Inductor and Inductance MCQ Quiz - Objective Question with Answer for Inductor and Inductance - Download Free PDF

Explanation:

Energy Stored in the Magnetic Field of a Solenoid

Definition: The energy stored in the magnetic field of a solenoid is the energy associated with the magnetic field created by the current flowing through the solenoid. This energy can be calculated using the formula for magnetic energy density and the volume of the solenoid.

Given Data:

• Length of the solenoid (L) = 40 cm = 0.4 m
• Diameter of the solenoid = 4 cm = 0.04 m
• Radius of the solenoid (r) = Diameter / 2 = 0.04 / 2 = 0.02 m
• Number of turns (N) = 100
• Current through the solenoid (I) = 20 A

Formula for Energy Stored in the Magnetic Field:

The energy stored in the magnetic field of a solenoid is given by:

U = (1/2) × L × I²

where:

• U = Energy stored (in Joules)
• L = Inductance of the solenoid (in Henry)
• I = Current through the solenoid (in Amperes)

Step 1: Calculate the Inductance (L) of the Solenoid

The inductance of a solenoid is given by the formula:

L = (μ₀ × N² × A) / l

where:

• L = Inductance (in Henry)
• μ₀ = Permeability of free space = 4π × 10^-7 H/m
• N = Number of turns
• A = Cross-sectional area of the solenoid = π × r² (in m²)
• l = Length of the solenoid (in meters)

Substitute the values:

A = π × r² = π × (0.02)² = 3.14 × 0.0004 = 0.001256 m²

L = (4π × 10^-7 × 100² × 0.001256) / 0.4

L = (4π × 10^-7 × 10000 × 0.001256) / 0.4

L = (1.577 × 10^-2) / 0.4

L = 3.9425 × 10^-2 H

Step 2: Calculate the Energy Stored (U)

U = (1/2) × L × I²

Substitute the values:

U = (1/2) × 3.9425 × 10^-2 × (20)²

U = 0.5 × 3.9425 × 10^-2 × 400

U = 0.5 × 15.77

U = 7.885 Joules

Therefore, the energy stored in the magnetic field of the solenoid is approximately 0.789 Joules.

Correct Option Analysis:

The correct option is:

Option 2: 0.789 Joule

This matches the calculated result, confirming that the energy stored in the magnetic field of the solenoid is approximately 0.789 Joules.

Important Information

To further understand the analysis, let’s evaluate the other options:

Option 1: 0.705 Joule

This value is incorrect because it results from either an error in the calculation of the inductance or the energy formula. The actual energy stored is higher than this value.

Option 3: 0.587 Joule

This value is also incorrect and is significantly lower than the actual calculated energy stored. This could result from an incorrect assumption about the cross-sectional area or current value.

Option 4: 0.658 Joule

This value is close to the correct value but still incorrect. It may result from a rounding or calculation error in one of the intermediate steps.

Conclusion:

The correct energy stored in the magnetic field of the solenoid, based on the given data and formula, is 0.789 Joules. This matches Option 2. The other options deviate due to calculation errors, incorrect assumptions, or approximations.

Explanation:

The magnetic field at a point on the axis of a current-carrying circular loop is given by the Biot-Savart law. For a large loop with radius b and current I₁ , the magnetic field at a point on the axis of the loop a distance z from the center of the loop is:

\( B(z) = \frac{\mu_0 I_1 b^2}{2 (b^2 + z^2)^{3/2}}.\)

This magnetic field is directed along the axis of the loop (the common axis of both loops).

The magnetic flux \(\Phi_2 \) through the small loop (radius a ) due to the magnetic field B(z) created by the large loop is the product of the magnetic field at the position of the small loop and the area of the small loop:

\( \Phi_2 = B(z) \cdot A_2 = \frac{\mu_0 I_1 b^2}{2 (b^2 + z^2)^{3/2}} \cdot \pi a^2.\)

Here,\( A_2 = \pi a^2 \) is the area of the small loop.

