Mutual Induction MCQ Quiz in বাংলা - Objective Question with Answer for Mutual Induction - বিনামূল্যে ডাউনলোড করুন [PDF]

Calculation:

We know that,

\(M = -\cfrac{e_{2}}{di_{1} / dt} = - \cfrac{e_{1}}{di_{2}/dt}\),

Also,

\(e_{1} = -L_{1} \cfrac{di_{1}}{dt}\) and \(e_{2} = -L_{2} \cfrac{di_{2}}{dt}\),

\(\Rightarrow M^{2} = \cfrac{e_{1}e_{2}}{\left(\cfrac{di_{1}}{dt}\right)\left(\cfrac{di_{2}}{dt}\right)} = L_{1}L_{2}\),

\(\Rightarrow M = \sqrt{L_{1}L_{2}}\)

Concept:

• Self-inductance: The property of the current-carrying coil to resists the change of current flowing through it.
  • This occurs mainly due to the self-induced emf produced in the coil itself.

​\(⇒ L=N\frac{ϕ}{I}\)   

Where N = number of turns in the coil, L = self-inductance of the coil, ϕ = flux associated with the coil and I = current in the coil

• Mutual - inductance: Current is induced in secondary coil kept in proximity with the coil carrying current.
  • This property of inducing a current in the secondary coil by changing the flux of the secondary coil is called mutal inductance.

​\(⇒ M =N_{1}\frac{\phi }{I_{2}}\)    

M is mutual inductance between Coil 2 and 1, N₁ is the number of turns in coil 1, ϕ is magnetic flux induced in coil 2.

• The induced emf in the secondary coil is given as

\(E = M\frac{di }{dt}\)

Calculation:

Given Mutual inductance M = 1. 25 Henery

Rate of change of current \(\frac{di }{dt} = 80 \) A

Emf induced E = 1.25 H × 80 A/s = 100 Volt

So, induced emf is 100 V.

CONCEPT:

Mutual-inductance

• When two coils are brought in proximity with each other the magnetic field in one of the coils tends to link with the other. If this magnetic field of the first coil is changed then the magnetic flux associated with the second coil changes and this leads to the generation of voltage in the second coil.
• This property of a coil that affects or changes the current and voltage in a secondary coil is called mutual inductance.
• The SI unit of the mutual inductance is Henry.
• Let two coils of the number of turns N1 and N2 are placed close to each other. Then the mutual inductance of coil 1 with respect to coil 2 is given as,

\(⇒ M_{12}=N_{1}\frac{ϕ_{1}}{I_{2}}\)

Where N1 = number of turns in coil 1, ϕ1 = flux linked with coil 1, and I2 current in coil 2

Mutual inductance for two co-axial solenoids:

• Let two long solenoids S1 and S2 of equal length are placed co-axially as shown in the figure.
• The solenoid S1 is placed inside the solenoid S2.
• The mutual inductance of both the solenoids will be equal and it is given as,

\(⇒ M_{12} = M_{21} = μ_on_1n_2πr_{1}^{2}l\)

For medium of relative permeability μr,

\(⇒ M_{12} = M_{21} = \mu_rμ_on_1n_2πr_{1}^{2}l\)

Where n1 = number of turns per unit length of solenoid 1, n2 = number of turns per unit length of solenoid 2, r1 = radius of the inner solenoid, and l = length of both the solenoids

EXPLANATION:

• We know that if there are two solenoids of equal length and one solenoid is placed coaxially inside the other solenoid then the mutual inductance of solenoid 1 with respect to solenoid 2 will be equal to the mutual inductance of solenoid 2 with respect to solenoid 1.
• The mutual inductance of both the solenoids is given as,

\(⇒ M_{12} = M_{21} = μ_on_1n_2πr_{1}^{2}l\)     -----(1)

Where n1 = number of turns per unit length of solenoid 1, n2 = number of turns per unit length of solenoid 2, r1 = radius of the inner solenoid, and l = length of both the solenoids

• By equation 1 it is clear that the mutual inductance of both the solenoids depends on the n1, n2, r1, and l. Hence, option 1 is correct.

