Electromagnetic Waves · Physics · NEET
MCQ (Single Correct Answer)
The following table presents the part of the electromagnetic spectrum and their corresponding major applications.
| Part of the electromagnetic spectrum | Applications | ||
| P. | Microwave | I. | For purifying the water |
| Q. | UV rays | II. | For warming the food |
| R. | Gamma rays | III. | For AM and FM communication systems |
| S. | Radio wave | IV. | For treating the Cancer cells |
An electromagnetic wave travelling in a lossless dielectric medium having a dielectric constant, $\varepsilon_r=9$, has the electric field, $E_x=E_0 \sin \left(k z-2 \pi \times 10^6 t\right) \vee m^{-1}$ where $E_0$ is the amplitude and $k$ is the wave vector. Among the following options, the incorrect choice is
$$ \text { Match List I with List II: } $$
| List-I (Electromagnetic wave) |
List-II (Production) |
||
|---|---|---|---|
| A. | Microwave | I. | Electrons in atoms emit light when they move from a higher energy level to a lower energy level |
| B. | Visible light | II. | Radioactive decay of nucleus |
| C. | Gamma rays | III. | Vibration of atoms and molecules |
| D. | Infra-red rays | IV. | Klystron valve or magnetron valve |
The electric field in a plane electromagnetic wave is given by $$ E_z=60 \cos \left(5 x+1.5 \times 10^9 t\right) \mathrm{V} / \mathrm{m}$$ Then expression for the corresponding magnetic field is (here subscripts denote the direction of the field) :
The electromagnetic radiation which has the smallest wavelength are
If the ratio of relative permeability and relative permittivity of a uniform medium is $$1: 4$$. The ratio of the magnitudes of electric field intensity $$(E)$$ to the magnetic field intensity $$(H)$$ of an EM wave propagating in that medium is (Given that $$\sqrt{\frac{\mu_0}{\varepsilon_0}}=120 \pi$$):
The property which is not of an electromagnetic wave travelling in free space is that:
In a plane electromagnetic wave travelling in free space, the electric field component oscillates sinusoidally at a frequency of $$2.0 \times 10^{10} \mathrm{~Hz}$$ and amplitude $$48 ~\mathrm{Vm}^{-1}$$. Then the amplitude of oscillating magnetic field is : (Speed of light in free space $$=3 \times 10^{8} \mathrm{~m} \mathrm{~s}^{-1}$$ )
The magnetic field of a plane electromagnetic wave is given by $$\overrightarrow B = 3 \times {10^{ - 8}}\cos (1.6 \times {10^3}x + 48 \times {10^{10}}t)\widehat j$$, then the associated electric field will be :
The ratio of the magnitude of the magnetic field and electric field intensity of a plane electromagnetic wave in free space of permeability $${\mu _0}$$ and permittivity $${\varepsilon _0}$$ is (Given that c - velocity) of light in free space
Match List - I with List - II
| List - I (Electromagnetic waves) |
List - II (Wavelength) |
||
|---|---|---|---|
| (a) | AM radio waves | (i) | $${10^{ - 10}}$$ m |
| (b) | Microwaves | (ii) | $${10^2}$$ m |
| (c) | Infrared radiations | (iii) | $${10^{ - 2}}$$ m |
| (d) | X-rays | (iv) | $${10^{ - 4}}$$ m |
Choose the correct answer from the options given below
When light propagates through a material medium of relative permittivity $$\varepsilon $$r and relative permeability $$\mu$$r, the velocity of light, v is given by (c-velocity of light in vacuum)
E = $$\widehat i$$ 40 cos (kz $$-$$ 6 $$ \times $$ 108 t) where E, z and t are in volt/m, meter and seconds respectively. The value of wave vector k is
(1) the wavelength $$\lambda $$ is 188.4 m.
(2) the wave number k is 0.33 rad/m.
(3) the wave amplitude is 10 V/m.
(4) the wave is propagating along +x direction.
Which one of the following pairs of statements is correct ?
$${E_y} = 2.5{N \over C}\cos \left[ {\left( {2\pi \times {{10}^6}{{rad} \over m}} \right)t - \left( {\pi \times {{10}^{ - 2}}{{rad} \over s}} \right)x} \right];$$
Ez = 0.