CUET UG Physics Booster Test 2-Mutual and Self-Inductance -

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CUET UG Physics Booster Test 2-Mutual and Self-Inductance

⏱️ 30 Minutes

❓ 20 Questions

📌 Answers are locked once submitted — results and explanations appear at the end.

CUET UG Physics Booster Test 2-Mutual and Self-Inductance

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QUESTION 1 OF 20

In a closely wound coil of N turns experiencing a varying current

Answer Review

1 In a closely wound coil of N turns experiencing a varying current

A the flux linkage NΦ_B decreases as current increases

B the same magnetic flux is linked with all the turns and the total flux linkage is proportional to the current

C only half of the turns experience flux linkage

D the flux through the coil is independent of the geometry

💡 Short Explanation

�� Closely wound turns link the same flux. �� Flux linkage = NΦ. �� Flux linkage ∝ current.

✅ Correct Answer Explanation

In a closely wound coil, each turn links approximately the same magnetic flux Φ. → Therefore total flux linkage is NΦ. → Since magnetic field and flux are proportional to current: NΦ ∝ I → This leads to the definition: NΦ = LI

❓ Why Other Options Are Incorrect

�� Option A → Flux linkage increases with current.
�� Option C → All turns participate in flux linkage.
�� Option D → Flux depends on geometry and material.

🎯 Strategy

Used

�� Conceptual Matching

Application:

�� Use the definition of flux linkage in a coil.

Final Logic:

�� Same flux through all turns gives NΦ ∝ I.

🧠 Memory Trick

"All Turns, Same Flux."

📚 Class XII – Physics Part I – Chapter: Electromagnetic Induction – Topic: Concept of Inductance

2 Inductance of a circuit depends primarily on which of the following factors?

A Geometry, Intrinsic material properties

B Voltage, External resistance

C Charge separation, Magnetic pole strength

D Electrostatic flux, Temperature

💡 Short Explanation

�� Shape matters. �� Medium matters. �� Current does not define L.

✅ Correct Answer Explanation

Inductance depends on: • Number of turns • Coil dimensions • Magnetic permeability of the medium → These are geometric and material properties.

❓ Why Other Options Are Incorrect

�� Option B → Voltage and resistance do not determine L.
�� Option C → Magnetic pole strength is not used.
�� Option D → Electrostatic flux is unrelated.

🎯 Strategy

Used

�� Elimination

Application:

�� Identify parameters appearing in inductance formulas.

Final Logic:

�� Geometry and medium determine inductance.

🧠 Memory Trick

"Shape + Material = L."

📚 Class XII – Physics Part I – Chapter: Electromagnetic Induction – Topic: Inductance

3 Identify the incorrect statement about the henry unit definition:

A It is the SI unit of inductance.

B It is denoted by the symbol H.

C It is named in honour of Joseph Henry who independently discovered electromagnetic induction.

D It has the dimensions of [M L² T⁻³ A⁻¹].

💡 Short Explanation

�� Henry is SI unit of inductance. �� Symbol is H. �� Dimension is [ML²T⁻²A⁻²].

✅ Correct Answer Explanation

Inductance dimensions are: [ML²T⁻²A⁻²] → Option D gives dimensions of voltage/current-related quantities incorrectly. → Therefore, D is the incorrect statement.

❓ Why Other Options Are Incorrect

�� Option A → Correct.
�� Option B → Correct notation.
�� Option C → Historically correct.

🎯 Strategy

Used

�� Dimensional Analysis

Application:

�� Recall dimensions of inductance.

Final Logic:

�� Henry has dimensions [ML²T⁻²A⁻²].

🧠 Memory Trick

"Henry = Flux ÷ Current."

📚 Class XII – Physics Part I – Chapter: Electromagnetic Induction – Topic: Unit and Dimensions of Inductance

4 Match List I with List II

List I	List II
1. Magnetic Flux	a. [M L² T⁻² A⁻²]
2. EMF	b. [M L² T⁻² A⁻¹]
3. Mutual Inductance	c. [M L² T⁻³ A⁻¹]
4. Self Inductance	d. [M L² T⁻² A⁻²]

A 1-b, 2-c, 3-a, 4-d

B 1-a, 2-b, 3-c, 4-d

C 1-c, 2-b, 3-a, 4-d

D 1-b, 2-a, 3-c, 4-d

💡 Short Explanation

�� Flux → Weber dimensions. �� EMF → Voltage dimensions. �� Both inductances have same dimensions.

