JEE Advanced Previous Year Questions (2018 - 2023): Electromagnetic Induction | Physics for JEE Main & Advanced PDF Download

2023

Q1: A thin conducting rod MN of mass  20 gm, length  25 cm and resistance  10 Ω is held on frictionless, long, perfectly conducting vertical rails as shown in the figure. There is a uniform magnetic field B₀ = 4 T directed perpendicular to the plane of the rod-rail arrangement. The rod is released from rest at time t = 0 and it moves down along the rails. Assume air drag is negligible. Match each quantity in List-I with an appropriate value from List-II, and choose the correct option.
[Given: The acceleration due to gravity   g = 10 m s⁻² and e⁻¹ = 0.4 ] 

(a) P→5,Q→2,R→3,S→1
(b) P→3,Q→1,R→4,S→5
(c) P→4,Q→3,R→1,S→2
(d) P→3,Q→4,R→2,S→5                      [JEE Advanced 2023 Paper 1]
Ans: (d)
From force equation

At t = 0.2 sec

(P) Now at t=0.2sec
The magnitude of the induced emf = E = Bvℓ

(Q) At t = 0.2sec, the magnitude of magnetic force = BIℓsin⁡θ

(R) At t  = 0.2sec, the power dissipated as heat 

(S) Magnitude of terminal velocity
At terminal velocity, the net force become zero

Hence, Answer is (D)

2022

Q1: Consider an LC circuit, with inductance L = 0.1H and capacitance   C = 10⁻³ F, kept on a plane. The area of the circuit is   1 m². It is placed in a constant magnetic field of strength B₀ which is perpendicular to the plane of the circuit. At time t = 0, the magnetic field strength starts increasing linearly as  with β= 0.04Ts⁻¹. 
The maximum magnitude of the current in the circuit is _________ mA.   [JEE Advanced 2022 Paper 1]
Ans: 3.98 to 4.02
Emf induced in the circuit is

So the circuit can be rearranged as

Using Kirchhoff's law we can write

Using SHM concept we can write 

So,

2021

Q1: In a circuit, a metal filament lamp is connected in series with a capacitor of capacitance C μF across a 200 V, 50 Hz supply. The power consumed by the lamp is 500 W while the voltage drop across it is 100 V. Assume that there is no inductive load in the circuit. Take rms values of the voltages. The magnitude of the phase-angle (in degrees) between the current and the supply voltage is φ. Assume, π√3 ≈ 5.
The value of C is ____________.     [JEE Advanced 2021 Paper 2]
Ans: 100
for lamp,

Q2: In a circuit, a metal filament lamp is connected in series with a capacitor of capacitance C μF across a 200 V, 50 Hz supply. The power consumed by the lamp is 500 W while the voltage drop across it is 100 V. Assume that there is no inductive load in the circuit. Take rms values of the voltages. The magnitude of the phase-angle (in degrees) between the current and the supply voltage is φ. Assume, π√3 ≈ 5.
The value of φ is ____________.     [JEE Advanced 2021 Paper 2]
Ans: 100
for lamp,

2020

Q1: The inductors of two LR circuits are placed next to each other, as shown in the figure. The values of the self-inductance of the inductors, resistors, mutual-inductance and applied voltages are specified in the given circuit. After both the switches are closed simultaneously, the total work done by the batteries against the induced EMF in the inductors by the time the currents reach their steady state values is _________ mJ.                      [JEE Advanced 2020 Paper 2]

Ans: 55
Mutual inductance is producing flux in same direction as self-inductance.

= 55 mJ

2019

Q1: A 10 cm long perfectly conducting wire PQ is moving with a velocity I cm/s on a pair of horizontal rails of zero resistance. One side of the rails is connected to an inductor L = 1 mH and a resistance R = 1Ω as shown in figure. The horizontal rails, L and R lie in the same plane with a uniform magnetic field B = 1 T perpendicular to the plane. If the key S is closed at certain instant, the current in the circuit after 1 millisecond is x × 10^-3 A, where the value of x is _____
[Assume the velocity of wire PQ remains constant (1 cm/s) after key S is closed. Given e^-1 = 0.37, where e is base of the natural logarithm]                              [JEE Advanced 2019 Paper 2]

Ans: 0.63
Motional emf,

i = 0.63 mA

Q2: A conducting wire of parabolic shape, initially y = x², is moving with velocity  in a non-uniform magnetic field , as shown in figure. If V₀, B₀, L and β are positive constants and Δϕ is the potential difference developed between the ends of the wire, then the correct statement(s) is/are 

(a) |Δϕ|=

(c) |Δϕ| =

Motional emf across the length dy is,

emf in loop is proportional to L for given value of β,

The length of projection of the wire Y = X of length √2L on the y-axis is thus, the answer remain unchanged.
Therefore, correct options are (a), (b) and (d) 

2018

(a)
(b)
(c)
(d)                         [JEE Advanced 2018 Paper 1]
Ans: (b) & (d)

From principle of superposition,

I is maximum when dI / dt = 0, which gives