CONCEPT:

Mutual-inductance

• When two coils are brought in proximity with each other the magnetic field in one of the coils tends to link with the other. If this magnetic field of the first coil is changed then the magnetic flux associated with the second coil changes and this leads to the generation of voltage in the second coil.
• This property of a coil that affects or changes the current and voltage in a secondary coil is called mutual inductance.
• The SI unit of the mutual inductance is Henry.
• Let two coils of the number of turns N1 and N2 are placed close to each other. Then the mutual inductance of coil 1 with respect to coil 2 is given as,

\(⇒ M_{12}=N_{1}\frac{ϕ_{1}}{I_{2}}\)

Where N1 = number of turns in coil 1, ϕ1 = flux linked with coil 1, and I2 current in coil 2

Mutual inductance for two co-axial solenoids:

• Let two long solenoids S1 and S2 of equal length are placed co-axially as shown in the figure.
• The solenoid S1 is placed inside the solenoid S2.
• The mutual inductance of both the solenoids will be equal and it is given as,

\(⇒ M_{12} = M_{21} = μ_on_1n_2πr_{1}^{2}l\)

For medium of relative permeability μr,

\(⇒ M_{12} = M_{21} = \mu_rμ_on_1n_2πr_{1}^{2}l\)

Where n1 = number of turns per unit length of solenoid 1, n2 = number of turns per unit length of solenoid 2, r1 = radius of the inner solenoid, and l = length of both the solenoids

EXPLANATION:

• We know that if there are two solenoids of equal length and one solenoid is placed coaxially inside the other solenoid then the mutual inductance of solenoid 1 with respect to solenoid 2 will be equal to the mutual inductance of solenoid 2 with respect to solenoid 1.
• The mutual inductance of both the solenoids is given as,

\(⇒ M_{12} = M_{21} = μ_on_1n_2πr_{1}^{2}l\)     -----(1)

Where n1 = number of turns per unit length of solenoid 1, n2 = number of turns per unit length of solenoid 2, r1 = radius of the inner solenoid, and l = length of both the solenoids

• By equation 1 it is clear that the mutual inductance of both the solenoids does not depend on the current in the solenoids.
• Therefore when the current in both the solenoids is doubled, the mutual inductance of both the solenoids S1 and S2 will remain unchanged. Hence, option 3 is correct.

Calculation: 

The mutual inductance \( M \) between two coils is given as 2 H.

If the current in the primary coil changes from 10 A to 0 A in 0.1 seconds, the change in current \( \Delta I \) is:

\(\Delta I = 10 \, \text{A} - 0 \, \text{A} = 10 \, \text{A}\)

The time interval \( \Delta t \) for this change is 0.1 s.

The induced emf \( \mathcal{E} \) in the secondary coil can be calculated using the formula for mutual induction:

\(\mathcal{E} = -M \frac{\Delta I}{\Delta t}\)

Substituting the given values:

\(\mathcal{E} = -2 \, \text{H} \times \frac{10 \, \text{A}}{0.1 \, \text{s}}\)

\(\mathcal{E} = -2 \, \text{H} \times 100 \, \text{A/s}\)

\(\mathcal{E} = -200 \, \text{V}\)

Since we are typically interested in the magnitude of the induced emf, we have:

\(\mathcal{E} = 200 \, \text{V}\)

∴ The correct answer is option 2.