✅ Correct Answer Explanation

1 → b because Flux = [ML²T⁻²A⁻¹] → 2 → c because EMF = [ML²T⁻³A⁻¹] → 3 → a because Mutual inductance = [ML²T⁻²A⁻²] → 4 → d because Self inductance has the same dimensions.

❓ Why Other Options Are Incorrect

�� Option B → Flux and inductance mismatched.
�� Option C → Flux and emf interchanged.
�� Option D → EMF dimension incorrect.

🎯 Strategy

Used

�� Dimensional Analysis

Application:

�� Match physical quantities with standard dimensions.

Final Logic:

�� Flux, EMF and inductance dimensions uniquely identify the matching.

🧠 Memory Trick

"Flux-A⁻¹, EMF-T⁻³, L-A⁻²."

📚 Class XII – Physics Part I – Chapter: Electromagnetic Induction – Topic: Dimensions and Units

5 If a varying current I₂ flows in coil 2, it induces an emf in a neighbouring coil 1. Thus, if the mutual inductance is M₁₂, the induced emf ε₁ will be

A - M₁₂ (dI₂ / dt)

B - M₁₂ (dΦ₁ / dt)

C M₁₂ (dI₁ / dt)

D - L (dI₂ / dt)

💡 Short Explanation

�� Mutual induction formula. �� Lenz's law introduces minus sign. �� Depends on current change in coil 2.

✅ Correct Answer Explanation

Induced emf due to mutual induction is: ε₁ = -M₁₂(dI₂/dt) → Negative sign arises from Lenz's law.

❓ Why Other Options Are Incorrect

�� Option B → Uses incorrect derivative.
�� Option C → Depends on wrong current.
�� Option D → Uses self-inductance instead of mutual inductance.

🎯 Strategy

Used

�� Formula Recognition

Application:

�� Recall standard mutual induction equation.

Final Logic:

�� ε = -M(dI/dt).

🧠 Memory Trick

"Mutual → M dI/dt."

📚 Class XII – Physics Part I – Chapter: Electromagnetic Induction – Topic: Mutual Induction

6 If two concentric circular coils, one of small radius r₁ = 0.02 m and the other of large radius r₂ = 0.2 m, are placed co-axially with centers coinciding, what will be their mutual inductance M₁₂? (μ₀ = 4π × 10⁻⁷ T m A⁻¹)

A 3.94 × 10⁻⁹ H

B 7.89 × 10⁻⁹ H

C 1.97 × 10⁻⁹ H

D 5.56 × 10⁻⁹ H

💡 Short Explanation

�� M = μ₀πr₁²/2r₂. �� Smaller coil determines flux area. �� Substitute values.

✅ Correct Answer Explanation

For concentric circular coils: M = μ₀πr₁² / 2r₂ → Substituting: μ₀ = 4π×10⁻⁷ r₁ = 0.02 m r₂ = 0.2 m → M ≈ 3.94 × 10⁻⁹ H

❓ Why Other Options Are Incorrect

�� Option B → Approximately double.
�� Option C → Half-value error.
�� Option D → Calculation error.

🎯 Strategy

Used

�� Substitution

Application:

�� Apply standard mutual inductance formula.

Final Logic:

�� Numerical substitution gives 3.94 × 10⁻⁹ H.

🧠 Memory Trick

"Small Coil Controls Flux."

📚 Class XII – Physics Part I – Chapter: Electromagnetic Induction – Topic: Mutual Inductance of Circular Coils

7 Consider two long co-axial solenoids S1 and S2. Which of the following statements are correct regarding the magnetic flux and mutual inductance between the solenoids?
Statements:
1. Total number of turns in the inner solenoid is n₁l.
2. Flux linked with the outer solenoid S₂ due to current I₁ in S₁ is assumed to be confined solely inside S₁.
3. The magnetic field μ₀n₁I₁ exists uniformly across the entire cross-section of S₂.
4. The flux linkage with solenoid S₂ is N₂Φ₂ = M₂₁I₁.

