A 1 m long metal rod AB completes the circuit as shown in figure. The area of circuit is perpendicular to the magnetic field of 0.10 T. If the resistance of the total circuit is 2Omega then the force needed to move the rod towards right with constant speed (v) of 1.5 m/s is ____ N.

### Related Formula E = B l v ### Core Logic To maintain a constant speed, the external force applied must balance the opposing magnetic force generated by the induced current. ### Step 1: Calculate External Force Given values: B = 0.10 T l = 1 m v = 1.5 m/s R = 2, Omega ### Pattern Recognition Standard "sliding rod on rails" problem. The required mechanical force to maintain terminal velocity is always F = (B2 L2 v/R). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Electromagnetic Induction

A 1 m long metal rod AB completes the circuit as shown in figure. The area of circuit is perpendicular to the magnetic field of 0.10 T. If the resistance of the total circuit is 2Omega then the force

### Related Formula E = B l v ### Core Logic To maintain a constant speed, the external force applied must balance the opposing magnetic force generated by the induced current. ### Step 1: Calculate External Force Given values: B = 0.10 T l = 1 m v = 1.5 m/s R = 2, Omega ### Pattern Recognition Standard "sliding rod on rails" problem. The required mechanical force to maintain terminal velocity is always F = (B2 L2 v/R). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics:

A 1 m long metal rod AB completes the circuit as shown in figure. The area of circuit is perpendicular to the magnetic field of 0.10 T. If the resistance of the total circuit is 2Omega then the force needed to move the rod towards right with constant speed (v) of 1.5 m/s is ____ N.

### Related Formula E = B l v ### Core Logic To maintain a constant speed, the external force applied must balance the opposing magnetic force generated by the induced current. ### Step 1: Calculate External Force Given values: B = 0.10 T l = 1 m v = 1.5 m/s R = 2, Omega ### Pattern Recognition Standard "sliding rod on rails" problem. The required mechanical force to maintain terminal velocity is always F = (B2 L2 v/R). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Electromagnetic Induction

A 1 m long metal rod AB completes the circuit as shown in figure. The area of circuit is perpendicular to the magnetic field of 0.10 T. If the resistance of the total circuit is 2Omega then the force

### Related Formula E = B l v ### Core Logic To maintain a constant speed, the external force applied must balance the opposing magnetic force generated by the induced current. ### Step 1: Calculate External Force Given values: B = 0.10 T l = 1 m v = 1.5 m/s R = 2, Omega ### Pattern Recognition Standard "sliding rod on rails" problem. The required mechanical force to maintain terminal velocity is always F = (B2 L2 v/R). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics:

A 1 m long metal rod AB completes the circuit as shown in figure. The area of circuit is perpendicular to the magnetic…

Year	2026	2025	2024	Total
Questions	12	6	10	28

Year

2026

2025

2024

Total

Questions

Solution & Explanation

Related Formula

$E = B l v$ $E = B l v$

i = (E)/(R)

FB = i l B = (B² l² v)/(R)

Core Logic

To maintain a constant speed, the external force applied must balance the opposing magnetic force generated by the induced current.

Fₑₓₜ = FB

Step 1: Calculate External Force

Given values: $B = 0.10 T$ $B = 0.10\text{ T}$ $l = 1 m$ $l = 1\text{ m}$ $v = 1.5 m/s$ $v = 1.5\text{ m/s}$ $R = 2 Ω$ $R = 2\, \Omega$

Fₑₓₜ = (B² l² v)/(R)

Fₑₓₜ = ((0.10)² \times (1)² \times 1.5)/(2)

Fₑₓₜ = (0.01 \times 1.5)/(2) = (0.015)/(2) = 0.0075

Fₑₓₜ = 7.5 \times 10⁻³ N

Motional EMF solution diagram for Q41 - JEE Main 2026 Morning — A conducting rod AB moves on a U-shaped rail in a perpendicular magnetic field.

