PYQSidePowered byPrepblink ExamsMy practice

Electromagnetic Waves

JEE Main 2025 (Online) 29th January Evening ShiftNumericalMedium📈 Topic in 21% papers+4
Ques. 1JEE Main 2025, 29 January (Evening Shift)

A parallel plate capacitor consisting of two circular plates of radius 10 cm is being charged by a constant current of 0.15 A. If the rate of change of potential difference between the plates is 7 108 V/s then the integer value of the distance between the parallel plates is (Take, 0=9 10-12 Fm, = 227) .

Answer

1320

Solution

I= dQdt=C dVdt with C= 0 r2d ⇒ d= 0 r2I dVdt

[figure]

= 9 10-12 227 (0.1)2 7 1080.15= 198 10-60.15

=1320 10-6 m

∴ d=1320 m

Answer: 1320

Keep practising — Electromagnetic Waves

Ques. 2JEE Main 2025, 28 January (Morning Shift)

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
Ques. 3JEE Main 2025, 28 January (Evening Shift)

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 ) ]
Ques. 4JEE Main 2025, 24 January (Evening Shift)

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

  1. 4 < 3 < 2 < 1
  2. 4 < 2 < 3 < 1
  3. 3 < 4 < 2 < 1
  4. 4 < 3 < 1 < 2
Ques. 5JEE Main 2025, 23 January (Morning Shift)

The electric field of an electromagnetic wave in free space is E=57 [7.5 106t-5 10-3(3x+4y) ](4 i-3 j) N/C. The associated magnetic field in Tesla is

  1. B= 573 108 [7.5 106t-5 10-3(3x+4y) ]( k)
  2. B= 573 108 [7.5 106t-5 10-3(3x+4y) ](5 k)
  3. B=- 573 108 [7.5 106t-5 10-3(3x+4y) ]( k)
  4. B=- 573 108 [7.5 106t-5 10-3(3x+4y) ](5 k)
Ques. 6JEE Main 2025, 23 January (Evening Shift)

A plane electromagnetic wave of frequency 20 MHz travels in free space along the +x direction. At a particular point in space and time, the electric field vector of the wave is Ey=9.3 V m-1. Then, the magnetic field vector of the wave at that point is

  1. Bz=1.55 10-8 T
  2. Bz=6.2 10-8 T
  3. Bz=3.1 10-8 T
  4. Bz=9.3 10-8 T
Ques. 7JEE Main 2025, 23 January (Evening Shift)

A time varying potential difference is applied between the plates of a parallel plate capacitor of capacitance 2.5 . The dielectric constant of the medium between the capacitor plates is 1 . It produces an instantaneous displacement current of 0.25 mA in the intervening space between the capacitor plates, the magnitude of the rate of change of the potential difference will be V s-1.

Ques. 8JEE Main 2025, 22 January (Evening Shift)

A parallel plate capacitor of area A=16 cm2 and separation between the plates 10 cm , is charged by a DC current. Consider a hypothetical plane surface of area A0=3.2 cm2 inside the capacitor and parallel to the plates. At an instant, the current through the circuit is 6A. At the same instant the displacement current through A0 is mA .

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