A 1, 3, 4

B 1, 2, 4

C 2, 3

D 1, 2, 3, 4

💡 Short Explanation

�� N = nl. �� Flux effectively confined to inner solenoid. �� Flux linkage relation is correct.

✅ Correct Answer Explanation

Statement 1 is correct because total turns = n₁l. → Statement 2 is correct in the ideal long-solenoid approximation. → Statement 3 is incorrect because the field is not uniform across the entire cross-section of S₂. → Statement 4 is correct from the definition of mutual inductance.

❓ Why Other Options Are Incorrect

�� Option A → Contains incorrect Statement 3.
�� Option C → Includes incorrect Statement 3.
�� Option D → Contains incorrect Statement 3.

🎯 Strategy

Used

�� Elimination

Application:

�� Check long-solenoid approximations.

Final Logic:

�� Statement 3 is false; 1, 2 and 4 are correct.

🧠 Memory Trick

"Flux Mostly Inside Inner."

📚 Class XII – Physics Part I – Chapter: Electromagnetic Induction – Topic: Co-axial Solenoids

8 The calculation of mutual inductance for long co-axial solenoids is,

A strictly requires knowing the resistance of the coils

B assumes the inner solenoid is perfectly spherical

C requires neglecting edge effects and assuming a uniform magnetic field

D is independent of the lengths of the solenoids

💡 Short Explanation

�� Long-solenoid approximation. �� Uniform field assumption. �� Edge effects ignored.

✅ Correct Answer Explanation

Mutual inductance derivation assumes: • Long solenoids • Uniform magnetic field • Negligible edge effects → Hence Option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → Resistance is irrelevant.
�� Option B → No spherical assumption.
�� Option D → Length appears in mutual inductance formula.

🎯 Strategy

Used

�� Conceptual Elimination

Application:

�� Recall assumptions used in derivation.

Final Logic:

�� Long-solenoid model requires uniform field.

🧠 Memory Trick

"Long Solenoid → Ignore Ends."

📚 Class XII – Physics Part I – Chapter: Electromagnetic Induction – Topic: Mutual Inductance Derivation

9 Identify the correct statements about the equality of mutual inductance M₁₂ = M₂₁
Statements:
1. It has only been proven for long co-axial solenoids and does not apply generally.
2. It allows calculation of mutual inductance even when calculating flux linkage directly is very difficult.
3. It is highly useful when the inner solenoid is much shorter than the outer one.
4. It shows that the induced emf in coil 1 due to coil 2 is fundamentally different in magnitude from the reverse situation.

A 1, 2

B 2, 3

C 3, 4

D 1, 2, 3

💡 Short Explanation

�� Reciprocity theorem is general. �� Useful in difficult calculations. �� Especially useful for unequal solenoids.

✅ Correct Answer Explanation

Statement 2 is correct because reciprocity avoids difficult flux calculations. → Statement 3 is correct because one direction may be much easier to compute. → Statement 1 is incorrect because reciprocity is general. → Statement 4 is incorrect because M₁₂ = M₂₁.

❓ Why Other Options Are Incorrect

�� Option A → Contains incorrect Statement 1.
�� Option C → Contains incorrect Statement 4.
�� Option D → Contains incorrect Statement 1.

🎯 Strategy

Used

�� Elimination

Application:

�� Identify statements consistent with reciprocity theorem.

Final Logic:

�� Only Statements 2 and 3 are correct.

🧠 Memory Trick

"Difficult One? Use Reciprocity."

📚 Class XII – Physics Part I – Chapter: Electromagnetic Induction – Topic: Reciprocity Theorem

10 In a situation where a short inner solenoid is placed well inside a long outer solenoid

A calculating M₁₂ is difficult because the inner solenoid is immersed in a non-uniform field

B calculating M₂₁ is extremely difficult as the field due to the inner solenoid varies across the outer solenoid

C the reciprocity theorem fails to apply

D the mutual inductance becomes infinite

💡 Short Explanation

�� Short solenoid produces non-uniform field. �� Flux calculation becomes difficult. �� Reciprocity theorem remains valid.

