JEE  >  Physics For JEE  >  JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1

JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE

1 Crore+ students have signed up on EduRev. Have you?

Q 1. A LCR circuit behaves like a damped harmonic oscillator. Comparing it with a physical spring-mass damped oscillator having damping constant b, the correct equivalence would be    [2020]
JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE     

 Ans: (4)
Solution:
Given LCR circuit
JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE
We know that equation of motion of damped oscillation is
JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE    (1)
Now in the LCR circuit, voltage drop along the circuit is
JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE
We know that
JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE
So
JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE
JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE     ..(2)
Comparing Eqs. (1) and (2), we get
m = L , b= R, K = 1/C
JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE

Q 2. An emf of 20 V is applied at time t = 0 to a circuit containing in series 10 mH inductor and 5 Ω resistor. The ratio of the currents at time t =∞ and t = 40 s is close to (Take e2 = 7.389)
(1) 1.06
(2) 1.15
(3) 1.46
(4) 0.84

Ans: (1)
Solution: 

Given that, L = 10 mH = 10 × 10−3 H, R = 5 Ω, V = 20 V.
We know that, current at instant time t is given by
JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE
where
I0 = V/R = 20/5 = 4A
And
JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE
So,
JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE 
I = 4[1 - exp( - 500t )]
at t =∞, we have

I = 4[1 - exp ( -∞] = 4(1-0) = 4
at t =40, we have
I40 = 4[1 - exp (-500 x 400)] = 4[1-exp (-20000)]
= 4[1-(e2)-10000] = 4[1 - (7.389)-10000]
So,
JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE
Therefore I/I40 is slightly greater than 1.

Q 3. As shown in the figure, a battery of emf ε is connected to an inductor L and resistance R in series. The switch is closed at t = 0. The total charge that flows from the battery, between t = 0 and t = tc (tc is the time constant of the circuit) is    [2020]
JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE
JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE
JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE
Ans:
(1)
Solution:
Charge flow for time t1 to t2 is given by
JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE
Here t1 = 0, t2 = tc
So,
JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE
Current flow in the circuit at any instant t is given by
JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE
Here I0 = ε/R, τ = tc
JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE
So,
JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE
JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE

Q 4. In LC circuit the inductance L = 40 mH and capacitance C = 100 μF. If a voltage V(t) = 10 sin (314t )  is applied to the circuit, the current in the circuit is given as    [2020]
(1) 0.52cos(314t)
(2) 10cos(314t)
(3) 5.2cos(314t)
(4) 0.52sin(314t)
Ans: 
(1)
Solution:
Given that LC series circuit having
L = 40mH = 40 x 10-3 H
C = 100μF = 100 x 10-6 F,
V(t) = 10sin (314t)
So,
ω = 314
X= ωL = 314 x 40 x 10-3  = 12.56 Ω
JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE
Z = XC - XL = 31.84 - 12.56 = 19.28 Ω
Since, XC > XL current lead the voltage by phase ϕ = π/2
Now, maximum current is given by
JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE
Therefore, current in the circuit is given by
JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE

Q 5. If the magnetic field in a plane electromagnetic wave is given by JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEEthen what will be expression for electric field?    [2020]
JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE
Ans:
(2)
Solution:
Given that  JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE 
Standard equation of magnetic field of EM wave is
JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE
So,
B0 = 3 x 10-8, k = 1.6 x 103 , ω = 48 x 1010, ϕ = 0
we know that
E0 = Bx c
= 3 x 10-8 x 3 x 108 = 9
Standard equation of electric field of EM wave is
JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE
Therefore,
JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE

Q 6. The electric field of a plane electromagnetic wave is given by
JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE

At t = 0, a positively charged particle is at the point(x ,y ,z ) = (0,0,π/k ). If its instantaneous velocity at (t = 0 ) is JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE the force acting on it due to the wave is    [2020]
(1) Parallel to JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE
(2) zero
(3) antiparallel to JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE
(4) parallel to JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE
Ans: 
(3)
Solution:
Given that,
JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE
at t = 0, position of charged particle (x,y,z) JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE
Instantaneous velocity, JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE
We know force is given by JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE
at t = 0, JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE
We know that speed of electromagnetic waves is given by
JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE
So, at t = 0,
JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE
JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE
Therefore,
JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE
JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE
So, force is antiparallel to  JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE

