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Electromagnetic Induction appeared 28 times across 3 years — 3.2% of Physics. This question is from Mutual Induction and Lenz's Law.

Year 2026 2025 2024 Total
Questions 12 6 10 28

Three identical coils C₁, C₂ and C₃ are closely placed such that they share a common axis. C₂ is exactly midway. C₁ carries current I in anti-clockwise direction while C₃ carries current I in clockwise direction. An induced current flows through C₂ will be in clockwise direction when
Electromagnetic Induction diagram for Q45 - JEE Main 2026 January Morning
Three coaxial identical coils with opposite current directions.

Solution & Explanation

Related Formula
Bₙₑₜ = BC₂ - BC₁, ε = -(dφ)/(dt)
Core Logic

Solution diagram for Q45 - JEE Main 2026 Morning
Three coaxial identical coils with opposite current directions.

Applying Lenz's law and magnetic field superposition: for induced current in C₂ to be clockwise, the net magnetic flux through C₂ must change accordingly. When C₁ moves towards C₂ and C₃ moves away from C₂, the net field change induces the specified clockwise current.

Pattern Recognition

Sees: Coaxial current-carrying coils in relative motion. Shortcut: Analyze net magnetic field variation at middle coil C₂ using Lenz's law. Check: Matches option (2). ✓

Chapter Mix

Class 12 Physics: Electromagnetic Induction

Reference Study Guides

More Electromagnetic Induction Previous-Year Questions — Page 6

Q53 jee_main_2024_31_jan_evening Faraday's Law
The magnetic flux φ (in weber) linked with a closed circuit of resistance 8 Ω varies with time (in seconds) as φ = 5t² - 36t + 1. The induced current in the circuit at t = 2 s is ________ A.
Numerical Answer. Answer: 2 to 2

Solution

Related Formula
ε = -(dφ)/(dt) I = (|ε|)/(R)
Core Logic

Calculate the time derivative of the magnetic flux to find the induced EMF. Evaluate it at the requested time, and then apply Ohm's law to find the current magnitude.

Step 1: Calculate Induced EMF
ε = -(d)/(dt) (5t² - 36t + 1) ε = -(10t - 36)

At t = 2 s:

ε = - (10 × 2 - 36) ε = -(20 - 36) = 16 V
Step 2: Calculate Induced Current
I = (ε)/(R) = (16)/(8) = 2 A
Pattern Recognition

Flux polynomials (At² - Bt + C) instantly trigger a simple derivative test. Remember to drop the negative sign for final current magnitude unless direction is specifically asked.

Chapter Mix

Class 12 Physics: Electromagnetic Induction

Q jee_main_2024_31_jan_morning Mutual Inductance
A small square loop of wire of side is placed inside a large square loop of wire of side L (L = ²). The loops are coplanar and their centers coincide. The value of the mutual inductance of the system is √(x) × 10⁻⁷ H, where x =
Numerical Answer. Answer: 128 to 128

Solution

Related Formula
M = (φ₂)/(i₁) Bstraight wire segment = (μ₀ i)/(4π d) ( θ₁ + θ₂)
Core Logic

Mutual Inductance diagram for Q57 - JEE Main 2024 Morning
Mutual Inductance diagram for Q57 - JEE Main 2024 Morning

Assume a current i flows through the larger square loop of side L. The magnetic field generated by it at its center acts as a uniform field across the very small inner loop of side .

The magnetic field at the center of the large square loop (distance d = L/2 from each side, angles 45^°):

B = 4 × [ (μ₀ i)/(4π (L/2)) ( 45^° + 45^°) ] B = (μ₀ i)/(π (L/2)) ( 2√(2) ) B = 2√(2) μ₀ iπ L
Step 2: Mutual Inductance Calculation

Flux linkage for the inner loop:

φ = B · ² φ = 2√(2) μ₀ iπ L ²

Given L = ²:

φ = 2√(2) μ₀ iπ ( ²) ² = 2√(2) μ₀ iπ

Mutual inductance M:

M = (φ)/(i) = 2√(2) μ₀π

Using μ₀ = 4π × 10⁻⁷:

M = 2√(2) × 4π × 10⁻⁷π M = 8√(2) × 10⁻⁷ H M = √(128) × 10⁻⁷ H

Comparing with √(x) × 10⁻⁷, we get x = 128.

Chapter Mix

Class 12 Physics: Electromagnetic Induction

Q44 jee_main_2024_31_jan_morning Faraday's Law
A coil is placed perpendicular to a magnetic field of 5000 ~T. When the field is changed to 3000 ~T in 2s, an induced emf of 22 ~V is produced in the coil. If the diameter of the coil is 0.02 ~m, then the number of turns in the coil is:
  • A. 7
  • B. 70
  • C. 35
  • D. 140

Solution

Related Formula
ε = N | (Δφ)/(Δ t) | Δφ = (Δ B) A θ
Core Logic

Given data: Initial Magnetic Field, Bᵢ = 5000 T Final Magnetic Field, Bf = 3000 T Time interval, Δ t = 2 s Diameter, d = 0.02 m ⇒ r = 0.01 m Induced emf, ε = 22 V

Change in magnetic field magnitude |Δ B| = 5000 - 3000 = 2000 T. Area of the coil A = π r² = π (0.01)² = 10⁻⁴π m².

Step 2: Equation Evaluation
Δφ = |Δ B| A = (2000) π (0.01)² = 0.2π

Using Faraday's Law:

22 = N ( (0.2π)/(2) )

22 = N (0.1π) Taking π ≈ 22/7:

22 = N ( 0.1 × (22)/(7) ) 1 = (N)/(70) ⇒ N = 70
Chapter Mix

Class 12 Physics: Electromagnetic Induction

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JEE Physics: Waves (+15.5%) | Electrostatics: Concentric Shells (-29.7%) | Modern Physics: Photoelectric Clones (+34.2%) | Mathematics: Definite Integrals (+18.1%) | Chemistry: Coordination Splitting (-11.4%) | JEE Physics: Waves (+15.5%) | Electrostatics: Concentric Shells (-29.7%) | Modern Physics: Photoelectric Clones (+34.2%) | Mathematics: Definite Integrals (+18.1%) | Chemistry: Coordination Splitting (-11.4%)