✅ Correct Answer Explanation

Field produced by the short inner solenoid varies significantly over the outer solenoid. → Direct flux integration becomes difficult. → Therefore, M₂₁ is difficult to compute directly.

❓ Why Other Options Are Incorrect

�� Option A → M₁₂ is easier to calculate.
�� Option C → Reciprocity remains valid.
�� Option D → Mutual inductance remains finite.

🎯 Strategy

Used

�� Contextual Matching

Application:

�� Analyze field distribution of the short solenoid.

Final Logic:

�� Non-uniform field complicates direct computation.

🧠 Memory Trick

"Short Solenoid → Complex Field."

📚 Class XII – Physics Part I – Chapter: Electromagnetic Induction – Topic: Reciprocity Theorem

11 Identify the correct statements regarding Self-induction:
1. It is the phenomenon of inducing an emf in an isolated coil due to flux change from varying current in the same coil.
2. The induced emf aids the change in current.
3. Flux linkage is inversely proportional to current.
4. The proportionality constant L is called self-inductance.

A 1, 4

B 2, 3

C 1, 2, 4

D 3, 4

💡 Short Explanation

�� Self-induction occurs in the same coil. �� Self-induced emf opposes change. �� L is self-inductance.

✅ Correct Answer Explanation

Statement 1 is correct because self-induction refers to emf induced in a coil due to changing current in the same coil. → Statement 2 is incorrect because the induced emf opposes the change in current according to Lenz's law. → Statement 3 is incorrect because flux linkage is directly proportional to current: NΦ = LI → Statement 4 is correct because L is called self-inductance.

❓ Why Other Options Are Incorrect

�� Option B → Statements 2 and 3 are incorrect.
�� Option C → Contains incorrect Statement 2.
�� Option D → Contains incorrect Statement 3.

🎯 Strategy

Used

�� Elimination

Application:

�� Identify statements violating Lenz's law and NΦ = LI.

Final Logic:

�� Only Statements 1 and 4 are correct.

🧠 Memory Trick

"Self → Same Coil."

📚 Class XII – Physics Part I – Chapter: Electromagnetic Induction – Topic: Self-Induction

12 If the self-induced emf in a coil opposes the change in current, it is mathematically linked to the time derivative of current. Thus, if a coil has self-inductance L, the induced emf ε will be

A L (dΦ/dt)

B - L (dI/dt)

C - I (dL/dt)

D LI

💡 Short Explanation

�� Self-induction law. �� Negative sign from Lenz's law. �� Depends on rate of current change.

✅ Correct Answer Explanation

Self-induced emf is given by: ε = -L(dI/dt) → The negative sign indicates opposition to the change producing it.

❓ Why Other Options Are Incorrect

�� Option A → Not the self-induction formula.
�� Option C → Assumes changing inductance.
�� Option D → Represents flux linkage relation indirectly, not emf.

🎯 Strategy

Used

�� Formula Recognition

Application:

�� Recall standard self-induction equation.

Final Logic:

�� ε = -L(dI/dt).

🧠 Memory Trick

"Lenz Gives Minus."

📚 Class XII – Physics Part I – Chapter: Electromagnetic Induction – Topic: Self-Induced EMF

13 The self-induced back emf in a circuit is,
Choose correct:

A (A)

B (B)

C (C)

D (D)

💡 Short Explanation

�� Back emf opposes change. �� Stabilizes flux linkage. �� Follows Lenz's law.

✅ Correct Answer Explanation

Since magnetic flux is proportional to current, opposing changes in current means opposing changes in flux. → Back emf therefore tends to maintain existing flux linkage.

❓ Why Other Options Are Incorrect

�� Option A → Back emf resists current growth.
�� Option C → Not its primary role.
�� Option D → Does not increase battery voltage.

🎯 Strategy

Used

�� Contextual Matching

Application:

�� Relate back emf to Lenz's law.

Final Logic:

�� Back emf opposes flux change.

🧠 Memory Trick

"Back EMF Resists Change."

📚 Class XII – Physics Part I – Chapter: Electromagnetic Induction – Topic: Back EMF

14 Identify the incorrect statement about electromagnetic inertia:

A Self-inductance physically plays the role of inertia.

B It is the electromagnetic analogue of mass in mechanics.

C It opposes the growth and decay of current in the circuit.