Step 3: Induced emf in the Small Loop

According to Faraday's law of induction, the induced emf in the small loop is related to the rate of change of the magnetic flux through the loop:

\(\mathcal{E}_2 = -\frac{d\Phi_2}{dt}.\)

Since the flux \(\Phi_2\) depends on the current I_1 in the large loop, which may change with time, the induced emf in the small loop will be:

\( \mathcal{E}_2 = -\frac{d}{dt} \left( \frac{\mu_0 I_1 b^2 \pi a^2}{2 (b^2 + z^2)^{3/2}} \right).\)

Using the fact that \(\frac{dI_1}{dt} = k\) (where k is the rate of change of the current in the large loop), we get:

\(\mathcal{E}_2 = -\frac{\mu_0 k b^2 \pi a^2}{2 (b^2 + z^2)^{3/2}}.\)

Step 4: Definition of Mutual Inductance

The mutual inductance M is defined by the relationship:

\(\mathcal{E}_2 = -M \frac{dI_1}{dt}.\)

Thus, the mutual inductance M between the two loops is:

\(M = \frac{\mu_0 \pi a^2 b^2}{2 (b^2 + z^2)^{3/2}}.\)

The correct answer is 4):\(M = \frac{\mu_0 \pi a^2 b^2}{2 (b^2 + z^2)^{3/2}}.\)

CONCEPT:

Self-Induction

• Whenever the electric current passing through a coil changes, the magnetic flu× 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 –

1. \(L=\frac{{{\mu }_{o}}{{N}^{2}}A}{l}\)
2. N\(\phi \)=LI ; ϕ" id="MathJax-Element-1-Frame" role="presentation" style="position: relative;" tabindex="0">ϕϕ $\phi$ = LIN" id="MathJax-Element-2-Frame" role="presentation" style="position: relative;" tabindex="0">LINLIN $\frac{LI}{N}$    

Where, μo = Absolute permeability, N = Number of turns, l = length of the solenoid, R= resistance of the coil , and A = Area of the solenoid, \(\phi \)= flu×, I= Current flowing through the coil.

CALCULATION:

Given - Number of turns (N) =1000, Inductance(L) =  1 mH, current(I) = 50 mA

\(\phi \)= \( \frac{LI}{N}\)=\( \frac{10^{-3}\cdot50\cdot10^{-3}}{1000}\)= 5 × 10-8 Wb.

Concept:

Inductor:

• A passive two-terminal electrical component that stores energy in a magnetic field when electric current flows through it.
• When current flows through the inductor a time-varying magnetic field induces an EMF in the conductor (Faraday's law of induction).
• According to Lenz's law, the induced voltage opposes the change in current that created it. So inductor opposes any changes in current through them.
• It is also known as coil, choke, or reactor.
• The energy stored in an inductor is given by

E = (½)LI2

Where E = Energy stored in Inductor (Joule)

L = Inductance of the inductor (H)

I = current trough inductor (A)

Calculation:

Given:

E = 2 Joules

I = 0.2 A

\(E = \frac{1}{2}Li^2\)

\(2 = \frac{1}{2}\times L\times0.2^2\)

L = 100 H

Concept:

Emf induced in a coil or inductor is given by

\(e = -L{di\over dt}\)

Where,

L = Inductance

di = final value of current - initial value of current

dt = final time - initial time

Concept:

Coefficient of Coupling (k):
The coefficient of coupling (k) between two coils is defined as the fraction of magnetic flux produced by the current in one coil that links the other.

Two coils have self-inductance L1 and L2, then mutual inductance M between them then Coefficient of Coupling (k) is given by 

\(k=\frac{M}{√ {L_1L_2}}\)

Where,

\(M=\frac{N_1N_2\mu_o \mu_rA}{ l}\)

\(L_1=\frac{N_1^2\mu_o \mu_rA}{ l}\)

\(L_2=\frac{N_2^2\mu_o \mu_rA}{ l}\)

N1 and N2 is the number of turns in coil 1 and coil 2 respectively
A is the cross-section area
l is the length

Calculation:

Given,

L1 = 0.2 H

L2 = 2.45 H

M = 0.14 H

From the above concept,

\(k=\frac{M}{√ {L_1L_2}}\)

= 0.14/(√0.2 × √2.45)

k = 0.2

Inductor

An inductor is an electrical element that stores energy in the form of a magnetic field.

An inductor opposes the change in current passing through it.

The S.I. unit of inductance is henry (H) and is denoted by the symbol L.

If current 'I' flows through the inductor, then the induced voltage is given by:

\(V_L=L{di\over dt}\)

Additional Information A capacitor stores energy in the form of electrical energy.

If the load is purely Inductive:

 Applying Ohm's Law to the circuit

\(i=\frac{1}{L}\int{V_msin(ω t)dω t}\)

After solving it,

\(i=\frac{-V_m}{ω L}cos(ω t)=\frac{-V_mcosω t}{X_L}\) .... (1)

Since, i = Imsinωt

Hence,

\(I=\frac{V}{X_L}\) .... (2)

Here, we represent V and I as RMS values for further solutions.