Q 7. A plane electromagnetic wave of frequency 25 GHz is propagating in vacuum along the z-direction. At a particular point in space and time, the magnetic field is given by JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE. The corresponding electric field JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE is (speed of light c = 3 x 108 m/s)    [2020]
JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE
JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE
JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE

Ans: (4)
Solution:
given that,
JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE
We know that
JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE
JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE

Q 8. The electric fields of two plane electromagnetic plane waves in vacuum are given by JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE.
At t = 0, a particle of charge q is at origin with a velocity JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE, (c is the speed of light in vacuum). The instantaneous force experienced by the particle is    [2020]
JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE
Ans: 
(4)
Solution:
Given that
JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE
For given electric fields of electromagnetic waves, there corresponding magnetic fields are
JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE
Force on a charge particle in these two electromagnetic waves is given by
JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE
JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE
JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE

Q 9. A plane electromagnetic wave is propagating along the direction JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE with its polarization along the direction JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE The correct form of the magnetic field of the wave would be (here B0 is an appropriate constant)    [2020]
JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE
JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE
Ans:
(1)
Solution:
Direction of propagation of EM wave is
JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE
Direction of polarization of electric filed is
JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE
So, direction of polarization of magnetic field is
JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE
Therefore, direction of polarization of magnetic field is
JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE

Q 10. A long solenoid of radius R carries a time (t) dependent current I(t) = I0t(1-t) with a ring of radius 2R is placed coaxially near its middle. During the time interval 0 ≤ t ≤ 1, the induced current (IR) and the induced EMF(VR) in the ring changes as    [2020]
(1) Direction of IR remains unchanged and VR is maximum at t = 0.5 s
(2) At t = 0.25 s direction of IR reserves and VR is maximum
(3) Direction of IR remains unchanged and VR is zero at t = 0.25 s
(4) At t = 0.5 s direction of IR reverses and VR is zero
Ans:
(4)
Solution:
Given that 
I(t) = I0t(1-t)
Induced emf is given by
JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE
where ϕ is flux
Solenoid magnetic flux is given by
JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE
So,
JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE
For t = 0.5 s,  VR = 0
So emf will be zero at t = 0.5 s
At t = 0.5 s, direction of IR reverses and VR is zero.

Q 11. A planar loop of wire rotates in a uniform magnetic field. Initially, at t = 0, the plane of the loop is perpendicular to the magnetic field. If it rotates with a period of 10 s about an axis in its plane then the magnitude of induced emf will be maximum and minimum, respectively at     [2020]
(1) 2.5 s and 7.5 s
(2) 2.5 s and 5.0 s
(3) 5.0 s and 7.5 s
(4) 5.0 s and 10.0 s
Ans: 
(2)
Solution:
We know, magnetic flux is given by
JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE
Where B is magnetic field and A is area
And emf induced is given by
JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE
JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE
emf will be maximum when JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE
and emf will be minimum when ωt = π, 2π, ....
So, magnitude of induced emf will be maximum and minimum at 2.5 s and 5.0 s respectively.

Q 12. At time t = 0, magnetic field of 1000 Gauss is passing perpendicularly through the area defined by the closed loop shown in the figure. If the magnetic field reduces linearly to 500 Gauss, in the next 5 s, the induced EMF in the loop is
JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE    [2020]
(1) 56 μV
(2) 28 μV
(3) 48 μV
(4) 36 μV
Ans:
(1)
Solution:
Given that
B1 = 1000 Gauss = 1000 x 10-4 T
B2 = 500 Gauss = 500 x 10-4 T
t= 5s
JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE
Induced EMF is given by
JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE

