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

More Electromagnetic Induction Previous-Year Questions

Q28 jee_main_2026_21_jan_morning Faraday's Law

A conducting circular loop of area

1.0 m²

is placed perpendicular to a magnetic field which varies as

B = (100 t) Tesla

. If the resistance of the loop is

100 Ω

, then the average thermal energy dissipated in the loop in one period is ____ J.

A. $(π)/(2)$ $\frac{\pi}{2}$
B. $2π$ $2\pi$
C. $π$ $\pi$
D. $π²$ $\pi^{2}$

Solution

Related Formula

φ = B \cdot A

E = -(dφ)/(dt)

P = (E²)/(R)

Core Logic

Given area of the loop, $A = 1 m²$ $A = 1\text{ m}^2$ and magnetic field $B = (100t)$ $B = \sin(100t)$. The magnetic flux passing through the loop is:

φ = B \cdot A = (100t) \times 1 = (100t)

Induced EMF $E = | (dφ)/(dt) | = 100 (100t)$

Step 1: Calculating Power and Energy

Instantaneous power $P = (E²)/(R) = (100² ²(100t))/(100) = 100 ²(100t)$ .

Thermal energy dissipated in one time period $T$ $T$:

Q = \int₀T P dt = \int₀T 100 ²(100t) dt

The angular frequency $ω = 100 rad/s$ $\omega = 100\text{ rad/s}$, so time period $T = (2π)/(ω) = (2π)/(100) = (π)/(50) sec$ .

Q = 100 \int₀π/50 ²(100t) dt = 100 \int₀π/50 (1 + (200t))/(2) dt

Q = 50 [ t + ( (200t))/(200) ]₀π/50

Q = 50 ( (π)/(50) - 0 ) = π Joules

Pattern Recognition

For a sinusoidal signal, the integral of $²(ω t)$ $\cos^2(\omega t)$ over one full period $T$ $T$ is always $T/2$ $T/2$. Thus, $\int P dt = Pmax \times (T)/(2)$ .

Chapter Mix

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

Q29 jee_main_2026_21_jan_evening LC Oscillations

A capacitor C is first charged fully with potential difference of

V₀

and disconnected from the battery. The charged capacitor is connected across an inductor having inductance L. In

t s

25% of the initial energy in the capacitor is transferred to the inductor. The value of t is ________ s.

A. $π\sqrt(LC)3$ $\frac{\pi\sqrt{LC}}{3}$
B. $π\sqrt(LC)6$ $\frac{\pi\sqrt{LC}}{6}$
C. $π\sqrt(LC)2$ $\frac{\pi\sqrt{LC}}{2}$
D. $π\sqrt((LC)/(2))$ $\pi\sqrt{\frac{LC}{2}}$

Solution

Related Formula

For LC oscillations, charge varies as:

Q(t) = Q₀ (ω t)

Where $ω = 1\sqrt(LC)$ $\omega = \frac{1}{\sqrt{LC}}$

Energy in capacitor:

UC = (Q²)/(2C)

Core Logic

Since 25% of the initial energy is transferred to the inductor, the remaining energy in the capacitor is 75% of its initial value.

UCf = 75% of UCᵢ

(Qf²)/(2C) = (3)/(4) \times (Q₀²)/(2C)

Qf² = (3)/(4) Q₀²

Qf = \sqrt(3)2 Q₀

Step 1: Finding Time

Using the equation for charge variation:

Q₀ (ω t) = \sqrt(3)2 Q₀

(ω t) = \sqrt(3)2

ω t = (π)/(6)

Step 2: Final Conclusion

Substitute $ω = 1\sqrt(LC)$ $\omega = \frac{1}{\sqrt{LC}}$:

1\sqrt(LC) t = (π)/(6)

t = π \sqrt(LC)6

Pattern Recognition

Energy is proportional to charge squared. $75%$ $75\%$ energy remaining means charge is $\sqrt(0.75) = \sqrt(3)2$ $\sqrt{0.75} = \frac{\sqrt{3}}{2}$ of the original. Cosine of $π/6$ $\pi/6$ yields this exact ratio.