Given:

V = 200 V

I = 10 A

From equation (1),

X_L = V/I = 200/10 = 20 Ω = ωL

ω = 2πf = 100π

\(L=\frac{20}{100\pi}=6.36\times 10^{-2}=63.6\ mH\)

Concept:

Emf induced in a coil or inductor is given by

\(e = -L{di\over dt}\)

Where,

L = Inductance

di = final value of current - initial value of current

dt = final time - initial time

The correct answer is option 3:(a solenoid bent in circular shape and ends are joined)

Concept:

• A toroid is a coil of insulated or enameled wire wound on a d-shaped form made of powdered iron.
• A toroid is used as an inductor in electronic circuits, especially at low frequencies where comparatively large inductances are necessary
• The toroid is solenoid is a coil of insulated wire wound on a rod-shaped form made of solid iron, solid steel, or powdered iron. Hence we can consider Toroid as a Ring-shaped closed solenoid.

• A toroid has more inductance, for a given number of turns, than a solenoid with a core of the same material and similar size. This makes it possible to construct high-inductance coils of reasonable physical size and mass.
• Toroidal coils of a given inductance can carry more current than solenoidal coils of similar size, because larger-diameter wires can be used, and the total amount of wire is less, reducing the resistance. In a toroid, all the magnetic flux is contained in the core material.
• This is because the core has no ends from which flux might leak off. The confinement of the flux prevents external magnetic fields from affecting the behaviour of the toroid, and also prevents the magnetic field in the toroid from affecting other components in a circuit.

Inductance

Inductance is the tendency of an electrical conductor to oppose a change in the electric current flowing through it.

The inductance is given by:

\(L={ϕ \over I}\)

where, L = Inductance

ϕ = Flux linkage

I = Current

From the above expression, it is found that the inductance is directly proportional to the flux linkage.

The flux linkage is maximum for ferromagnetic materials such as iron, cobalt, and nickel.

Hence, the iron core will have the highest inductance.

From the given figure we have

\(i\left( t \right) = \frac{{\left( {5\ A} \right)}}{T}t\)

Voltage across inductor \(= {V_L} = L.\frac{{di\left( t \right)}}{{dt}}\)

\(\therefore {V_L} = L.\frac{d}{{dt}}\left( {\frac{{5\ A}}{T}t} \right) = \frac{{5AL}}{T}\)

So the shape of the waveform is a step signal.

Hence, the signal can be drawn as,

CONCEPT:

Self-Induction

• Whenever the electric current passing through a coil changes, the magnetic flu× 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 –

1. \(L=\frac{{{\mu }_{o}}{{N}^{2}}A}{l}\)
2. N\(\phi \)=LI ; ϕ" id="MathJax-Element-1-Frame" role="presentation" style="position: relative;" tabindex="0">ϕϕ $\phi$ = LIN" id="MathJax-Element-2-Frame" role="presentation" style="position: relative;" tabindex="0">LINLIN $\frac{LI}{N}$    

Where, μo = Absolute permeability, N = Number of turns, l = length of the solenoid, R= resistance of the coil , and A = Area of the solenoid, \(\phi \)= flu×, I= Current flowing through the coil.

CALCULATION:

Given - Number of turns (N) =1000, Inductance(L) =  1 mH, current(I) = 50 mA

\(\phi \)= \( \frac{LI}{N}\)=\( \frac{10^{-3}\cdot50\cdot10^{-3}}{1000}\)= 5 × 10-8 Wb.

Concept:

The inductance of a coil is given by,

\(L=\frac{{{\mu }_{0}}{{\mu }_{r}}{{N}^{2}}A}{l}\)

μ₀ is magnetic permeability of free space

μ_r is relative permeability

N is number if turns

A is cross-sectional area

l is length

Calculation:

Given that, the number of turns of the coil is doubled

Length is halved

The cross-sectional area of the core remains constant

So, new inductance \({{L}_{new}}=\frac{{{\mu }_{0}}{{\mu }_{r}}{{\left( 2N \right)}^{2}}A}{l/2}\)

\({{L}_{new}}=8\frac{{{\mu }_{0}}{{\mu }_{r}}{{N}^{2}}A}{l}\)

CONCEPT:

• Inductor: A device or component of a circuit that has significant self-inductance and stores energy in a magnetic field. (fig.)

The formula used to find the energy stored in a magnetic field is

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

Where L is the inductance and I is current in the Inductor.

EXPLANATION:

The energy stored in an inductor of self-inductance L henry carrying a current of I ampere is:

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