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

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Questions 12 6 10 28

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 Type:
Enter a numerical value Answer: 2 to 2 +4 marks

Solution & Explanation

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

Reference Study Guides

More Electromagnetic Induction Previous-Year Questions — Page 5

Q42 jee_main_2024_27_jan_morning Faraday's Law
A rectangular loop of length 2.5 m and width 2 m is placed at 60° to a magnetic field of 4 T. The loop is removed from the field in 10 sec. The average emf induced in the loop during this time is:
  • A. -2 V
  • B. +2 V
  • C. +1 V
  • D. -1 V

Solution

Related Formula
emf = -(Δφ)/(Δ t)
Core Logic

Initial magnetic flux is given by φᵢ = B A θ, where A = 2.5 × 2 = 5 m², B = 4 T, and θ = 60° (angle aligned with the axis mapping context rules in the problem source text).

φᵢ = 4 × 5 × (60°) = 20 × 0.5 = 10 Wb

Final flux after removal φf = 0.

Step 1: Compute Induced EMF
emf = -(φf - φᵢ)/(Δ t) = -(0 - 10)/(10) = +1 V
Pattern Recognition

Removal fields yield positive flux variants under canonical sign configurations due to the absolute reduction profile mapped by Lenz/Faraday relationships.

Chapter Mix

Class 12 Physics: Electromagnetic Induction

Q53 jee_main_2024_27_jan_morning Mutual Induction
Two coils have mutual inductance 0.002 H. The current changes in the first coil according to the relation i = i₀ ω t, where i₀ = 5 A and \omega = 50π rad/s. The maximum value of emf in the second coil is (π)/(α) V. The value of α is ______.
Numerical Answer. Answer: 2 to 2

Solution

Related Formula
emf = -M (di)/(dt)
Core Logic

Differentiating the current expression with respect to time:

(di)/(dt) = (d)/(dt)(i₀ ω t) = i₀ ω ω t

Hence, the expression for induced emf is:

emf = -M i₀ ω ω t
Step 1: Isolate Maximum value

The peak magnitude occurs when ω t = 1:

emfmax = M i₀ ω

Substitute given numerical values:

emfmax = 0.002 × 5 × 50π = 0.5π = (π)/(2) V
Step 2: Match with target variable

Comparing (π)/(2) with (π)/(α) gives:

α = 2

Pattern Recognition

Harmonic driving functions induce derivative cosine arrays whose peak amplitudes scale strictly as the product of primary parameters: M · i₀ · ω.

Chapter Mix

Class 12 Physics: Electromagnetic Induction

Q jee_main_2024_29_jan_morning Faraday's and Lenz's Law
A square loop of side 10 ~cm and resistance 0.7 Ω is placed vertically in east-west plane. A uniform magnetic field of 0.20 ~T is set up across the plane in north east direction. The magnetic field is decreased to zero in 1 ~s at a steady rate. Then, magnitude of induced emf is x × 10⁻³ ~V. The value of x is
Numerical Answer. Answer: 2 to 2

Solution

Related Formula

According to Faraday's Law of Electromagnetic Induction, the magnitude of induced EMF (e) is:

e = (Δ φ)/(Δ t)

where magnetic flux φ is defined via dot product:

φ = B · A = B A θ
Core Logic

Let the East-West plane lie vertical along the x-z plane. The normal vector to the loop points along the North direction (along j):

A = (0.1 ~m)² j = 0.01 j ~m²

The uniform magnetic field points North-East, meaning it is at an angle of 45^° to the North vector direction:

B = B 45^° i + B 45^° j = 0.2√(2) i + 0.2√(2) j

Vector reference showing orientation of loop area and North-East magnetic field vectors for Q53
Vector reference showing orientation of loop area and North-East magnetic field vectors for Q53

Step 1: Calculate Initial Flux
$φᵢ = B · A = ( 0.2√(2) ) × 0.01 = 2 × 10⁻³√(2) = √(2) × 10⁻³ ~Wb
Step 2: Find Induced EMF

Since the field goes steadily to zero in

Step 2: Find Induced EMF

Since the field goes steadily to zero in $\Delta t = 1 \mathrm{~s}, the final flux value\phi_f = 0:

e = (|0 - φᵢ|)/(1) = √(2) × 10⁻³ ~V
Step 3: Extract x

Comparing this with the target expression

Step 3: Extract x

Comparing this with the target expression $\sqrt{x} \times 10^{-3} \mathrm{~V}:

$x = 2

Pattern Recognition

Be careful when identifying angles between directional plane descriptors. An East-West plane has a normal axis directed North-South. A North-East field makes a clean

Pattern Recognition

Be careful when identifying angles between directional plane descriptors. An East-West plane has a normal axis directed North-South. A North-East field makes a clean $45^\circ$ angle relative to this structural normal line.