Pattern Recognition

Standard "sliding rod on rails" problem. The required mechanical force to maintain terminal velocity is always $F = (B² L² v)/(R)$ $F = \frac{B^2 L^2 v}{R}$.

Chapter Mix

Class 12 Physics: Electromagnetic Induction

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 $t = 2 s$ $t = 2 \text{ s}$:

ε = - (10 \times 2 - 36)

ε = -(20 - 36) = 16 V

Step 2: Calculate Induced Current

I = (ε)/(R) = (16)/(8) = 2 A

Pattern Recognition

Flux polynomials ( $At² - Bt + C$ $At^2 - Bt + C$) 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

\sqrt(x) \times 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

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

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

B = 4 \times [ (μ₀ i)/(4π (L/2)) ( 45^° + 45^°) ]

B = (μ₀ i)/(π (L/2)) ( 2\sqrt(2) )

B = 2\sqrt(2) μ₀ iπ L

Step 2: Mutual Inductance Calculation

Flux linkage for the inner loop:

φ = B \cdot ²

φ = 2\sqrt(2) μ₀ iπ L ²

Given $L = ²$ $L = \ell^2$:

φ = 2\sqrt(2) μ₀ iπ ( ²) ² = 2\sqrt(2) μ₀ iπ

Mutual inductance $M$ $M$:

M = (φ)/(i) = 2\sqrt(2) μ₀π

Using $μ₀ = 4π \times 10⁻⁷$ $\mu_0 = 4\pi \times 10^{-7}$:

M = 2\sqrt(2) \times 4π \times 10⁻⁷π

M = 8\sqrt(2) \times 10⁻⁷ H

M = \sqrt(128) \times 10⁻⁷ H

Comparing with $\sqrt(x) \times 10⁻⁷$ $\sqrt{x} \times 10^{-7}$, we get $x = 128$ $x = 128$.

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

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. $7$ $7$
B. $70$ $70$
C. $35$ $35$
D. $140$ $140$

Solution

Related Formula

ε = N | (Δφ)/(Δ t) |

Δφ = (Δ B) A θ

Core Logic

Given data: Initial Magnetic Field, $Bᵢ = 5000 T$ $B_i = 5000\mathrm{\,T}$ Final Magnetic Field, $Bf = 3000 T$ $B_f = 3000\mathrm{\,T}$ Time interval, $Δ t = 2 s$ $\Delta t = 2\mathrm{\,s}$ Diameter, $d = 0.02 m \Rightarrow r = 0.01 m$ Induced emf, $ε = 22 V$ $\varepsilon = 22\mathrm{\,V}$

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) )

$22 = N (0.1π)$ $22 = N (0.1\pi)$ Taking $π \approx 22/7$ $\pi \approx 22/7$:

22 = N ( 0.1 \times (22)/(7) )

1 = (N)/(70) \Rightarrow N = 70

Chapter Mix

Class 12 Physics: Electromagnetic Induction

Solution & Explanation

Related Formula

Core Logic

Step 1: Calculate External Force

Pattern Recognition

Chapter Mix

More Electromagnetic Induction Previous-Year Questions — Page 6

Solution

Related Formula

Core Logic

Step 1: Calculate Induced EMF

Step 2: Calculate Induced Current

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 2: Mutual Inductance Calculation

Chapter Mix

Solution

Related Formula

Core Logic

Step 2: Equation Evaluation

Chapter Mix

More Electromagnetic Induction Questions — jee_main_2026_21_jan_morning

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

Solution & Explanation

Related Formula

Core Logic

Step 1: Calculate External Force

Pattern Recognition

Chapter Mix

More Electromagnetic Induction Previous-Year Questions — Page 6

Solution

Related Formula

Core Logic

Step 1: Calculate Induced EMF

Step 2: Calculate Induced Current

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 2: Mutual Inductance Calculation

Chapter Mix

Solution

Related Formula

Core Logic

Step 2: Equation Evaluation

Chapter Mix

More Electromagnetic Induction Questions — jee_main_2026_21_jan_morning

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