Q 13. In a fluorescent lamp choke (a small transformer), 100 V of reverse voltage is produced when the choke current changes uniformly from 0.25 A to 0 in a duration of 0.025 ms. The self-inductance of the choke (in mH) is estimated to be ________.
Ans:
(10)
Solution:
Given that
V =100V, I1 = 0.25A, I2 = 0 A, t = 0.025 ms
Voltage across inductor is given by
JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE
JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE

Q 14. A small circular loop of conducting wire has radius 'a' and carries current I. It is placed in a uniform magnetic field B perpendicular to its plane such that when rotated slightly about its diameter and released, it starts performing simple harmonic motion of time period T. If the mass of the loop is m  then    [2020]
JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE
JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE
Ans:
(3)
Solution:
Magnetic moment of circular loop carrying current I and radius a is given by
JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE

Where  JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE is the area vector of the circular loop.
Given conditions is shown in the following figure
JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE
Torque on a current carrying circular loop in a magnetic field is given by
JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE
And
JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE
Where  JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE angular acceleration of the circular loop.
So,      Iα =  MBsin θ
For small θ , we have
JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE   
We know that, for SHM
α  =-ω2θ
So,
JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE
JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE
We know that
T = 2π/ω
JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE

15. A conducting circular loop made of a thin wire, has area 3.5 × 10−3 m2 and resistance 10 Ω. It is placed perpendicular to a time dependent magnetic field B(t) = (0.4 T) sin (50πt). The field is uniform in space. Then, the net charge flowing through the loop during t = 0 s and t = 10 ms is close to     [2019]
(1) 14 mC
(2) 7 mC
(3) 21 mC
(4) 6 mC
Ans:
(1)
Solution:
We have, A = 3.5 × 10−3 m2, R = 10 Ω
JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE
=(1/10) x 3.5 x 10-3 x 0.4
= 1.4 × 10−4 = 14 mC

Q 16. A solid metal cube of edge length 2 cm is moving in a positive y-direction at a constant speed of 6 m/s. There is a uniform magnetic field of 0.1 T in the positive z-direction. The potential difference between the two faces of the cube perpendicular to the x-axis, is    [2019]
(1) 12 mV
(2) 6 mV
(3) 1 mV
(4) 2 mV
Ans:
(1)
Solution:
Potential difference between two faces = E × l
= (v × B)l
= 6 × 0.1 × 2 × 10–2
= 12 mV

Q 17. There are two long coaxial solenoids of same length l. The inner and outer coils have radii r1 and r2 and number of turns per unit length n1 and n2, respectively. The ratio of mutual inductance to the self-inductance of the inner-coil is    [2019]
JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE
Ans:
(4)
Solution:
Self-Inductance is
L = μ0n12πr12l       [ϕ = Li]
Mutual inductance is
M = μ0n1n2πr12l [ϕ1 = Mi2]
JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE

Q 18. A copper wire is wound on a wooden frame whose shape is that of an equilateral triangle. If the linear dimension of each side of the frame is increased by a factor of 3, keeping the number of turns of the coil per unit length of the frame the same, then the self-inductance of the coil    [2019]
(1) decreases by a factor of 9.
(2) increases by a factor of 27.
(3) increases by a factor of 3.
(4) decreases by a factor of 9√3
Ans:
(1)
Solution:
The self-inductance of the coil is
JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE
⇒ N/l = constant
⇒ N ∝ l
⇒ L ∝ lA
⇒ L ∝ la2
JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE
Therefore, self-inductance will increase by a factor of 3.

Q 19. A 10 m long horizontal wire extends from North East to South West. It is falling with a speed of 5.0 m/s, at right angles to the horizontal component of the Earth’s magnetic field, of 0.3 × 10−4 Wb/m2. The value of the induced emf in wire is    [2019]
(1) 1.5 × 10−3 V
(2) 1.1 × 10−3 V
(3) 2.5 × 10−3 V
(4) 0.3 × 10−3 V
Ans:
(2)
Solution:
Induced emf = Bvl sin θ
Since, θ = 45°
JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE

Q 20. A power transmission line feeds input power at 2300 V to a step-down transformer with its primary windings having 4000 turns. The output power is delivered at 230 V by the transformer. If the current in the primary of the transformer is 5 A and its efficiency is 90%, the output current would be    [2019]
(1) 50 A
(2) 45 A
(3) 35 A
(4) 25 A
Ans:
(2)
Solution:
JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE

Q 21. A series AC circuit containing an inductor (20 mH), a capacitor (120 μF) and a resistor (60 Ω) is driven by an AC source of 24 V/50 Hz. The energy dissipated in the circuit in 60 s is     [2019]
(1) 5.65 × 102 J

(2) 2.26 × 103 J
(3) 5.17 × 102 J
(4) 3.39 × 103 J
Ans:
(3)
Solution:
Given R = 60 Ω, f = 50 Hz, C = 120 μF = 120 × 10−6 F
JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE
Inductive Reactance
XL = ωL
= 2πfL
= 2 × π × 50 × 20 × 10−3 
= 6.28 Ω
Now,
XC – XL = 26.52 – 6.28
= 20.24
XC – XL ≈ 20
Impedance
JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE
JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE
Power consumed by the AC
JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE
Hence, energy dissipated in the circuit in 60s = 5.17 × 102 J

Q 22.  The self-induced emf of a coil is 25 volts. When the current in it is changed at uniform rate from 10 A to 25 A in 1 s, the change in the energy of the inductance is    [2019]
(1) 740 J
(2) 437.5 J
(3) 540 J
(4) 637.5 J
Ans:
(2)
Solution:
Induced emf,
JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE
JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE
Therefore, change in energy of inductance is
JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE

Q 23. In the circuit shown, the switch S1 is closed at time t = 0 and the switch S2 is kept open. At some later time "t0", the switch S1 is opened and S2, is closed. The behavior of the current I as a function of time t is given by    [2019]
JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE
(1)
JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE
(2)
JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE
(3)
JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE
(4)
JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE

Ans: (N*)
*Disputed question – As more than one option is correct [options (1), (2), (4) are correct
Solution:
We have,
JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE
Now, integrating both the sides, we get
JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE
JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE
JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE during decay
JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE

Q 24. In the figure shown, a circuit contains two identical resistors with resistance R = 5Ω and an inductance with L = 2 mH. An ideal battery of 15 V is connected in the circuit. What will be the current through the battery long after the switch is closed?
JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE    [2019]
(1) 5.5 A
(2) 7.5 A
(3) 3 A
(4) 6 A
Ans: 
(4)
Solution:
JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE
Let I be the current in the circuit.
Ideal inductor will behave like zero resistance long time after switch is closed.
I = 2E/R
⇒ I = (2 x 15) / 5
⇒ I = 6A

Q 25. In the circuit shown here, JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE and R1 = 10 Ω. Current in L–R1 path is I1 and in C–R2 path it is I2. The voltage of AC source is given by V = 200√2 sin(100 t) volts. The phase difference between I1 and I2 is
JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE    [2019]

(1) 60°
(2) 30°
(3) 90°
(4) 0°
Ans:
(3)
Solution:
Given V = 200√2 sin100t
ω = 100
XL = ωL
JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE
JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE
Thus, θ1 is close to 90°
JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE
JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE
⇒θ2 = 60°
So, the phase difference = 90° + 60° = 150°
If R2 is 20 kΩ, then the phase difference = 60° + 30° = 90°.

Q 26. A plane electromagnetic wave of frequency 50 MHz travels in free space along the positive x direction. At a particular point in space and time, JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE The corresponding magnetic field JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE at that point will be    [2019]
(1) 18.9 x 10-8 JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE
(2) 2.1 x 10-8 JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE
(3) 6.3 x 10-8 JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE
(4) 18.9 x 108 JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE
Ans: 
(2)
Solution:
Let JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE be the magnetic field, at a particular point JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE
So,
JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE
where c is the speed of light.
JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE
Unit vector
JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE

Q 27. The energy associated with electric field is ( UE) and with magnetic field is (UB) for an electromagnetic wave in free space. Then    [2019]
(1)  UB = UB/2
(2) UE > UB
(3) UE < UB
(4) UE = UB
Ans:
(4)
Solution:
The average electric energy density is
JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE
The average magnetic energy density is
JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE
In electromagnetic wave, the electric and magnetic field vary sinusoidally in free space so, in above expression E and B are replaced by their rms values.
Therefore,
JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE

Since energy density of electric and magnetic field is equal.
Hence, UB = UE

Q 28. If the magnetic field of a plane electromagnetic wave is given by ( the speed of light = 3 × 10m/s)JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEEthen the maximum electric field associated  with it is     [2019]
(1) 6 × 104 N/C
(2) 3 × 104 N/C
(3) 4 × 104 N/C
(4) 4.5 × 104 N/C
Ans:
(2)
Solution:
speed of light is
c = 3 × 108 m/s
Magnetic field is
JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE
Maximum magnetic field, B0 = 100 × 10−6 T
Therefore, electric field is
E0 = c × B0 
= 3 × 108 × 100 × 10−6
= 3 × 104 N/C

Q 29. The electric field of a plane polarized electromagnetic wave in free space at time t = 0 is given by an expression JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE The magnetic field JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE is given by (c is the velocity of light)    [2019]
(1) JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE
(2) JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE
(3) JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE
(4) JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE
Ans: 
(2)
Solution:
Electric field is
JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE
since, JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE is direction of waves
So, direction of JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE will be along
JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE
Let JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE and unit vector in direction of propagation of EM wave is
JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE
JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE
Therefore, JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE

Q 30. An electromagnetic wave of intensity 50 W/m2 enters in a medium of refractive index ‘ n’ without any loss. The ratio of the magnitudes of electric fields, and the ratio of the magnitudes of magnetic fields of the wave before and after entering into the medium are respectively, given by    [2019]
JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE
Ans:
(3)
Solution:
As we know that,
JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE
For transparent medium
JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE
From Eq. (1) and Eq. (2), we have
JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE
We know that, intensity in term of electric field is
JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE
From Eq. (3) and Eq. (4), we have
JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE
Similarly, in terms of magnetic field
JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE
From Eq. (5) and Eq. (6), we get
JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE

Q 31.  A 27 mW laser beam has a cross-sectional area of 10 mm2. The magnitude of the maximum electric field in this electromagnetic wave is given by
(Given: Permittivity of space ε0 = 9x10-12 SI units; Speed of light c = 3 × 108 m/s.)    [2019]
(1) 2 kV/m
(2) 0.7 kV/m
(3) 1 kV/m
(4) 1.4 kV/m
Ans:
(4)
Solution:
Intensity of electromagnetic wave is given by
JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE
JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE
E = √2 x 103 V/m
E =1.4 kV/m

Q 32. In an a.c. circuit, the instantaneous e.m.f. and current are given by e = 100 sin 30 t JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE . In one cycle of a.c., the average power consumed by the circuit and the wattless current are, respectively:    [2018]
(1) 50, 10
(2) JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE

(3) JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE
(4) 50, 0
Ans: 
(2)
Solution:
JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE
JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE
JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE
JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE

Q 33. For an RLC circuit driven with voltage of amplitudeJEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE and frequency JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE, the current exhibits resonance. The quality factor, Q is given by    [2018]
(1) JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE

(2) JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE

(3) JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE

(4) JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE
Ans:
(1)
Solution:
JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE
Alternate solution
JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE is the only dimensionless  quantity, hence must be the quality factor.

The document JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE is a part of the JEE Course Physics For JEE.
All you need of JEE at this link: JEE
257 videos|633 docs|256 tests
Download as PDF

Download free EduRev App

Track your progress, build streaks, highlight & save important lessons and more!

Related Searches

practice quizzes

,

past year papers

,

Important questions

,

Objective type Questions

,

Extra Questions

,

mock tests for examination

,

Viva Questions

,

video lectures

,

study material

,

JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE

,

pdf

,

shortcuts and tricks

,

ppt

,

Free

,

JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE

,

Exam

,

Sample Paper

,

JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 1 - Notes | Study Physics For JEE - JEE

,

MCQs

,

Semester Notes

,

Previous Year Questions with Solutions

,

Summary

;