Electromagnetic Waves Numerical Class-12 Nootan ISC Physics Solution Ch-13. Step by step solutions of Kumar and Mittal Physics of Nageen Prakashan as council latest prescribe guideline for upcoming exam. Visit official Website CISCE for detail information about ISC Board Class-12 Physics.
Electromagnetic Waves Numerical Class-12 Nootan ISC Physics Solution Ch-13
| Board | ISC |
| Class | 12 |
| Subject | Physics |
| Book | Nootan |
| Chapter-13 | Electromagnetic Waves |
| Topics | Numericals on Electromagnetic Waves |
| Academic Session | 2025-2026 |
Numericals on Electromagnetic Waves
Class-12 Nootan ISC Physics Solution Ch-13
Que-2: Electromagnetic waves enter a medium of relative permeability 9 and relative permittivity 100. Find the speed of the waves in the medium.
Ans- v = c/√μr. cr = 3 x 10^8/√9 x 100
= 1 x 10^7 m/s
Que-3: A radar transmitter generates waves of frequency 3 x 10^9 Hz. What is the wavelength?
Ans- λ = c/f = 3 x 10^8 / 3 x 10^9
= 10^-1 m = 10 cm
Que-4: A radio station broadcasts at 30 m. Find the frequency of broadcast.
Ans- f = c/λ = 3 x 10^8 / 30
= 10 x 10^6 Hz = 10 MHz
Que-5: A plane electromagnetic wave has a maximum electric field of 3.20 x 10^-4 V m^-1 What is maximum magnetic field?
Ans- Bo = Eo/c = 3.20 x 10^-4 / 3 x 10^8
= 1.07 x 10^-12 T
Que-6: In a plane EM wave, the magnetic field oscillates sinusoidally with a frequency of 2 x 10^10 Hz and an amplitude of 46 Wb m^-2 Find the wavelength of the wave and the amplitude of the oscillating electric field.
Ans- λ = c/f = 3 x 10^8 / 2 x 10^10
= 1.5 x 10^-2 m
Eo = Boc = 46 x 3 x 10^8
= 138 x 10^8 V/m
Que-7: In a plane EM wave, the electric field oscillates sinusoidally with a frequency of 2 x 10^10 Hz and an amplitude of 48V m^-1 Find the wavelength and the amplitude of the oscillating magnetic field.
Ans- λ = c/f = 3 x 10^8 / 2 x 10^10
= 1.5 x 10^-2 m
Bo = Eo/c = 48 / 3 x 10^8
= 0.16 x 10^-6 T
= 0.16 μT
Que-8: A laser beam has intensity 2.5 x 10^14 W m^-2 Find the amplitudes of electric and magnetic fields in the beam.
Ans- (1). Calculate the Electric Field Amplitude (Eo): I = (1/2) x c x ε₀ x Eo^2
Rearrange for Eo : Eo = √(2I / (c ε₀))
Substitute the values:
Eo = √(2 * 2.5 x 10^14 / (3 x 10^8 * 8.854 x 10^-12))
Eo ≈ 4.34 x 10^8 V/m
(2) Calculate the Magnetic Field Amplitude (Bo): Bo = Eo / c
Substitute the values:
Bo = (4.34 x 10^8) / (3 x 10^8)
Bo ≈ 1.45 T
Que-9: In a plane electromagnetic wave, the amplitude of the magnetic field is 5.0 x 10^-6 T. Find the amplitude of the electric field and the total average energy density of the wave.
Ans- Eo = Bo.c = 5 x 10^-6 x 3 x 10^8
= 1.5 x 10^3 V/m
Total Energy density
= ue + um = (1/2 Eo E^2 + 1/2 B^2/μo) 1/2
= B^2/2μo = (5 x 10^-6)^2 / 2 x (4π x 10^-7)
= 1.0 x 10^-5 J m^-3
Que-10: In a plane electromagnetic wave the electric field varies with time having an amplitude of 1.0 V m^-1 Find the average energy density of magnetic field and that of electric field.
Ans- ue = 1/4 ∈o Eo
= 1/4 x 8.85 x 10^-12 x 1
= 2.2 x 10^-12 J m^-3 each.
— : End of Electromagnetic Waves Numerical Class-12 Nootan ISC Physics Solution Ch-13. :–
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