D It relies on the presence of a second neighbouring coil to manifest.

💡 Short Explanation

�� Self-inductance exists in a single coil. �� No second coil required. �� Electromagnetic inertia arises from self-induction.

✅ Correct Answer Explanation

Self-inductance is a property of a single circuit. → Back emf appears whenever current changes in that circuit. → Therefore, a second neighbouring coil is unnecessary.

❓ Why Other Options Are Incorrect

�� Option A → Correct analogy.
�� Option B → Standard mechanical analogy.
�� Option C → Correct effect of inductance.

🎯 Strategy

Used

�� Elimination

Application:

�� Distinguish self-induction from mutual induction.

Final Logic:

�� Electromagnetic inertia arises from self-induction alone.

🧠 Memory Trick

"Self Means One Coil."

📚 Class XII – Physics Part I – Chapter: Electromagnetic Induction – Topic: Electromagnetic Inertia

15 If a long solenoid of cross-sectional area 1.0 × 10⁻³ m² and length 0.5 m has 1000 turns, what will be its self-inductance? (μ₀ = 4π × 10⁻⁷ T m A⁻¹)

A 2.51 × 10⁻³ H

B 1.26 × 10⁻³ H

C 5.02 × 10⁻³ H

D 3.14 × 10⁻⁴ H

💡 Short Explanation

�� Use L = μ₀N²A/l. �� Substitute values. �� Obtain 2.51 × 10⁻³ H.

✅ Correct Answer Explanation

L = μ₀N²A/l = (4π×10⁻⁷)(1000)²(10⁻³)/(0.5) = 2.51 × 10⁻³ H

❓ Why Other Options Are Incorrect

�� Option B → Half the correct value.
�� Option C → Double the correct value.
�� Option D → Major calculation error.

🎯 Strategy

Used

�� Substitution

Application:

�� Apply solenoid self-inductance formula.

Final Logic:

�� Direct substitution gives 2.51 × 10⁻³ H.

🧠 Memory Trick

"L ∝ N²A/l."

📚 Class XII – Physics Part I – Chapter: Electromagnetic Induction – Topic: Self-Inductance of Solenoid

16 Core permeability and self-inductance relation:

A High relative permeability, High self-inductance

B High relative permeability, Low self-inductance

C Low relative permeability, High self-inductance

D Zero relative permeability, Infinite self-inductance

💡 Short Explanation

�� L ∝ μᵣ. �� Larger permeability increases flux linkage. �� Therefore L increases.

✅ Correct Answer Explanation

L = μᵣμ₀n²Al → Since L is directly proportional to μᵣ, increasing relative permeability increases self-inductance.

❓ Why Other Options Are Incorrect

�� Option B → Opposite dependence.
�� Option C → Opposite dependence.
�� Option D → Physically incorrect.

🎯 Strategy

Used

�� Direct Formula Recognition

Application:

�� Observe proportionality with μᵣ.

Final Logic:

�� Higher μᵣ ⇒ Higher L.

🧠 Memory Trick

"More μ, More L."

📚 Class XII – Physics Part I – Chapter: Electromagnetic Induction – Topic: Magnetic Core Effects

17 Choose the correct statements about the work done against back emf
Statements:
1. Work needs to be done against the back emf to establish a current.
2. The rate of work done is given by |dW/dt| = εI.
3. If resistive losses are ignored, dW/dt = LI(dI/dt).
4. The total work done is dissipated entirely into the surroundings.

A 1, 3

B 1, 2, 3

C 2, 4

D 1, 2, 3, 4

💡 Short Explanation

�� Work is required against back emf. �� Power = εI. �� Energy is stored magnetically.

✅ Correct Answer Explanation

Statement 1 is correct because current establishment requires work against back emf. → Statement 2 is correct: P = εI → Statement 3 is correct because: ε = L(dI/dt) Therefore: P = LI(dI/dt) → Statement 4 is incorrect because energy is stored in the magnetic field.

❓ Why Other Options Are Incorrect

�� Option A → Omits correct Statement 2.
�� Option C → Contains incorrect Statement 4.
�� Option D → Contains incorrect Statement 4.

🎯 Strategy

Used

�� Elimination

Application:

�� Identify the statement inconsistent with magnetic energy storage.

