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

Q41 jee_main_2026_23_january_evening Faraday's Law of Induction

A circular loop of radius 7 cm is placed in uniform magnetic field of 0.2 T directed perpendicular to plane of loop. The loop is converted into a square loop in 0.5 s. The EMF induced in the loop is ____ mV.

A. $6.6$ $6.6$
B. $13.2$ $13.2$
C. $8.25$ $8.25$
D. $1.32$ $1.32$

Solution

Related Formula

EMF = (Δ φ)/(Δ t) = B (A₁ - A₂)/(Δ t)

Core Logic

The magnetic field is constant, but the area changes as the circle deforms into a square. Length of the wire is conserved. Circumference of circular loop = $2π r \approx 14π cm$ $2\pi r \approx 14\pi \, \mathrm{cm}$.

Side length of the new square loop:

4a = 14π a = (7π)/(2) cm

Step 1: Calculate Change in Area

Δ A = Acircle - Asquare

A₁ = π r² = π (7)² = 49π cm²

A₂ = a² = ( (7π)/(2) )² = (49π²)/(4) cm²

Δ A = ( 49π - (49π²)/(4) ) \times 10⁻⁴ m²

Step 2: Calculate Flux Change and EMF

Δ φ = B Δ A = 0.2 ( 49π - (49π²)/(4) ) \times 10⁻⁴

Using $π \approx 3.1415$ $\pi \approx 3.1415$: $49π \approx 153.938$ $49\pi \approx 153.938$ $(49π²)/(4) \approx 120.89$ $\frac{49\pi^2}{4} \approx 120.89$ $Δ A \approx 33.04 \times 10⁻⁴ m²$

Wait, taking standard approximations:

Δ φ = 0.2 \times 33.07 \times 10⁻⁴ = 6.614 \times 10⁻⁴ Wb

EMF = (Δ φ)/(Δ t) = 6.614 \times 10⁻⁴0.5 = 13.228 \times 10⁻⁴ V

EMF = 1.32 mV

Pattern Recognition

When a loop is deformed into a different shape, perimeter is conserved. For a given perimeter, a circle always encloses maximum area. Subtracting the new area from the old area immediately gives the $Δ A$ $\Delta A$ driving the induced EMF.

Chapter Mix

Class 12 Physics: Electromagnetic Induction

Q42 jee_main_2026_23_january_evening Mutual Inductance

Suppose a long solenoid of 100 cm length, radius 2 cm having 500 turns per unit length, carries a current

I = 10 (ω t)

A, where

ω = 1000 rad/s

. A circular conducting loop (B) of radius 1 cm coaxially slid through the solenoid at a speed

v = 1 cm/s

. The r.m.s. current through the loop when the coil B is inserted 10 cm inside the solenoid is

α/\sqrt(2)

. The value of

α

is ____. [Resistance of the loop

= 10Ω

]

A. $197$ $197$
B. $80$ $80$
C. $280$ $280$
D. $100$ $100$

Solution

Related Formula

ε = - A (dB)/(dt)

$B = μ₀ n I$ $B = \mu_0 n I$

irms = εrmsR

Core Logic

Since the loop B is well inside the long solenoid (10 cm deep into a 100 cm length), the magnetic field is uniform across its cross-section and approximately equal to the ideal solenoid field $B = μ₀ n I$ $B = \mu_0 n I$. Here $n = 500$ $n = 500$ turns/m. The sliding velocity $v$ $v$ just means it's moving through a uniform field region, so motional EMF does not contribute to the flux change; only the transformer EMF (time-varying $B$ $B$ field) creates the induced current.