Chapter Mix

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

Q29 jee_main_2026_22_january_morning Motional EMF and Terminal Speed

XPQY is a vertical smooth long loop having a total resistance R where PX is parallel to QY and separation between them is l. A constant magnetic field B perpendicular to the plane of the loop exists in the entire space. A rod CD of length L (L > l) and mass m is made to slide down from rest under the gravity as shown in figure. The terminal speed acquired by the rod is \_\_\_\_ m/s. (g = acceleration due to gravity)

Electromagnetic Induction diagram for Q29 - JEE Main 2026 January Morning — Vertical smooth long loop with sliding rod under gravity and magnetic field.

A. $2mgRB²l²$ $\frac{2\mathrm{mgR}}{\mathrm{B}^2l^2}$
B. $8mgRB²l²$ $\frac{8\mathrm{mgR}}{\mathrm{B}^2l^2}$
C. $2 mgRB ^ 2 L ^ 2$
D. $mgRB²l²$ $\frac{\mathrm{mgR}}{\mathrm{B}^2l^2}$

Solution

Related Formula

e = Bvl, i = (e)/(R)

Core Logic

Solution diagram for Q29 - JEE Main 2026 Morning — Vertical smooth long loop with sliding rod under gravity and magnetic field.

At equilibrium (for terminal velocity):

mg = iBl mg = ((Bvl)/(R))Bl

v = (mgR)/(B²l²)

Pattern Recognition

Sees: Sliding rod in magnetic field reaching terminal speed. Shortcut: Equate gravitational force with magnetic force $iBl$ $iBl$ at terminal velocity. Check: Matches option (4). ✓

Chapter Mix

Class 12 Physics: Electromagnetic Induction

Q45 jee_main_2026_22_january_morning Mutual Induction and Lenz's Law

Three identical coils

C₁

C₂

and

C₃

are closely placed such that they share a common axis.

C₂

is exactly midway.

C₁

carries current I in anti-clockwise direction while

C₃

carries current I in clockwise direction. An induced current flows through

C₂

will be in clockwise direction when

Electromagnetic Induction diagram for Q45 - JEE Main 2026 January Morning — Three coaxial identical coils with opposite current directions.

A. $C₁$ $C_{1}$ and $C₃$ $C_{3}$ move with equal speeds away from $C₂$ $C_{2}$
B. $C₁$ $C_{1}$ moves towards $C₂$ $C_{2}$ and $C₃$ $C_{3}$ moves away from $C₂$ $C_{2}$
C. $C₁$ $\mathrm{C}_1$ moves away from $C₂$ $\mathrm{C}_2$ and $C₃$ $\mathrm{C}_3$ moves towards $C₂$ $\mathrm{C}_2$
D. $C₁$ $C_{1}$ and $C₃$ $C_{3}$ move with equal speeds towards $C₂$ $C_{2}$

Solution

Related Formula

Bₙₑₜ = BC₂ - BC₁, ε = -(dφ)/(dt)

Core Logic

Solution diagram for Q45 - JEE Main 2026 Morning — Three coaxial identical coils with opposite current directions.

Applying Lenz's law and magnetic field superposition: for induced current in $C₂$ $C_2$ to be clockwise, the net magnetic flux through $C₂$ $C_2$ must change accordingly. When $C₁$ $C_1$ moves towards $C₂$ $C_2$ and $C₃$ $C_3$ moves away from $C₂$ $C_2$, the net field change induces the specified clockwise current.

Pattern Recognition

Sees: Coaxial current-carrying coils in relative motion. Shortcut: Analyze net magnetic field variation at middle coil $C₂$ $C_2$ using Lenz's law. Check: Matches option (2). ✓

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

Solution

Related Formula

Core Logic

Step 1: Calculating Power and Energy

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Finding Time

Step 2: Final Conclusion

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Pattern Recognition

Chapter Mix

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