Concept Used:

The magnetic field inside the solenoid is given by the formula:

B = μ₀ n I = μ₀ (N / L) I

where:

N is the number of turns of the solenoid

L is the length of the solenoid

I is the current flowing through the solenoid

μ₀ is the permeability of free space

The magnetic flux (Φ) through the smaller coil is given by:

Φ = B A = μ₀ (N / L) I A

where:

A is the area of the smaller coil

The induced emf (e) in the smaller coil is given by Faraday's Law of electromagnetic induction:

e = - (dΦ / dt)

Substituting the expression for magnetic flux (Φ), we get:

e = - μ₀ (N / L) A (dI / dt)

Calculation:

The current is reversed in 0.2s, so the rate of change of current (dI / dt) is calculated as:

dI / dt = (2 - (-2)) / 0.2 = 20 A / s

Substituting the values into the formula for emf:

e = - (4π × 10^-7) × (3000 / 0.2) × 1.4 × 10^-3 × 20

e = - 1.32 × 10^-1 V

Therefore, the induced emf produced is:

e = 1.32 × 10^-1 V

Concept Used:

The mutual inductance between two loops depends on the magnetic flux through one loop due to the current in the other loop. The general formula for mutual inductance (M) is given by:

M = Nφ / I

Where: N = number of turns in the loop φ = magnetic flux through the loop I = current in the loop

Solution:

The mutual inductance is given by:

M = (μ₀ I R²) / (2 (8R³)³/2) × cos 45°

Now, we can simplify the expression by considering the geometry of the setup. Substituting the values for the magnetic flux and simplifying:

μ₀ a² / (8R × 2¹/2) = R × 2²/2

Therefore, the value of p is 7.

Conclusion:
The correct answer is: p = 7.

The coupled flux of two coils system is used to define the mutual inductance between the coils. The mutual inductance between the coils is \(M_{21}=\dfrac{N_2\phi_{21}}{I_1}\)

So it is defined as the proportionality between the emf generated in coil 2 due to the current flows in coil 1. Thus It depends on the relative position and orientation of two coils.

Concept:

• Mutual inductance (M) between two coils quantifies the ability of one coil to induce an electromotive force (emf) in the other due to a changing current.
• It depends on the geometric arrangement of the coils, not on the rate of change of current.

Factors Affecting Mutual Inductance:

• Relative position and orientation of coils: Coils placed closer and aligned properly will have higher mutual inductance.
• Number of turns in the coils: More turns in a coil increase the linkage of magnetic flux.
• Permeability of the core material: A ferromagnetic core increases mutual inductance.

Explanation of Options:

• Rate of change of current in A (Incorrect): Mutual inductance is a geometric property and does not depend on how fast the current changes.
• Rate of change of current in A and B (Incorrect): While emf depends on the rate of change of current, mutual inductance does not.
• Material of the wire (Incorrect): The material of the wire does not significantly affect mutual inductance.
• Relative position and orientation of coils (Correct Answer): The distance and alignment of the coils directly influence their mutual inductance.
• Direction of flow of current in B (Incorrect): The direction of current does not affect the value of mutual inductance, only the polarity of induced emf.

The mutual inductance between coils A and B depends on the relative position and orientation of A and B.

Given:

A square loop of side length a is moving away from an infinitely long current-carrying conductor at a constant speed v. Let x be the instantaneous distance between the conductor and the side AB of the loop. The goal is to determine how the mutual inductance M between the loop and the conductor changes with time.

Concept:

• The mutual inductance M depends on the magnetic flux through the loop caused by the current in the long conductor.
• For an infinitely long conductor, the magnetic field B at a distance x from the conductor is proportional to 1/x.
• The magnetic flux through the square loop is proportional to B \cdot \text{Area of the loop}, and since \(B \propto 1/x \)and the area remains constant, the flux linkage does not depend on x.
• Thus, the mutual inductance M, which is proportional to the flux linkage, remains constant regardless of the distance x.

Calculation:

Using the relation for mutual inductance:

M = constant, because the flux linkage and geometric configuration between the loop and conductor remain constant.

Conclusion:

∴ The mutual inductance M remains constant with time, as shown in option 3 (horizontal line).