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Electromagnetic Induction appeared 28 times across 3 years — 3.2% of Physics. This question is from Motional EMF and Terminal Speed.

Year	2026	2025	2024	Total
Questions	12	6	10	28

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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)
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 & Explanation

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

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

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

Reference Study Guides

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

in

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

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