A conducting moves on two conducting rails as shown in the figure. A constant magnetic field B into the page. The…

Year	2026	2025	2024	Total
Questions	12	6	10	28

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. $-2 V$ $-2\text{ V}$
B. $+2 V$ $+2\text{ V}$
C. $+1 V$ $+1\text{ V}$
D. $-1 V$ $-1\text{ V}$

Solution

Related Formula

emf = -(Δφ)/(Δ t)

Core Logic

Initial magnetic flux is given by $φᵢ = B A θ$ $\phi_i = B A \cos\theta$, where $A = 2.5 \times 2 = 5 m²$ $A = 2.5 \times 2 = 5\text{ m}^2$, $B = 4 T$ $B = 4\text{ T}$, and $θ = 60°$ $\theta = 60^{\circ}$ (angle aligned with the axis mapping context rules in the problem source text).

φᵢ = 4 \times 5 \times (60°) = 20 \times 0.5 = 10 Wb

Final flux after removal $φf = 0$ $\phi_f = 0$.

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 $ω t = 1$ $\cos\omega t = 1$:

emfmax = M i₀ ω

Substitute given numerical values:

emfmax = 0.002 \times 5 \times 50π = 0.5π = (π)/(2) V

Step 2: Match with target variable

Comparing $(π)/(2)$ $\frac{\pi}{2}$ with $(π)/(α)$ $\frac{\pi}{\alpha}$ gives:

$α = 2$ $\alpha = 2$

Pattern Recognition

Harmonic driving functions induce derivative cosine arrays whose peak amplitudes scale strictly as the product of primary parameters: $M \cdot i₀ \cdot ω$ $M \cdot i_0 \cdot \omega$.

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 \times 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 ( $e$ $e$) is:

e = (Δ φ)/(Δ t)

where magnetic flux $φ$ $\phi$ is defined via dot product:

φ = B \cdot 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 $j$ $\hat{j}$):

A = (0.1 ~m)² j = 0.01 j ~m²

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

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

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

Step 1: Calculate Initial Flux

$φᵢ = B \cdot A = ( 0.2\sqrt(2) ) \times 0.01 = 2 \times 10⁻³\sqrt(2) = \sqrt(2) \times 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 $\Delta t = 1 \mathrm{~s}, the final flux value$, the final flux value $\phi_f = 0:

e = (|0 - φᵢ|)/(1) = \sqrt(2) \times 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 $η$ $\eta$ 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: $Vₚ = 2.3 ~kV = 2300 ~V$ $Iₚ = 5 ~A$ $I_p = 5 \mathrm{~A}$

Pᵢₙ = 2300 \times 5 ~W

Step 2: Calculate Output Power and Current

Efficiency $η = 90% = 0.9$ $\eta = 90\% = 0.9$

Pout = η \times Pᵢₙ = 0.9 \times 2300 \times 5

Since $Pout = Vₛ Iₛ$ $P_{\text{out}} = V_s I_s$ and $Vₛ = 230 ~V$ $V_s = 230 \mathrm{~V}$:

Vₛ Iₛ = 230 \times Iₛ = 0.9 \times 2300 \times 5

Iₛ = (0.9 \times 2300 \times 5)/(230) = 0.9 \times 10 \times 5

Iₛ = 9 \times 5 = 45 ~A

Pattern Recognition

Transformer equations are direct: $η Vₚ Iₚ = Vₛ Iₛ$ $\eta V_p I_p = V_s I_s$. The number of turns ( $3000$ $3000$) 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 \times 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 ( $Bv$ $B_v$). 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 \times 10⁻⁴ ~T) \times (1)/(2) = 0.25 \times 10⁻⁴ ~T = (1)/(4) \times 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) \times 10⁻⁴) (40π) (0.8)²

ε = (1)/(8) \times 10⁻⁴ \times 40π \times 0.64

ε = 5π \times 10⁻⁴ \times 0.64

ε = 3.2π \times 10⁻⁴ ~V

ε = 32π \times 10⁻⁵ ~V

Step 3: Extract N

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

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 $Bv$ $B_v$ for horizontal spinners.

Chapter Mix

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

Solution & Explanation

Related Formula

Core Logic

Step 1: Geometric Analysis

Pattern Recognition

Chapter Mix

Reference Study Guides

More Electromagnetic Induction Previous-Year Questions — Page 5

Solution

Related Formula

Core Logic

Step 1: Compute Induced EMF

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Isolate Maximum value

Step 2: Match with target variable

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Calculate Initial Flux

Step 2: Find Induced EMF

Step 2: Find Induced EMF

Step 3: Extract x

Step 3: Extract x

Pattern Recognition

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Calculate Input Power

Step 2: Calculate Output Power and Current

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Calculate Field and Omega

Step 2: Calculate Induced EMF

Step 3: Extract N

Pattern Recognition

Chapter Mix

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