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

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 ∫_a^b (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

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.

Calculation:

ϕ₁ = L₁ I₁ + M₁₂ I₂

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

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

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.

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.

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

\(⇒ e =- L\frac{{di}}{{dt}}\)

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 

\(L = \frac{{{μ _o}{N^2}A}}{l}\)

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.

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

X_L = 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 X_L = 2πfL. If frequency f will decrease, the impedance will get reduced and the bulb will glow more.

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,

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

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

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

\(⇒L_2=\frac{L_1}{2}\)​

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

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 (ϕ_B/ 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.

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.

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.