Step 1: Calculate Induced EMF

I = I₀ ω t (dI)/(dt) = I₀ ω ω t

ε = A (dB)/(dt) = A μ₀ n (dI)/(dt) = (π r²) μ₀ n (I₀ ω ω t)

Step 2: Calculate RMS Current

Peak EMF $ε₀ = π r² μ₀ n I₀ ω$ .

i₀ = (ε₀)/(R) = (π r² μ₀ n I₀ ω)/(R)

irms = i₀\sqrt(2) = π r² μ₀ n I₀ ω\sqrt(2) R

Step 3: Numerical Substitution

irms = π \times (10⁻²)² \times (4π \times 10⁻⁷) \times 500 \times 10 \times 1000\sqrt(2) \times 10

irms = π \times 10⁻⁴ \times 4π \times 10⁻⁷ \times 5 \times 10⁶\sqrt(2) \times 10

irms = 20π² \times 10⁻⁶ \times 10\sqrt(2) \times 10 = 20π²\sqrt(2) \times 10⁻⁶ A

Using $π² \approx 9.87$ $\pi^2 \approx 9.87$, $20π² \approx 197.39$ $20\pi^2 \approx 197.39$. Thus, $irms \approx 197\sqrt(2)$ . $α = 197$ $\alpha = 197$.

Pattern Recognition

Motion parallel to uniform field lines inside an ideal solenoid induces no motional EMF. Only the time derivative of the AC current $(dI)/(dt)$ $\frac{dI}{dt}$ contributes to Faraday induction.

Chapter Mix

Class 12 Physics: Electromagnetic Induction Class 12 Physics: Alternating Current

Q8 jee_main_2025_02_april_evening Self Induction and Energy Stored

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 :

A. $(B² A l)/(μ₀)$ $\frac{B^2 A l}{\mu_0}$
B. $(B² A l)/(2 μ₀)$ $\frac{B^2 A l}{2 \mu_0}$
C. $B² A l$ $B^2 A l$
D. $(B² A l)/(4 μ₀)$ $\frac{B^2 A l}{4 \mu_0}$

Solution

Related Formula

Magnetic Energy Density (energy per unit volume):

um = (B²)/(2 μ) = (B²)/(2 μᵣ μ₀)

Total Energy stored:

U = um \times V = um \times (A l)

Core Logic

We are given:

Relative permeability $μᵣ = 2$ $\mu_r = 2$
Permeability of the filled core medium $μ = μᵣ μ₀ = 2 μ₀$ $\mu = \mu_r \mu_0 = 2 \mu_0$

Substitute $μᵣ = 2$ $\mu_r = 2$ into the energy density expression:

um = (B²)/(2 (2 μ₀)) = (B²)/(4 μ₀)

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

U = um \cdot V = (B²)/(4 μ₀) A l

Pattern Recognition

Sees: Magnetic energy stored in a solenoid with medium relative permeability. Trap: Placing the relative permeability $μᵣ$ $\mu_r$ in the numerator of the formula instead of the denominator. Shortcut: Magnetic energy density is always inversely proportional to permeability. With medium $μ = 2μ₀$ $\mu = 2\mu_0$, energy density is halved compared to free space, giving $(B²)/(4μ₀)$ $\frac{B^2}{4\mu_0}$. Multiply by volume $Al$ $Al$ to get $(B² Al)/(4μ₀)$ $\frac{B^2 Al}{4\mu_0}$.

Chapter Mix

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

Q jee_main_2025_04_april_morning Motional Electromotive Force

Conductor wire ABCDE with each arm

10~cm

in length is placed in magnetic field of

1\sqrt(2)~Tesla

, perpendicular to its plane. When conductor is pulled towards right with constant velocity of

10~cm/s

, induced emf between points A and E is ________ mV.

Conductor wire path geometry layout for Q25 - JEE Main 2025 Morning — The figure outlines a segmented conductive wire trail pulled laterally through an orthogonal inward magnetic field matrix region.

Numerical Answer. Answer: 10 to 10

Solution

Related Formula

Motional electromotive force formula:

ε = B v leff

where $leff$ $l_{\text{eff}}$ is the perpendicular component of the straight-line displacement vector connecting the endpoints $A$ $A$ and $E$ $E$ ( $lAE$ $l_{AE}$) relative to velocity $v$ $\vec{v}$.

Core Logic

In a uniform magnetic field, the motional EMF induced in any arbitrary conductor wire depends solely on the straight-line displacement vector connecting its endpoints, rather than the detailed path:

ε = ( v \times B) \cdot leff

Effective vector length translation resolution mapping for Q25 - JEE Main 2025 Morning — The figure outlines a segmented conductive wire trail pulled laterally through an orthogonal inward magnetic field matrix region.

