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Electromagnetic Waves

JEE Main 2024 (Online) 9th April Morning ShiftMCQEasy📈 Topic in 64% papers+4−1
Ques. 1JEE Main 2024, 9 April (Morning Shift)

A plane EM wave is propagating along x direction. It has a wavelength of 4 mm. If electric field is in y direction with the maximum magnitude of 60 V m-1, the equation for magnetic field is :

  1. Bz = 2 10-7 [ 2 (x - 3 108 t ) ] k T
  2. Bz = 2 10-7 [ 2 103 (x - 3 108 t ) ] k T
  3. Bz = 60 [ 2 (x - 3 108 t ) ] k T
  4. Bx = 60 [ 2 (x - 3 108 t ) ] i T

Answer

(2) Bz = 2 10-7 [ 2 103 (x - 3 108 t ) ] k T

Solution

Amplitude: B0 = E0c = 603 108 = 2 10-7 T

Wave number: k = 2 = 2 4 10-3 = 2 103 m-1

Since E B points along the propagation, j k = i, so B is along z.

Hence Bz = B0 [k(x - ct)] k = 2 10-7 [ 2 103 (x - 3 108 t ) ] k T

Answer: (2)

Keep practising — Electromagnetic Waves

Ques. 2JEE Main 2024, 9 April (Evening Shift)

The magnetic field in a plane electromagnetic wave is By = (3.5 10-7 ) (1.5 103 x + 0.5 1011 t ) T. The corresponding electric field will be :

  1. Ez = 105 (1.5 103 x + 0.5 1011 t ) V m-1
  2. Ey = 10.5 (1.5 103 x + 0.5 1011 t ) V m-1
  3. Ey = 1.17 (1.5 103 x + 0.5 1011 t ) V m-1
  4. Ez = 1.17 (1.5 103 x + 0.5 1011 t ) V m-1
Ques. 3JEE Main 2024, 8 April (Morning Shift)

Average force exerted on a non-reflecting surface at normal incidence is 2.4 10-4 N. If 360 W/cm2 is the light energy flux during span of 1 hour 30 minutes, then the area of the surface is:

  1. 20 m2
  2. 0.2 m2
  3. 0.1 m2
  4. 0.02 m2
Ques. 4JEE Main 2024, 6 April (Morning Shift)

Electromagnetic waves travel in a medium with speed of 1.5 108 m s-1. The relative permeability of the medium is 2.0. The relative permittivity will be:

  1. 4
  2. 1
  3. 2
  4. 5
Ques. 5JEE Main 2024, 6 April (Evening Shift)

In the given electromagnetic wave Ey = 600 ( t - kx) Vm-1, intensity of the associated light beam is (in W/m2) : (Given 0 = 9 10-12 C2 N-1 m-2)

  1. 486
  2. 729
  3. 243
  4. 972
Ques. 6JEE Main 2024, 5 April (Morning Shift)

An alternating voltage of amplitude 40 V and frequency 4 kHz is applied directly across the capacitor of 12 . The maximum displacement current between the plates of the capacitor is nearly :

  1. 10 A
  2. 8 A
  3. 13 A
  4. 12 A
Ques. 7JEE Main 2024, 5 April (Evening Shift)

Match List I with List II : Choose the correct answer from the options given below :

  1. (A)-(I), (B)-(III), (C)-(II), (D)-(IV)
  2. (A)-(III), (B)-(II), (C)-(IV), (D)-(I)
  3. (A)-(IV), (B)-(III), (C)-(II), (D)-(I)
  4. (A)-(II), (B)-(I), (C)-(IV), (D)-(III)
Ques. 8JEE Main 2024, 4 April (Morning Shift)

The electric field in an electromagnetic wave is given by E= i40 (t-z/c) NC-1. The magnetic field induction of this wave is (in SI unit) :

  1. B= j 40c (t-z/c)
  2. B= i 40c (t-z/c)
  3. B= j40 (t-z/c)
  4. B= k 40c (t-z/c)
Ques. 9JEE Main 2024, 4 April (Evening Shift)

Arrange the following in the ascending order of wavelength: Choose the most appropriate answer from the options given below

  1. 1 < 2 < 3 < 4
  2. 2 < 1 < 4 < 3
  3. 4 < 3 < 2 < 1
  4. 4 < 3 < 1 < 2
Ques. 10JEE Main 2024, 1 February (Morning Shift)

A parallel plate capacitor has a capacitance C=200 pF. It is connected to 230 V ac supply with an angular frequency 300 rad/s. The rms value of conduction current in the circuit and displacement current in the capacitor respectively are :

  1. 14.3 and 143
  2. 13.8 and 13.8
  3. 13.8 and 138
  4. 1.38 and 1.38
Ques. 11JEE Main 2024, 1 February (Evening Shift)

If frequency of electromagnetic wave is 60 MHz and it travels in air along z direction then the corresponding electric and magnetic field vectors will be mutually perpendicular to each other and the wavelength of the wave (in m) is :

  1. 2.5
  2. 5
  3. 10
  4. 2

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