,

\(⇒ e =- L\frac{{di}}{{dt}}=-(5\times (-2) )=10\, V\)

When the length of the solenoid is doubled without any change in the number of turns and the area of the coil.  Then its self-inductance will 

  1. Nine times
  2. Half times
  3. Doubled
  4. Unchanged

Answer (Detailed Solution Below)

Option 2 : Half times

Self-Induction Question 11 Detailed Solution

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CONCEPT:

Self-Induction

  • Whenever the electric current passing through a coil changes, the magnetic flux linked with it will also change.
  • As a result of this, in accordance with Faraday’s laws of electromagnetic induction, an emf is induced in the coil which opposes the change that causes it.
  • This phenomenon is called ‘self-induction’ and the emf induced is called back emf, current so produced in the coil is called induced current.
  • Self-inductance of a solenoid is given by –

\(⇒ L=\frac{{{\mu }_{o}}{{N}^{2}}A}{l}\)

Where μo = Absolute permeability, N = Number of turns, l = length of the solenoid, and A = Area of the solenoid.

EXPLANATION:

Given - l2 = 2l1

  • Self-inductance of a solenoid is given by:

\(⇒ L=\frac{{{\mu }_{o}}{{N}^{2}}A}{l}\)

  • According to the question, the length of the solenoid is doubled without any change in the number of turns and the area of the coil

\(⇒ L\propto \frac{1}{l}\)

⇒ L1l1 = L2l2

\(⇒ \frac{{{L}_{1}}}{{{L}_{2}}}=\frac{l_2}{l_1}=\frac{2l_1}{l_1}=2\)

\(⇒L_2=\frac{L_1}{2}\)

The energy stored in a 50 mH inductor carrying a current of 4 A will be

  1. 0.4 J
  2. 4.0 J
  3. 0.8 J
  4. 0.04 J

Answer (Detailed Solution Below)

Option 1 : 0.4 J

Self-Induction Question 12 Detailed Solution

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CONCEPT:

  • The magnetic potential energy of inductor (U): In building a steady current in the circuit, the source emf has to do work against of self-inductance of the coil and whatever energy consumed for this work stored in a magnetic field of coil this energy called as magnetic potential energy (U) of the coil.

\(U=\frac{1}{2}L{{I}^{2}}\)

Where L = Self-inductance and I = current

EXPLANATION:

Given that:

L = 50 mH = 50 × 10-3 H and I = 4 A

The magnetic potential energy of inductor (U) is:

\(U=\frac{1}{2}\times 50\times {{10}^{-3}}\times {{\left( 4 \right)}^{2}}\)

\(U=\frac{1}{2}\times 50\times {{10}^{-3}}\times 16\)

U = 400 × 10-3 H

U = 0.4 J

A long solenoid has 500 turns. When a current of 2 amperes is passed through it, the resulting magnetic flux linked with each turn of the solenoid is 4 × 10–3 Wb. The self-inductance of the solenoid is -

  1. 1.0 henry
  2. 4.0 henry
  3. 2.5 henry
  4. 2.0 henry

Answer (Detailed Solution Below)

Option 1 : 1.0 henry

Self-Induction Question 13 Detailed Solution

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Concept:

Solenoid: The solenoid is a long cylindrical coil of wire consisting of a large number of turns bound together very tightly.

Magnetic flux: The Number of magnetic lines of forces set up in a magnetic circuit is called magnetic flux. It is analogous to an electric current, I is an electric circuit. Its SI unit is Weber

The Self-Inductance of the solenoid is given by:

 L = N (ϕB/ i)

L = self-inductance. N = number of turns. ,ϕB = Magnetic flux linked with each turn of the solenoid, i =  current flowing  through the solenoid

F1 Jitendra Madhu 26.10.20 D8

Explanations:​

Given that:

N = 500 , ϕB  =  4 × 10-3 , i = 2 Ampere

L = (500 × 4 × 10-3) / 2 

L =  1.0 henry.

hence option 1 is correct.

The self inductance of a coil can be changed by-

  1. changing electric current in the coil
  2. changing number of turns per unit length of coil
  3. changing conductance of the coil
  4. All of the above

Answer (Detailed Solution Below)

Option 2 : changing number of turns per unit length of coil

Self-Induction Question 14 Detailed Solution

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CONCEPT:

  • Self-Induction: Whenever the electric current passing through a coil changes, the magnetic flux linked with it will also change.
    • As a result of this, in accordance with Faraday’s laws of electromagnetic induction, an emf is induced in the coil which opposes the change that causes it.
    • This phenomenon is called ‘self-induction’ and the emf induced is called back emf, current so produced in the coil is called induced current.

EXPLANATION:

The self-inductance of a solenoid/coil is given by:

\(L = \frac{{{\mu _o}{N^2}A}}{l}\)

Dependence of self-inductance (L):

  • Self-inductance does not depend upon current flowing or change in current flowing and temperature but it depends upon:
    • Number of turns ( N )
    • Area of cross-section ( A )
    • Permeability of medium (μo).
  • Thus the self-inductance of a coil can be changed by the number of turns per unit length of the coil.
  • Hence option 2 is correct.

The self-inductance of the coil depends on ______. 

  1. the temperature of the coil 
  2. the current flowing through the coil 
  3. the area of the coil  
  4. the induced EMF developed in the coil 

Answer (Detailed Solution Below)

Option 3 : the area of the coil  

Self-Induction Question 15 Detailed Solution

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Explanation:

Self inductance is defines as the induction of a voltage in a current-carrying wire when the current in the wire itself is changing. Self-inductance of a coil is numerically equal to the amount of magnetic flux linked with the coil when unit current flows through the coil.

According to Faraday's law,

  • \(E=-L\frac{dI}{dt}\)

where, E is E.M.F. of and L is self inductance and \(\frac{dI}{dt}\) is rate of change of current.

 

Also, \(E=-\frac{d\phi}{dt}\)

  • \(\implies-\frac{d\phi}{dt}=-L\frac{dI}{dt}=\frac{d(NBA)}{dt}=L\frac{dI}{dt}\)

 

  • \(\frac{d(NBA)}{dt}=-L\frac{dI}{dt}\)

 

  • \(\frac{d(NBA)}{dt}=L\frac{dI}{dt}\)

So, self inductance is directly proportional to area of the coil,\(L\propto BA\)

Hence, the correct answer is Option-3-the area of the coil.

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