Step 1: Compute Effective Length

From the geometry of the symmetric wire segments oriented at $45°$ $45^{\circ}$ to the horizontal:

leff = 2 \times (10 45°) = 2 \times 10 \times 1\sqrt(2) = 10\sqrt(2) cm = 0.1\sqrt(2) m

Step 2: Calculate Induced EMF

Substitute the given parameters into the motional EMF expression:

ε = B v leff

ε = ( 1\sqrt(2)) \times (0.1 m/s) \times (0.1\sqrt(2) m)

ε = 1\sqrt(2) \times 0.1 \times 0.1\sqrt(2) = 0.01 V = 10 mV

Pattern Recognition

In a uniform magnetic field, motional EMF is path-independent. Replace any zig-zag or curved conductor with an equivalent straight line joining the two endpoints perpendicular to the velocity vector.

Evaluation Rubric / Model Answer

10

Chapter Mix

Class 12 Physics: Electromagnetic Induction

Q1 jee_main_2025_28_jan_evening Motional EMF

A uniform magnetic field of

0.4 T

acts perpendicular to a circular copper disc

20 ~cm

in radius. The disc is having a uniform angular velocity of

10 π

rad

s⁻¹

about an axis through its centre and perpendicular to the disc. What is the potential difference developed between the axis of the disc and the rim? (

π = 3.14

)

A. $0.0628 V$ $\text{0.0628 V}$
B. $0.5024 V$ $\text{0.5024 V}$
C. $0.2512 V$ $\text{0.2512 V}$
D. $0.1256 V$ $\text{0.1256 V}$

Solution

Related Formula

The induced electromotive force (EMF) developed between the center and the rim of a rotating disc in a perpendicular magnetic field is given by:

E = (1)/(2) B ω R²

Core Logic

Given parameters from the problem statement [cite: 655, 657, 658]:

Magnetic field, $B = 0.4 T$ $B = 0.4 \text{ T}$
Radius of the disc, $R = 20 cm = 0.2 m$ $R = 20 \text{ cm} = 0.2 \text{ m}$
Angular velocity, \omega = 10\pi \text{ rad s}^{-1}

Substituting the values into the governing formula:

E = (1)/(2) \times 0.4 \times (10 \times 3.14) \times (0.2)²

E = 0.2 \times 31.4 \times 0.04

E = 0.2512 V

Step 1: Evaluation

The potential difference developed between the axis of the disc and the rim is precisely $0.2512 V$ $0.2512 \text{ V}$.

Pattern Recognition

For any rotating conductor of length $R$ $R$ or a continuous disc rotating about its center in a perpendicular magnetic field, the induced EMF is mathematically equivalent to a single radial rod sweeping the area, leading directly to the formula $(1)/(2)Bω R²$ $\frac{1}{2}B\omega R^2$.

Chapter Mix

Class 12 Physics: Electromagnetic Induction

Solution & Explanation

Related Formula

Core Logic

Step 1: Geometric Analysis

Pattern Recognition

Chapter Mix

Reference Study Guides

More Electromagnetic Induction Previous-Year Questions — Page 3

Solution

Related Formula

Core Logic

Step 1: Calculate Change in Area

Step 2: Calculate Flux Change and EMF

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Calculate Induced EMF

Step 2: Calculate RMS Current

Step 3: Numerical Substitution

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Compute Effective Length

Step 2: Calculate Induced EMF

Pattern Recognition

Evaluation Rubric / Model Answer

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Evaluation

Pattern Recognition

Chapter Mix

More Electromagnetic Induction Questions — jee_main_2025_28_jan_evening

Select Your Target Exam

Full Mock Test Hub

Topic-wise Practice Hub

Exam Analysis Tools

Chapter Weightage & Trends

Topic-wise Practice Hub

JEE Main Exam Papers

Chapter Weightage Board

Formula Recognition

Exam Questions

Select an Exam

Welcome back