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
Answer (Detailed Solution Below)
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
Answer (Detailed Solution Below)
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:
Answer (Detailed Solution Below)
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
Answer (Detailed Solution Below)
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
Answer (Detailed Solution Below)
Self-Induction Question 6 Detailed Solution
Download Solution PDFCONCEPT:
- 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:
Answer (Detailed Solution Below)
Self-Induction Question 7 Detailed Solution
Download Solution PDFCONCEPT:
- 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 __________.
Answer (Detailed Solution Below)
Self-Induction Question 8 Detailed Solution
Download Solution PDFCONCEPT:
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
\(L = \frac{{{μ _o}{N^2}A}}{l}\)
EXPLANATION:
- Self-inductance of a solenoid is given by
- 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:
Answer (Detailed Solution Below)
Self-Induction Question 9 Detailed Solution
Download Solution PDFCONCEPT:
- 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
Answer (Detailed Solution Below)
Self-Induction Question 10 Detailed Solution
Download Solution PDFCONCEPT:
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
Answer (Detailed Solution Below)
Self-Induction Question 11 Detailed Solution
Download Solution PDFCONCEPT:
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
Answer (Detailed Solution Below)
Self-Induction Question 12 Detailed Solution
Download Solution PDFCONCEPT:
- 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 JA 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 -
Answer (Detailed Solution Below)
Self-Induction Question 13 Detailed Solution
Download Solution PDFConcept:
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
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-
Answer (Detailed Solution Below)
Self-Induction Question 14 Detailed Solution
Download Solution PDFCONCEPT:
- 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 ______.
Answer (Detailed Solution Below)
Self-Induction Question 15 Detailed Solution
Download Solution PDFExplanation:
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.