PYQSide ExamsMy practice

Electromagnetic Waves · Question 18

JEE Main 2026 (Online) 21st January Morning ShiftMCQEasy📈 Topic in 65% papers+4−1
1

The electric field in a plane electromagnetic wave is given by : Ey = 69 [0.6 103x - 1.8 1011t ] V/m. The expression for magnetic field associated with this electromagnetic wave is T.

  1. Bz = 2.3 10-7 [0.6 103x - 1.8 1011t ]
  2. Bz = 2.3 10-7 [0.6 103x + 1.8 1011t ]
  3. By = 2.3 10-7 [0.6 103x - 1.8 1011t ]
  4. By = 69 [0.6 103x + 1.8 1011t ]

Answer

(1) Bz = 2.3 10-7 [0.6 103x - 1.8 1011t ]

Solution

Phase (kx - t) wave travels along +x; speed = k = 1.8 10110.6 103 = 3 108 m/s

B0 = E0c = 693 108 = 2.3 10-7 T

B is perpendicular to both E (y-direction) and the direction of propagation (x), and E B is along +x:

j k = i B is along z, in phase with E

Bz = 2.3 10-7 [0.6 103x - 1.8 1011t ] T

Answer: (1)

JEE Main 2026 (Online) 21st January Morning Shift

Keep practising — Electromagnetic Waves

2

An electromagnetic wave of frequency 100 MHz propagates through a medium of conductivity, = 10 mho/m. The ratio of maximum conduction current density to maximum displacement current density is . [Take 14 0 = 9 109 N m2/C2]

JEE Main 2026 (Online) 21st January Evening Shift
3

If 0 denotes the permittivity of free space and E is the flux of the electric field through the area bounded by the closed surface, then dimensions of ( 0 d Edt ) are that of :

  1. electric charge
  2. electric field
  3. electric current
  4. electric potential
JEE Main 2025 (Online) 7th April Morning Shift
4

The unit of 2I 0 c is : (I = intensity of an electromagnetic wave, c = speed of light)

  1. Vm
  2. NC-1
  3. NC
  4. Nm
JEE Main 2025 (Online) 7th April Evening Shift
5

If an optical medium possesses a relative permeability of 10 and relative permittivity of 10.0885, then the velocity of light is greater in vacuum than that in this medium by times. ( 0 = 4 10-7 H/m, 0 = 8.85 10-12 F/m, c = 3 108 m/s)

JEE Main 2025 (Online) 4th April Evening Shift
6

The radiation pressure exerted by a 450 W light source on a perfectly reflecting surface placed 2 m away from it, is

  1. 3 10-8 Pascals
  2. 0
  3. 1.5 10-8 Pascals
  4. 6 10-8 Pascals
JEE Main 2025 (Online) 3rd April Morning Shift
7

Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R). In the light of the above statements, choose the most appropriate answer from the options given below :

  1. Both (A) and (R) are true and (R) is the correct explanation of (A)
  2. Both (A) and (R) are true but (R) is not the correct explanation of (A)
  3. (A) is false but (R) is true
  4. (A) is true but (R) is false
JEE Main 2025 (Online) 29th January Morning Shift
8

Due to presence of an em-wave whose electric component is given by E = 100 ( t - kx) NC-1 a cylinder of length 200 cm holds certain amount of em-energy inside it. If another cylinder of same length but half diameter than previous one holds same amount of em-energy, the magnitude of the electric field of the corresponding em-wave should be modified as

  1. 50 ( t - kx) NC-1
  2. 400 ( t - kx) NC-1
  3. 200 ( t - kx) NC-1
  4. 25 ( t - kx) NC-1
JEE Main 2025 (Online) 28th January Morning Shift
9

The magnetic field of an E.M. wave is given by B = ( 32 i + 12 j )30 [ (t - zc ) ] (S.I. Units). The corresponding electric field in S.I. units is:

  1. E = ( 12 i + 32 j )30c [ (t + zc ) ]
  2. E = ( 12 i - 32 j )30c [ (t - zc ) ]
  3. E = ( 32 i - 12 j )30c [ (t + zc ) ]
  4. E = ( 34 i + 14 j )30c [ (t - zc ) ]
JEE Main 2025 (Online) 28th January Evening Shift

All Electromagnetic Waves PYQs →

PDFDownload Electromagnetic Waves PYQs as PDF Free, with answer key and a QR code to every solution — print it or study offline⬇
All JEE Main Physics chapters