Chapter Mix

Class 12 Physics: Electromagnetic Induction

Q51 jee_main_2024_30_january_evening Transformer Efficiency
A power transmission line feeds input power at 2.3 ~kV to a step down transformer with its primary winding having 3000 turns. The output power is delivered at 230 ~V by the transformer. The current in the primary of the transformer is 5 A and its efficiency is 90%. The winding of transformer is made of copper. The output current of transformer is ________ A.
Numerical Answer. Answer: 45 to 45

Solution

Related Formula
η = PoutPᵢₙ Pᵢₙ = Vₚ Iₚ Pout = Vₛ Iₛ
Core Logic

The efficiency η of a transformer is the ratio of output power to input power. We can use this to find the secondary (output) current.

Step 1: Calculate Input Power

Given: Vₚ = 2.3 ~kV = 2300 ~V Iₚ = 5 ~A

Pᵢₙ = 2300 × 5 ~W
Step 2: Calculate Output Power and Current

Efficiency η = 90% = 0.9

Pout = η × Pᵢₙ = 0.9 × 2300 × 5

Since Pout = Vₛ Iₛ and Vₛ = 230 ~V:

Vₛ Iₛ = 230 × Iₛ = 0.9 × 2300 × 5 Iₛ = (0.9 × 2300 × 5)/(230) = 0.9 × 10 × 5 Iₛ = 9 × 5 = 45 ~A
Pattern Recognition

Transformer equations are direct: η Vₚ Iₚ = Vₛ Iₛ. The number of turns (3000) is distractor data not needed unless calculating the secondary turns.

Chapter Mix

Class 12 Physics: Alternating Current

Q60 jee_main_2024_30_jan_morning Motional EMF in Rotating Blades
A ceiling fan having 3 blades of length 80 ~cm each is rotating with an angular velocity of 1200 ~rpm. The magnetic field of earth in that region is 0.5 ~G and angle of dip is 30°. The emf induced across the blades is N π × 10⁻⁵ ~V. The value of N is _______.
Numerical Answer. Answer: 32 to 32

Solution

Related Formula
ε = (1)/(2) Bv ω ² Bv = B δ
Core Logic

For a horizontal fan rotating in the Earth's magnetic field, the blades cut only the vertical component of the magnetic field (Bv). The number of blades is a distractor, as they are all in parallel, meaning the EMF developed across one blade is the same as the EMF across the whole fan setup.

Step 1: Calculate Field and Omega

Vertical component of magnetic field:

Bv = B (30^°) = (0.5 × 10⁻⁴ ~T) × (1)/(2) = 0.25 × 10⁻⁴ ~T = (1)/(4) × 10⁻⁴ ~T

Angular velocity in rad/s:

ω = 2π f = 2π ((1200)/(60)) = 40π ~rad/s
Step 2: Calculate Induced EMF

Length of blade = 80 ~cm = 0.8 ~m.

ε = (1)/(2) Bv ω ² ε = (1)/(2) ((1)/(4) × 10⁻⁴) (40π) (0.8)² ε = (1)/(8) × 10⁻⁴ × 40π × 0.64 ε = 5π × 10⁻⁴ × 0.64 ε = 3.2π × 10⁻⁴ ~V ε = 32π × 10⁻⁵ ~V
Step 3: Extract N

Given ε = Nπ × 10⁻⁵ ~V, we can directly see that N = 32.

Pattern Recognition

Motional EMF for rotating rods acts like a battery. Multiple identical blades radiating from the center to the rim behave like multiple identical batteries in parallel; the total voltage does not stack. Isolate Bv for horizontal spinners.

Chapter Mix

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

More Electromagnetic Induction Questions — jee_main_2024_31_jan_evening

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