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Electromagnetic Induction appeared 28 times across 3 years — 3.2% of Physics. This question is from Self Induction and Energy Stored.

Year 2026 2025 2024 Total
Questions 12 6 10 28

A solenoid having area A and length l is filled with a material having relative permeability 2. The magnetic energy stored in the solenoid is :

Solution & Explanation

Related Formula
  • Magnetic Energy Density (energy per unit volume):
um = (B²)/(2 μ) = (B²)/(2 μᵣ μ₀)
  • Total Energy stored:
U = um × V = um × (A l)
Core Logic

We are given:

  • Relative permeability μᵣ = 2
  • Permeability of the filled core medium μ = μᵣ μ₀ = 2 μ₀
  • Substitute μᵣ = 2 into the energy density expression:

um = (B²)/(2 (2 μ₀)) = (B²)/(4 μ₀)

Multiply by the total volume of the solenoid (V = A l):

U = um · V = (B²)/(4 μ₀) A l
Pattern Recognition

Sees: Magnetic energy stored in a solenoid with medium relative permeability. Trap: Placing the relative permeability μᵣ in the numerator of the formula instead of the denominator. Shortcut: Magnetic energy density is always inversely proportional to permeability. With medium μ = 2μ₀, energy density is halved compared to free space, giving (B²)/(4μ₀). Multiply by volume Al to get (B² Al)/(4μ₀).

Chapter Mix

Class 12 Physics: Electromagnetic Induction Class 12 Physics: Magnetism and Matter

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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