Final Logic:

�� Statement 4 is false.

🧠 Memory Trick

"Work Stored, Not Lost."

📚 Class XII – Physics Part I – Chapter: Electromagnetic Induction – Topic: Magnetic Energy Storage

18 Match List I with List II based on mechanical and electrical analogues

List I	List II
1. Mass (m)	a. Current (I)
2. Velocity (v)	b. Back EMF (ε)
3. Kinetic Energy (½mv²)	c. Magnetic Energy (½LI²)
4. Force opposing acceleration	d. Self-Inductance (L)

A 1-a, 2-d, 3-c, 4-b

B 1-d, 2-a, 3-c, 4-b

C 1-b, 2-c, 3-d, 4-a

D 1-d, 2-c, 3-a, 4-b

💡 Short Explanation

�� Mass ↔ Inductance. �� Velocity ↔ Current. �� Kinetic energy ↔ Magnetic energy.

✅ Correct Answer Explanation

1 → d because mass corresponds to self-inductance. → 2 → a because velocity corresponds to current. → 3 → c because kinetic energy corresponds to magnetic energy. → 4 → b because back emf opposes current change like force opposes acceleration.

❓ Why Other Options Are Incorrect

�� Option A → Mass and current mismatched.
�� Option C → Multiple incorrect analogies.
�� Option D → Velocity and magnetic energy mismatched.

🎯 Strategy

Used

�� Option Grouping

Application:

�� Match standard mechanical-electrical analogies.

Final Logic:

�� Mass:L :: Velocity:I.

🧠 Memory Trick

"Mass→L, Speed→I."

📚 Class XII – Physics Part I – Chapter: Electromagnetic Induction – Topic: Electromagnetic Inertia

19 If the magnetic energy stored in a solenoid is (½)LI², and its volume is Al, the magnetic energy per unit volume (u_B) can be rewritten using the magnetic field B. Thus, if a magnetic field B exists in the solenoid, the energy density u_B will be

A B²/2μ₀

B B/2μ₀

C B²μ₀/2

D B²/μ₀

💡 Short Explanation

�� Standard magnetic energy density. �� Depends on B². �� Inversely proportional to μ₀.

✅ Correct Answer Explanation

Magnetic energy density is: u_B = B²/(2μ₀) → Derived from magnetic energy stored in a solenoid divided by its volume.

❓ Why Other Options Are Incorrect

�� Option B → Missing square of B.
�� Option C → Incorrect dependence on μ₀.
�� Option D → Missing factor ½.

🎯 Strategy

Used

�� Formula Recognition

Application:

�� Recall standard energy-density formula.

Final Logic:

�� u_B = B²/(2μ₀).

🧠 Memory Trick

"B-square over 2μ."

📚 Class XII – Physics Part I – Chapter: Electromagnetic Induction – Topic: Magnetic Energy Density

20 In a comparison between the magnetic energy density of a solenoid and the electrostatic energy density of a parallel plate capacitor

A only the magnetic energy is proportional to the square of the field strength

B both energy densities are inversely proportional to the square of their respective field strengths

C both cases demonstrate that energy is proportional to the square of the field strength

D electrostatic energy relies on free space permeability μ₀

💡 Short Explanation

�� u_B ∝ B². �� u_E ∝ E². �� Both are quadratic field relations.

✅ Correct Answer Explanation

Magnetic energy density: u_B = B²/(2μ₀) → Electrostatic energy density: u_E = ½ε₀E² → Both expressions show energy density proportional to the square of the field magnitude.

❓ Why Other Options Are Incorrect

�� Option A → Electrostatic energy is also proportional to field squared.
�� Option B → Opposite of actual dependence.
�� Option D → Electrostatic energy uses ε₀, not μ₀.

🎯 Strategy

Used

�� Option Grouping

Application:

�� Compare the two energy-density formulas.

Final Logic:

�� Both contain squared field terms.

🧠 Memory Trick

"Field² Stores Energy."

📚 Class XII – Physics Part I – Chapter: Electromagnetic Induction – Topic: Magnetic and Electrostatic Energy Density

CUET UG Physics Booster Test 2-Mutual and Self-Inductance

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